Rajnikant Sinha Real and Complex Analysis

Rajnikant Sinha

Real and Complex AnalysisVolume 1

Real and Complex Analysis

Rajnikant Sinha

Real and Complex AnalysisVolume 1

123

Rajnikant SinhaVaranasi, Uttar Pradesh, India

ISBN 978-981-13-0937-3 ISBN 978-981-13-0938-0 (eBook)https://doi.org/10.1007/978-981-13-0938-0

Library of Congress Control Number: 2018945438

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Preface

The book is an introduction to real and complex analysis that will be useful toundergraduate students of mathematics and engineering. It is designed to equip thereader with tools that will help them to understand the concepts of real analysis andcomplex analysis. In addition, it contains the essential topics of analysis that areneeded for the study of functional analysis. Its guiding principle is to help develop thenecessary concepts rigorously with enough detail and with the minimum prerequi-sites. Further, I have endeavored to make this book both accessible and readable. Thisbook contains complete solutions to almost all the problems discussed within. Thiswill be beneficial to readers only if used correctly: readers are encouraged to look atthe solution to a problem only after trying to solve the problem.

Certainly, at times, the reader may find the proofs excruciatingly detailed, but itis better to be detailed than concise. Furthermore, omitting the detailed calculationcan sometimes be perplexing for beginners. I have tried to make it a readable textthat caters to a broad audience. This approach should certainly benefit beginnerswho have not yet tussled with the subject in a serious way.

This book contains several useful theorems and their proofs in the realm of realand complex analysis. Most of these theorems are the works of some the greatmathematicians of the 19th and 20th centuries. In alphabetical order, some of theseinclude: Arzela, Ascoli, Baire, Banach, Carathéodory, Cauchy, Dirichlet, Egoroff,Fatou, Fourier, Fubini, Hadamard, Jordan, Lebesgue, Liouville, Minkowski,Mittag-Leffler, Morera, Nikodym, Ostrowski, Parseval, Picard, Plancherel, Poisson,Radon, Riemann, Riesz, Runge, Schwarz, Taylor, Tietze, Urysohn, Weierstrass, andYoung. I have spent several years providing their proofs in unprecedented detail.

There are plenty of superb texts on real and complex analysis, but there is adearth of books that blend real analysis with complex analysis. Libraries alreadycontain several excellent reference books on real and complex analysis, whichinterested students can consult for a deeper understanding. It was not my intentionto replace such books. This book is written under the assumption that studentsalready know the fundamentals of advanced calculus. The proofs of various named

v

theorems should be considered to be at the core of the book by any reader who isserious about learning the subject.

The book is divided into two volumes. Volume 1 contains three chapters:Lebesgue integration, Lp-spaces and Fourier transforms. In Chap. 1, we begin withthe definition of an exponential function and prove that it maps the set of allcomplex numbers onto the set of all nonzero complex numbers. After that, wedevelop the Lebesgue theory of abstract integration of complex-valued functions.Next, we prove the Riesz representation theorem in enough detail and use it toanswer the question: is every set of n-tuples of real numbers Lebesgue measurablein R

n? The theme of Chap. 2 is Lp-spaces. First of all, we introduce convexfunctions and then prove the Riesz–Fischer theorem. In the end, we derive someproperties of Banach algebra. We have shown that Lp-spaces is an example of aBanach space. In Chap. 3, we introduce total variation, and prove the Radon–Nikodym theorem. Fubini theorem and the change-of-variable theorem are alsoproved in this chapter. Finally, we discuss the Plancherel theorem on Fouriertransforms.

I am particularly indebted to Walter Rudin and Paul Richard Halmos for theirletters discussing academic questions. By great good fortune, some colleagues ofmine were able to join in with this enterprise a few years ago, some of whom haveprovided a meticulous reading of the manuscript from a user’s viewpoint. I extendmy great thanks to all of them for their expert services.

While studying this book, I hope that readers will experience the thrill of creativeeffort and the joy of achievement.

Varanasi, India Rajnikant Sinha

vi Preface

Contents

1 Lebesgue Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.3 Integration of Positive Functions . . . . . . . . . . . . . . . . . . . . . . . . 511.4 Integration of Complex-Valued Functions . . . . . . . . . . . . . . . . . 851.5 Sets of Measure Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 981.6 Preliminaries to Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1191.7 Preliminaries to Riesz Representation Theorem. . . . . . . . . . . . . . 1391.8 Riesz Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1671.9 Borel Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1771.10 Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1871.11 Existence of Non-Lebesgue Measurable Sets . . . . . . . . . . . . . . . 216

2 Lp-Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2372.1 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2372.2 The Lp-Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2602.3 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2852.4 Orthogonal Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2992.5 Riesz-Fischer Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3152.6 Baire’s Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3372.7 Hahn-Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3532.8 Banach Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

3 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3913.1 Total Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3913.2 Radon–Nikodym Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4113.3 Bounded Linear Functionals on Lp . . . . . . . . . . . . . . . . . . . . . . . 4313.4 Lebesgue Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4573.5 Metric Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4823.6 Vitali–Caratheodory Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 510

vii

3.7 Change-of-Variables Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 5353.8 Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5443.9 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5663.10 Distribution Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5783.11 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5873.12 Inversion Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6143.13 Plancherel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637

viii Contents

About the Author

Rajnikant Sinha is Former Professor of Mathematics at Magadh University, BodhGaya, India. As a passionate mathematician, he has published numerous interestingresearch findings in international journals and books, including Smooth Manifolds(Springer) and the contributed book Solutions to Weatherburn’s Elementary VectorAnalysis. His research focuses on topological vector spaces, differential geometryand manifolds.

ix

Chapter 1Lebesgue Integration

Inadequacies of the Riemann integration, and difficulties in handling limit processesin it were largely overcome with the advent of abstract integration. Next, uponapplying a remarkable result—the Riesz representation theorem—Lebesgue mea-sure in Euclidean space is introduced. Just like the real number system is a com-pletion, in a certain sense, of the rational number system, Lebesgue integration is acompletion, in a certain sense, of Riemann integration. These phenomena will bevividly demonstrated in this somewhat long chapter. We begin with the definitionof exponential function, and prove that it maps the set of all complex numbers ontothe set of all nonzero complex numbers. After that, we develop the Lebesgue theoryof abstract integration of complex-valued functions. Next we prove the Rieszrepresentation theorem in sufficient detail and use it to answer the question: Is everyset of n-tuples of real numbers is Lebesgue measurable in Rn?

1.1 Exponential Function

Since exponential function will occur quite frequently in later chapters, it seemsprudent to lay a good foundation for this at the earliest opportunity. This section isdevoted to this end.

Note 1.1 For every complex number z,

limn!1

znþ 1

nþ 1ð Þ!znn!

�� ¼ lim

n!1zj j

nþ 1¼ zj j lim

n!11

nþ 1

� �¼ zj j � 0 ¼ 0\1;

so, by the ratio test of convergence, the series 1þ zþ z22! þ z3

3! þ � � � is absolutelyconvergent for every complex number z.

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Conclusion 1.2 For every complex number z, the series 1þ zþ z22! þ z3

3! þ � � � isabsolutely convergent.

Notation The sum of the series 1þ zþ z22! þ z3

3! þ � � � is denoted by exp zð Þ. Inshort, for every z 2 C,

exp zð Þ �X1n¼0

zn

n!:

Note 1.3 Let us take any z;w 2 C. By Conclusion 1.2, 1þ zþ z22! þ z3

3! þ � � � isabsolutely convergent, and 1þwþ w2

2! þ w3

3! þ � � � is absolutely convergent, andhence, by Mertens’ theorem, the sum of their Cauchy product

1þ zþwð Þþ z2

2!þ zwþ w2

2!

� �þ z3

3!þ z2

2!wþ z

w2

2!þ w3

3!

� �þ � � �

¼ 1þ zþwð Þþ 12!

z2 þ 2zwþw2� �þ 13!

z3 þ 3z2wþ 3zw2 þw3� �þ � � �

¼ 1þ zþwð Þþ zþwð Þ22!

þ zþwð Þ33!

þ � � �

is

1þ zþ z2

2!þ z3

3!þ � � �

� �1þwþ w2

2!þ w3

3!þ � � �

� �

Thus

exp zþwð Þ ¼ 1þ zþwð Þþ zþwð Þ22!

þ zþwð Þ33!

þ � � �

¼ 1þ zþ z2

2!þ z3

3!þ � � �

� �1þwþ w2

2!þ w3

3!þ � � �

� �¼ exp zð Þð Þ exp wð Þð Þ:

Observe that

exp 0ð Þ ¼ 1þ 0þ 02

2!þ 03

3!þ � � � ¼ 1;

and

exp 1ð Þ ¼ 1þ 1þ 12

2!þ 13

3!þ � � � ¼ 1þ 1þ 1

2!þ 1

3!þ � � � ¼ e:

2 1 Lebesgue Integration

Thus

exp 0ð Þ ¼ 1; and exp 1ð Þ ¼ e:

Notation For every z 2 C, it is customary to denote exp zð Þ by ez.

Conclusion 1.4 For every z;w 2 C, exp zð Þð Þ exp wð Þð Þ ¼ ezew ¼ ezþw. Also,e0 ¼ 1, and e1 ¼ e.

Note 1.5 Let A be a nonempty bounded subset of C. So, there exists a positive realnumber R such that for every z 2 A, zj j\R. For every z 2 A, zn

n!

�� Rn

n! , and, byConclusion 1.2,

P1n¼0

Rn

n! is convergent, so, by Weierstrass M-test,P1

n¼0znn! is uni-

formly convergent on A.

Conclusion 1.6 For every nonempty bounded subset A of C,P1

n¼0znn! is uniformly

convergent on A.

Note 1.7 We want to show that the map exp : C ! C is continuous.For this purpose, let us take any z0 2 C. We have to show that the map

z 7! exp zð Þ is continuous at z0.Clearly, the open disk

D z0; 1ð Þ � z : z 2 C and z� z0j j\1f gð Þ

is a nonempty bounded subset of C, and z0 2 D z0; 1ð Þ. Now, by Conclusion 1.6,P1n¼0

znn! is uniformly convergent on D z0; 1ð Þ. SinceP1

n¼0znn! is uniformly convergent

on D z0; 1ð Þ, z0 2 D z0; 1ð Þ, and each map z 7! 1n! z

n is continuous at z0, by a theorem(cf. [5], Theorem 7.12), the map z 7! P1

n¼0znn! ¼ exp zð Þð Þ is continuous at z0, and

hence z 7! exp zð Þ is continuous at z0.Conclusion 1.8 The map exp : C ! C is continuous.

Note 1.9 We want to show that for every z 2 C, ez 6¼ 0.If not, otherwise, suppose that there exists z 2 C such that ez ¼ 0. This would

lead us to arrive at a contradiction. Here

1 ¼ e0 ¼ ez�z ¼ eze�z ¼ 0e�z ¼ 0;

gives a contradiction. Thus, exp : C ! C0 � C� 0f gð Þð Þ.Conclusion 1.10 exp : C ! C0 � C� 0f gð Þð Þ.Note 1.11 For every complex number z,

limn!1

znnþ 1ð Þ!zn�1

n!

�� ¼ lim

n!1zj j

nþ 1¼ zj j lim

n!11

nþ 1

� �¼ zj j � 0 ¼ 0\1;

1.1 Exponential Function 3

so, by the ratio test of convergence, the series 1þ z2! þ z2

3! þ � � � is absolutelyconvergent for every complex number z.

Conclusion 1.12 For every complex number z, the series 1þ z2! þ z2

3! þ � � � isabsolutely convergent.

Note 1.13 Let A be a nonempty bounded subset of C. So, there exists a positivereal number R such that for every z 2 A, zj j\R. For every z 2 A, and, for everypositive integer n,

zn�1

n!

�� Rn�1

n!;

and, by Conclusion 1.12, 1þ R2! þ R2

3! þ � � � is convergent, so, by Weierstrass M-

test, 1þ z2! þ z2

3! þ � � � is uniformly convergent on A.

Conclusion 1.14 For every nonempty bounded subset A of C,P1

n¼0zn

nþ 1ð Þ! isuniformly convergent on A.

Note 1.15 We want to show that the map

z 7! 1þ z2!

þ z2

3!þ � � �

� �

from C to C is continuous. For this purpose, let us take any z0 2 C. We have toshow that the map

z 7! 1þ z2!

þ z2

3!þ � � �

� �

is continuous at z0. Clearly, the open disk D z0; 1ð Þ � z : z 2 C and z� z0j j\1f gð Þis a nonempty bounded subset of C, and z0 2 D z0; 1ð Þ. Now, by Conclusion 1.14,1þ z

2! þ z23! þ � � � is uniformly convergent on D z0; 1ð Þ. Since 1þ z

2! þ z23! þ � � � is

uniformly convergent on D z0; 1ð Þ; z0 2 D z0; 1ð Þ; and each map z 7! 1nþ 1ð Þ! z

n is

continuous at z0; by a theorem (cf. [5], Theorem 7.12), the map

z 7! 1þ z2!

þ z2

3!þ � � �

� �

is continuous at z0:


Conclusion 1.16 The map

z 7! 1þ z2!

þ z2

3!þ � � �

� �

from C to C is continuous.

Note 1.17 We shall try to show that

limz!0

exp zð Þð Þ � 1z

¼ 1:

Observe that, for every z 2 C� 0f gð Þ;


¼ 1z

1þ zþ z2

2!þ z3

3!þ � � �

� �� 1

� �

¼ 1z

zþ z2

2!þ z3

3!þ � � �

� �

¼ 1þ z2!

þ z2

3!þ � � � ;

so, for every z 2 C� 0f g;


¼ 1þ z2!

þ z2

3!þ � � � :

By Conclusion 1.16, the map

z 7! 1þ z2!

þ z2

3!þ � � �

� �

from C to C is continuous, and hence

limz!0

1þ z2!

þ z2

3!þ � � �

� �¼ 1þ 0

2!þ 02

3!þ � � �

� �¼ 1:

Since

limz!0

1þ z2!

þ z2

3!þ � � �

� �¼ 1;

and, for every z 2 C� 0f gð Þ;



¼ 1þ z2!

þ z2

3!þ � � � ;

we have

limz!0


¼ 1:

Conclusion 1.18 limz!0exp zð Þð Þ�1

z ¼ 1:

Note 1.19 We want to show that, for every z 2 C;

limw!0

exp zþwð Þ � exp zð Þw

¼ exp zð Þ:

Let us fix any z 2 C: Now, by Conclusion 1.18,

LHS ¼ limw!0


¼ limw!0

exp zð Þð Þ exp wð Þð Þ � exp zð Þw

¼ limw!0

exp zð Þð Þ exp wð Þð Þ � 1w

¼ exp zð Þð Þ limw!0

exp wð Þð Þ � 1w

� �¼ exp zð Þð Þ1 ¼ exp zð Þ ¼ RHS:

Conclusion 1.20 For every z 2 C;

limw!0


¼ exp zð Þ:

Note 1.21 Clearly, the restriction expjR: R ! R0 � R� 0f gð Þ: For every x[ 0;

expjR� �

xð Þ ¼ exp xð Þ ¼ 1þ xþ x2

2!þ x3

3!þ � � � [ 0:

Also, for every x[ 0;

expjR� � �xð Þ� �

expjR� �

xð Þ� � ¼ exp �xð Þð Þ exp xð Þð Þ ¼ exp �xþ xð Þ ¼ exp 0ð Þ ¼ 1;

so, for every x[ 0; expjR� � �xð Þ ¼ 1

expjRð Þ xð Þ [ 0: Thus, expjR: R ! 0;1ð Þ:We shall try to show that expjR: R ! 0;1ð Þ is strictly increasing.Let us take any x 2 R: Since, by Conclusion 1.20,


expjR� �0

xð Þ ¼ limh ! 0

h 2 R

expjR� �

xþ hð Þ � expjR� �

xð Þh

¼ limh ! 0

h 2 R

exp xþ hð Þ � exp xð Þh

¼ limh ! 0

h 2 C

exp xþ hð Þ � exp xð Þh

¼ exp xð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ expjR� �

xð Þ 2 0;1ð Þ;

we have expjR� �0

xð Þ[ 0: It follows that expjR: R ! 0;1ð Þ is strictly increasing.

Conclusion 1.22 expjR: R ! 0;1ð Þ is strictly increasing.

Note 1.23 We shall try to show thata. limt!þ1

t2RexpjR� �

tð Þ ¼ þ1; b. lims!�1s2R

expjR� �

sð Þ ¼ 0;

For a: For every t[ 0; we have

expjR� �

tð Þ ¼ exp tð Þ ¼ 1þ tþ t2

2!þ t3

3!þ � � � [ t;

so, for every t[ 0; we have t\ expjR� �

tð Þ; and hence,limt!þ1

t2RexpjR� �

tð Þ ¼ þ1:

For b: For every t[ 0;

expjR� � �tð Þ ¼ exp �tð Þ ¼ 1

exp tð Þ :

Since, for every t[ 0; expjR� � �tð Þ ¼ 1

exp tð Þ ; and by (a), limt!1t2R

expjR� �

tð Þ ¼þ1; we have

LHS ¼ lims ! �1s 2 R

expjR� �

sð Þ ¼ limt ! þ1t 2 R

expjR� � �tð Þ

¼ limt ! þ1t 2 R

1exp tð Þ ¼ lim

t ! þ1t 2 R

1expjR� �

tð Þ ¼ 0 ¼ RHS:


Conclusion 1.24 a. limt!þ1t2R

expjR� �

tð Þ ¼ þ1; b. lims!�1s2R

expjR� �

sð Þ ¼ 0:

Note 1.25 For every complex number z;

limn!1

�1ð Þn z2n2nð Þ!

�1ð Þn�1 z2 2n�1ð Þ2 n�1ð Þð Þ!

��

¼ limn!1

zj j22nð Þ 2n� 1ð Þ

¼ zj j2� limn!1

12nð Þ 2n� 1ð Þ

¼ zj j2�0 ¼ 0\1;

and

limn!1

�1ð Þn z2nþ 1

2nþ 1ð Þ!�1ð Þn�1 z2n�1

2n�1ð Þ!

��

¼ limn!1

zj j22nþ 1ð Þ 2nð Þ

¼ zj j2� limn!1

12nþ 1ð Þ 2nð Þ

¼ zj j2�0 ¼ 0\1;

so, by the ratio test of convergence, the series 1þ � z22!

� þ z4

4! þ � � � ; and

zþ � z33!

� þ z5

5! þ � � � are absolutely convergent for every complex number z:

Let A be a nonempty bounded subset of C: So, there exists a positive realnumber R such that for every z 2 A; zj j\R: Now, for every z 2 A;

�1ð Þn z2n

2nð Þ!��

�� R2n

2nð Þ! ;

and 1þ R2

2! þ R4

4! þ � � � is convergent, so, by Weierstrass M-test,

1þ � z22!

� þ z4

4! þ � � � is uniformly convergent on A: Similarly,

zþ � z33!

� þ z5

5! þ � � � is uniformly convergent on A: We want to show that the map

z 7! 1þ � z2

2!

� �þ z4

4!þ � � �

� �


from C to C is continuous. For this purpose, let us take any z0 2 C: We have toshow that the map

z 7! 1þ � z2

2!

� �þ z4

4!þ � � �

� �

is continuous at z0: Clearly, the open disk D z0; 1ð Þ � z : z 2 C and z� z0j j\1f gð Þis a nonempty bounded subset of C; and z0 2 D z0; 1ð Þ: Since

1þ � z2

2!

� �þ z4

4!þ � � �

is uniformly convergent on D z0; 1ð Þ; z0 2 D z0; 1ð Þ; and each map z 7! �1ð Þn z2n2nð Þ! is

continuous at z0; by a theorem (cf. [5], Theorem 7.12), the map

z 7! 1þ � z2

2!

� �þ z4

4!þ � � �

� �

is continuous at z0: Thus, the map

z 7! 1þ � z2

2!

� �þ z4

4!þ � � �

� �

from C to C is continuous. Similarly, the map

z 7! zþ � z3

3!

� �þ z5

5!þ � � �

� �

from C to C is continuous.

Conclusion 1.26 z 7! 1þ � z22!

� þ z4

4! þ � � ��

and z 7! zþ � z33!

� þ z5

5! þ � � ��

are continuous functions from C to C.

Note 1.27 We shall try to show that, for every real t;

exp itð Þ ¼ exp �itð Þ:


Let us fix any real t: Here,

exp itð Þ ¼ 1þ itþ itð Þ22!

þ itð Þ33!

þ � � � ¼ 1þ it � t2

2!� i

t3

3!þ � � �

¼ 1þ 0þ � t2

2!

� �þ 0þ � � �

� �þ 0þ itþ 0þ �i

t3

3!

� �þ � � �

� �

¼ 1þ 0þ � t2

2!

� �þ 0þ � � �

� �þ i 0þ tþ 0þ � t3

3!

� �þ � � �

� �

¼ 1þ � t2

2!

� �þ t4

4!þ � � �

� �þ i tþ � t3

3!

� �þ t5

5!þ � � �

� �;

so

LHS ¼ exp itð Þ ¼ 1þ � t2

2!

� �þ t4

4!þ � � �

� �� i tþ � t3

3!

� �þ t5

5!þ � � �

� �:

Similarly,

RHS ¼ exp �itð Þ ¼ exp i �tð Þð Þ

¼ 1þ � �tð Þ22!

!þ �tð Þ4

4!þ � � �

!þ i �tð Þþ � �tð Þ3

3!

!þ �tð Þ5

5!þ � � �

!

¼ 1þ � t2

2!

� �þ t4

4!þ � � �

� �� i tþ � t3

3!

� �þ t5

5!þ � � �

� �:

Hence exp itð Þ ¼ exp �itð Þ:Conclusion 1.28 For every real t; exp itð Þ ¼ exp �itð Þ:Definition For every z 2 C;

1þ � z2

2!

� �þ z4

4!þ � � �

is denoted by cos z: Thus, cos : C ! C: Clearly the restriction cosjR: R ! R:


Note 1.29 We want to show that, for every z 2 C; cos z ¼ 12 exp izð Þþ exp �izð Þð Þ:

Here,

RHS ¼ 12

exp izð Þþ exp �izð Þð Þ

¼ 12

1þ izþ izð Þ22!

þ izð Þ33!

þ � � � !

þ 1þ �izð Þþ �izð Þ22!

þ �izð Þ33!

þ � � � ! !

¼ 12

2þ 2izð Þ22!

þ 2izð Þ44!

þ � � � !

¼ 1þ izð Þ22!

þ izð Þ44!

þ � � �

¼ 1� z2

2!þ z4

4!� � � �

¼ cos z ¼ LHS:

Conclusion 1.30 For every z 2 C; cos z ¼ 12 exp izð Þþ exp �izð Þð Þ:

Definition For every z 2 C;

zþ � z3

3!

� �þ z5

5!þ � � �

is denoted by sin z: Thus, sin : C ! C: Clearly, the restriction sinjR: R ! R:

Note 1.31 We want to show that, for every z 2 C; sin z ¼ 12i exp izð Þ � exp �izð Þð Þ:

Here,

RHS ¼ 12i

exp izð Þ � exp �izð Þð Þ

¼ 12i

1þ izþ izð Þ22!

þ izð Þ33!

þ � � � !

� 1þ �izð Þþ �izð Þ22!

þ �izð Þ33!

þ � � � ! !

¼ 12i

2 izð Þþ 2izð Þ33!

þ 2izð Þ55!

þ � � � !

¼ z 1þ izð Þ23!

þ izð Þ45!

þ � � � !

¼ z� z3

3!þ z5

5!� � � �

¼ sin z ¼ LHS:

Conclusion 1.32 For every z 2 C; sin z ¼ 12i exp izð Þ � exp �izð Þð Þ:

Note 1.33 We shall try to show that, for every real t;a. exp itð Þ ¼ cos tð Þþ i sin tð Þ; b. cos tð Þ2 þ sin tð Þ2¼ 1:


For a: As in Note 1.27,

LHS ¼ exp itð Þ ¼ 1þ � t2

2!

� �þ t4

4!þ � � �

� �þ i tþ � t3

3!

� �þ t5

5!þ � � �

� �¼ cos tð Þþ i sin tð Þ ¼ RHS:

For b: By Note 1.27,

LHS ¼ cos tð Þ2 þ sin tð Þ2¼ cos tð Þþ i sin tð Þð Þ cos tð Þþ i � sin tð Þð Þ¼ cos tð Þþ i sin tð Þð Þ cos tð Þþ i sin tð Þð Þ�ð Þ ¼ exp itð Þð Þ exp itð Þð Þ�ð Þ¼ exp itð Þð Þ exp �itð Þð Þ ¼ exp it � itð Þ ¼ exp 0ð Þ ¼ 1 ¼ RHS:

Conclusion 1.34 For every real t; a. exp itð Þ ¼ cos tð Þþ i sin tð Þ; b.cos tð Þ2 þ sin tð Þ2¼ 1:

Notation Here, it is customary to write (b) as: cos2 tþ sin2 t ¼ 1:It is easy to see that, for every z;w 2 C;

(i) cos �zð Þ ¼ cos z;(ii) sin �zð Þ ¼ � sin z;(iii) cos2zþ sin2z ¼ 1;(iv) cos z� wð Þ ¼ cos z cos w� sin z sin w;(v) sin z� wð Þ ¼ sin z cos w� cos z sin w;(vi) sin 2z ¼ 2 sin z cos z;(vii) cos 2z ¼ 2 cos2z� 1 ¼ 1� 2 sin2z;(viii) exp izð Þ ¼ cos zþ i sin z:

Note 1.35 We shall try to show that, for every real t;

a. sinjR� �0

tð Þ ¼ cosjR� �

tð Þ;b. cosjR� �0

tð Þ ¼ � sinjR� �

tð Þ;c. sinjR; cosjR, expjR are continuous functions,d. expjR: R ! 0;1ð Þ is 1-1,e. expjR maps R onto 0;1ð Þ:

For a: For this purpose, let us fix any real t0: We have to show that

sinjR� �0

t0ð Þ ¼ cosjR� �

t0ð Þ:

We want to apply a theorem (cf. [5], Theorem 7.17). Here, each term of

tþ � t3

3!

� �þ t5

5!þ � � �

represents a differentiable function. Further, the series of their derivatives is


1þ � t2

2!

� �þ t4

4!þ � � � ¼ cosjR

� �tð Þ� �

;

which converges uniformly on the compact set t0 � 1; t0 þ 1½ �; by Conclusion 1.26.Again, by Conclusion 1.26,

t0 þ � t303!

� �þ t50

5!þ � � �

is convergent. Now, by a theorem, the map

t 7! tþ � t3

3!

� �þ t5

5!þ � � �

� �¼ sinjR� �

tð Þ� �is differentiable at t0; and

sinjR� �0

t0ð Þ ¼ 1þ � t202!

� �þ t40

4!þ � � � ¼ cosjR

� �t0ð Þ:

For b: Its proof is similar to that of (a).For c: Since sinjR is differentiable over R; sinjR is continuous over R: Similarly,

cosjR is continuous over R: By Conclusions 1.8, z 7! ez from C to C0 is continuous,so its restriction expjR: R ! 0;1ð Þ is continuous.

For d: By Conclusion 1.22, expjR: R ! 0;1ð Þ is strictly increasing, so expjR:R ! 0;1ð Þ is 1-1.

For e: Let us take any t 2 0;1ð Þ:Case I: when 1\t: Here

expjR� �

0ð Þ ¼ exp 0ð Þ ¼ 1\t\1þ tþ t2

2!þ � � � ¼ exp tð Þ ¼ expjR

� �tð Þ:

Since expjR� �

0ð Þ\t\ expjR� �

tð Þ; and expjR is a continuous function, by theintermediate value theorem, there exists r 2 R such that expjR

� �rð Þ ¼ t:

Case II: when 0\t\1: Here 1\ 1t : So, by Case I, there exists r 2 R such that

expjR� �

rð Þ ¼ 1t ; and hence,

t ¼ 1expjR� �

rð Þ ¼1

exp rð Þ ¼ exp �rð Þ ¼ expjR� � �rð Þ:

Thus, expjR� � �rð Þ ¼ t:

Case III: when t ¼ 1: Here

expjR� �

0ð Þ ¼ exp 0ð Þ ¼ 1 ¼ t:

So, in all cases, there exists r 2 R such that expjR� �

rð Þ ¼ t: Hence expjR mapsR onto 0;1ð Þ:


Conclusion 1.36 For every real t;

a. sinjR� �0

tð Þ ¼ cosjR� �

tð Þ;b. cosjR� �0


tð Þ;c. sinjR; cosjR, expjR are continuous functions,d. expjR: R ! 0;1ð Þ is 1-1,e. expjR maps R onto 0;1ð Þ:

Note 1.37 Observe that

cosjR� �

2ð Þ ¼ cos 2 ¼ 1þ � 22

2!

� �þ 24

4!þ � 26

6!

� �þ � � �

¼ 1þ � 22

2!

� �þ 24

4!

� �� 26

6!� 28

8!

� �� 210

10!� 212

12!

� ��

¼ 1þ � 22

2!

� �þ 24

4!

� �� 26

6!1� 22

7 8

� �� 210

10!1� 22

11 12

� ��

\ 1þ � 22

2!

� �þ 24

4!

� �¼ �1þ 2

3¼ � 1

3\0:

Thus, cosjR� �

2ð Þ\0: Also,

cosjR� �

1ð Þ ¼ cos 1 ¼ 1þ � 12

2!

� �þ 14

4!þ � 16

6!

� �þ � � �

¼ 1þ � 12!

� �þ 1

4!þ � 1

6!

� �þ � � �

¼ 1þ � 12!

� �� þ 1

4!� 16!

� �þ 1

8!� 110!

� �þ � � �

[ 1þ � 12!

� �� ¼ 1

2[ 0:

Since cosjR� �

2ð Þ\0\ cosjR� �

1ð Þ; and, by Conclusion 1.36(c), cosjR is con-tinuous over R; by the intermediate value theorem, there exists p0 2 1; 2ð Þ such thatcos p0 ¼ð Þ cosjR

� �p0ð Þ ¼ 0:

Conclusion 1.38 There exists p0 2 1; 2ð Þ such that cos p0 ¼ 0:

Note 1.39 Let A � p : p[ 0 and cos p ¼ 0f g:By Conclusion 1.38, A is nonempty. Clearly, 0 is a lower bound of A: It follows

that inf A exists and 0� inf A:


Problem 1.40 inf A 6¼ 0:

(Solution If not, otherwise, let inf A ¼ 0: We have to arrive at a contradiction.Since inf A ¼ 0; there exists a sequence tnf g in A such that limn!1 tn ¼ inf Að Þ ¼0: Since each tn 2 A; each cos tn becomes 0: By Conclusion 1.36(c), cosjR iscontinuous and limn!1 tn ¼ 0; so

0 ¼ limn!1 0 ¼ lim

n!1 cos tn ¼ limn!1 cosjR

� �tnð Þ ¼ cosjR

� �0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ cos 0 ¼ 1;

which is a contradiction. ■)Since inf A 6¼ 0; and 0� inf A; we have

0\ inf A ¼ inf p : p[ 0 and cos p ¼ 0f g:Conclusion 1.41 0\ inf p : p[ 0 and cos p ¼ 0f g:Definition The positive real number inf p : p[ 0 and cos p ¼ 0f g is denoted by p

2 :

Thus

p2� inf p : p[ 0 and cos p ¼ 0f g:

Note 1.42 We want to show thata. 0\p\4; b. p is the smallest positive real number such that cos p2 ¼ 0:For a: By Conclusion 1.41, 0\ p

2 : Also, by Conclusion 1.38,

p2¼ inf p : p[ 0 and cos p ¼ 0f g\2:

Thus 0\p\4:For b: We shall try to show that p

2 2 p : p[ 0 and cos p ¼ 0f g: Put

A � p : p[ 0 and cos p ¼ 0f g:

Since inf A ¼ p2 ; there exists a sequence tnf g in A such that limn!1 tn ¼

inf Að Þ ¼ p2 : Since each tn 2 A; each cos tn is 0: By Conclusion 1.36(c), cosjR is

continuous, and limn!1 tn ¼ p2 ; so

0 ¼ limn!1 0 ¼ lim

n!1 cos tn ¼ limn!1 cosjR

� �tnð Þ ¼ cosjR

� � p2

� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ cos

p2:


Since 0\ p2 ; and cos p2 ¼ 0; we have p

2 2 p : p[ 0 and cos p ¼ 0f g: Since

inf p : p[ 0 and cos p ¼ 0f g ¼ p22 p : p[ 0 and cos p ¼ 0f g;

we have

min p : p[ 0 and cos p ¼ 0f g ¼ p2:

Thus, p is the smallest positive real number such that cos p2 ¼ 0:

Conclusion 1.43 a. 0\p\4; b. p is the smallest positive real number such thatcos p2 ¼ 0:

Note 1.44 We want to show that, for every t 2 0; p4� �

; 12\ cos t� 1:

Let us fix any t 2 0; p4� �

: Here, by Conclusion 1.43(a), t 2 0; p4� � 0; 1ð Þ; so

t2 2 0; 1ð Þ; and hence,

cos t ¼ 1þ � t2

2!

� �þ t4

4!þ � t6

6!

� �þ � � �

¼ 1� t2

2!

� �þ t4

4!1� t2

5 6

� �þ t8

8!1� t2

9 10

� �þ � � �

[ 1� t2

2!

� �[

12:

Thus, 12\ cos t: Since cos2 t� cos2 tþ sin2 t ¼ 1; we have cos t� 1:

Conclusion 1.45 For every t 2 0; p4� �

; 12\ cos t� 1:

Note 1.46 We want to show that t 7! cosjR� �

tð Þ is strictly decreasing over 0; p4� �

:

Let us take any t 2 0; p4� �

: Here, by Conclusion 1.43(a), t 2 0; p4� � 0; 1ð Þ; so

t2 2 0; 1ð Þ; and hence

cosjR� �0


tð Þ ¼ � sin tð Þ ¼ � tþ � t3

3!

� �þ t5

5!þ � � �

� �

¼ �tþ t3

3!

� �� t5

5!1� t2

6 7

� �� t9

9!1� t2

10 11

� �� \ �tþ t3

3!

� �:

Now, since x 7! �xþ x33!

� is a strictly decreasing polynomial over 0; 1ð Þ; for

every t 2 0; p4� �

; we have

cosjR� �0

tð Þ\ �tþ t3

3!

� �� xþ x3

3!

� ��x¼0

¼ 0;


and hence for every t 2 0; p4� �

; cosjR� �0

tð Þ\0: It follows that t 7! cosjR� �

tð Þ isstrictly decreasing over 0; p4

� �:

Conclusion 1.47 t 7! cosjR� �

tð Þ is strictly decreasing over 0; p4� �

:

Note 1.48 We want to show that, for every t 2 0; p4� �

; 0\ sin t:If not, otherwise, suppose that there exists t 2 0; p4

� �such that sin t� 0:We have

to arrive at a contradiction. Since sin t� 0; we have

0� � sin t ¼ � sinjR� �

tð Þ� � ¼ cosjR� �0

tð Þ:

Since t 2 0; p4� �

; by Conclusion 1.47, cosjR� �0

tð Þ\0: This is a contradiction.

Conclusion 1.49 For every t 2 0; p4� �

; 0\ sin t:

Note 1.50 We shall try to show:

a. cos p4 ¼ 1ffiffi2

p ;

b. sin p4 ¼ 1ffiffi

2p ;

c. sin p2 ¼ 1;

d. sin p ¼ 0;e. cos p ¼ �1;f. sin 2p ¼ 0;g. cos 2p ¼ 1;h. For every integer n;

(i) sin np ¼ 0;(ii) cos 2np ¼ 1;(iii) e2npi ¼ 1:

For a: Here, 0 ¼ cos p2 ¼ cos 2 p4

� � ¼ 2 cos p4� �2�1; so cos p4 ¼ 1ffiffi

2p or cos p4 ¼

� 1ffiffi2

p : There exists a convergent sequence tnf g in 0; p4� �

such that limn!1 tn ¼ p4 :

Since cosjR is continuous over R; we have limn!1 cos tn ¼ cos p4 : Since each tn isin 0; p4� �

; by Conclusion 1.45, each cos tn [ 12 ; and hence

12� lim

n!1 cos tn�

¼ cosp4:

It follows that 12 � cos p4 : Since

12 � cos p4 ; and cos p4 ¼ 1ffiffi

2p or cos p4 ¼ � 1ffiffi

2p

� ;

we have cos p4 ¼ 1ffiffi2

p :

For b: Here, 0 ¼ cos p2 ¼ cos 2 p4

� � ¼ 1� 2 sin p4

� �2; so sin p

4 ¼ 1ffiffi2

p or sin p4 ¼

� 1ffiffi2

p : There exists a convergent sequence tnf g in 0; p4� �

such that limn!1 tn ¼ p4 :

Since sinjR is continuous over R; we have limn!1 sin tn ¼ sin p4 : Since each tn is in


0; p4� �

; by Conclusion 1.49, sin tn [ 0; and hence 0� limn!1 sin tn ¼ sin p4

� �: It

follows that 0� sin p4 : Since 0� sin p

4 ; and sin p4 ¼ 1ffiffi

2p or sin p

4 ¼ � 1ffiffi2

p�

; we have

sin p4 ¼ 1ffiffi

2p :

For c: Here

LHS ¼ sinp2¼ sin 2

p4

� ¼ 2 sin

p4cos

p4¼ 2 � 1ffiffiffi

2p � 1ffiffiffi

2p ¼ 1 ¼ RHS:

For d: Here

LHS ¼ sin p ¼ sin 2p2

� ¼ 2 sin

p2cos

p2¼ 2 1ð Þ 0ð Þ ¼ 0 ¼ RHS:

Similarly, all other proofs can be supplied.

Conclusion 1.51

a. cos p4 ¼ 1ffiffi2

p ;

b. sin p4 ¼ 1ffiffi

2p ;

c. sin p2 ¼ 1;

d. sin p ¼ 0;e. cos p ¼ �1;f. sin 2p ¼ 0;g. cos 2p ¼ 1;h. For every integer n;

(i) sin np ¼ 0;(ii) cos 2np ¼ 1;(iii) e2npi ¼ 1:

Note 1.52 We shall try to prove:a. For every z 2 C; if z

2pi is an integer then ez ¼ 1; b. If ez ¼ 1 then z2pi is an

integer.For a: Let z � xþ iy; where x; y 2 R: Let z

2pi ¼ n 2 Z: Now, by Conclusion1.51,

LHS ¼ ez ¼ e2pin ¼ 1 ¼ RHS:

For b: Let ez ¼ 1; where z � xþ iy; and x; y 2 R: Here

1 ¼ ez ¼ exþ iy ¼ exeiy ¼ ex cos yð Þþ i sin yð Þð Þ;


so,

1 ¼ 1j j ¼ ex cos yð Þþ i sin yð Þð Þj j ¼ exj j cos yð Þþ i sin yð Þj j¼ exj j

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficos2 yþ sin2 y

q¼ exj j

ffiffiffi1

p¼ exj j ¼ ex;

and hence, ex ¼ 1: Since ex ¼ 1; and 1 ¼ ex cos yð Þþ i sin yð Þð Þ; we have

cos yð Þþ i sin yð Þ ¼ 1 ¼ 1þ i0;

and hence, cos y ¼ 1 and sin y ¼ 0: Since ex ¼ 1 ¼ e0; and by Conclusion 1.36,expjR: R ! 0;1ð Þ is 1-1, we have x ¼ 0: Now

z2pi

¼ xþ iy2pi

¼ 0þ iy2pi

¼ y2p

:

It suffices to show that y2p is an integer. If not, otherwise, let y

2p � nþ r; wheren 2 Z; and r 2 0; 1ð Þ: We have to arrive at a contradiction. Here

1 ¼ ez ¼ exþ iy ¼ e0þ iy ¼ eiy

¼ ei2p nþ rð Þ ¼ ei2npþ i2rp

¼ ei2npei2rp ¼ 1ei2rp ¼ ei2rp

¼ cos 2rpþ i sin 2rp:

Since, cos 2rpþ i sin 2rp ¼ 1 ¼ 1þ i0; we have cos 2rp ¼ 1; and sin 2rp ¼0: Since cos 2rp ¼ 1; we have

0 ¼ 1� cos 2rp ¼ 2 sin2 rp;

and hence sin rp ¼ 0: Since sin rp ¼ 0; we have 2 sin rp2 cos

rp2 ¼ 0: It follows that

sin rp2 ¼ 0 or cos rp2 ¼ 0: Since r 2 0; 1ð Þ; we have 0\ rp

2 \p2 ; and therefore, by

Conclusion 1.43, cos rp2 6¼ 0: Since cos rp2 6¼ 0; and sin rp2 ¼ 0 or cos rp2 ¼ 0

� �; we

have sin rp2 ¼ 0: Now, since 0 ¼ 2 sin rp

4 cosrp4 ; we have sin rp

4 ¼ 0 or cos rp4 ¼ 0:Since 0\ rp

4 \p4 ; we have, by Conclusion 1.45, 1

2\ cos rp4 � 1: Since12\ cos rp4 � 1 and sin rp

4 ¼ 0 or cos rp4 ¼ 0� �

; we have sin rp4 ¼ 0: Since 0\ rp

4 \p4 ;

we have, by Conclusion 1.49, 0\ sin rp4 ; and hence sin rp

4 6¼ 0; a contradiction.

Conclusion 1.53 a. For every z 2 C; if z2pi is an integer then ez ¼ 1; b. If ez ¼ 1

then z2pi is an integer.

Note 1.54 By Conclusions 1.45 and 1.49, cos; sin are positive over 0; p4� �

: Now letus take any t 2 p

4 ;p2

� �: It follows that p

2 � t� � 2 0; p4

� �; and hence

0\ cosp2� t

� |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} ¼ cos

p2cos tþ sin

p2sin t ¼ 0 cos tþ 1 sin t ¼ sin t:


Thus, sin is positive over 0; p2� �

: Similarly cos is positive over 0; p2� �

; sin ispositive over p

2 ; p� �

; cos is negative over p2 ; p� �

; sin is negative over p; 3p2� �

; cos isnegative over p; 3p2

� �; sin is negative over 3p

2 ; 2p� �

; and cos is positive over3p2 ; 2p� �

:

Conclusion 1.55 sin is positive over 0; p2� �

; cos is positive over 0; p2� �

; sin ispositive over p

2 ; p� �

; cos is negative over p2 ; p� �

; sin is negative over p; 3p2� �

; cos isnegative over p; 3p2

� �; sin is negative over 3p

2 ; 2p� �

; and cos is positive over3p2 ; 2p� �

:

Note 1.56 We shall try to prove: The map t 7! eit from R to the unit circlez : z 2 C and zj j ¼ 1f g is onto.By Conclusion 1.34, t 7! eit is a mapping from R to the unit circle

z : z 2 C and zj j ¼ 1f g: It remains to show that this map is onto. For this purpose,let us take any xþ iyð Þ 2 C such that x; y 2 R; and

ffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p¼ 1: We have to find

a real number t such that cos tþ i sin t ¼ xþ iy; that is, cos t ¼ x; and sin t ¼ y:

Sinceffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 þ y2

p¼ 1; we have

xj j2¼ x2 � x2 þ y2 ¼ 1;

and hence xj j � 1: Since xj j � 1; we have x 2 �1; 1½ �: Similarly, y 2 �1; 1½ �:Case I: when 0\y\1: Since x 2 �1; 1½ �; cosjR: R ! R is continuous,

cosjR� �

0ð Þ ¼ cos 0 ¼ 1; and cosjR� �

pð Þ ¼ cos p ¼ �1; by the intermediate valuetheorem, there exists t 2 0; p½ � such that cos t ¼ cosjR

� �tð Þ ¼ x: Here t 2 0; p½ �; so

by Conclusion 1.55, 0� sin t; and hence sin tj j ¼ sin t: Here 0\y\1; andx2 þ y2 ¼ 1; so

y ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� cosjR

� �tð Þ� �2q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� cos tð Þ2

q¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffisin tð Þ2

q¼ sin tj j ¼ sin t:

Thus, cos tþ i sin t ¼ xþ iy:Case II: when �1\y\0: Since x 2 �1; 1½ �; cosjR: R ! R is continuous,

cosjR� �

2pð Þ ¼ cos 2p ¼ 1; and cosjR� �

pð Þ ¼ cos p ¼ �1; by the intermediatevalue theorem, there exists t 2 p; 2p½ � such that cos t ¼ cosjR

� �tð Þ ¼ x: Here t 2

p; 2p½ �; so by Conclusion 1.55, sin t� 0; and hence sin tj j ¼ � sin t: Here�1\y\0; and x2 þ y2 ¼ 1; so

�y ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� cosjR

� �tð Þ� �2q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� cos tð Þ2

q¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffisin tð Þ2

q¼ sin tj j ¼ � sin t:


Thus, in both cases, there exists a real t such that cos tþ i sin t ¼ xþ iy: Also,cos 0þ i sin 0 ¼ 1þ i0; cos p2 þ i sin p

2 ¼ 0þ i1; cos pþ i sin p ¼ �1þ i0; andcos 3p2 þ i sin 3p

2 ¼ 0þ i �1ð Þ:Thus, the map t 7! eit from R to the unit circle z : z 2 C and zj j ¼ 1f g is onto.

Conclusion 1.57 The map t 7! eit from R to the unit circle z : z 2 C and zj j ¼ 1f gis onto.

Note 1.58 We shall try to prove: The map z 7! ez from C to C0 � C� 0f gð Þ isonto.

By Conclusion 1.10, z 7! ez is a mapping from C to C� 0f g: We have to showthat this map is onto. For this purpose, let us take a nonzero complex number z: Itfollows that zj j 2 0;1ð Þ; and z

zj j 2 z : z 2 C and zj j ¼ 1f g: Since zj j 2 0;1ð Þ; and,by Conclusion 1.36, expjR: R ! 0;1ð Þ is onto, there exists x 2 R such thatex ¼ð Þ expjR

� �xð Þ ¼ zj j: Since z

zj j 2 z : z 2 C and zj j ¼ 1f g; and, by Conclusion

1.57, the map t 7! eit from R to w : w 2 C and wj j ¼ 1f g is onto, there exists y 2 R

such that eiy ¼ zzj j : Since eiy ¼ z

zj j ; and ex ¼ zj j; we have

exþ iy ¼ exeiy ¼ zj j zzj j ¼ z;

where xþ iyð Þ 2 C: Thus, e xþ iyð Þ ¼ z: It follows that the map z 7! ez from C to C0

� C� 0f gð Þð Þ is onto.Conclusion 1.59 The map z 7! ez from C to C0 � C� 0f gð Þ is onto.

1.2 Measurable Functions

If we compare measure theory with topology, we notice an analogy: the counter-parts of measure spaces, measurable sets and measurable functions are topologicalspaces, open sets and continuous functions respectively. However, the concept ofmeasure has no counterpart in topology.

Definition Let X be a nonempty set. Let ℳ be any collection of subsets of X: If

1. X 2 ℳ;2. if A 2 ℳ; then the complement Ac 2 ℳ;3. ℳ is closed with respect to countable union, in the sense that, if A1;A2; . . . are

in ℳ then A1 [A2 [ � � � is in ℳ;

then we say that ℳ is a r-algebra in X: Here members of ℳ are called themeasurable sets, and X is called the measurable space.


Lemma 1.60 Let X be a nonempty set. Let ℳ be a r-algebra in X: Then

1. the empty set ; is in ℳ;2. ℳ is closed with respect to countable intersection,3. ℳ is closed with respect to finite union,4. ℳ is closed with respect to finite intersection,5. ℳ is closed with respect to difference,6. ℳ is closed with respect to symmetric difference.

Proof

1. Since X 2 ℳ; ; ¼ð ÞXc is in ℳ; and hence ; is in ℳ:2. Let A1;A2; . . . be in ℳ: We have to show that A1 \A2 \ � � � is in ℳ: Since

A1;A2; . . . are in ℳ; then Ac1;A

c2; . . . are in ℳ; and hence

A1 \A2 \ � � �ð Þc¼ð ÞAc1 [Ac

2 [ � � �

is in ℳ; which in turn implies that A1 \A2 \ � � �ð Þc is in ℳ: It follows that

A1 \A2 \ � � � ¼ð Þ A1 \A2 \ � � �ð Þcð Þc

is in ℳ; and hence A1 \A2 \ � � � is in ℳ:3. Let A1;A2; . . .;An be in ℳ: We have to show that A1 [A2 [ � � � [An is in ℳ:

Since A1;A2; . . .;An; ;; ;; . . . are in ℳ;

A1 [A2 [ � � � [An ¼ð ÞA1 [A2 [ � � � [An [;[ ;[ � � �

is in ℳ; and hence, A1 [A2 [ � � � [An is in ℳ:4. Its proof is similar to Proof 3.5. Let A;B 2 ℳ: We have to show that A� B 2 ℳ: Since B 2 ℳ; we have

Bc 2 ℳ: Since A;Bc 2 ℳ; by Proof 4, A� B ¼ð ÞA\ Bcð Þ 2 ℳ; and hence,A� B 2 ℳ:

6. Let A;B 2 ℳ: We have to show that A D B 2 ℳ: Since A;B 2 ℳ; we have, byProof 5, A� Bð Þ 2 ℳ: Similarly, B� Að Þ 2 ℳ: Since A� Bð Þ 2 ℳ; andB� Að Þ 2 ℳ; we have, by Proof 3,

ADB ¼ð Þ A� Bð Þ [ B� Að Þ 2 ℳ;

and hence ADB 2 ℳ: ■

Definition Let X be a nonempty set. Let ℳ be a r-algebra in X: Let Y be atopological space. Let f : X ! Y :

If, for every open set V in Y ; f�1 Vð Þ 2 ℳ; then we say that f : X ! Y is ameasurable function.


Lemma 1.61 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let Y ; Z betopological spaces. Let f : X ! Y be a measurable function. Let g : Y ! Z be acontinuous function. Then their composite g � f : X ! Z is a measurable function.

Proof For this purpose, let us take any open set W in Z: We have to show that

f�1 g�1 Wð Þ� � ¼ f�1� �g�1� �

Wð Þ� � ¼ f�1� � � g�1� �� Wð Þ ¼ g � fð Þ�1 Wð Þ 2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is, f�1 g�1 Wð Þð Þ 2 ℳ: Since g : Y ! Z; and W is an open set in Z; g�1 Wð Þ isopen in Y : Since g�1 Wð Þ is open in Y ; and f : X ! Y is a measurable function,f�1 g�1 Wð Þð Þ 2 ℳ: ■

Lemma 1.62

1. Let X be a nonempty set. Let ℳ be a r-algebra in X: Let u : X ! R be ameasurable function. Let v : X ! R be a measurable function. Let Y be atopological space. Let U : R2 ! Y be a continuous map. Then the map

h : x 7!U u xð Þ; v xð Þð Þ

from X to Y is a measurable function.2. Let X be a nonempty set. Let ℳ be a r-algebra in X: Let u : X ! C be a

measurable function. Let v : X ! C be a measurable function. Let Y be atopological space. Let U : C2 ! Y be a continuous map. Then the map h :x 7!U u xð Þ; v xð Þð Þ from X to Y is a measurable function.

Proof

1. We first try to show that the map g : x 7! u xð Þ; v xð Þð Þ from X to R2 is a mea-surable function.

For this purpose, let us take any nonempty open set V of R2: We have to showthat g�1 Vð Þ 2 ℳ: Since V is a nonempty open set of R2; there exists a countablecollection of open rectangles

a1; b1ð Þ c1; d1ð Þ; a2; b2ð Þ c2; d2ð Þ; . . .

such that

V ¼ a1; b1ð Þ c1; d1ð Þð Þ [ a2; b2ð Þ c2; d2ð Þð Þ [ � � � :

1.2 Measurable Functions 23

Here

g�1 Vð Þ ¼ g�1 a1; b1ð Þ c1; d1ð Þð Þ [ a2; b2ð Þ c2; d2ð Þð Þ [ � � �ð Þ¼ g�1 a1; b1ð Þ c1; d1ð Þð Þ� �[ g�1 a2; b2ð Þ c2; d2ð Þð Þ� �[ � � � :

Problem 1:63 g�1 a1; b1ð Þ c1; d1ð Þð Þ ¼ u�1 a1; b1ð Þð Þð Þ \ v�1 c1; d1ð Þð Þð Þ:(Solution Let x 2 LHS: It follows that

u xð Þ; v xð Þð Þ ¼ð Þg xð Þ 2 a1; b1ð Þ c1; d1ð Þ;

and hence u xð Þ 2 a1; b1ð Þ; and v xð Þ 2 c1; d1ð Þ: This shows that x 2 u�1 a1; b1ð Þð Þ;and x 2 v�1 c1; d1ð Þð Þ; and hence x 2 RHS: Thus, LHS RHS: Now let y 2 RHS:It follows that y 2 u�1 a1; b1ð Þð Þ; and y 2 v�1 c1; d1ð Þð Þ: Now u yð Þ 2 a1; b1ð Þ; andv yð Þ 2 c1; d1ð Þ; and hence

g yð Þ ¼ð Þ u yð Þ; v yð Þð Þ 2 a1; b1ð Þ c1; d1ð Þ:

Therefore, g yð Þ 2 a1; b1ð Þ c1; d1ð Þ: This shows that y 2 LHS: Thus, RHS LHS: This proves LHS ¼ RHS: ■)

Since u : X ! R is a measurable function, and a1; b1ð Þ is open in R;

u�1 a1; b1ð Þð Þ 2 ℳ: Similarly, v�1 c1; d1ð Þð Þ 2 ℳ: Now, by Lemma 1.60,

g�1 a1; b1ð Þ c1; d1ð Þð Þ ¼� �u�1 a1; b1ð Þð Þ� �\ v�1 c1; d1ð Þð Þ� � 2 ℳ;

and hence g�1 a1; b1ð Þ c1; d1ð Þð Þ 2 ℳ: Similarly, g�1 a2; b2ð Þ c2; d2ð Þð Þ 2ℳ; etc. Since

g�1 a1; b1ð Þ c1; d1ð Þð Þ 2 ℳ; g�1 a2; b2ð Þ c2; d2ð Þð Þ 2 ℳ; . . .;

and ℳ is a r-algebra in X;

g�1 Vð Þ ¼� �g�1 a1; b1ð Þ c1; d1ð Þð Þ� �[ g�1 a2; b2ð Þ c2; d2ð Þð Þ� �[ � � �

is in ℳ; and therefore g�1 Vð Þ 2 ℳ:

Thus, we have shown that x 7! u xð Þ; v xð Þð Þ from X to R2 is a measurablefunction. Since x 7! u xð Þ; v xð Þð Þ from X to R2 is a measurable function, and U :

R2 ! Y is a continuous map, by Lemma 1.61, their composite

x 7!U u xð Þ; v xð Þð Þ ¼ h xð Þð Þ

is a measurable function, and hence h : x 7!U u xð Þ; v xð Þð Þ from X to Y is a mea-surable function.

2. Its proof is similar to Proof 1. ■


Lemma 1.64 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let u : X ! R

be a measurable function. Let v : X ! R be a measurable function. Then

uþ ivð Þ : t 7! u tð Þð Þþ i v tð Þð Þ

from X to C is a measurable function.

Proof Observe that the map U : x; yð Þ 7! xþ iyð Þ from R2 to C is continuous. Now,by Lemma 1.62,

x 7!U u xð Þ; v xð Þð Þ ¼ u xð Þð Þþ i v xð Þð Þ ¼ uþ ivð Þ xð Þð Þ

is a measurable function, and hence uþ ivð Þ is a measurable function. ■

Lemma 1.65 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let u : X ! R;and v : X ! R: Let the function

uþ ivð Þ : t 7! u tð Þð Þþ i v tð Þð Þ

from X to C be measurable. Then

1. the map u is a measurable function,2. the map v is a measurable function,3. the map uþ ivj j : t 7! u tð Þð Þþ i v tð Þð Þj j from X to R is a measurable function.

Proof

1. Observe that the map g : xþ iyð Þ 7! x from C to R is continuous. Now, byLemma 1.61,

t 7! g u tð Þð Þþ i v tð Þð Þð Þ ¼ u tð Þð Þ

is a measurable function, and hence u is a measurable function.2. Its proof is similar to Proof 1.3. Observe that the map g : xþ iyð Þ 7! xþ iyj j from C to R is continuous. Now, by

Lemma 1.61,

t 7! g u tð Þð Þþ i v tð Þð Þð Þ ¼ u tð Þð Þþ i v tð Þð Þj j ¼ uþ ivj j tð Þð Þ

is a measurable function, and hence uþ ivj j is a measurable function. ■

Lemma 1.66 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let f : X ! C

be a measurable function. Let g : X ! C be a measurable function. Then

1. f þ gð Þ : t 7! f tð Þþ g tð Þð Þ from X to C is a measurable function,2. f � gð Þ : t 7! f tð Þð Þ g tð Þð Þ from X to C is a measurable function.


Proof

1. We know that þ : z;wð Þ 7! zþwð Þ from C2 to C is continuous. Now, byLemma 1.62,

t 7! þ f tð Þ; g tð Þð Þ ¼ f tð Þþ g tð Þ ¼ f þ gð Þ tð Þð Þ

is a measurable function, and hence f þ gð Þ is a measurable function.2. We know that the product � : z;wð Þ 7! z � wð Þ from C2 to C is continuous. Now,

by Lemma 1.62,

t 7! � f tð Þ; g tð Þð Þ ¼ f tð Þð Þ � g tð Þð Þ ¼ f � gð Þ tð Þð Þ

is a measurable function, and hence f � gð Þ is a measurable function. ■

Lemma 1.67 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let E 2 ℳ: LetvE : X ! R be the function defined as follows: for every x 2 X;

vE xð Þ � 1 ifx 2 E0 ifx 62 E:

�Here vE is called the characteristic function of E:ð Þ

Then vE is a measurable function.

Proof Let U be any nonempty open subset of R: We have to show thatvEð Þ�1 Uð Þ 2 ℳ:Case I: when 1 2 U and 0 2 U: Here

vEð Þ�1 Uð Þ ¼ vEð Þ�1 1ð Þ�

[ vEð Þ�1 0ð Þ�

¼ E [ Ecð Þ ¼ X 2 ℳ:

Case II: when 1 2 U and 0 62 U: Here


¼ E 2 ℳ:

Case III: when 0 2 U and 1 62 U: Here


¼ Ec 2 ℳ:

Case IV: when 1 62 U and 0 62 U: Here

vEð Þ�1 Uð Þ ¼ ; 2 ℳ:

So, in all cases, vEð Þ�1 Uð Þ 2 ℳ: ■

Lemma 1.68 Let X be a nonempty set. Let ℳ be a r-algebra in X: Let f : X ! C

be a measurable function. Then there exists a function a : X ! C such that


1. a : X ! C is a measurable function,2. aj j ¼ 1;3. f ¼ a � fj j:

Proof Case I: when f xð Þ is nonzero for every x 2 X: Let us define a : X ! C asfollows: for every t 2 X; a tð Þ � 1

f tð Þj j f tð Þð Þ: Since the map z 7! 1zj j zð Þ is continuous

from C0 to C0; and f : X ! C0 is a measurable function, by Lemma 1.61 the map

t 7! 1f tð Þj j f tð Þð Þ ¼ a tð Þð Þ

is a measurable function, and hence a : X ! C is a measurable function. Thus, 1holds. 2, 3 are clear.

Case II: when f xð Þ ¼ 0 for some x 2 X: Since C0 is an open subset of C; andf : X ! C is a measurable function, we have f�1 C0ð Þ 2 ℳ: It follows thatf�1 C0ð Þð Þc2 ℳ: Put

ℳ� � A\ f�1 C0ð Þ� �: A 2 ℳ

� :

Problem 1:69 ℳ� is a r-algebra in f�1 C0ð Þ:(Solution

1. Since f�1 C0ð Þ 2 ℳ;

f�1 C0ð Þ ¼� �f�1 C0ð Þ� �\ f�1 C0ð Þ� � 2 ℳ�;

and hence f�1 C0ð Þð Þ 2 ℳ�:2. Take any A\ f�1 C0ð Þð Þ 2 ℳ�; where A 2 ℳ: We have to show that

f�1 C0ð Þ� �� A ¼� �f�1 C0ð Þ� �� A\ f�1 C0ð Þ� �� 2 ℳ�;

that is, f�1 C0ð Þð Þ � Að Þ 2 ℳ: Here A; f�1 C0ð Þð Þ 2 ℳ; so by Lemma 1.60,f�1 C0ð Þð Þ � Að Þ 2 ℳ:

3. Let

A1 \ f�1 C0ð Þ� �; A2 \ f�1 C0ð Þ� �

; . . .

be in ℳ�; where A1;A2; . . . are in ℳ: We have to show that

A1 [A2 [ � � �ð Þ \ f�1 C0ð Þ� � ¼� �A1 \ f�1 C0ð Þ� �� [ A2 \ f�1 C0ð Þ� �� [ � � �

is in ℳ�; that is

A1 [A2 [ � � �ð Þ \ f�1 C0ð Þ� � 2 ℳ�:

Since A1;A2; . . . are inℳ; and ℳ is a r-algebra, we have A1 [A2 [ � � �ð Þ 2 ℳ;and hence


A1 [A2 [ � � �ð Þ \ f�1 C0ð Þ� � 2 ℳ�:

Thus, ℳ� is a r-algebra in f�1 C0ð Þ: ■)

Problem 1:70 ℳ� ℳ:

(Solution Let A\ f�1 C0ð Þð Þ 2 ℳ�; where A 2 ℳ: We have to show thatA\ f�1 C0ð Þð Þ 2 ℳ: Since A 2 ℳ; and f�1 C0ð Þ 2 ℳ; by Lemma 1.60,A\ f�1 C0ð Þð Þ 2 ℳ: ■)

Problem 1:71 The restriction f jf�1 C0ð Þ : f�1 C0ð Þ ! C0 is a measurable function.

(Solution Let us take any nonempty open subset U of C0: We have to show that

f�1 Uð Þ ¼� �f jf�1 C0ð Þ� �1

Uð Þ 2 ℳ�;

that is, f�1 Uð Þ 2 ℳ�: Since U is open in C0; and C0 is open in C; U is open in C:

Since U is open in C; and f : X ! C is a measurable function, we have f�1 Uð Þ 2ℳ; and hence

f�1 Uð Þ ¼ f�1 U \C0ð Þ ¼ f�1 Uð Þ� �\ f�1 C0ð Þ� � 2 ℳ�|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :It follows that f�1 Uð Þ 2 ℳ�: ■)Now, by Case I, there exists a function b : f�1 C0ð Þ ! C such that

1′. b : f�1 C0ð Þ ! C is a measurable function,2′. bj j ¼ 1;

3′. f jf�1 C0ð Þ�

¼ b � f jf�1 C0ð Þ� �� :

Let us define a function a : X ! C as follows: for every t 2 X;

a tð Þ � b tð Þ if t 2 f�1 C0ð Þ1 if t 62 f�1 C0ð Þ:

�

Clearly,

a�1 1ð Þ f�1 C0ð Þ� �c¼ f�1 C0ð Þcð Þ ¼ f�1 0ð Þ:

We have to show:

a. aj j ¼ 1;b. f ¼ a � fj j;c. a : X ! C is a measurable function.


For a: Let us take any t 2 X: We have to show that a tð Þj j ¼ð Þ aj j tð Þ ¼ 1; that is,a tð Þj j ¼ 1:Case I: when t 2 f �1 C0ð Þ: Here a tð Þ ¼ b tð Þ: Now by 20,

LHS ¼ a tð Þj j ¼ b tð Þj j ¼ bj j tð Þ ¼ 1 ¼ RHS:

Case II: when t 62 f�1 C0ð Þ: Here

LHS ¼ a tð Þj j ¼ 1j j ¼ 1 ¼ RHS:

This proves a.For b: Let us take any t 2 X: We have to show that

f tð Þ ¼ a � fj jð Þ tð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a tð Þð Þ fj j tð Þð Þ ¼ a tð Þð Þ f tð Þj j;

that is, f tð Þ ¼ a tð Þð Þ f tð Þj j:Case I: when t 2 f �1 C0ð Þ: Here a tð Þ ¼ b tð Þ: Now by 30;

RHS ¼ a tð Þð Þ f tð Þj jð Þ¼ b tð Þð Þ f tð Þj jð Þ¼ b tð Þð Þ f jf�1 C0ð Þ

� tð Þ

�� ¼ b tð Þð Þ f jf�1 C0ð Þ

� �� tð Þ� ¼ b � f jf�1 C0ð Þ

� �� tð Þ

¼ f jf�1 C0ð Þ�

tð Þ¼ f tð Þ ¼ LHS:

Case II: when t 62 f�1 C0ð Þ. Since t 62 f�1 C0ð Þ, we have f tð Þ 62 C0, and hence,f tð Þ ¼ 0. Now

RHS ¼ a tð Þð Þ f tð Þj jð Þ ¼ a tð Þð Þ 0j j ¼ 0 ¼ f tð Þ ¼ LHS:

This proves b.For c: Let us take any nonempty open subset U of C: We have to show that

a�1 Uð Þ 2 ℳ: Observe that

a�1 Uð Þ ¼ a�1 Uð Þ \X ¼ a�1 Uð Þ \ f�1 C0ð Þ [ f�1 0ð Þ� �¼ a�1 Uð Þ \ f�1 C0ð Þ� �[ a�1 Uð Þ \ f�1 0ð Þ� �

:


It suffices to show that(ci) a�1 Uð Þ \ f�1 C0ð Þð Þ 2 ℳ; and (cii) a�1 Uð Þ \ f�1 0ð Þð Þ 2 ℳ:For (ci):

Problem 1:72 a�1 Uð Þ \ f�1 C0ð Þ ¼ b�1 Uð Þ:(Solution Let us take any t 2 LHS; that is t 2 a�1 Uð Þ \ f�1 C0ð Þ: We shall try toshow that t 2 RHS; that is, t 2 b�1 Uð Þ; that is, b tð Þ 2 U: Since t 2a�1 Uð Þ \ f�1 C0ð Þ; we have t 2 a�1 Uð Þ; and t 2 f�1 C0ð Þ: It follows that a tð Þ 2 U;and a tð Þ ¼ b tð Þ; and hence, b tð Þ 2 U: Thus, LHS RHS:

Next, let t 2 RHS; that is, t 2 b�1 Uð Þ; that is, b tð Þ 2 U: We have to show thatt 2 LHS: Since U is a subset of C; and b : f�1 C0ð Þ ! C; we havet 2ð Þb�1 Uð Þ f�1 C0ð Þ, and hence t 2 f�1 C0ð Þ: Since t 2 f�1 C0ð Þ; we havea tð Þ ¼ b tð Þ 2 Uð Þ; and hence, t 2 a�1 Uð Þ: Since t 2 a�1 Uð Þ; and t 2 f�1 C0ð Þ; wehave t 2 a�1 Uð Þ \ f�1 C0ð Þ ¼ LHS:

Thus, RHS LHS: Since RHS LHS; and LHS RHS; so LHS ¼ RHS: ■)Since b : f�1 C0ð Þ ! C is a measurable function, and U is an open subset of C;

we have b�1 Uð Þ 2 ℳ�: Since

a�1 Uð Þ \ f�1 C0ð Þ ¼ b�1 Uð Þ 2 ℳ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ℳ;

we have a�1 Uð Þ \ f�1 C0ð Þ 2 ℳ:

For (cii): We have to show that a�1 Uð Þ \ f�1 0ð Þð Þ 2 ℳ:Case I: when 1 62 U:

Problem 1:73 a�1 Uð Þ \ f�1 0ð Þð Þ ¼ b�1 Uð Þ or ;:(Solution If not, otherwise, let a�1 Uð Þ \ f�1 0ð Þ 6¼ b�1 Uð Þ; and a�1 Uð Þ \ f�1 0ð Þ 6¼ ;: We have to arrive at a contradiction. Since a�1 Uð Þ \ f�1 0ð Þ 6¼ ;; there existst 2 a�1 Uð Þ \ f�1 0ð Þ; and hence t 2 a�1 Uð Þ; and t 2 f�1 0ð Þ: Since t 2 f�1 0ð Þ; wehave t 62 f�1 C0ð Þ; and hence, U3ð Þa tð Þ ¼ 1; a contradiction. ■)

Since b : f�1 C0ð Þ ! C is a measurable function, and U is open in C; we haveb�1 Uð Þ 2 ℳ� ℳð Þ; and hence, b�1 Uð Þ 2 ℳ: Since b�1 Uð Þ 2 ℳ; ; 2 ℳ; and

a�1 Uð Þ \ f�1 0ð Þ� � ¼ b�1 Uð Þ or ;� �;

we have a�1 Uð Þ \ f�1 0ð Þ 2 ℳ:Case II: when 1 2 U: Since 1 2 U; we have

f�1 0ð Þ a�1 1ð Þ a�1 Uð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl};


and hence

a�1 Uð Þ \ f�1 0ð Þ ¼ f�1 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f�1 C0ð Þ� �c2 ℳ:

Thus, in all cases, a�1 Uð Þ \ f�1 0ð Þ 2 ℳ: ■

Lemma 1.74 Let X be any nonempty set. Let E be any nonempty collection ofsubsets of X: Then there exists a r-algebra ℳ in X such that

1. E ℳ; 2. If ℳ1 is a r-algebra in X satisfying E ℳ1; then ℳ ℳ1:(In short, there exists a smallest r-algebra in X containing E:)

Proof Let F be the collection of all r-algebras in X that contain E: Clearly, thepower set P Xð Þ of X is a r-algebras in X; and E is contained in P Xð Þ: It followsthat P Xð Þ 2 F Thus, F is nonempty.

Problem 1:75 \F is a r-algebra in X:

(Solution

1. We have to show that ; is in \F . For this purpose, let us take any ℳ 2 F . Wehave to show that ; 2 ℳ: Since ℳ 2 F , ℳ is a r-algebra in X; and hence; 2 ℳ:

2. Let us take any A in \F . We have to show that Ac is in \F . For this purpose,let us take any ℳ 2 F . We have to show that Ac 2 ℳ: Since A is in \F , andℳ 2 F , we have A 2 ℳ: Sinceℳ 2 F ,ℳ is a r-algebra in X: Sinceℳ is a r-algebra in X; and A 2 ℳ; we have Ac 2 ℳ:

3. Let us take any A1;A2; . . . in \F . We have to show that A1 [A2 [ � � � is in\F . For this purpose, let us take any ℳ 2 F . We have to show thatA1 [A2 [ � � �ð Þ 2 ℳ: Since each An is in \F , and ℳ 2 F , each An 2 ℳ:Since ℳ 2 F , ℳ is a r-algebra in X: Since ℳ is a r-algebra in X; and eachAn 2 ℳ; we have A1 [A2 [ � � �ð Þ 2 ℳ: ■)

It suffices to show: 1. E (\F ), and 2. If ℳ1 is a r-algebra in X satisfyingE ℳ1; then (\F ) ℳ1:

For 1: Since each member of F contains E; \F contains E:For 2: Let ℳ1 be a r-algebra in X satisfying E ℳ1: We have to show that

(\F ) ℳ1: Since ℳ1 is a r-algebra in X satisfying E ℳ1; ℳ1 is in F , andhence, (\F ) ℳ1: ■

Lemma 1.76 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let Y be anynonempty set. Let f : X ! Y : Then A : A Y and f�1 Að Þ 2 ℳ

� is a r-algebra in

Y :

Proof

1. Here ; Y : Sinceℳ is a r-algebra in X; we have f�1 ;ð Þ ¼ð Þ; 2 ℳ; and hencef�1 ;ð Þ 2 ℳ: Since ; Y ; and f�1 ;ð Þ 2 ℳ; we have


; 2 A : A Y and f�1 Að Þ 2 ℳ�

:

2. Let

B 2 A : A Y and f�1 Að Þ 2 ℳ�

:

We have to show that

Bc 2 A : A Y and f�1 Að Þ 2 ℳ�

:

Since

B 2 A : A Y and f�1 Að Þ 2 ℳ�

;

we have

B Y and f�1 Bð Þ 2 ℳ:

Since B Y ; we have Bc Y : Since f�1 Bð Þ 2 ℳ; and ℳ is a r-algebra in X;we have

f�1 Bcð Þ ¼� �f�1 Bð Þ� �c2 ℳ;

and hence, f�1 Bcð Þ 2 ℳ: Since Bc Y ; and f�1 Bcð Þ 2 ℳ; we have

Bc 2 A : A Y and f�1 Að Þ 2 ℳ�

:

3. For each n ¼ 1; 2; . . .; let

Bn 2 A : A Y and f�1 Að Þ 2 ℳ�

:


B1 [B2 [ � � �ð Þ 2 A : A Y and f�1 Að Þ 2 ℳ�

;

that is,

B1 [B2 [ � � �ð Þ Y and f�1 B1 [B2 [ � � �ð Þ 2 ℳ:

Since each

Bn 2 A : A Y and f�1 Að Þ 2 ℳ�

;


so each Bn Y and each f�1 Bnð Þ 2 ℳ: Since each Bn Y ; we haveB1 [B2 [ � � �ð Þ Y : Since each f�1 Bnð Þ 2 ℳ; and ℳ is a r-algebra in X;we have

f�1 B1 [B2 [ � � �ð Þ ¼� �f�1 B1ð Þ [ f�1 B2ð Þ [ � � �� 2 ℳ;

and hence, f�1 B1 [B2 [ � � �ð Þ 2 ℳ: Thus,

A : A Y and f�1 Að Þ 2 ℳ�

is a r-algebra in Y : ■

Lemma 1.77 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let

f : X ! �1;1½ � � R[ �1;1f gð Þ:

For every a 2 R; let f�1 a;1ð �ð Þ 2 ℳ: Then f : X ! �1;1½ � is a measurablefunction.

Proof Let a 2 R:

Problem 1:78 f�1 a;1½ �ð Þ 2 ℳ:

(Solution By the assumption, for every n ¼ 1; 2; . . ., each f�1 a� 1n ;1

� �� 2 ℳ:

Since each f�1 a� 1n ;1

� �� 2 ℳ; and ℳ is a r-algebra in X;

f�1 a;1½ �ð Þ ¼ f�1 a� 11;1

� �[ a� 1

2;1

� �[ � � �

� �¼

� �

f�1 a� 11;1

� �� [ f�1 a� 1

2;1

� �� [ � � �

is in ℳ; and hence, f�1 a;1½ �ð Þ 2 ℳ: ■)

Problem 1:79 f�1 �1; a½ Þð Þ 2 ℳ:

(Solution Since f�1 a;1½ �ð Þ 2 ℳ; and ℳ is a r-algebra in X;

f �1 �1; a½ Þð Þ ¼ f�1 a;1½ �cð Þ ¼ f�1 a;1½ �ð Þ� �c2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence, f�1 �1; a½ Þð Þ 2 ℳ: ■)

As above, it is clear that f�1 �1; a½ �ð Þ 2 ℳ: Thus, for every real a;

f�1 �1; a½ Þð Þ; f�1 �1; a½ �ð Þ; f�1 a;1ð �ð Þ; f�1 a;1½ �ð Þ 2 ℳ:


Now, let a\b:

Problem 1:80 f�1 a; bð Þð Þ 2 ℳ:

(Solution Here f�1 a;1ð �ð Þ 2 ℳ; f�1 �1; b½ Þð Þ 2 ℳ; and ℳ is a r-algebra in X;so

f�1 a; bð Þð Þ ¼ f�1 a;1 \� ½ �1; bð Þð Þ ¼ f�1 a;1ð �ð Þ \ f�1 �1; b½ Þð Þ 2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence, f�1 a; bð Þð Þ 2 ℳ: ■)

Similarly,

f�1 a; bð �ð Þ; f�1 a; b½ Þð Þ; f�1 a; b½ �ð Þ 2 ℳ:

Also, for every a 2 R; f�1 að Þ 2 ℳ: It is easy to see thatf�1 �1ð Þ; f�1 1ð Þ 2 ℳ:

Now, let G be a nonempty set, which is open in R[ �1;1f g: It suffices toshow that f�1 Gð Þ 2 ℳ:

Case I: when �1 2 G; and 1 62 G: Since �1 2 G; and G is open inR[ �1;1f g; there exists a real number a such that �1; a½ Þ �1; a½ � G:Since a;1ð � is open in �1;1½ �; so �1; a½ � ¼ð Þ �1;1½ � � a;1ð � is closed in�1;1½ �; and hence �1; a½ � is closed in �1;1½ �: Since �1; a½ � is closed in�1;1½ �; and G is open in �1;1½ �; so G� �1; a½ � is open in �1;1½ �; andhence,

G� �1; a½ �ð Þ \ �1;1ð Þ

is open in �1;1ð Þ: Since �1 2 G; and 1 62 G; we have G� �1; a½ �ð Þ �1;1ð Þ; and hence G� �1; a½ �ð Þ \ �1;1ð Þ ¼ G� �1; a½ �: Since

G� �1; a½ �ð Þ \ �1;1ð Þ ¼ G� �1; a½ �;

and

G� �1; a½ �ð Þ \ �1;1ð Þ

is open in �1;1ð Þ; G� �1; a½ � is open in �1;1ð Þ: Since G� �1; a½ � is openin R; there exist real numbers a1; b1; a2; b2; . . . such that a1\b1; a2\b2; . . .; and

G� �1; a½ � ¼ a1; b1ð Þ [ a2; b2ð Þ [ � � � :


Thus

G ¼ �1; a½ � [ G� �1; a½ �ð Þ ¼ �1; a½ � [ a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ:

It follows that

f�1 Gð Þ ¼ f�1 �1; a½ � [ a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ¼ f�1 �1; a½ �ð Þ [ f�1 a1; b1ð Þð Þ [ f�1 a2; b2ð Þð Þ [ � � � :

Since

f�1 �1; a½ �ð Þ; f�1 a1; b1ð Þð Þ; f�1 a2; b2ð Þð Þ; . . .

are in ℳ; and ℳ is a r-algebra in X;

f�1 Gð Þ ¼� �f�1 �1; a½ �ð Þ [ f�1 a1; b1ð Þð Þ [ f�1 a2; b2ð Þð Þ [ � � �

is in ℳ; and hence, f�1 Gð Þ 2 ℳ:Case II: when 1 2 G; and �1 62 G. This case is similar to case I.Similarly, all other cases can be dealt with. ■

Definition Let X be a topological space with topology O: By Lemma 1.74, thereexists a r-algebra B in X such that

1. O B; and 2. If ℳ is a r-algebra in X satisfying O ℳ; then B ℳ:Members of B are called Borel sets in X:Example: Let a; b 2 R: Let a\b: Clearly, a; b½ Þ is a Borel set in R: (Reason:

since each a� 1n ; b

� �is open in R; so

a; b½ Þ ¼ð Þ a� 11; b

� �[ a� 1

2; b

� �[ a� 1

3; b

� �[ � � �

is a Borel set in R; and hence a; b½ Þ is a Borel set in R:)

Lemma 1.81 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let Y be atopological space. Let f : X ! Y be a measurable function. Let E be a Borel set inY : Then f�1 Eð Þ 2 ℳ:

Proof Let O be the topology of Y : Let B be the r-algebra in Y such that1. O B; and 2. If ℳ is a r-algebra in Y satisfying O ℳ; then B ℳ:Now since E is a Borel set in Y ; so E 2 B: By Lemma 1.76,


is a r-algebra in Y :


Problem 1:82 O A : A Y and f�1 Að Þ 2 ℳ�

:

(Solution Let U 2 O: We have to show that U Y and f�1 Uð Þ 2 ℳ: Since U 2O; and O is the topology of Y ; we have U Y : Since U 2 O; and f : X ! Y is ameasurable function, we have f�1 Uð Þ 2 ℳ: ■)

Since


is a r-algebra in Y satisfying

O A : A Y and f�1 Að Þ 2 ℳ�

;

so by 2,

E 2ð ÞB A : A Y and f�1 Að Þ 2 ℳ�

;

and hence f�1 Eð Þ 2 ℳ: ■

Definition Let X be a topological space with topology O: Let B be the r-algebra inX such that

1. O B2. If ℳ is a r-algebra in X satisfying O ℳ; then B ℳ: (In short, B is the

smallest r-algebra in X that contains O; that is, B is the collection of all Borelsets in X:)

Let Y be a topological space with topology O1: Let f : X ! Y : If for everyU 2 O1; f�1 Uð Þ 2 B; then we say that f : X ! Y is a Borel mapping.

Lemma 1.83 Let X be a topological space with topology O: Let Y be a topologicalspace with topology O1: Let f : X ! Y be continuous. Then f : X ! Y is a Borelmapping.

Proof Let B be the r-algebra in X such that

1. O B;2. If ℳ is a r-algebra in X satisfying O ℳ; then B ℳ:

Let U 2 O1: We have to show that f�1 Uð Þ 2 B: Since U 2 O1; and f : X ! Yis continuous, we have f�1 Uð Þ 2 O Bð Þ; and hence f�1 Uð Þ 2 B: ■

Lemma 1.84 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let Y be atopological space with topology O1: Let Z be a topological space with topology O2:Let f : X ! Y be a measurable function. Let g : Y ! Z be a Borel mapping. Thenthe composite


g � f : X ! Z

is a measurable function.

Proof Let B be the r-algebra in Y such that

1. O1 B;2. If ℳ is a r-algebra in Y satisfying O1 ℳ; then B ℳ:

Let U 2 O2: We have to prove that

f�1 g�1 Uð Þ� � ¼ g � fð Þ�1 Uð Þ 2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is f�1 g�1 Uð Þð Þ 2 ℳ: Since g : Y ! Z is a Borel mapping, and U 2 O2; wehave g�1 Uð Þ 2 B; and hence g�1 Uð Þ is a Borel set in Y : Since g�1 Uð Þ is a Borel setin Y ; and f : X ! Y is a measurable function, by Lemma 1.81 we havef�1 g�1 Uð Þð Þ 2 ℳ: ■

Lemma 1.85 Let X be any nonempty set. Let ℳ be a r-algebra in X: For everyn ¼ 1; 2; . . .; let fn : X ! �1;1½ � be a measurable function. Let us define g1 :X ! �1;1½ � as follows: for every x 2 X;

g1 xð Þ � sup f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g:

Let us define g2 : X ! �1;1½ � as follows: for every x 2 X;

g2 xð Þ � sup f2 xð Þ; f3 xð Þ; f4 xð Þ; . . .f g; etc:

Then g1 : X ! �1;1½ �; g2 : X ! �1;1½ �; and so on are measurablefunctions.

Proof We want to show that g2 : X ! �1;1½ � is a measurable function. For thispurpose, let us take any a 2 R: By Lemma 1.77, it suffices to show thatg�12 a;1ð �ð Þ 2 ℳ: Observe that

g�12 a;1ð �ð Þ ¼ x : g2 xð Þ 2 a;1ð �f g ¼ x : a\g2 xð Þf g

¼ x : a\ sup f2 xð Þ; f3 xð Þ; . . .f gf g¼ x : there exists n� 2 such that a\fn xð Þf g¼ x : there exists n� 2 such that fn xð Þ 2 a;1ð �f g¼ x : there exists n� 2 such that x 2 f�1

n a;1ð �ð Þ� ¼ f�1

2 a;1ð �ð Þ [ f�13 a;1ð �ð Þ [ � � � :

Since a;1ð � is open in �1;1½ �; and f2 : X ! �1;1½ � is a measurablefunction, we have f�1

2 a;1ð �ð Þ 2 ℳ: Similarly,


f�13 a;1ð �ð Þ 2 ℳ; f�1

4 a;1ð �ð Þ 2 ℳ; . . .:

Since

f�12 a;1ð �ð Þ; f�1

3 a;1ð �ð Þ; f�14 a;1ð �ð Þ; . . .

are in ℳ; and ℳ is a r-algebra in X; we have

g�12 a;1ð �ð Þ ¼� �

f�12 a;1ð �ð Þ [ f�1

3 a;1ð �ð Þ [ � � �

is in ℳ; and hence g�12 a;1ð �ð Þ 2 ℳ:

Thus, g2 : X ! �1;1½ � is a measurable function. Similarly, g3 : X !�1;1½ � is a measurable function, and so on. ■

Lemma 1.86 Let X be any nonempty set. Let ℳ be a r-algebra in X: For everyn ¼ 1; 2; . . .; let fn : X ! �1;1½ � be a measurable function. Let us define h1 :X ! �1;1½ � as follows: for every x 2 X;

h1 xð Þ � inf f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g:

Let us define h2 : X ! �1;1½ � as follows: for every x 2 X;

h2 xð Þ � inf f2 xð Þ; f3 xð Þ; f4 xð Þ; . . .f g; etc:

Then h1 : X ! �1;1½ �; h2 : X ! �1;1½ �; etc. are measurable functions.

Proof Its proof is similar to Lemma 1.85. ■

Lemma 1.87 Let X be any nonempty set. Let ℳ be a r-algebra in X: For everyn ¼ 1; 2; . . .; let fn : X ! �1;1½ � be a measurable function. Let us define h1 :X ! �1;1½ � as follows: for every x 2 X;

h1 xð Þ � inf f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g:


h2 xð Þ � inf f2 xð Þ; f3 xð Þ; f4 xð Þ; . . .f g; etc:

Let us define lim infn!1 fnð Þ : X ! �1;1½ � as follows: for every x 2 X;

lim infn!1 fn

� xð Þ � sup h1 xð Þ; h2 xð Þ; h3 xð Þ; . . .f g:

Then lim infn!1 fnð Þ : X ! �1;1½ � is a measurable function.

Proof By Lemma 1.86, h1 : X ! �1;1½ �; h2 : X ! �1;1½ �; etc. are measur-able functions. Again, by Lemma 1.85,


lim infn!1 fn

� : X ! �1;1½ �

is measurable function. ■

Lemma 1.88 Let X be any nonempty set. Let ℳ be a r-algebra in X: For everyn ¼ 1; 2; . . .; let

fn : X ! �1;1½ �

be a measurable function. Let us define h1 : X ! �1;1½ � as follows: for everyx 2 X;

h1 xð Þ � sup f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g:


h2 xð Þ � sup f2 xð Þ; f3 xð Þ; f4 xð Þ; . . .f g; etc.

Let us define

lim supn!1

fn

� �: X ! �1;1½ �

as follows: for every x 2 X;

lim supn!1

fn

� �xð Þ � inf h1 xð Þ; h2 xð Þ; h3 xð Þ; . . .f g:

Then lim supn!1 fnð Þ : X ! �1;1½ � is a measurable function.


Lemma 1.89 Let X be any nonempty set. Let ℳ be a r-algebra in X: For everyn ¼ 1; 2; . . .; let fn : X ! C be a measurable function. For every x 2 X; let fn xð Þf gbe convergent. Let limn!1 fnð Þ : X ! C be the function defined as follows: orevery x 2 X;

limn!1 fn�

xð Þ � limn!1 fn xð Þð Þ:

Then limn!1 fnð Þ : X ! C is a measurable function.

Proof For every x 2 X; and for every n ¼ 1; 2; . . .; put fn xð Þ � un xð Þþ i vn xð Þð Þ;where un xð Þ; vn xð Þ 2 R: For every x 2 X; let


limn!1 fn�

xð Þ � u xð Þþ i v xð Þð Þ:

Thus, for every n ¼ 1; 2; . . .; un : X ! R �1;1½ �ð Þ; vn : X !R �1;1½ �ð Þ; u : X ! R �1;1½ �ð Þ; and v : X ! R �1;1½ �ð Þ: Sinceeach fn : X ! C is a measurable function, by Lemma 1.65, each un is a measurablefunction and each vn is a measurable function. Since for every x 2 X; fn xð Þ ¼un xð Þþ i vn xð Þð Þ; where un xð Þ; vn xð Þ 2 R; and fn xð Þf g converges to u xð Þþ i v xð Þð Þ,for every x 2 X; un xð Þf g converges to u xð Þ, and hence, for every x 2 X;

u xð Þ ¼ limn!1 un xð Þð Þ ¼ lim inf

n!1 un xð Þð Þ ¼ lim infn!1 un

� xð Þ:

It follows that u ¼ lim infn!1 un: By Lemma 1.87, lim infn!1 unð Þ : X !�1;1½ � is a measurable function. Since lim infn!1 unð Þ : X ! �1;1½ � is ameasurable function, and u ¼ lim infn!1 un; so u : X ! R is a measurable func-tion. Similarly, v : X ! R is a measurable function. Since u : X ! R; and v : X !R are measurable functions, by Lemma 1.64

uþ ivð Þ : t 7! u tð Þð Þþ i v tð Þð Þ ¼ limn!1 fn�

tð Þ�

from X to C is a measurable function, and hence, limn!1 fnð Þ : X ! C is a mea-surable function. ■


f : X ! �1;1½ �; and g : X ! �1;1½ �

be measurable functions. Let us define max f ; gf gð Þ : X ! �1;1½ � as follows: forevery x 2 X;

max f ; gf gð Þ xð Þ � max f xð Þ; g xð Þf g:

Then max f ; gf g is a measurable function.

Proof By Lemma 1.85, the map

x 7! sup f xð Þ; g xð Þ; g xð Þ; . . .f g ¼ max f xð Þ; g xð Þf g ¼ max f ; gf gð Þ xð Þð Þ

is a measurable function, so max f ; gf g is a measurable function. ■


f : X ! �1;1½ �; and g : X ! �1;1½ �

be measurable functions. Let us define min f ; gf gð Þ : X ! �1;1½ � as follows: forevery x 2 X;


min f ; gf gð Þ xð Þ � min f xð Þ; g xð Þf g:

Then min f ; gf g is a measurable function.


Lemma 1.92 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let f :X ! �1;1½ � be a measurable function. Let us define �fð Þ : X ! �1;1½ � asfollows: for every x 2 X; �fð Þ xð Þ � � f xð Þð Þ: Then �fð Þ is a measurable function.

Proof Let us take any a 2 R: By Lemma 1.77, it suffices to show that�fð Þ�1 a;1ð �ð Þ 2 ℳ: Here

�fð Þ�1 a;1ð �ð Þ ¼ x : �fð Þ xð Þ 2 a;1ð �f g ¼ x : a\ �fð Þ xð Þf g¼ x : a\� f xð Þð Þf g ¼ x : f xð Þ\ �að Þf g¼ x : f xð Þ 2 �1;�a½ Þf g ¼ f�1 �1;�a½ Þð Þ:

Since �1;�a½ Þ is open in �1;1½ �; and f : X ! �1;1½ � is a measurablefunction, so

�fð Þ�1 a;1ð �ð Þ ¼�

f�1 �1;�a½ Þð Þ 2 ℳ;

and hence, �fð Þ�1 a;1ð �ð Þ 2 ℳ: ■

Lemma 1.93 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let f :X ! �1;1½ � be a measurable function. Let us define f� : X ! 0;1½ � as follows:for every x 2 X; f�ð Þ xð Þ � � min f xð Þ; 0f gð Þ: Let us define f þ : X ! 0;1½ � asfollows: for every x 2 X; f þð Þ xð Þ � max f xð Þ; 0f g: Then f�; f þ are measurablefunctions.

Here f� is called the negative part of f and f þ is called the positive part of f :ð Þ

Proof Clearly the constant function 0 : x 7! 0 from X to �1;1½ � is a measurablefunction. In addition, it is clear that f� ¼ � min f ; 0f gð Þ: Since f ; 0 are measurablefunctions, by Lemma 1.91 min f ; 0f g is a measurable function. Since min f ; 0f g is ameasurable function, by Lemma 1.92 f� ¼ð Þ � min f ; 0f gð Þð Þ is a measurablefunction, and hence f� is a measurable function. Similarly, f þ is a measurablefunction. ■


f : X ! 0;1½ �; g : X ! 0;1½ �; h : X ! 0;1½ �:

Let f ¼ g� h: Then1. f þ � g; and 2. f� � h: (Here 1�1 � 0:)


Proof

1. Let us take any x 2 X: We have to show that f þ xð Þ� g xð Þ; that is,max f xð Þ; 0f g� g xð Þ; that is, f xð Þ� g xð Þ and 0� g xð Þð Þ: Now since 0� g xð Þ; itremains to for us to show that f xð Þ� g xð Þ:Case I: when g xð Þ ¼ 1: Here f xð Þ�1 ¼ g xð Þð Þ; so f xð Þ� g xð Þ:Case II: when g xð Þ 6¼ 1: Since 0� h; so 0� h xð Þ; and hence � h xð Þð Þ� 0: Thus

f xð Þ ¼ð Þg xð Þ � h xð Þð Þ� g xð Þþ 0 ¼ g xð Þð Þ:

It follows that f xð Þ� g xð Þ: Thus f þ � g:

2. Let us take any x 2 X: We have to show that f�ð Þ xð Þ� h xð Þ; that is,� min f xð Þ; 0f gð Þ� h xð Þ; that is, � h xð Þð Þ�min f xð Þ; 0f g; that is,� h xð Þð Þ� f xð Þ and� h xð Þð Þ� 0ð Þ, that is, � h xð Þð Þ� f xð Þ and 0� h xð Þ). Nowsince 0� h xð Þ; it remains for us to show that � h xð Þð Þ� f xð Þ:Case I: when h xð Þ ¼ 1: Here �1� f xð Þ; so � h xð Þð Þð Þ� f xð Þ:Case II: when h xð Þ 6¼ 1: Since 0� g; we have 0� g xð Þ; and hence

� h xð Þð Þ ¼ð Þ0� h xð Þð Þ� g xð Þ � h xð Þð Þ ¼ f xð Þð Þ:

It follows that � h xð Þð Þ� f xð Þ:Thus, f� � h: ■

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let s : X ! C

be any function. If there exist finite-many distinct a1; . . .; an in C such that s Xð Þ ¼a1; . . .; anf g; then we say that s is a simple function. By the nonnegative simple

function s we mean that s Xð Þ 0;1½ Þ:Problem 1.95 s is a measurable function if and only if s�1 a1ð Þ; . . .; s�1 anð Þ 2 ℳ:

(Solution Let s be a measurable function. We shall try to show that s�1 a1ð Þ 2 ℳ:Since a1; . . .; an are distinct members of C; there exists a real number r[ 0 suchthat the open disk D a1; rð Þ does not contain a2; . . .; an: Since D a1; rð Þ is an opendisk with center a1; D a1; rð Þ is an open neighborhood of a1: Since D a1; rð Þ is anopen neighborhood of a1; and s : X ! C is a measurable function,

s�1 a1ð Þ ¼� �s�1 D a1; rð Þð Þ 2 ℳ;

and hence s�1 a1ð Þ 2 ℳ: Similarly, s�1 a2ð Þ 2 ℳ; etc.Conversely, let s�1 a1ð Þ; . . .; s�1 anð Þ 2 ℳ: We have to show that s : X ! C is a

measurable function. For this purpose, let us take any open set V in C: We have toshow that s�1 Vð Þ 2 ℳ: Since s is a simple function, either s�1 Vð Þ ¼ ;ð Þ or s�1 Vð Þis a union of some members of s�1 a1ð Þ; . . .; s�1 anð Þ: If s�1 Vð Þ ¼ ; 2 ℳð Þ; thens�1 Vð Þ 2 ℳ: So we consider only the case when s�1 Vð Þ is a union of somemembers of


s�1 a1ð Þ; . . .; s�1 anð Þ� :

Since s�1 Vð Þ is a union of some members of s�1 a1ð Þ; . . .; s�1 anð Þ� ; and

s�1 a1ð Þ; . . .; s�1 anð Þ are members of the r-algebra ℳ; s�1 Vð Þ 2 ℳ: ■)

Problem 1.96 If s is a measurable function, then

s ¼ a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ:

(Solution Let x 2 X: We have to show that

s xð Þ ¼ a1 v s�1 a1ð Þð Þ xð Þ�

þ � � � þ an v s�1 anð Þð Þ xð Þ�

:

Since s Xð Þ ¼ a1; . . .; anf g; and a1; . . .; an are distinct members of C;

s�1 a1ð Þ; . . .; s�1 anð Þ� is a partition of X: Since s�1 a1ð Þ; . . .; s�1 anð Þ�

is a partition of X; and x 2 X; x iscontained in exactly one member of s�1 a1ð Þ; . . .; s�1 anð Þ�

: For simplicity, sup-pose that

x 2 s�1 a1ð Þ; x 62 s�1 a2ð Þ; . . .; x 62 s�1 anð Þ:

It follows that

v s�1 a1ð Þð Þ xð Þ ¼ 1; v s�1 a2ð Þð Þ xð Þ ¼ 0; . . .; v s�1 anð Þð Þ xð Þ ¼ 0:

Since x 2 s�1 a1ð Þ; we have s xð Þ ¼ a1:

RHS ¼ a1 v s�1 a1ð Þð Þ xð Þ�

þ a2 v s�1 a2ð Þð Þ xð Þ�

þ � � � þ an v s�1 anð Þð Þ xð Þ�

¼ a1 1ð Þþ a2 0ð Þþ � � � þ an 0ð Þ ¼ a1 ¼ s xð Þ ¼ LHS:

∎)Observe that

0; 1 123

� �; 1 1

23; 2 1

23

� �; 2 1

23; 3 1

23

� �; . . .;

�

26 � 1� � 1

23; 26 1

23

� �; 26 1

23;1

� ��


is a partition of 0;1½ �: Also,

0; 1 123

� �; 1 1

23; 2 1

23

� �; 2 1

23; 3 1

23

� �; . . .;

�

26 � 1� � 1

23; 26 1

23

� �; 26 1

23;1

� ��

is a collection of Borel sets in 0;1½ �:Let X be any nonempty set. Let ℳ be a r-algebra in X: It is easy to see that

i. If s : X ! 0;1½ Þ; and t : X ! 0;1½ Þ are simple functions, then sþ tð Þ :x 7! s xð Þþ t xð Þð Þ from X to 0;1½ Þ; and s � tð Þ : x 7! s xð Þð Þ t xð Þð Þ from X to0;1½ Þ are simple functions;

ii. If s : X ! 0;1½ Þ; and t : X ! 0;1½ Þ are measurable functions, then sþ tð Þ :x 7! s xð Þþ t xð Þð Þ from X to 0;1½ Þ; and s � tð Þ : x 7! s xð Þð Þ t xð Þð Þ from X to0;1½ Þ are measurable functions.

Let us define a function u3 : 0;1 !� ½0;1½ Þ as follows: For every t 2 0;1½ �;

u3 tð Þ �

0 if t 2 0; 1 123

� �1 1

23 if t 2 1 123 ; 2 1

23� �

2 123 if t 2 2 1

23 ; 3 123

� �...

26 � 1� � 1

23 if t 2 26 � 1� � 1

23 ; 26 1

23� �

26 123 if t 2 26 1

23 ;1� �

:

8>>>>>>><>>>>>>>:

Since the collection of all Borel sets in 0;1½ � is a r-algebra in 0;1½ �; and

u3 0;1½ �ð Þ ¼ 0; 1 123

; 2 123

; . . .; 26 � 1� � 1

23; 26 1

23

� ��

has only finite-many elements, u3 is a simple function. Also,

u3ð Þ�1 0ð Þ ¼ 0; 1 123

� �� ; u3ð Þ�1 1 1

23

� �¼ 1 1

23; 2 1

23

� �� ;

u3ð Þ�1 2 123

� �¼ 2 1

23; 3 1

23

� �� ; . . .;

u3ð Þ�1 26 � 1� � 1

23

� �¼ 26 � 1

� � 123

; 26 123

� �� ;

u3ð Þ�1 26 123

� �¼ 26 1

23;1

� ��


are Borel sets in 0;1½ �:Problem 1.97 u3 is a Borel mapping.

(Solution For this purpose, let us take any open set V in 0;1½ Þ: We have to showthat u3ð Þ�1 Vð Þ is a Borel set in 0;1½ �: Since u3 is a simple function, either

u3ð Þ�1 Vð Þ ¼ ;�

or u3ð Þ�1 Vð Þ is a union of some members of

u3ð Þ�1 0ð Þ; u3ð Þ�1 1 123

� �; u3ð Þ�1 2 1

23

� �; . . .;

�

u3ð Þ�1 26 � 1� � 1

23

� �; u3ð Þ�1 26 1

23

� ��:

If u3ð Þ�1 Vð Þ ¼ ;; then u3ð Þ�1 Vð Þ is a Borel set in 0;1½ �: So, we consider thecase when u3ð Þ�1 Vð Þ is a union of some members of

u3ð Þ�1 0ð Þ; u3ð Þ�1 1 123

� �; u3ð Þ�1 2 1

23

� �; . . .;

�

u3ð Þ�1 26 � 1� � 1

23

� �; u3ð Þ�1 26 1

23

� ��:

Since u3ð Þ�1 Vð Þ is a union of some members of

u3ð Þ�1 0ð Þ; u3ð Þ�1 1 123

� �; u3ð Þ�1 2 1

23

� �; . . .;

�

u3ð Þ�1 26 � 1� � 1

23

� �; u3ð Þ�1 26 1

23

� ��;

and

u3ð Þ�1 0ð Þ; u3ð Þ�1 1 123

� �; u3ð Þ�1 2 1

23

� �; . . .;

�

u3ð Þ�1 26 � 1� � 1

23

� �; u3ð Þ�1 26 1

23

� ��

are Borel sets in 0;1½ �; u3ð Þ�1 Vð Þ is a Borel set in 0;1½ �: ■)If we draw the graph of u3; the straight line t 7! t; and the straight line

t 7! t � 123

� �; it is easy to observe that, for every t in 0; 23½ �;

t � 123

� �\u3 tð Þ� t:


Let us define a function u4 : 0;1 !� ½0;1½ Þ as follows: For every t 2 0;1½ �;

u4 tð Þ �

0 if t 2 0; 1 124

� �1 1

24 if t 2 1 124 ; 2 1

24� �

2 124 if t 2 2 1

24 ; 3 124

� �...

28 � 1ð Þ 124 if t 2 28 � 1ð Þ 1

24 ; 28 1

24� �

28 124 if t 2 28 1

24 ;1� �

:

8>>>>>>><>>>>>>>:

As above, u4 is a simple Borel mapping, and for every t in 0; 24½ �;t � 1

24� �

\u4 tð Þ� t: Similar definitions and results can be obtained foru1;u2;u5;u6; etc. If we draw the graphs of u1;u2;u3;u4; . . .; it is easy to observethat for every t in 0;1½ �;

0�u1 tð Þ�u2 tð Þ�u3 tð Þ�u4 tð Þ� � � � :

Since, for every t in 0; 2n½ �;

t � 12n

� �\un tð Þ� t and lim

n!1 t � 12n

� �¼ t;

we have, for every t in 0;1½ Þ; limn!1 un tð Þ ¼ t: Also, since

un 1ð Þ ¼ 22n 12n

¼ 2n ! 1 as n ! 1ð Þ;

for every t in 0;1½ �; limn!1 un tð Þ ¼ t:Let X be any nonempty set. Let ℳ be a r-algebra in X: Let f : X ! 0;1½ � be a

measurable function. We have shown above that each un : 0;1 !� ½0;1½ Þ is asimple Borel mapping, so, by Lemma 1.84, each composite un � fð Þ : X ! 0;1½ Þis a simple measurable function. Since for every t in 0;1½ �;

0�u1 tð Þ�u2 tð Þ�u3 tð Þ�u4 tð Þ� � � � ;

it follows that, for every x in X;

0�u1 f xð Þð Þ�u2 f xð Þð Þ�u3 f xð Þð Þ�u4 f xð Þð Þ� � � � ;

and hence, for every x in X;

0� u1 � fð Þ xð Þ� u2 � fð Þ xð Þ� u3 � fð Þ xð Þ� u4 � fð Þ xð Þ� � � � :

Since for every t in 0;1½ �; limn!1 un tð Þ ¼ t; we have, for every x in X;limn!1 un f xð Þð Þ ¼ f xð Þ: Thus, for every x in X; limn!1 sn xð Þ ¼ f xð Þ; where sn �un � f : Thus, we get the following.


Lemma 1.98 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let f :X ! 0;1½ � be a measurable function. Then there exists a sequence snf g of simplemeasurable functions, sn : X ! 0;1½ Þ; such that, for every x in X;

0� s1 xð Þ� s2 xð Þ� s3 xð Þ� s4 xð Þ� � � � ;

and limn!1 sn xð Þ ¼ f xð Þ:Further, if E is a nonempty subset of X such that f is bounded on E; then

limn!1 sn ¼ f uniformly on E:Proof of the remaining part Let E be a nonempty subset of X such that f is

bounded on E: We have to show that limn!1 un � fð Þ ¼ f uniformly on E: Since fis bounded on E; there exists a positive integer N such that for every n�N; and forevery x 2 E; f xð Þ� 2n: Now, since for every positive integer n; and for every t in0; 2n½ �;

t � 12n

� �\un tð Þ� t;

we have for every n�N; and, for every x 2 E;

f xð Þ � 12n

� �\un f xð Þð Þ� f xð Þ \ f xð Þþ 1

2n

� �� :

Thus, for every n�N; and, for every x 2 E;

un f xð Þð Þ � f xð Þj j\212n

� �;

that is, for every n�N; and, for every x 2 E; sn xð Þ � f xð Þj j\ 12n�1 : Now, since

limn!1 12n�1 ¼ 0; we have limn!1 sn ¼ f uniformly on E: ■

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a function. By a positive measure (or, simply, measure) l onℳ we meanthat

1. l is countably additive, in the sense that, if A1;A2;A3; . . .f g is a countablecollection of members in ℳ such that i 6¼ j implies Ai \Aj ¼ ;; then

l A1 [A2 [A3 [ � � �ð Þ ¼ l A1ð Þþ l A2ð Þþ l A3ð Þþ � � � ;

2. there exists A in ℳ such that l Að Þ\1:


Lemma 1.99 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l :ℳ ! 0;1½ � be a positive measure on ℳ: Then

1. l ;ð Þ ¼ 0;2. if A1; . . .;Anf g is a finite collection of members in ℳ such that i 6¼ j implies

Ai \Aj ¼ ;; then

l A1 [ � � � [Anð Þ ¼ l A1ð Þþ � � � þ l Anð Þ;

3. if A;B 2 ℳ; and A B; then l Að Þ� l Bð Þ;4. if A1;A2;A3; . . .f g is a countable collection of members in ℳ such that A1

A2 A3 � � � ; then

limn!1 l Anð Þ ¼ l A1 [A2 [A3 [ � � �ð Þ;

5. if A1;A2;A3; . . .f g is a countable collection of members in ℳ such that � � � A3 A2 A1; and l A1ð Þ\1; then

limn!1 l Anð Þ ¼ l A1 \A2 \A3 \ � � �ð Þ:

Proof

1. Since l : ℳ ! 0;1½ � is a positive measure onℳ; there exists A inℳ such thatl Að Þ\1: Since l is countably additive, we have

l Að Þ ¼ð Þl A[;[ ;[ ;[ � � �ð Þ ¼ l Að Þþ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � � :

Since

l Að Þ ¼ l Að Þþ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � � ;

and l Að Þ\1; we have 0 ¼ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � � : This shows thatl ;ð Þ ¼ 0:

2. Let A1; . . .;Anf g be a finite collection of members in ℳ such that i 6¼ j )Ai \Aj ¼ ;: Since l is countably additive, and A1; . . .;Anf g is a finite collectionof members in ℳ satisfying i 6¼ j ) Ai \Aj ¼ ;; we have

l A1 [ � � � [An [;[;[ ;[ � � �ð Þ ¼ l A1ð Þþ � � � þ l Anð Þþ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � � :

Now since,

l A1 [ � � � [An [;[ ;[;[ � � �ð Þ ¼ l A1 [ � � � [Anð Þ ¼ LHS


and

l A1ð Þþ � � � þ l Anð Þþ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � �¼ l A1ð Þþ � � � þ l Anð Þþ 0þ 0þ 0þ � � �¼ l A1ð Þþ � � � þ l Anð Þ ¼ RHS;

we have LHS ¼ RHS:3. Let A;B 2 ℳ; and A B: Since ℳ is a r-algebra, and A;B 2 ℳ; we have

B� Að Þ 2 ℳ: Now, since A B; A; B� Að Þf g is a finite collection of membersin ℳ satisfying A\ B� Að Þ ¼ ;: It follows, from Lemma 1.99(2), that

l Bð Þ ¼ l A[ B� Að Þð Þ ¼ l Að Þþ l B� Að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � l Að Þþ 0 ¼ l Að Þ:

Thus, l Að Þ� l Bð Þ:4. Let A1;A2;A3; . . .f g be a countable collection of members in ℳ such that

A1 A2 A3 � � � : It follows, from Lemma 1.99(3), that

l A1ð Þ� l A2ð Þ� l A3ð Þ� � � � :

Since ℳ is a r-algebra, and each An 2 ℳ; we have A1 [A2 [A3 [ � � �ð Þ 2 ℳ:Since each An 2 ℳ; A1 [A2 [A3 [ � � �ð Þ 2 ℳ; and An A1 [A2 [A3 [ � � �ð Þ;by Lemma 1.99(3),

l Anð Þ� l A1 [A2 [A3 [ � � �ð Þ:

Thus, l A1ð Þ; l A2ð Þ; l A3ð Þ; . . .f g is a monotonically increasing sequence ofmembers in 0;1½ �; and each l Anð Þ� l A1 [A2 [A3 [ � � �ð Þ: It follows thatlimn!1 l Anð Þ exists and limn!1 l Anð Þ� l A1 [A2 [A3 [ � � �ð Þ: We have toshow that

limn!1 l Anð Þ ¼ l A1 [A2 [A3 [ � � �ð Þ:

If not, otherwise, let limn!1 l Anð Þ\l A1 [A2 [A3 [ � � �ð Þ:We have to arrive ata contradiction. Since

limn!1 l Anð Þ\l A1 [A2 [A3 [ � � �ð Þ;

and l A1ð Þ; l A2ð Þ; l A3ð Þ; . . .f g is a monotonically increasing sequence ofmembers in 0;1½ �; it follows that l A1ð Þ; l A2ð Þ; l A3ð Þ; . . .f g is a monotonicallyincreasing sequence of members in 0;1½ Þ: Since l is countably additive,


l A1 [ A2 � A1ð Þ [ A3 � A2ð Þ [ A4 � A3ð Þ [ � � �ð Þ¼ l A1ð Þþ l A2 � A1ð Þþ l A3 � A2ð Þþ l A4 � A3ð Þþ � � � :

Since

l A1 [ A2 � A1ð Þ [ A3 � A2ð Þ [ A4 � A3ð Þ [ � � �ð Þ ¼ l A1 [A2 [A3 [ � � �ð Þ

and

l A1ð Þþ l A2 � A1ð Þþ l A3 � A2ð Þþ l A4 � A3ð Þþ � � �¼ l A1ð Þþ l A2 � A1ð Þþ l A1ð Þð Þ � l A1ð Þð Þþ l A3 � A2ð Þþ l A2ð Þð Þ � l A2ð Þð Þþ l A4 � A3ð Þþ l A3ð Þð Þ � l A3ð Þð Þþ � � �

¼ l A1ð Þþ l A2 � A1ð Þ [A1ð Þ � l A1ð Þð Þþ l A3 � A2ð Þ [A2ð Þ � l A2ð Þð Þþ l A4 � A3ð Þ [A3ð Þ � l A3ð Þð Þþ � � �

¼ l A1ð Þþ l A2ð Þ � l A1ð Þð Þþ l A3ð Þ � l A2ð Þð Þþ l A4ð Þ � l A3ð Þð Þþ � � �¼ lim

n!1 l A1ð Þ;l A1ð Þþ l A2ð Þ � l A1ð Þð Þ;l A1ð Þþ l A2ð Þ � l A1ð Þð Þþ l A3ð Þ � l A2ð Þð Þ; . . .f g¼ lim

n!1 l A1ð Þ;l A2ð Þ; l A3ð Þ;l A4ð Þ; . . .f g ¼ limn!1 l Anð Þ;

we have

l A1 [A2 [A3 [ � � �ð Þ ¼ limn!1 l Anð Þ:

This is a contradiction.5. Let A1;A2;A3; . . .f g be a countable collection of members in ℳ such that

� � � A3 A2 A1; and l A1ð Þ\1: It follows, from Lemma 1.99(3), that

� � � � l A3ð Þ� l A2ð Þ� l A1ð Þ\1:

Since ℳ is a r-algebra, and each An 2 ℳ; we have A1 \A2 \A3 \ � � �ð Þ 2 ℳ:Since each An 2 ℳ; A1 \A2 \A3 \ � � �ð Þ 2 ℳ; and An A1 \A2 \A3 \ � � �ð Þ;by Lemma 1.99(3), for every positive integer n; l A1 \A2 \A3 \ � � �ð Þ � l Anð Þ:Thus, l A1ð Þ; l A2ð Þ; l A3ð Þ; . . .f g is a monotonically decreasing sequence ofmembers in 0;1½ Þ; and each l Anð Þ� l A1 \A2 \A3 \ � � �ð Þ:It follows that limn!1 l Anð Þ exists and

limn!1 l Anð Þ� l A1 \A2 \A3 \ � � �ð Þ:


limn!1 l Anð Þ ¼ l A1 \A2 \A3 \ � � �ð Þ:

Since l is countably additive,


l A1 \A2 \A3 \ � � �ð Þ [ A1 � A2ð Þ [ A2 � A3ð Þ [ A3 � A4ð Þ [ � � �ð Þ¼ l A1 \A2 \A3 \ � � �ð Þ þ l A1 � A2ð Þþ l A2 � A3ð Þþ l A3 � A4ð Þþ � � �

Now since,

l A1 \A2 \A3 \ � � �ð Þ [ A1 � A2ð Þ [ A2 � A3ð Þ [ A3 � A4ð Þ [ � � �ð Þ ¼ l A1ð Þ

and

l A1 \A2 \A3 \ � � �ð Þ þ l A1 � A2ð Þþ l A2 � A3ð Þþ l A3 � A4ð Þþ � � �¼ l A1 \A2 \A3 \ � � �ð Þ þ l A1 � A2ð Þþ l A2ð Þð Þ � l A2ð Þð Þþ l A2 � A3ð Þþ l A3ð Þð Þ � l A3ð Þð Þþ � � �

¼ l A1 \A2 \A3 \ � � �ð Þ þ l A1 � A2ð Þ [A2ð Þ � l A2ð Þð Þþ l A2 � A3ð Þ [A3ð Þ � l A3ð Þð Þþ � � �

¼ l A1 \A2 \A3 \ � � �ð Þ þ l A1ð Þ � l A2ð Þð Þþ l A2ð Þ � l A3ð Þð Þþ � � �¼ l A1 \A2 \ � � �ð Þþ lim

n!1 l A1ð Þ � l A2ð Þ; l A1ð Þ � l A2ð Þð Þþ l A2ð Þ � l A3ð Þð Þ; . . .f g¼ l A1 \A2 \A3 \ � � �ð Þ þ lim

n!1 l A1ð Þ � l A2ð Þ; l A1ð Þ � l A3ð Þ; . . .f g¼ l A1 \A2 \A3 \ � � �ð Þ þ l A1ð Þ � lim

n!1 l Anð Þ;

we have

l A1 \A2 \A3 \ � � �ð Þ ¼ limn!1 l Anð Þ: ∎

1.3 Integration of Positive Functions

The abstract integration theory presented here is largely a work of Lebesgue. Hedeveloped his theory in a step-by-step manner, from simple function to “slightlymore complicated” functions. We shall present all proofs in a sufficiently elaboratemanner, such that no important points are overlooked.

Definition Let us ‘extend’ the definition of ¼ (equal to) over 0;1½ Þ to 0;1½ � asfollows: For every a 2 0;1½ Þ, we define that a ¼ 1 is false, and 1 ¼ a is false.Also, we define that 1 ¼ 1 is true.

Let us ‘extend’ the definition of þ (addition) over 0;1½ Þ to 0;1½ � as follows:For every a 2 0;1½ �,


aþ1 � 1; and 1þ a � 1:

Thus,

þ : 0;1½ � 0;1½ � ! 0;1½ �;

that is, þ is a binary operation over 0;1½ �:Problem 1.100 þ is commutative over 0;1½ �:(Solution Let us take any a; b 2 0;1½ �: We have to show that aþ b ¼ bþ a:

Case I: when a ¼ 1: LHS ¼ aþ b ¼ 1þ b ¼ 1 ¼ bþ1 ¼ bþ a ¼ RHS:Case II: when b ¼ 1: LHS ¼ aþ b ¼ aþ1 ¼ 1 ¼ 1þ a ¼ bþ a ¼ RHS:Case III: when a 6¼ 1; and b 6¼ 1: This case is trivial. ■)

Problem 1.101 þ is associative over 0;1½ �:(Solution Let us take any a; b; c 2 0;1½ �: We have to show thataþ bð Þþ c ¼ aþ bþ cð Þ:Case I: when a ¼ 1: LHS ¼ aþ bð Þþ c ¼ 1þ bð Þþ c ¼ 1þ c ¼ 1 ¼

1þ bþ cð Þ ¼ aþ bþ cð Þ ¼ RHS:Case II: when b ¼ 1: LHS ¼ aþ1ð Þþ c ¼ 1þ c ¼ 1 ¼ aþ1 ¼

aþ 1þ cð Þ ¼ aþ bþ cð Þ ¼ RHS:Case III: when c ¼ 1: LHS ¼ aþ bð Þþ1 ¼ 1 ¼ aþ1 ¼ aþ bþ1ð Þ ¼

aþ bþ cð Þ ¼ RHS:Case IV: when 6¼ 1; b 6¼ 1, and c 6¼ 1: This case is trivial. ■)It is clear that 0 serves the purpose of the additive identity in 0;1½ �: Let us

‘extend’ the definition of � (multiplication) over 0;1½ Þ to 0;1½ � as follows: Forevery a 2 0;1ð �,

a � 1 � 1; and 1 � a � 1:

Further, we define that 0 � 1 � 0; and 1 � 0 � 0: Thus, for every a 2 0;1½ �,0 � a ¼ 0 ¼ a � 0: Here, � : 0;1½ � 0;1½ � ! 0;1½ �; that is, � is a binary operationover 0;1½ �:Problem 1.102 � is commutative over 0;1½ �:(Solution Let us take any a; b 2 0;1½ �: We have to show that a � b ¼ b � a:

When a ¼ 0; LHS ¼ a � b ¼ 0 � b ¼ 0 ¼ b � 0 ¼ b � a ¼ RHS: Similarly, whenb ¼ 0; a � b ¼ b � a: So, it suffices to show that a � b ¼ b � a for every a; b 2 0;1ð �:

Case Ia: when a ¼ 1: LHS ¼ a � b ¼ 1 � b ¼ 1 ¼ b � 1 ¼ b � a ¼ RHS:Case Ib: when b ¼ 1: This case is similar to the case Ia.Case II: when a 2 0;1ð Þ; and b 2 0;1ð Þ: This is a trivial case. ■)

Problem 1.103 � is associative over 0;1½ �:(Solution Let us take any a; b; c 2 0;1½ �: We have to show thata � bð Þ � c ¼ a � b � cð Þ:


When a ¼ 0;LHS ¼ a � bð Þ � c ¼ 0 � bð Þ � c ¼ 0 � c ¼ 0 ¼ 0 � b � cð Þ ¼ a � b � cð Þ ¼ RHS:

The cases when b ¼ 0 or c ¼ 0. So, it suffices to show that a � bð Þ � c ¼a � b � cð Þ; for every a; b; c 2 0;1ð �:

Case Ia: when a ¼ 1: LHS ¼ a � bð Þ � c ¼ 1 � bð Þ � c ¼ 1 � c ¼ 1 ¼1 � b � cð Þ ¼ a � b � cð Þ ¼ RHS:

Case Ib: when b ¼ 1: This case is similar to the case Ia.Case Ic: when c ¼ 1: This case is similar to the case Ia.Case II: when a 2 0;1ð Þ; b 2 0;1ð Þ; and c 2 0;1ð Þ: This is a trivial case. ■)It is clear that 1 serves the purpose of the multiplicative identity in 0;1½ �: Also,

Problem 1.104 � distributes over þ in 0;1½ �:(Solution Let us take any a; b; c 2 0;1½ �: We have to show thata � bþ cð Þ ¼ a � bð Þþ a � cð Þ:

Case I: when a ¼ 0: LHS ¼ a � bþ cð Þ ¼ 0 � bþ cð Þ ¼ 0 ¼ 0þ 0 ¼0 � bð Þþ 0 � cð Þ ¼ a � bð Þþ a � cð Þ ¼ RHS:Case II: when b ¼ 0: LHS ¼ a � bþ cð Þ ¼ a � 0þ cð Þ ¼ a � c ¼ 0þ a � cð Þ ¼

a � 0ð Þþ a � cð Þ ¼ a � bð Þþ a � cð Þ ¼ RHS:Case III: when c ¼ 0: This case is similar to case II.Now, it suffices to show that a � bþ cð Þ ¼ a � bð Þþ a � cð Þ for every

a; b; c 2 0;1ð �:Case IV(a): when a 2 0;1ð Þ; and c ¼ 1: LHS ¼ a � bþ cð Þ ¼ a � bþ1ð Þ ¼

a � 1 ¼ 1 ¼ a � bð Þþ1

¼ a � bð Þþ a � 1ð Þ ¼ a � bð Þþ a � cð Þ ¼ RHS:

Case IV(b): when a 2 0;1ð Þ; and b ¼ 1: This case is similar to case IVa.Case IV(c): when a 2 0;1ð Þ; b 2 0;1ð Þ; and c 2 0;1ð Þ: This is a trivial case.Case V: when a ¼ 1: LHS ¼ a � bþ cð Þ ¼ 1 � bþ cð Þ ¼ 1 ¼ 1þ1 ¼

1 � bð Þþ 1 � cð Þ ¼ a � bð Þþ a � cð Þ ¼ RHS: ■)

Problem 1.105 Let a 2 0;1½ Þ; and b; c 2 0;1½ �: Let aþ b ¼ aþ c: Then b ¼ c:

(Solution Case I: when b 2 0;1½ Þ: Since a; b 2 0;1½ Þ; aþ c ¼ð Þaþ b 2 0;1½ Þ;and hence aþ c 2 0;1½ Þ: Since aþ c 2 0;1½ Þ; and a 2 0;1½ Þ; we have c 6¼ 1:Since c 6¼ 1; and c 2 0;1½ �; we have c 2 0;1½ Þ: Since a; b; c 2 0;1½ Þ; andaþ b ¼ aþ c; we have b ¼ c:

Case II: when b ¼ 1: Since b ¼ 1; and a 2 0;1½ Þ; aþ c ¼ð Þaþ b ¼ 1; andhence aþ c ¼ 1: Since aþ c ¼ 1; a 2 0;1½ Þ; we have c ¼ 1 ¼ bð Þ; and henceb ¼ c:

Thus, in all cases, b ¼ c: ■)

Problem 1.106 Let a 2 0;1ð Þ; and b; c 2 0;1½ �: Let a � b ¼ a � c: Then b ¼ c:

(Solution Case I: when b 2 0;1½ Þ: Since a 2 0;1ð Þ; and b 2 0;1½ Þ; a � c ¼ð Þa �b 2 0;1½ Þ; and hence a � c 2 0;1½ Þ: Since a � c 2 0;1½ Þ; and a 2 0;1ð Þ; we

1.3 Integration of Positive Functions 53

have c 6¼ 1: Since c 6¼ 1; and c 2 0;1½ �; we have c 2 0;1½ Þ: Since a 2 0;1ð Þ;b; c 2 0;1½ Þ; and a � b ¼ a � c; we have b ¼ c:

Case II: when b ¼ 1: Since b ¼ 1; and a 2 0;1ð Þ; a � c ¼ð Þa � b ¼ 1; andhence a � c ¼ 1: Since a � c ¼ 1; a 2 0;1ð Þ; we have c ¼ 1 ¼ bð Þ; and henceb ¼ c:

Thus, in all cases, b ¼ c: ■)

Lemma 1.107 For every n ¼ 1; 2; . . .; let an; bn 2 0;1½ �: Let a1 � a2 � a3 � � � � ;and b1 � b2 � b3 � � � � : Let limn!1 an ¼ a; and limn!1 bn ¼ b: (Since eachan; bn 2 0;1½ �; each an þ bnð Þ 2 0;1½ �:) Then

limn!1 an þ bnð Þ ¼ aþ bð Þ:

Proof Case I: when a ¼ 1; and b 2 0;1½ Þ: Since 0� a1 � a2 � a3 � � � � ; andlimn!1 an ¼ a ¼ 1ð Þ; either there exists a positive integer n0 such that an0 ¼ 1;or a1; a2; a3; . . .f g is a set of real numbers that is not bounded above.

Case I(a): when there exists a positive integer n0 such that an0 ¼ 1: Since0� b1 � b2 � b3 � � � � �1; and limn!1 bn ¼ b 2 0;1½ Þ; each bn 2 0;1½ Þ: Sincean0 ¼ 1; and a1 � a2 � a3 � � � � ; we have an ¼ 1 for every n� n0: Since an ¼ 1for every n� n0; and each bn 2 0;1½ Þ; an þ bn ¼ 1 for every n� n0; and hence

limn!1 an þ bnð Þ ¼ 1 ¼ 1þ b ¼ aþ bð Þ:

Case I(b): when a1; a2; a3; . . .f g is a set of real numbers that is not boundedabove. Since 0� b1 � b2 � b3 � � � � �1; and limn!1 bn ¼ b 2 0;1½ Þ; each bn 20;1½ Þ: Now, since a1; a2; a3; . . .f g is a set of real numbers that is not boundedabove, a1 þ b1; a2 þ b2; a3 þ b3; . . .f g is a set of real numbers that is not boundedabove. Also, a1 þ b1 � a2 þ b2 � a3 þ b3 � � � � : It follows that

limn!1 an þ bnð Þ ¼ 1 ¼ 1þ b ¼ aþ bð Þ:

Thus, limn!1 an þ bnð Þ ¼ aþ bð Þ:Case II: when b ¼ 1; and a 2 0;1½ Þ: This case is similar to the case I.Case III: when a 2 0;1½ Þ; and b 2 0;1½ Þ: Since 0� b1 � b2 � b3 � � � � �1;

and limn!1 bn ¼ b 2 0;1½ Þ; each bn 2 0;1½ Þ: Similarly, each an 2 0;1½ Þ: Sinceeach an 2 0;1½ Þ; each bn 2 0;1½ Þ;

limn!1 an ¼ a; lim

n!1 bn ¼ b; a 2 0;1½ Þ; and b 2 0;1½ Þ;

it is known that

limn!1 an þ bnð Þ ¼ aþ bð Þ:

We have seen that, in all cases, limn!1 an þ bnð Þ ¼ aþ bð Þ: ∎


Lemma 1.108 For every n ¼ 1; 2; . . .; let an; bn 2 0;1½ �: Let a1 � a2 � a3 � � � � ;and b1 � b2 � b3 � � � � : Let limn!1 an ¼ a; and limn!1 bn ¼ b: Since eachan; bn 2 0;1½ �; each an � bnð Þ 2 0;1½ �: Then

limn!1 an � bnð Þ ¼ a � bð Þ:

Proof Case I: when a ¼ 0: Since 0� a1 � a2 � a3 � � � � ; and limn!1 an ¼a ¼ 0ð Þ; we have each an ¼ 0; and hence each an � bnð Þ ¼ 0: This shows thatlimn!1 an � bnð Þ ¼ 0 ¼ 0 � b ¼ a � bð Þ; and hence,


Case II: when b ¼ 0: This case is similar to case I.Case III: when a ¼ 1; and b 2 0;1ð Þ: Since a1 � a2 � a3 � � � � ; and

limn!1 an ¼ a ¼ 1ð Þ; either there exists a positive integer n0 such that an0 ¼ 1;or a1; a2; a3; . . .f g is a set of real numbers that is not bounded above.

Case III(a): when there exists a positive integer n0 such that an0 ¼ 1: Since0� b1 � b2 � b3 � � � � �1; and limn!1 bn ¼ b 2 0;1ð Þ; each bn 2 0;1½ Þ:Since each bn 2 0;1½ Þ; and limn!1 bn ¼ b 2 0;1ð Þ; there exists a positive inte-ger n1 � n0 such that bn 2 0;1ð Þ for every n� n1: Since an0 ¼ 1; n1 � n0; anda1 � a2 � a3 � � � � ; we have an ¼ 1 for every n� n1: Since an ¼ 1 for everyn� n1; and bn 2 0;1ð Þ for every n� n1; we have an � bnð Þ ¼ 1 for every n� n1;and hence

limn!1 an � bnð Þ ¼ 1 ¼ 1 � b ¼ a � bð Þ:

Thus, limn!1 an � bnð Þ ¼ a � bð Þ:Case III(b): when a1; a2; a3; . . .f g is a set of real numbers that is not bounded

above. Since 0� b1 � b2 � b3 � � � � �1; and limn!1 bn ¼ b 2 0;1ð Þ; each bn 20;1½ Þ: Since each bn 2 0;1½ Þ; and limn!1 bn ¼ b 2 0;1ð Þ; there exists a posi-tive integer n0 [ 1 such that bn 2 0;1ð Þ for every n� n0: Since a1; a2; a3; . . .f g isa set of real numbers that is not bounded above,

a1; . . .; an0�1; an0 � bn0 ; an0 þ 1 � bn0 ; an0 þ 2 � bn0 ; an0 þ 2 � bn0 ; . . .f g

is a set of real numbers that is not bounded above. Since 0\bn0 � bn0 þ 1\1; and

an0 þ 1 2 0;1½ Þ; an0 þ 1 � bn0 � an0 þ 1 � bn0 þ 1\1:

Since

a1; . . .; an0�1; an0 � bn0 ; an0 þ 1 � bn0 ; an0 þ 2 � bn0 ; an0 þ 2 � bn0 ; . . .f g

is a set of real numbers that is not bounded above, and


an0 þ 1 � bn0 � an0 þ 1 � bn0 þ 1\1;

a1; . . .; an0�1; an0 � bn0 ; an0 þ 1 � bn0 þ 1; an0 þ 2 � bn0 ; an0 þ 3 � bn0 ; . . .f g

is a set of real numbers that is not bounded above. Similarly,

a1; . . .; an0�1; an0 � bn0 ; an0 þ 1 � bn0 þ 1; an0 þ 2 � bn0 þ 2; an0 þ 3 � bn0 ; . . .f g

is a set of real numbers that is not bounded above, and

a1; . . .; an0�1; an0 � bn0 ; an0 þ 1 � bn0 þ 1; an0 þ 2 � bn0 þ 2; an0 þ 3 � bn0 þ 3; . . .f g

is a set of real numbers that is not bounded above, etc. It follows that

limn!1 an � bnð Þ ¼ 1 ¼ 1 � b ¼ a � bð Þ:

Thus, limn!1 an � bnð Þ ¼ a � bð Þ:Case IV: when b ¼ 1; and a 2 0;1ð Þ: This case is similar to Case III.Case V: when a ¼ 1; and b ¼ 1: Since 0� a1 � a2 � a3 � � � � ; and

limn!1 an ¼ a ¼ 1ð Þ; either there exists a positive integer n0 such that an0 ¼ 1;or a1; a2; a3; . . .f g is a set of real numbers that is not bounded above. Similarly,either there exists a positive integer n1 such that bn1 ¼ 1; or b1; b2; b3; . . .f g is a setof real numbers that is not bounded above.

Case V(a): when there exists a positive integer n0 such that an0 ¼ 1; and thereexists a positive integer n1 such that bn1 ¼ 1: Clearly, an � bn ¼ 1 for everyn�max n0; n1f g; and hence

limn!1 an � bnð Þ ¼ 1 ¼ 1 �1 ¼ a � bð Þ:

Thus,


Case V(b): when there exists a positive integer n0 such that an0 ¼ 1; andb1; b2; b3; . . .f g is a set of real numbers that is not bounded above. It follows that

there exists a positive number n1 � n0 such that 0\bn for every n� n1; and hencean � bn ¼ 1 � bn ¼ 1ð Þ for every n� n1: Hence,

limn!1 an � bnð Þ ¼ 1 ¼ 1 �1 ¼ a � bð Þ:

Thus, limn!1 an � bnð Þ ¼ a � bð Þ:Case V(c): when there exists a positive integer n1 such that bn1 ¼ 1; and

a1; a2; a3; . . .f g is a set of real numbers that is not bounded above. This case issimilar to case Vb.


Case V(d): when a1; a2; a3; . . .f g is a set of real numbers that is not boundedabove, and b1; b2; b3; . . .f g is a set of real numbers that is not bounded above. Itfollows that a1 � b1; a2 � b2; a3 � b3; . . .f g is a set of real numbers that is not boundedabove, and hence

limn!1 an � bnð Þ ¼ 1 ¼ 1 �1 ¼ a � bð Þ:

Thus,


Case VI: when a 2 0;1ð Þ; and b 2 0;1ð Þ: Since 0� a1 � a2 � a3 � � � � ; andlimn!1 an ¼ a 2 0;1ð Þ; each an 2 0;1½ Þ: Similarly, each bn 2 0;1½ Þ: Sinceeach an 2 0;1½ Þ; each bn 2 0;1½ Þ; limn!1 an ¼ a; and limn!1 bn ¼ b; it isknown that


We have seen that, in all cases, limn!1 an � bnð Þ ¼ a � bð Þ: ∎

Lemma 1.109 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letf : X ! 0;1½ � be a measurable function, and g : X ! 0;1½ � be a measurablefunction. Then f þ gð Þ : x 7! f xð Þþ g xð Þð Þ from X to 0;1½ � is a measurablefunction.

Proof Let us take any a 2 0;1½ Þ: By Lemma 1.77, it suffices to show thatf þ gð Þ�1 a;1ð �ð Þ 2 ℳ: Observe that

f þ gð Þ�1 a;1ð �ð Þ ¼ x : f þ gð Þ xð Þ 2 a;1ð �f g ¼ x : a\ f þ gð Þ xð Þ�1f g¼ x : 0�ð Þa\f xð Þþ g xð Þ�1f g¼ x : 0�ð Þa\f xð Þþ g xð Þ\1f g[ x : f xð Þ ¼ 1f g[ x : g xð Þ ¼ 1f g¼ x : a� f xð Þ\g xð Þ\1 and f xð Þ 2 0;1ð Þf g[ x : f xð Þ ¼ 1f g[ x : g xð Þ ¼ 1f g¼ [ r2Q x : a� f xð Þ\r\g xð Þ\1 and f xð Þ 2 0;1ð Þf gð Þ [ x : f xð Þ ¼ 1f g[ x : g xð Þ ¼ 1f g

Now since Q; the set of all rational numbers, is countable, and ℳ is a r-algebra,it suffices to show that for every r 2 Q;

x : a� f xð Þ\r\g xð Þ\1 and f xð Þ 2 0;1ð Þf g 2 ℳ;

x : f xð Þ ¼ 1f g 2 ℳ; and x : g xð Þ ¼ 1f g 2 ℳ:

Observe that, for every r 2 Q; we have


x : a� f xð Þ\r\g xð Þ\1 and f xð Þ 2 0;1ð Þf g¼ x : a� r\f xð Þ and r\g xð Þ\1 and f xð Þ 2 0;1ð Þf g¼ f�1 a� r;1ð Þð Þ \ g�1 r;1ð Þð Þ:

Let us fix any r 2 Q: Since a� r;1ð Þ\ 0;1½ � is open in 0;1½ �; and f ismeasurable,

f�1 a� r;1ð Þð Þ ¼ f�1 a� r;1ð Þð Þ \X

¼ f�1 a� r;1ð Þð Þ \ f�1 0;1½ �ð Þ¼ f�1 a� r;1ð Þ\ 0;1½ �ð Þ 2 ℳ

and hence f�1 a� r;1ð Þð Þ 2 ℳ: Similarly, g�1 r;1ð Þð Þ 2 ℳ: Sincef�1 a� r;1ð Þð Þ 2 ℳ; g�1 r;1ð Þð Þ 2 ℳ; and ℳ is a r-algebra,

x : a� f xð Þ\r\g xð Þ\1 and f xð Þ 2 0;1ð Þf g¼ f�1 a� r;1ð Þð Þ \ g�1 r;1ð Þð Þ 2 ℳ;

and hence x : a� f xð Þ\r\g xð Þ\1 and f xð Þ 2 0;1ð Þf g 2 ℳ: Now we want toshow that x : f xð Þ ¼ 1f g 2 ℳ:

Since f is a measurable function, and ℳ is a r-algebra,

f�1 1ð Þ ¼ f�1 \ r2Q r;1ð �ð Þ ¼ \ r2Q f�1 r;1ð �ð Þ� � 2 ℳ;

and hence f�1 1ð Þ 2 ℳ: Similarly, g�1 1ð Þ 2 ℳ: ■

Lemma 1.110 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letf : X ! 0;1½ � be a measurable function, and g : X ! 0;1½ � be a measurablefunction. Then f � gð Þ : x 7! f xð Þð Þ g xð Þð Þ from X to 0;1½ � is a measurable function.

(Observe that, for every c 2 0;1½ Þ; the constant function c : x 7! c from X to0;1½ � is a measurable function. Now, by Lemma 1.110, if f : X ! 0;1½ � is ameasurable function, and c 2 0;1½ Þ; the product cfð Þ : x 7! cð Þ f xð Þð Þ from X to0;1½ � is a measurable function.)

Proof Let us take any a 2 0;1½ Þ: By Lemma 1.77, it suffices to show that

f � gð Þ�1 a;1ð �ð Þ 2 ℳ:


Observe that

f � gð Þ�1 a;1ð �ð Þ ¼ x : f � gð Þ xð Þ 2 a;1ð �f g¼ x : a\ f � gð Þ xð Þ�1f g ¼ x : 0�ð Þa\ f xð Þð Þ g xð Þð Þ�1f g¼ x : 0�ð Þa\ f xð Þð Þ g xð Þð Þ\1f g

[ x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g

¼ x :a

f xð Þ\g xð Þ\1 and 0\f xð Þ\1� �[ x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g

¼ [ r2Qþ x :a

f xð Þ\r\g xð Þ\1 and 0\f xð Þ\1� ��

[ x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g

Now since Qþ ; the set of all positive rational numbers, is countable, and ℳ is ar-algebra, it suffices to show that for every r 2 Qþ ;

f �1 ar;1

� � \ g�1 r;1ð Þð Þ

¼ x :ar\f xð Þ and r\g xð Þ\1

n o¼ x :

af xð Þ\r\g xð Þ\1 and 0\f xð Þ\1

� �2 ℳ;

and

x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g 2 ℳ:

For this purpose, let us fix any r in Qþ : Here, ar ;1� �

is open in 0;1½ �; andf : X ! 0;1½ � is a measurable function, so f�1 a

r ;1� �� 2 ℳ: Similarly,

g�1 r;1ð Þð Þ 2 ℳ: Since f�1 ar ;1� �� 2 ℳ; g�1 r;1ð Þð Þ 2 ℳ; and ℳ is a r-

algebra,

x :a

f xð Þ\r\g xð Þ\1 and 0\f xð Þ\1� �

¼ f�1 ar;1

� � \ g�1 r;1ð Þð Þ 2 ℳ;

and hence

x :a

f xð Þ\r\g xð Þ\1 and 0\f xð Þ\1� �

2 ℳ:


Now, it remains for us to show that

x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g 2 ℳ:

Here,

x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g ¼ f�1 1ð Þ\ g�1 1ð Þ:

Since f is a measurable function, and ℳ is a r-algebra,

f�1 1ð Þ ¼ f�1 \ r2Qþ r;1ð �� ¼ \ r2Qþ f�1 r;1ð �ð Þ� � 2 ℳ;

and hence f�1 1ð Þ 2 ℳ: Similarly, g�1 1ð Þ 2 ℳ: Since f�1 1ð Þ; g�1 1ð Þ 2 ℳ;and ℳ is a r-algebra, we have

x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g¼ f�1 1ð Þ\ g�1 1ð Þ 2 ℳ;

and hence x : f xð Þð Þ g xð Þð Þ ¼ 1; f xð Þ 2 0;1ð � and g xð Þ 2 0;1ð �f g 2 ℳ: ■

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let s : X ! 0;1½ Þ be a simple measurablefunction. Let E 2 ℳ: Since s : X ! 0;1½ Þ is a simple function, there existsfinite-many distinct a1; . . .; an in 0;1½ Þ such that s Xð Þ ¼ a1; . . .; anf g:

Now, since s : X ! 0;1½ Þ is a measurable function, we haves�1 a1ð Þ; . . .; s�1 anð Þ 2 ℳ; and

s ¼ a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ:

Since each s�1 aið Þ 2 ℳ; E 2 ℳ; and ℳ is a r-algebra, each s�1 aið Þð Þ \E 2ℳ; and hence, each l s�1 aið Þð Þ \Eð Þ 2 0;1½ �: Since each l s�1 aið Þð Þ \Eð Þ 20;1½ �; and each ai 2 0;1½ Þ; each product aið Þ l s�1 aið Þð Þ \Eð Þð Þ 2 0;1½ �: It fol-lows that

a1ð Þ l s�1 a1ð Þ� �\E� �� þ � � � þ anð Þ l s�1 anð Þ� �\E

� �� 2 0;1½ �:

Here

a1ð Þ l s�1 a1ð Þ� �\E� �� þ � � � þ anð Þ l s�1 anð Þ� �\E

� �� is denoted by

RE sdl; and is called the Lebesgue integral of s over E; with respect to

the measure l:Let X be any nonempty set. Letℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be a

positive measure on ℳ: Let E 2 ℳ: Let s : X ! 0;1½ Þ; and t : X ! 0;1½ Þ besimple measurable functions. Suppose that, for every x 2 X; s xð Þ� t xð Þ:


Problem 1.111RE sdl�

RE tdl:

(Solution Since s : X ! 0;1½ Þ is a simple function, there exists finite-many dis-tinct a1; . . .; an in 0;1½ Þ such that s Xð Þ ¼ a1; . . .; anf g: Now, since s : X ! 0;1½ Þis a measurable function, we have

s�1 a1ð Þ; . . .; s�1 anð Þ 2 ℳ:

Since t : X ! 0;1½ Þ is a simple function, there exists finite-many distinctb1; . . .; bm in 0;1½ Þ such that t Xð Þ ¼ b1; . . .; bmf g: Now, since t : X ! 0;1½ Þ is ameasurable function, we have

t�1 b1ð Þ; . . .; t�1 bmð Þ 2 ℳ:

For simplicity of argument, suppose that

s Xð Þ ¼ a1; a2; a3f g; and t Xð Þ ¼ b1; b2f g:


a1 l s�1 a1ð Þ� �\E� �� þ a2 l s�1 a2ð Þ� �\E

� �� þ a3 l s�1 a3ð Þ� �\E� ��

� b1 l t�1 b1ð Þ� �\E� �� þ b2 l t�1 b2ð Þ� �\E

� �� :

LHS ¼ a1 l s�1 a1ð Þ� �\E� �� þ a2 l s�1 a2ð Þ� �\E

� �� þ a3 l s�1 a3ð Þ� �\E� ��

¼ a1 l s�1 a1ð Þ \ t�1 b1ð Þ \E� �þ l s�1 a1ð Þ \ t�1 b2ð Þ \E

� �� þ a2 l s�1 a2ð Þ \ t�1 b1ð Þ \E

� �þ l s�1 a2ð Þ \ t�1 b2ð Þ \E� ��

þ a3 l s�1 a3ð Þ \ t�1 b1ð Þ \E� �þ l s�1 a3ð Þ \ t�1 b2ð Þ \E

� �� ¼ a1 l s�1 a1ð Þ \ t�1 b1ð Þ \E

� �� þ a1 l s�1 a1ð Þ \ t�1 b2ð Þ \E� ��

þ a2 l s�1 a2ð Þ \ t�1 b1ð Þ \E� �� þ a2 l s�1 a2ð Þ \ t�1 b2ð Þ \E

� �� þ a3 l s�1 a3ð Þ \ t�1 b1ð Þ \E

� �� þ a3 l s�1 a3ð Þ \ t�1 b2ð Þ \E� ��

� b1 l s�1 a1ð Þ \ t�1 b1ð Þ \E� �� þ b2 l s�1 a1ð Þ \ t�1 b2ð Þ \E

� �� þ b1 l s�1 a2ð Þ \ t�1 b1ð Þ \E

� �� þ b2 l s�1 a2ð Þ \ t�1 b2ð Þ \E� ��

þ b1 l s�1 a3ð Þ \ t�1 b1ð Þ \E� �� þ b2 l s�1 a3ð Þ \ t�1 b2ð Þ \E

� �� ¼ b1 l s�1 a1ð Þ \ t�1 b1ð Þ \E

� �þ l s�1 a2ð Þ \ t�1 b1ð Þ \E� �þl s�1 a3ð Þ \ t�1 b1ð Þ \E

� �� þ b2 l s�1 a1ð Þ \ t�1 b2ð Þ \E

� �þ l s�1 a2ð Þ \ t�1 b2ð Þ \E� �þ l s�1 a3ð Þ \ t�1 b2ð Þ \E

� �� ¼ b1 l t�1 b1ð Þ \E

� �� þ b2 l t�1 b2ð Þ \E� �� ¼ RHS:

∎)Let X be any nonempty set. Letℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be a

positive measure on ℳ: Let E 2 ℳ: Let s : X ! 0;1½ Þ be a simple measurablefunction. From the above discussion, we find that


supZE

tdl : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� s

8<:

9=;

¼ maxZE


8<:

9=;

¼ZE

sdl;

and hence

ZE

sdl ¼ supZE


8<:

9=;:

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let f : X ! 0;1½ � be a mea-surable function. By Lemma 1.98, there exists a sequence snf g of simple mea-surable functions sn : X ! 0;1½ Þ such that for every x in X;0� s1 xð Þ� s2 xð Þ� s3 xð Þ� � � �, and

s1 xð Þ� sup s1 xð Þ; s2 xð Þ; s3 xð Þ; . . .f g ¼ limn!1 sn xð Þ ¼ f xð Þ:

Hence 0� s1 � f : It follows that

ZE

tdl : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� f

8<:

9=;

is a nonempty subset of 0;1½ �; and hence

supZE


8<:

9=;

exists, and is a member of 0;1½ �: Here

supZE


8<:

9=;


is denoted byRE f dl; and is called the Lebesgue integral of f over E; with respect

to the measure l:

Lemma 1.112 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let f : X ! 0;1½ �; andg : X ! 0;1½ � be measurable functions. Let f � g: ThenZ

E

f dl�ZE

gdl:

Proof Since 0� f � g; we have

t : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� ff g s : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� gf g;

and hence

ZE


8<:

9=;

ZE

sdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� g

8<:

9=;:

It follows that

ZE

f dl ¼ supZEtdl : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� f

� �

� supZE

sdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� g

8<:

9=;

¼ZE

gdl;

and therefore,RE f dl�

RE gdl: ∎

Lemma 1.113 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let A;B 2 ℳ: Let A B: Let f :X ! 0;1½ � be a measurable function. ThenZ

A

f dl�ZB

f dl:


Proof Case I: when f is a simple function. It follows that there exist finite-manydistinct points a1; . . .; an in 0;1½ Þ such that f Xð Þ ¼ a1; . . .; anf g: Now, since f :X ! 0;1½ Þ is a measurable function, we have

f�1 a1ð Þ; . . .; f�1 anð Þ 2 ℳ:

Since A;B 2 ℳ; each f�1 aið Þ 2 ℳ; and ℳ is a r-algebra, each f�1 aið Þ \A;f�1 aið Þ \B 2 ℳ: Since A B; we have, for each i ¼ 1; . . .; n; f�1 aið Þ \A f�1 aið Þ \B; and hence, for each i ¼ 1; . . .; n;

l f�1 aið Þ \A� �� l f�1 aið Þ \B

� �:

Now since each ai 2 0;1½ Þ; for each i ¼ 1; . . .; n;

aið Þ l f�1 aið Þ \A� �� aið Þ l f�1 aið Þ \B

� �� ;

and henceZA

f dl ¼ a1ð Þ l f�1 a1ð Þ \A� �� þ � � � þ anð Þ l f�1 anð Þ \A

� �� a1ð Þ l f�1 a1ð Þ \B

� �� þ � � � þ anð Þ l f�1 anð Þ \B� ��

¼ZB

f dl:

Thus, ZA

f dl�ZB

f dl:

Case II: when f is not a simple function. By Case I,

ZA

f dl ¼ supZA

sdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� f

8<:

9=;

� supZB


8<:

9=;

¼ZB

f dl:


Thus, ZA

f dl�ZB

f dl:

So, in all cases,RA f dl�

RB f dl: ∎

Lemma 1.114 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let f : X ! 0;1½ � be ameasurable function. Let c 2 0;1½ Þ: ThenZ

E

cfð Þdl ¼ cZE

f dl:

Proof

Problem 1:115 For every simple measurable function s : X ! 0;1½ Þ;ZE

csð Þdl ¼ cZE

sdl:

(Solution Case I: when c ¼ 0: This case is trivial.Case II: when c 6¼ 0: There exists finite-many distinct a1; . . .; an in 0;1½ Þ such

that f Xð Þ ¼ a1; . . .; anf g: Now, since f : X ! 0;1½ Þ is a measurable function, wehave f�1 a1ð Þ; . . .; f�1 anð Þ 2 ℳ: Since f Xð Þ ¼ a1; . . .; anf g; we have

cfð Þ Xð Þ ¼ ca1; . . .; canf g:

Since, c 6¼ 0; for each i ¼ 1; . . .; n; we have cfð Þ�1 caið Þ ¼ f�1 aið Þ: We have toshow that

ca1ð Þ l cfð Þ�1 ca1ð Þ \E� �

þ � � � þ canð Þ l cfð Þ�1 canð Þ \E� �

¼ c a1ð Þ l f�1 a1ð Þ \E� �� þ � � � þ anð Þ l f �1 anð Þ \E

� �� :

LHS ¼ ca1ð Þ l cfð Þ�1 ca1ð Þ \E� �

þ � � � þ canð Þ l cfð Þ�1 canð Þ \E� �

¼ c a1ð Þ l cfð Þ�1 ca1ð Þ \E� �

þ � � � þ anð Þ l cfð Þ�1 canð Þ \E� � �

¼ c a1ð Þ l f�1 a1ð Þ \E� �� þ � � � þ anð Þ l f �1 anð Þ \E

� �� ¼ RHS:

∎)Let us observe that

t : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� cff g¼ cs : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� ff g:


Now,

LHS ¼ZE

cfð Þdl

¼ supZE

tdl : t : X ! 0;1½ Þ is a simple measurable function satisfying 0� t� cf

8<:

9=;

¼ supZE

csð Þdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� f

8<:

9=;

¼ sup cZE


8<:

9=;

¼ c supZE


8<:

9=;

0@

1A

¼ cZE

f dl ¼ RHS: ∎)

Lemma 1.116 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let c 2 0;1½ Þ: ThenZ

E

cdl ¼ cð Þ l Eð Þð Þ:

(It follows thatRE 0dl ¼ 0ð Þ l Eð Þð Þ ¼ 0ð Þ: Thus RE 0dl ¼ 0:)

Proof Here, the constant function c is a simple measurable function, so

LHS ¼ZE

cdl ¼ cð Þ l x : c xð Þ ¼ cf g\Eð Þð Þ

¼ cð Þ l X \Eð Þð Þ ¼ cð Þ l Eð Þð Þ ¼ RHS: ∎

Lemma 1.117 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ; and l Eð Þ ¼ 0: Letf : X ! 0;1½ � be a measurable function. Then

1.RE 1dl ¼ 0; 2.

RE f dl ¼ 0:


Proof

1. Here, the constant function 1 is a measurable function, and

s : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s�1f g¼ s : s : X ! 0;1½ Þ is a simple measurable functionf g:

Let us fix any simple measurable function s : X ! 0;1½ Þ: There existsfinite-many distinct a1; . . .; an in 0;1½ Þ such that f Xð Þ ¼ a1; . . .; anf g; andf�1 a1ð Þ; . . .; f�1 anð Þ 2 ℳ: Now, since each f�1 aið Þ \E E; we have

0�ð Þl f�1 aið Þ \E� �� l Eð Þ ¼ 0ð Þ;

and hence, each l f�1 aið Þ \Eð Þ ¼ 0: It follows that

ZE

sdl ¼Xni¼1

aið Þ l f�1 aið Þ \E� �� ¼Xn

i¼1

aið Þ 0ð Þ ¼ 0:

Thus, for every simple measurable function s : X ! 0;1½ Þ; RE sdl ¼ 0: Now,

LHS ¼ZE

1dl ¼ supZEsdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s�1

� �

¼ supZE

sdl : s : X ! 0;1½ Þ is a simple measurable function

8<:

9=; ¼ sup 0f g ¼ 0 ¼ RHS:

2. Since, for every x 2 X; f xð Þ�1 ¼ 1 xð Þð Þ; by Lemma 1.112,RE f dl�

RF 1dl: By 1,

0�ZE

f dl�ZE

1dl ¼ 0

|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};

soRE f dl ¼ 0: ∎

Lemma 1.118 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E;F 2 ℳ: Let E\F ¼ ;: Lets : X ! 0;1½ � be a simple measurable function. ThenZ

E[F

sdl ¼ZE

sdlþZF

sdl:


Proof There exists finite-many distinct a1; . . .; an in 0;1½ Þ such that f Xð Þ ¼a1; . . .; anf g; and f�1 a1ð Þ; . . .; f�1 anð Þ 2 ℳ: We have to prove that

Xni¼1

aið Þ l f�1 aið Þ \ E [Fð Þ� �� ¼Xni¼1

aið Þ l f�1 aið Þ \E� �� þ Xn

i¼1

aið Þ l f�1 aið Þ \F� ��

:

Since E \F ¼ ;; each f�1 aið Þ \Eð Þ \ f�1 aið Þ \Fð Þ ¼ ;:

LHS ¼Xni¼1

aið Þ l f�1 aið Þ \ E [Fð Þ� �� ¼Xni¼1

aið Þ l f�1 aið Þ \E� �[ f�1 aið Þ \F

� �� ¼Xni¼1

aið Þ l f�1 aið Þ \E� �þ l f�1 aið Þ \F

� �� ¼Xni¼1

aið Þ l f�1 aið Þ \E� �� þ aið Þ l f�1 aið Þ \F

� �� ¼Xni¼1

aið Þ l f�1 aið Þ \E� �� þ Xn

i¼1

aið Þ l f�1 aið Þ \F� ��

¼ RHS:

∎

Lemma 1.119 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let f : X ! 0;1½ � be ameasurable function. Then Z

X

vE � fð Þdl�ZE

f dl:

Proof Since, E 2 ℳ; the characteristic function vE : X ! 0;1½ � is a simplemeasurable function. Since vE : X ! 0;1½ � is a measurable function, and f : X !0;1½ � is a measurable function, by Lemma 1.110, their product vE � fð Þ : X !0;1½ � is a measurable function. Since vE � fð Þ : X ! 0;1½ � is a measurablefunction, and X 2 ℳ;

RX vE � fð Þdl exists. It is clear that vE � fð Þ� f : Since


ZX

vE � fð Þdl ¼Z

E [ Ecð Þ

vE � fð Þdl

¼ supZ

E [ Ecð Þ

sdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� vE � fð Þ

8><>:

9>=>;

¼ supZE

sdlþZEcð Þ


8><>:

9>=>;

� supZE

sdl : s : X ! 0;1½ Þ is a simple measurable function satisfying 0� s� vE � fð Þ8<:

9=;

þ supZEcð Þ


8><>:

9>=>;

¼ZE

vE � fð ÞdlþZEcð Þ

vE � fð Þdl;

we have ZX


vE � fð ÞdlþZEcð Þ

vE � fð Þdl:

Problem 1:120R

Ecð Þ vE � fð Þdl ¼ 0:

(Solution Let s : X ! 0;1½ Þ be a simple measurable function satisfying 0� s� vE � fð Þ:It suffices to show that

REcð Þ sdl ¼ 0: Since s : X ! 0;1½ Þ is a simple measurable

functionsatisfying0� s� vE � fð Þ; there exists finite-many distinct a1; . . .; an in0;1½ Þ such that vE � fð Þ Xð Þ ¼ a1; . . .; anf g; and

vE � fð Þ�1 a1ð Þ; . . .; vE � fð Þ�1 anð Þ 2 ℳ:

We have to prove that

Xni¼1

aið Þ l vE � fð Þ�1 aið Þ \ Ecð Þ� �

¼ 0:

Fix any i 2 1; . . .; nf g:Case I: when vE � fð Þ�1 aið Þ \ Ecð Þ 6¼ ;: There exists x 2 vE � fð Þ�1 aið Þ \ Ecð Þ: It

follows that x 62 E; and

0 ¼ 0 � f xð Þ ¼ð ÞvE xð Þ � f xð Þ ¼ ai:

This shows that aið Þ l vE � fð Þ�1 aið Þ \ Ecð Þ� �

¼ 0:


Case II:when vE � fð Þ�1 aið Þ \ Ecð Þ ¼ ;: It follows thatl vE � fð Þ�1 aið Þ \ Ecð Þ�

¼0; and hence, aið Þ l vE � fð Þ�1 aið Þ \ Ecð Þ

� � ¼ 0:

Thus, in all cases, each aið Þ l vE � fð Þ�1 aið Þ \ Ecð Þ� �

¼ 0; and hence

Xni¼1

aið Þ l vE � fð Þ�1 aið Þ \ Ecð Þ� �

¼ 0:

∎)Since

RX vE � fð Þdl� R

E vE � fð Þdlþ REcð Þ vE � fð Þdl; and R Ecð Þ vE � fð Þdl ¼ 0; we

have ZX


vE � fð Þdl:

Since vE � fð Þ� f ; by Lemma 1.112,

ZX

vE � fð Þdl�0@

1AZ

E


f dl;

and hence ZX


f dl:

∎

Lemma 1.121 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let f : X ! 0;1½ � be ameasurable function. Then Z

X

vE � fð Þdl ¼ZE

f dl:

Proof If not, otherwise, letRX vE � fð Þdl 6¼ RE f dl: We have to arrive at a contra-

diction. By Lemma 1.119,RX vE � fð Þdlk R

E f dl: SinceZX

vE � fð ÞdlkZE

f dl; andZX

vE � fð Þdl 6¼ZE

f dl;


it follows thatZX

vE � fð Þdl\ZE

f dl

¼ supZE


8<:

9=;;

and hence there exists a simple measurable function s : X !0;1½ Þ satisfying 0� s� f ; and

0�ð ÞZX

vE � fð Þdl\ZE

sdl:

Since s : X ! 0;1½ Þ is a simple measurable function; and 0\RE sdl; there

exist nonzero distinct real numbers a1; . . .; an such that s�1 a1ð Þ \E; . . .; s�1 anð Þ \Eare nonempty members of ℳ; andZ

E

sdl ¼ a1ð Þ l s�1 a1ð Þ \E� �� þ � � � þ anð Þ l s�1 anð Þ \E

� �� :

Since

s�1 a1ð Þ \E; . . .; s�1 anð Þ \E

are pairwise disjoint sets, and 0� s� f ; we have

a1vs�1 a1ð Þ \E þ � � � þ anvs�1 anð Þ \E � vE � f :

It follows thatZX

vE � fð Þdl� a1ð Þ l s�1 a1ð Þ \E� �\X� �� þ � � � þ anð Þ l s�1 anð Þ \E

� �\X� ��

¼ a1ð Þ l s�1 a1ð Þ \E� �� þ � � � þ anð Þ l s�1 anð Þ \E

� �� ¼ ZE

sdl;

and henceRE sdl�

RX vE � fð Þdl: This is a contradiction. ∎

Lemma 1.122 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let s : X ! 0;1½ Þ be a simplemeasurable function. Then the function E 7! R

E sdl from ℳ to 0;1½ � is a positivemeasure on ℳ:


Proof Since s : X ! 0;1½ Þ is a simple measurable function; there exist distinctreal numbers a1; . . .; an such that s�1 a1ð Þ; . . .; s�1 anð Þ are members of ℳ:

1. Countably additive: Let A1;A2;A3; . . .f g be a countable collection of membersin ℳ such that i 6¼ j ) Ai \Aj ¼ ;: We have to show that

ZA1 [A2 [ ��

sdl ¼ZA1

sdlþZA2

sdlþ � � � :

LHS ¼Z

A1 [A2 [ ��sdl ¼ a1ð Þ l s�1 a1ð Þ \ A1 [A2 [ � � �ð Þ� ��

þ � � � þ anð Þ l s�1 anð Þ \ A1 [A2 [ � � �ð Þ� �� ¼ a1ð Þ l s�1 a1ð Þ \A1

� �[ s�1 a1ð Þ \A2� �[ � � ��

þ � � � þ anð Þ l s�1 anð Þ \A1� �[ s�1 anð Þ \A2

� �[ � � �� ¼ a1ð Þ l s�1 a1ð Þ \A1

� �þ l s�1 a1ð Þ \A2� �þ � � ��

þ � � � þ anð Þ l s�1 anð Þ \A1� �þ l s�1 anð Þ \A2

� �þ � � �� ¼ a1ð Þ l s�1 a1ð Þ \A1

� �� þ � � � þ anð Þ l s�1 anð Þ \A1� ��

þ a1ð Þ l s�1 a1ð Þ \A2� �� þ � � � þ anð Þ l s�1 anð Þ \A2

� �� þ � � �

¼ZA1

sdlþZA2

sdlþ � � � ¼ RHS:

2. Since

Z;

sdl ¼ a1ð Þ l s�1 a1ð Þ \ ;� �� þ � � � þ anð Þ l s�1 anð Þ \ ;� �� ¼ a1ð Þ l ;ð Þð Þþ � � � þ anð Þ l ;ð Þð Þ ¼ a1ð Þ 0ð Þþ � � � þ anð Þ 0ð Þ ¼ 0\1;

we haveR; sdl\1: ∎

Lemma 1.123 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let s : X ! 0;1½ Þ; and t : X !0;1½ Þ be simple measurable functions. ThenZ

X

sþ tð Þdl ¼ZX

sdlþZX

tdl:


Proof Since s : X ! 0;1½ Þ is a simple measurable function, there exist distinctreal numbers a1; . . .; an such that s�1 a1ð Þ; . . .; s�1 anð Þ are members of ℳ: Sincet : X ! 0;1½ Þ is a simple measurable function, there exist distinct real numbersb1; . . .; bm such that t�1 b1ð Þ; . . .; t�1 bmð Þ are members of ℳ: For simplicity, letn ¼ 3; and m ¼ 2: Thus s�1 a1ð Þ; s�1 a2ð Þ; s�1 a3ð Þ�

is a partition of X; andt�1 b1ð Þ; t�1 b2ð Þ�

is a partition of X; and hence

s�1 a1ð Þ \ t�1 b1ð Þ; s�1 a2ð Þ \ t�1 b1ð Þ; s�1 a3ð Þ \ t�1 b1ð Þ;�s�1 a1ð Þ \ t�1 b2ð Þ; s�1 a2ð Þ \ t�1 b2ð Þ; s�1 a3ð Þ \ t�1 b2ð Þ

is a partition of X: Now, by Lemma 1.122,

ZX

sþ tð Þdl ¼X2j¼1

X3i¼1

Zs�1 aið Þ \ t�1 bjð Þ

sþ tð Þdl

0BB@

1CCA;

and

ZXsdlþ

ZX

tdl ¼X2j¼1

X3i¼1


sdl

0BB@

1CCAþ

X2j¼1

X3i¼1


tdl

0BB@

1CCA

¼X2j¼1

X3i¼1


sdlþZ

s�1 aið Þ \ t�1 bjð Þtdl

0BB@

1CCA

0BB@

1CCA:

Thus, it suffices to show that for every i 2 1; 2; 3f g; and j 2 1; 2f g;Z

s�1 aið Þ \ t�1 bjð Þsþ tð Þdl ¼


sdlþZ

s�1 aið Þ \ t�1 bjð Þtdl:


Observe that


sdlþZ


¼ aið Þ l s�1 aið Þ \ t�1 bj� �� þ Z


¼ aið Þ l s�1 aið Þ \ t�1 bj� �� þ bj

� �l s�1 aið Þ \ t�1 bj

� �� ¼ ai þ bj� �

l s�1 aið Þ \ t�1 bj� ��

:

Let us fix any i 2 1; 2; 3f g; and j 2 1; 2f g: We have to show that


sþ tð Þdl ¼ ai þ bj� �

l s�1 aið Þ \ t�1 bj� ��

:

For every x 2 s�1 aið Þ \ t�1 bj� �

; we have s xð Þ ¼ ai; and t xð Þ ¼ bj; and hence,for every x 2 s�1 aið Þ \ t�1 bj

� �;

sþ tð Þ xð Þ ¼ s xð Þþ t xð Þ ¼ ai þ bj:

It follows that


sþ tð Þdl ¼ ai þ bj� �

l s�1 aið Þ \ t�1 bj� ��

:

∎

Lemma 1.124 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letf : X ! �1;1½ �; and g : X ! �1;1½ � be measurable functions. Then

1. x : f xð Þ\g xð Þf g 2 ℳ; 2. x : f xð Þ ¼ g xð Þf g 2 ℳ; and 3.x : f xð Þ� g xð Þf g 2 ℳ:

Proof

1. Here


x : f xð Þ\g xð Þf g ¼ [ r2Qþ x : f xð Þ\r\g xð Þf g¼ [ r2Qþ x : f xð Þ\rf g\ x : r\g xð Þf gð Þ¼ [ r2Qþ x : f xð Þ\rf g\ g�1 r;1ð �ð Þ� �¼ [ r2Qþ x : r� f xð Þf gc \ g�1 r;1ð �ð Þ� �¼ [ r2Qþ f�1 r;1½ �ð Þ� �c \ g�1 r;1ð �ð Þ� �¼ [ r2Qþ f�1 \ n2N r � 1

n;1

� �� c

\ g�1 r;1ð �ð Þ� �

¼ [ r2Qþ \ n2Nf�1 r � 1n;1

� �� c

\ g�1 r;1ð �ð Þ� �

2 ℳ:

2. By 1,

x : f xð Þ ¼ g xð Þf g ¼ x : f xð Þ\g xð Þf g[ x : g xð Þ\f xð Þf gð Þc2 ℳ;

so x : f xð Þ ¼ g xð Þf g 2 ℳ:3. By 1,2,

x : f xð Þ� g xð Þf g ¼ x : f xð Þ\g xð Þf g[ x : f xð Þ ¼ g xð Þf g 2 ℳ;

so x : f xð Þ� g xð Þf g 2 ℳ:

∎

Theorem 1.125 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: For each n ¼ 1; 2; 3; . . .; let fn :X ! 0;1½ � be a measurable function. For every x 2 X; let f1 xð Þ� f2 xð Þ� � � � : Forevery x 2 X; let limn!1 fn xð Þ ¼ f xð Þ: Then1. f : X ! 0;1½ � is a measurable function,2. limn!1

RX fndl

� � ¼ RX f dl:Proof

1. Since, for every x 2 X; f1 xð Þ� f2 xð Þ� � � � ; and limn!1 fn xð Þ ¼ f xð Þ; it followsthat, for every x 2 X;

f xð Þ ¼ð Þ limn!1 fn xð Þ ¼ sup f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g;

and hence, by Lemma 1.85, f : X ! 0;1½ � is a measurable function.2. Since, for every x 2 X; f1 xð Þ� f2 xð Þ� � � � � f xð Þ; each fn is a measurable

function, and f is a measurable function, by Lemma 1.112,ZX

f1dl�ZX

f2dl� � � � �ZX

f dl;


and hence

limn!1

ZX

fndl

0@

1A�

ZX

f dl:

Now, it remains to show thatRX f dl� limn!1

RX fndl

� �:

Observe that

ZX

f dl ¼ supZX


8<:

9=;

For this purpose, let us take any simple measurable function s : X !0;1½ Þ satisfying 0� s� f : It suffices to show that

RX sdl� limn!1

RX fndl

� �:

Observe that

sup cZX

sdl : 0\c\1

8<:

9=; ¼

ZX

sdl

0@

1A � sup c : 0\c\1f gð Þ

¼ZX

sdl

0@

1A � 1 ¼

ZX

sdl;

Let us fix any real number c 2 0; 1ð Þ: It suffices to show that

ZX

csð Þdl� limn!1

ZX

fndl

0@

1A:

Since 0� s� f ; and 0\c\1; we have 0� cs� f : Since, for every x 2 X;f1 xð Þ� f2 xð Þ� � � � ; we have

x : csð Þ xð Þ� f1 xð Þf g x : csð Þ xð Þ� f2 xð Þf g x : csð Þ xð Þ� f3 xð Þf g � � � :Problem 1:126 x : csð Þ xð Þ� f1 xð Þf g [ x : csð Þ xð Þ� f2 xð Þf g [ x : csð Þ xð Þ� f3 xð Þf g[ � � � ¼ X:

(Solution Let us take any a 2 X: It is enough to show that there exists a positiveinteger n such that c s að Þð Þ� fn að Þ:

Case I: when f að Þ ¼ 0: Since 0� s að Þ� f að Þ ¼ 0ð Þ; we have s að Þ ¼ 0: Now,since

csð Þ að Þ ¼ c � s að Þð Þ ¼ c � 0 ¼ 0� f1 að Þ|fflfflfflfflffl{zfflfflfflfflffl};


we have

a 2 x : csð Þ xð Þ� f1 xð Þf g;

and hence

a 2 x : csð Þ xð Þ� f1 xð Þf g[ x : csð Þ xð Þ� f2 xð Þf g[ x : csð Þ xð Þ� f3 xð Þf g [ � � � :

Case II: when 0\f að Þ: Since 0� s að Þ� f að Þ ¼ sup f1 að Þ; f2 að Þ; . . .f g; and0\c\1;

0� csð Þ að Þ\1 sup f1 að Þ; f2 að Þ; . . .f gð Þ ¼ sup f1 að Þ; f2 að Þ; . . .f gð Þ;

and hence, there exists a positive integer n such that c s að Þð Þ� fn að Þ:∎)

By Lemma 1.122, function E 7! RE csð Þdl from ℳ to 0;1½ � is a positive

measure on ℳ: By Lemma 1.124,

x : csð Þ xð Þ� f1 xð Þf g; x : csð Þ xð Þ� f2 xð Þf g; x : csð Þ xð Þ� f3 xð Þf g; . . . 2 ℳ:

Now, by Lemma 1.99,

limn!1

ZX

fndl� limn!1

Zx: csð Þ xð Þ� fn xð Þf g

fndl

¼ limn!1

ZX

v x: csð Þ xð Þ� fn xð Þf g � fnð Þ�

dl

� limn!1

ZX

v x: csð Þ xð Þ� fn xð Þf g � csð Þ�

dl

¼ limn!1

Zx: csð Þ xð Þ� fn xð Þf g

csð Þdl

¼Z

[ n2N x: csð Þ xð Þ� fn xð Þf g

csð Þdl

¼ZX

csð Þdl;

so ZX

csð Þdl� limn!1

ZX

fndl:

∎


Theorem 1.125, known as the Lebesgue’s monotone convergence theorem, isdue to H. L. Lebesgue (28.06.1875–26.07.1941, French). His generalization of theRiemann integral revolutionized the field of integration. He also worked on theFourier series, complex analysis and topology. But his main work was on inte-gration theory.

Lemma 1.127 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! 0;1½ �; and g : X !0;1½ � be measurable functions. ThenZ

X

f þ gð Þdl ¼ZX

f dlþZX

gdl:

Proof By Lemma 1.98, there exists a sequence snf g of simple measurable func-tions sn : X ! 0;1½ Þ such that, for every x in X; 0� s1 xð Þ� s2 xð Þ� � � � ; andlimn!1 sn xð Þ ¼ f xð Þ: Also, there exists a sequence tnf g of simple measurablefunctions tn : X ! 0;1½ Þ such that, for every x in X; 0� t1 xð Þ� t2 xð Þ� � � � ; andlimn!1 tn xð Þ ¼ g xð Þ: It follows that sn þ tnf g is a sequence of simple measurablefunctions

sn þ tnð Þ : X ! 0;1½ Þ

such that, for every x in X; 0� s1 þ t1ð Þ xð Þ� s2 þ t2ð Þ xð Þ� � � � ; and

limn!1 sn þ tnð Þ xð Þ ¼ f þ gð Þ xð Þ:

It follows, by Theorem 1.125, that

ZX

f þ gð Þdl ¼ limn!1

ZX

sn þ tnð Þdl0@

1A ¼ lim

n!1

ZX

sndlþZX

tndl

0@

1A

¼ limn!1

ZX

sndl

0@

1Aþ lim

n!1

ZX

tndl

0@

1A ¼

ZX

f dlþZX

gdl:

HenceRX f þ gð Þdl ¼ RX f dlþ R

X gdl: ∎

Lemma 1.128 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: For each n ¼ 1; 2; 3; . . .; let fn :X ! 0;1½ � be a measurable function. For every x 2 X; let f1 xð Þþ f2 xð Þþ � � � ¼f xð Þ: Then1. f : X ! 0;1½ � is a measurable function,2.RX f1dlþ

RX f2dlþ � � � ¼ RX f dl:


Proof 1: For every positive integer n; and for every x 2 X; put

sn xð Þ � f1 xð Þþ � � � þ fn xð Þ:

Since, each fn : X ! 0;1½ � is a measurable function, by Lemma 1.109, eachsn : X ! 0;1½ � is a measurable function. Since, for every x 2 X;f1 xð Þþ f2 xð Þþ � � � ¼ f xð Þ; we have limn!1 sn xð Þ ¼ f xð Þ: Since each fn : X !0;1½ �; we have, for every x 2 X; s1 xð Þ� s2 xð Þ� � � � : Hence, by Theorem 1.125,f : X ! 0;1½ � is a measurable function, and

ZX

f dl ¼ limn!1

ZX

sndl

0@

1A ¼ lim

n!1

ZX

f1 þ � � � þ fnð Þdl0@

1A

¼ limn!1

ZX

f1dlþ � � � þZX

fndl

0@

1A ¼

ZX

f1dlþZX

f2dlþ � � � :

Thus, ZX

f1dlþZX

f2dlþ � � � ¼ZX

f dl:

∎Let us take 1; 2; 3; . . .f g for X; and the set of all subsets of 1; 2; 3; . . .f g as the r-

algebra ℳ in X: If E is any infinite subset of 1; 2; 3; . . .f g; then put l Eð Þ � 1: If Eis any finite subset of 1; 2; 3; . . .f g; then put

l Eð Þ � number of elements in Eð Þ:

Clearly, l is a positive measure on ℳ: Suppose that, for every i; j in1; 2; 3; . . .f g; let aij 2 0;1½ �: For every i; j in 1; 2; 3; . . .f g; put fi jð Þ � aij: Thus,

f1 : 1; 2; 3; . . .f g ! 0;1½ � is a measurable function, f2 : 1; 2; 3; . . .f g ! 0;1½ � is ameasurable function, etc. For every j 2 1; 2; 3; . . .f g;

X1i¼1

aij ¼ a1j þ a2j þ � � � ¼ f1 jð Þþ f2 jð Þþ � � � ¼ f jð Þ:

Observe that

ZX

f1dl ¼ f1 1ð Þð Þ 1ð Þþ f1 2ð Þð Þ 1ð Þþ � � � ¼ a11 þ a12 þ � � � ¼X1j¼1

a1j;


thusRX f1dl ¼P1

j¼1 a1j: Similarly,RX f2dl ¼P1

j¼1 a2j; etc. Next,ZX

f dl ¼ f 1ð Þð Þ 1ð Þþ f 2ð Þð Þ 1ð Þþ � � � ¼ f 1ð Þþ f 2ð Þþ � � �

¼X1j¼1

f jð Þ ¼X1j¼1

X1i¼1

aij

!;

so

ZX

f dl ¼X1j¼1

X1i¼1

aij

!:

By, Lemma 1.128,

X1i¼1

X1j¼1

aij

!¼X1i¼1

ZX

fidl

0@

1A ¼

ZX

f1dlþZX

f2dlþ � � �

¼ZX

f dl ¼X1j¼1

X1i¼1

aij

! !;

so

X1i¼1

X1j¼1

aij

!¼X1j¼1

X1i¼1

aij

!:

Corollary 1.129 For every i; j in 1; 2; 3; . . .f g; let aij 2 0;1½ �: Then

X1i¼1

X1j¼1

aij

!¼X1j¼1

X1i¼1

aij

!:

Lemma 1.130 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: For each n ¼ 1; 2; 3; . . .; let fn :X ! 0;1½ � be a measurable function. Then, by Lemma 1.87, lim infn!1 fnð Þ :X ! 0;1½ � is measurable function. Also,

ZX

lim infn!1 fn

� dl� lim inf

n!1

ZX

fndl

0@

1A:


Proof Let us define h1 : X ! 0;1½ � as follows: for every x 2 X; h1 xð Þ �inf f1 xð Þ; f2 xð Þ; f3 xð Þ; . . .f g: Let us define h2 : X ! 0;1½ � as follows: For everyx 2 X; h2 xð Þ � inf f2 xð Þ; f3 xð Þ; f4 xð Þ; . . .f g; etc. Clearly, for every x 2 X;h1 xð Þ� h2 xð Þ� � � � : By Lemma 1.86, h1 : X ! 0;1½ �; h2 : X ! 0;1½ �; . . . aremeasurable functions. Also, for every x 2 X;

lim infn!1 fn

� xð Þ ¼ sup h1 xð Þ; h2 xð Þ; h3 xð Þ; . . .f g ¼ lim

n!1 hn xð Þ:

Thus, for every x 2 X; limn!1 hn xð Þ ¼ lim infn!1 fnð Þ xð Þ: Now, byTheorem 1.125,

limn!1

ZX

hndl

0@

1A ¼

ZX

lim infn!1 fn

� dl:

Since, for every x 2 X; and, for every positive integer n;

hn xð Þ ¼ inf fn xð Þ; fnþ 1 xð Þ; fnþ 2 xð Þ; . . .f g� fn xð Þ;

we have, for every positive integer n;RX hndl�

RX fndl; and hence,

ZX

lim infn!1 fn

� dl ¼ lim

n!1

ZX

hndl

0@

1A� lim

n!1

ZX

fndl

0@

1A:

Thus,

ZX

lim infn!1 fn

� dl� lim

n!1

ZX

fndl

0@

1A:

∎Lemma 1.130, known as Fatou’s lemma, is due to P. J. L. Fatou (28.02.1878–

10.08.1929; French). He is known for major contributions to several branches ofanalysis. He was also an astronomer.

Lemma 1.131 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! 0;1½ � be a measurablefunction. Then the function E 7! R

E f dl from ℳ to 0;1½ � is a positive measure onℳ:


Proof

1. Countably additive: Let A1;A2;A3; . . .f g be a countable collection of membersin ℳ such that i 6¼ j ) Ai \Aj ¼ ;: We have to show thatZ

A1 [A2 [ ��f dl ¼

ZA1

f dlþZA2

f dlþ � � � :

By Lemma 1.121,

ZX

vA1 [A2 [ �� f� �

dl ¼Z

A1 [A2 [ ��f dl;

ZX

vA1� f� �

dl

¼ZA1

f dl;ZX

vA2� f� �

dl ¼ZA2

f dl; etc:

Since A1;A2;A3; . . .f g is a countable collection of members in ℳ such thati 6¼ j ) Ai \Aj ¼ ;; we have

vA1 [A2 [ �� f ¼ vA1� f þ vA2

� f þ � � � :

By Lemma 1.128,

RHS ¼ZA1

f dlþZA2

f dlþ � � � ¼ZX

vA1� f� �

dlþZX

vA2� f� �

dlþ � � �

¼ZX

vA1 [A2 [ �� f� �

dlZ

A1 [A2 [ ��f dl ¼ LHS;

so LHS ¼ RHS:

2. Since

Z;

f dl ¼ supZ;


8><>:

9>=>;

¼ sup 0f g ¼ 0;

we haveR; f dl ¼ 0: Hence, the function E 7! R

E f dl from ℳ to 0;1½ � is apositive measure on ℳ: ■

Lemma 1.132 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! 0;1½ � be a measurable


function. By Lemma 1.131, the function u : E 7! RE f dl from ℳ to 0;1½ � is a

positive measure on ℳ: Let g : X ! 0;1½ � be a measurable function. ThenZX

gdu ¼ZX

g � fð Þdl:

In short, we write du ¼ f dl:

Proof Case I: when g is a characteristic function, say vE; where E 2 ℳ: Here,

LHS ¼ZX

gdu ¼ZX

vEdu ¼ 1 � u E\Xð Þð Þþ 0 � u Ec \Xð Þð Þ

¼ u Eð Þ ¼ZE

f dl ¼ZX

vE � fð Þdl ¼ZX

g � fð Þdl ¼ RHS:

Case II: when g is a simple function, say

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ;

where each ai 2 0;1½ Þ; and each g�1 aið Þ 2 ℳ: On using Case I,

LHS ¼ZX

gdu ¼ZX

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ�

du

¼ZX

a1v g�1 a1ð Þð Þ�

duþ � � � þZX

anv g�1 anð Þð Þ�

du

¼ a1

ZX

v g�1 a1ð Þð Þ�

duþ � � � þ an

ZX

v g�1 anð Þð Þ�

du

¼ a1

ZX

v g�1 a1ð Þð Þ � f�

dlþ � � � þ an

ZX

v g�1 anð Þð Þ � f�

dl

¼ZX

a1 v g�1 a1ð Þð Þ � f�

þ � � � þ an v g�1 anð Þð Þ � f�

dl

¼ZX

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ�

� f�

dl ¼ZX

g � fð Þdl ¼ RHS:

Case III: when g is not a simple function. By Lemma 1.98, there exists asequence snf g of simple measurable functions sn : X ! 0;1½ Þ such that, for everyx in X; 0� s1 xð Þ� s2 xð Þ� � � � ; and limn!1 sn xð Þ ¼ g xð Þ: By Theorem 1.125,


ZX

gdu ¼ limn!1

ZX

sndu

0@

1A:

By Case II, for each positive integer n;ZX

sndu ¼ZX

sn � fð Þdl:

It follows that

ZX

gdu ¼ limn!1

ZX

sndu

0@

1A ¼ lim

n!1

ZX

sn � fð Þdl0@

1A:

Since, for every x in X; limn!1 sn xð Þ ¼ g xð Þ; we have, for every x in X;

limn!1 sn � fð Þ xð Þð Þ ¼ lim

n!1 f xð Þð Þ sn xð Þð Þ ¼ f xð Þð Þ g xð Þð Þ ¼ g � fð Þ xð Þ:

Thus, for every x in X;

limn!1 sn � fð Þ xð Þð Þ ¼ g � fð Þ xð Þ:

Since, for every x in X; 0� s1 xð Þ� s2 xð Þ� � � � ; and f : X ! 0;1½ �; we have,for every x in X; 0� s1 xð Þð Þ f xð Þð Þ� s2 xð Þð Þ f xð Þð Þ� � � � : Thus, for every x in X;0� s1 � fð Þ xð Þ� s2 � fð Þ xð Þ� � � � : Further, each sn � fð Þ : X ! 0;1½ � is a measur-able function. Now, by Theorem 1.125,

ZX

gdu ¼0@

1A lim

n!1

ZX

sn � fð Þdl0@

1A ¼

ZX

g � fð Þdl:

Thus, in all cases, ZX

gdu ¼ZX

g � fð Þdl:

∎


1.4 Integration of Complex-Valued Functions

From the properties of integration of positive functions, we enable ourselves tonaturally develop the abstract theory of complex functions. Here, the mostremarkable result is Lebesgue’s dominated convergence theorem.

Note 1.133 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let f : X ! C be a measurable function.By Lemma 1.65, fj j : X ! 0;1½ �; Re fð Þ : X ! �1;1½ �; and Im fð Þ : X !

�1;1½ � are measurable functions. Since fj j : X ! 0;1½ �; we haveRX fj jdl 2

0;1½ �: Since Re fð Þ : X ! �1;1½ � is a measurable function, by Lemma 1.93,Re fð Þð Þþ : X ! 0;1½ �; and Re fð Þð Þ�: X ! 0;1½ � are measurable functions.Similarly, Im fð Þð Þþ : X ! 0;1½ �; and Im fð Þð Þ�: X ! 0;1½ � are measurablefunctions. Observe that, for every x 2 X;

Re fð Þð Þþ� �xð Þ� Re fð Þð Þþ� �

xð Þþ Re fð Þð Þ�ð Þ xð Þ ¼ Re fð Þð Þþ þ Re fð Þð Þ�� xð Þ

¼ Re fð Þj j xð Þ ¼ Re fð Þð Þ xð Þj j ¼ Re f xð Þð Þj j � f xð Þj j ¼ fj j xð Þ;

so Re fð Þð Þþ � fj j: Since Re fð Þð Þþ : X ! 0;1½ �; and fj j : X ! 0;1½ � are mea-surable functions, and Re fð Þð Þþ � fj j; by Lemma 1.112,RX Re fð Þð Þþ dl� R

X fj jdl: Similarly,

ZX

Re fð Þð Þ�dl�ZX

fj jdl;ZX

Im fð Þð Þþ dl�ZX

fj jdl; andZX

Im fð Þð Þ�dl�ZX

fj jdl:

It follows that ifRX fj jdl\1; thenZ

X

Re fð Þð Þþ dl;ZX

Re fð Þð Þ�dl;ZX

Im fð Þð Þþ dl;ZX

Im fð Þð Þ�dl 2 0;1½ Þ:

Thus, ifRX fj jdl\1; then

ZX

Re fð Þð Þþ dl�ZX

Re fð Þð Þ�dl0@

1A;

ZX


Im fð Þð Þ�dl0@

1A 2 R:

1.4 Integration of Complex-Valued Functions 85

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let f : X ! C be a measurable function. LetRX fj jdl\1: We have seen that

ZX



1A;

ZX



1A

2 R;

and hence

ZX



1Aþ i

ZX



1A

2 C:

The complex number

ZX



1Aþ i

ZX



1A

is denoted byRXf dl:

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: The collection of all measurable functionsf : X ! C satisfying

RX fj jdl\1 is denoted by L1 lð Þ; and the members of L1 lð Þ

are called Lebesgue integrable functions with respect to l:If f 2 L1 lð Þ; then the nonnegative real number

RX fj jdl is also denoted by f1k k;

and is called L1-norm of f : Thus, for every f 2 L1 lð Þ;ZX

f dl ¼ZX



1Aþ i

ZX



1A:

Here, if Im fð Þ ¼ 0; then, by Lemma 1.116,ZX

f dl ¼ZX


Re fð Þð Þ�dl:


Hence, ZX

Re fð Þð Þdl ¼ZX


Re fð Þð Þ�dl:

Similarly, ZX

Im fð Þð Þdl ¼ZX


Im fð Þð Þ�dl:

Thus, ZX

f dl ¼ZX

Re fð Þð Þdlþ iZX

Im fð Þð Þdl:

Lemma 1.134 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! C; and g : X ! C bemeasurable functions. Let f ; g 2 L1 lð Þ: Let a; b 2 C: Then

1. af þ bgð Þ 2 L1 lð Þ;2.RX f þ gð Þdl ¼ RX f dlþ R

X gdl;3.RX �fð Þdl ¼ � RX f dl;

4. if a is real, thenRX afð Þdl ¼ a

RX f dl

� �;

5.RX ifð Þdl ¼ i

RX f dl

� �;

6.RX afð Þdl ¼ a

RX f dl

� �:

Proof

1. Since f : X ! C; and g : X ! C are measurable functions, and a; b 2 C; byLemma 1.66, af þ bgð Þ : X ! C is a measurable function. Now, we want toshow that

RX af þ bgj jdl\1: Since f 2 L1 lð Þ; RX fj jdl\1: Similarly,R

X gj jdl\1: Now, on using Lemma 1.114, and Lemma 1.127, we haveZX

aj j fj j þ bj j gj jð Þdl ¼ZX

aj j fj jdlþZX

bj j gj jdl

¼ZX

aj j fj jdlþ bj jZX

gj jdl0@

1A

¼ aj jZX

fj jdl0@

1Aþ bj j

ZX

gj jdl0@

1A\1;


and hence, ZX

aj j fj j þ bj j gj jð Þdl\1:

Since, for every x in X;

af þ bgj j xð Þ ¼ af þ bgð Þ xð Þj j ¼ a f xð Þð Þþ b g xð Þð Þj j � a f xð Þð Þj j þ b g xð Þð Þj j¼ aj j f xð Þj j þ b g xð Þð Þj j ¼ aj j f xð Þj j þ bj j g xð Þj j ¼ aj j fj j xð Þð Þþ bj j gj j xð Þð Þ¼ aj j fj jð Þ xð Þþ bj j gj jð Þ xð Þ ¼ aj j fj j þ bj j gj jð Þ xð Þ;

we have, for every x in X;

af þ bgj j xð Þ� aj j fj j þ bj j gj jð Þ xð Þ:

Since af þ bgð Þ : X ! C is a measurable function, by Lemma 1.65, af þ bgj j :X ! 0;1½ � is a measurable function. Since f : X ! C is a measurable function,fj j : X ! 0;1½ � is a measurable function. Similarly, gj j : X ! 0;1½ � is ameasurable function. Now, on using Lemma 1.110, aj j fj j : X ! 0;1½ �; andbj j gj j : X ! 0;1½ � are measurable functions, and hence, by Lemma 1.109,aj j fj j þ bj j gj jð Þ : X ! 0;1½ � is a measurable function. Since af þ bgj j : X !0;1½ �; aj j fj j þ bj j gj jð Þ : X ! 0;1½ � are measurable functions, and, for every xin X;

af þ bgj j xð Þ� aj j fj j þ bj j gj jð Þ xð Þ;

by Lemma 1.112,ZX

af þ bgj jdl�ZX

aj j fj j þ bj j gj jð Þdl \1ð Þ;

and hence ZX

af þ bgj jdl\1:

This shows that af þ bgð Þ 2 L1 lð Þ:2. Let us denote Re fð Þ by u1; Im fð Þ by u2; Re gð Þ by v1; and Im gð Þ by v2: It

follows that Re f þ gð Þ ¼ u1 þ v1; Im f þ gð Þ ¼ u2 þ v2: Here,

u1ð Þþ� u1ð Þ�� þ v1ð Þþ� v1ð Þ�� ¼ u1 þ v1ð Þ ¼ u1 þ v1ð Þþ� u1 þ v1ð Þ�;


so

u1 þ v1ð Þ� þ u1ð Þþ þ v1ð Þþ¼ u1 þ v1ð Þþ þ u1ð Þ� þ v1ð Þ�:

By Lemma 1.127,

ZX

u1þ v1ð Þ� þ u1ð Þþ þ v1ð Þþ� �dl ¼

ZX

u1þ v1ð Þ�ð ÞdlþZX

u1ð Þþ� �dlþ

ZX

v1ð Þþ� �dl;

and ZX

u1 þ v1ð Þ� þ u1ð Þþ þ v1ð Þþ� �dl

¼ZX

u1 þ v1ð Þþ þ u1ð Þ� þ v1ð Þ�� dl

¼ZX

u1 þ v1ð Þþ� �dlþ

ZX

u1ð Þ�ð ÞdlþZX

v1ð Þ�ð Þdl;

so ZX

u1 þ v1ð Þ�ð ÞdlþZX

u1ð Þþ� �dlþ

ZX

v1ð Þþ� �dl

¼ZX

u1 þ v1ð Þþ� �dlþ

ZX

u1ð Þ�ð ÞdlþZX

v1ð Þ�ð Þdl;

and hence,

ZX

u1dlþZX

v1dl ¼ZX

u1ð Þþ� �dl�

ZX

u1ð Þ�ð Þdl0@

1Aþ

ZX

v1ð Þþ� �dl�

ZX

v1ð Þ�ð Þdl0@

1A

¼ZX

u1þ v1ð Þþ� �dl�

ZX

u1 þ v1ð Þ�ð Þdl ¼ZX

u1 þ v1ð Þdl:

Thus, ZX

u1 þ v1ð Þdl ¼ZX

u1dlþZX

v1dl:


Similarly,

ZX

u2 þ v2ð Þdl ¼ZX

u2dlþZX

v2dl:

LHS ¼ZX

f þ gð Þdl ¼ZX

Re f þ gð Þð Þdlþ iZX

Im f þ gð Þð Þdl

¼ZX

u1 þ v1ð Þdlþ iZX

u2 þ v2ð Þdl

¼ZX

u1dlþZX

v1dl

0@

1Aþ i

ZX

u2dlþZX

v2dl

0@

1A

¼ZX

u1dlþ iZX

u2dl

0@

1Aþ

ZX

v1dlþ iZX

v2dl

0@

1A

¼ZX

Re fð Þð Þdlþ iZX

Im fð Þð Þdl0@

1Aþ

ZX

Re gð Þð Þdlþ iZX

Im gð Þð Þdl0@

1A

¼ZX

f dlþZX

gdl ¼ RHS:

3. Let us denote Re fð Þ by u1; Im fð Þ by u2: It follows that Re �fð Þ ¼ �u1; andIm fð Þ ¼ �u2: Also,

Re �fð Þð Þþ¼ �u1ð Þþ¼ u1ð Þ�¼ Re fð Þð Þ�;

and

Re �fð Þð Þ�¼ �u1ð Þ�¼ u1ð Þþ¼ Re fð Þð Þþ :

Thus,

Re �fð Þð Þþ¼ Re fð Þð Þ�; and Re �fð Þð Þ�¼ Re fð Þð Þþ :

Similarly,

Im �fð Þð Þþ ¼ Im fð Þð Þ�; and Im �fð Þð Þ�¼ Im fð Þð Þþ

LHS ¼ZX

�fð Þdl

¼ZX

Re �fð Þð Þþ dl�ZX

Re �fð Þð Þ�dl0@

1A


þ iZX

Im �fð Þð Þþ dl�ZX

Im �fð Þð Þ�dl0@

1A

¼ZX

Re fð Þð Þ�dl�ZX

Re fð Þð Þþ dl0@

1A

þ iZX

Im fð Þð Þ�dl�ZX

Im fð Þð Þþ dl0@

1A

¼ �ZX



1Aþ i

ZX



1A

0@

1A

¼ �ZX

f dl ¼ RHS:

4. Case I: when a� 0: By Lemma 1.114,ZX

a Re fð Þð Þ�ð Þdl ¼ aZX

Re fð Þð Þ�dl;ZX

a Re fð Þð Þþ� �dl ¼ a

ZX

Re fð Þð Þþ dl;ZX

a Im fð Þð Þ�ð Þdl ¼ aZX

Im fð Þð Þ�dl;ZX

a Im fð Þð Þþ� �dl ¼ a

ZX

Im fð Þð Þþ dl:

Since a� 0; it is clear that

Re afð Þð Þþ¼ a Re fð Þð Þð Þþ¼ a Re fð Þð Þþ� �;

and hence

Re afð Þð Þþ¼ a Re fð Þð Þþ� �:

Similarly,

Re afð Þð Þ� ¼ a Re fð Þð Þ�ð Þ; Im afð Þð Þþ¼ a Im fð Þð Þþ� �

; Im afð Þð Þ�¼ a Im fð Þð Þ�ð Þ:LHS ¼

ZX

afð Þdl

¼ZX

Re afð Þð Þþ dl�ZX

Re afð Þð Þ�dl0@

1A


þ iZX

Im afð Þð Þþ dl�ZX

Im afð Þð Þ�dl0@

1A

¼ZX

a Re fð Þð Þþ� �dl�

ZX

a Re fð Þð Þ�ð Þdl0@

1A

þ iZX

a Im fð Þð Þþ� �dl�

ZX

a Im fð Þð Þ�ð Þdl0@

1A

¼ aZX

Re fð Þð Þþ dl� aZX


1A

þ i aZX

Im fð Þð Þþ dl� aZX


1A

¼ aZX



1Aþ i

ZX



1A

0@

1A

¼ aZX

f dl

0@

1A ¼ RHS:

Case II: when a\0: Here, a ¼ � aj j: Now, by Case I,

ZX

afð Þdl ¼ZX

� aj jð Þfð Þdl ¼ZX

aj j �fð Þð Þdl ¼ aj jZX

�fð Þdl0@

1A;

so ZX

afð Þdl ¼ aj jZX

�fð Þdl0@

1A:

Since, by 3, ZX

�fð Þdl ¼ �ZX

f dl;

we haveZX

afð Þdl ¼ aj j �ZX

f dl

0@

1A ¼ � aj jð Þ

ZX

f dl ¼ aZX

f dl

0@

1A:


Thus, in all cases,RX afð Þdl ¼ a

RX f dl

� �:

5. Let us denote Re fð Þ by u1; Im fð Þ by u2: It follows that Re ifð Þ ¼ �u2; andIm ifð Þ ¼ u1; and hence, Re ifð Þð Þþ¼ �u2ð Þþ¼ u2ð Þ�; and Re ifð Þð Þ�¼�u2ð Þ�¼ u2ð Þþ : Thus, Re ifð Þð Þþ¼ u2ð Þ�; and Re ifð Þð Þ�¼ u2ð Þþ : Also,Im ifð Þð Þþ¼ u1ð Þþ ; and Im ifð Þð Þ�¼ u1ð Þ�:

LHS ¼ZX

ifð Þdl ¼ZX

Re ifð Þð Þþ dl�ZX

Re ifð Þð Þ�dl0@

1A

þ iZX

Im ifð Þð Þþ dl�ZX

Im ifð Þð Þ�dl0@

1A

¼ZX

u2ð Þ�dl�ZX

u2ð Þþ dl0@

1Aþ i

ZX

u1ð Þþ dl�ZX

u1ð Þ�dl0@

1A

¼ iZX

u1ð Þþ dl�ZX

u1ð Þ�dl0@

1Aþ i

ZX

u2ð Þþ dl�ZX

u2ð Þ�dl0@

1A

0@

1A

¼ iZX



1Aþ i

ZX



1A

0@

1A

¼ iZX

f dl ¼ RHS:

6. Let a ¼ a1 þ ia2; where a1; a2 2 R: By 2,RX a1f þ ia2fð Þdl ¼ RX a1f dlþR

X ia2f dl: By 4,RX a1f dl ¼ a1

RX f dl; and

RX a2 ifð Þdl ¼ a2

RX ifð Þdl: By 5,R

X ifð Þdl ¼ iRX f dl

� �:

Hence

LHS ¼ZX

afð Þdl ¼ZX

a1f þ ia2fð Þdl

¼ZX

a1f dlþZX

ia2f dl ¼ a1

ZX

f dlþ a2

ZX

ifð Þdl

¼ a1

ZX

f dlþ a2 iZX

f dl

0@

1A

0@

1A ¼ a1 þ ia2ð Þ

ZX

f dl

¼ aZX

f dl ¼ RHS:

∎


By Lemma 1.134(1), it is clear that L1 lð Þ is a complex linear space under thepointwise addition and pointwise scalar multiplication.

Lemma 1.135 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ Let f : X ! C be a measurablefunction. Let f 2 L1 lð Þ: Then1. fj j 2 L1 lð Þ; 2. RX f dl�� R

X fj jdl:

Proof

1. Since f 2 L1 lð Þ; f : X ! C is a measurable function satisfyingRX fj jdl\1:

Since f : X ! C is a measurable function, by Lemma 1.65, fj j : X !0;1½ Þ Cð Þ is a measurable function. Since fj jð Þj j ¼ fj j;Z

X

fj jð Þj jdl ¼ZX

fj jdl \1ð Þ;

and henceRX fj jð Þj jdl\1: Since fj j : X ! C is a measurable function, andR

X fj jð Þj jdl\1; by the definition of L1 lð Þ; we have fj j 2 L1 lð Þ:2. Case I: when

RX f dl ¼ 0: This case is trivial.

Case II: whenRX f dl 6¼ 0: It follows that

RX f dl

�� 2 C� 0f gð Þ: SinceRX f dl;

RX f dl

�� 2 C� 0f gð Þ; and C� 0f gð Þ is a multiplicative group, thereexists a 2 C� 0f gð Þ such that

ZX

f dl

�� ¼ a

ZX

f dl

0@

1A ¼

ZX

afð Þdl; by Lemma 1:134 6ð Þ0@

1A:

Since

ZX

f dl

�� ¼ a

ZX

f dl

0@

1A; and

ZX

f dl

�� 2 C� 0f gð Þ;

we have aj j ¼ 1: Since

ZX

f dl

�� ¼

ZX

afð Þdl

¼ZX



1Aþ i

ZX



1A;


and ZX

f dl

��;

ZX



1A;

ZX



1A 2 R;

ZX



1A ¼ 0;

we have

ZX

f dl

�� ¼

ZX


Re afð Þð Þ�dl�ZX

Re afð Þð Þþ dlþZX

Re afð Þð Þ�dl

¼ZX

Re afð Þð Þþ þ Re afð Þð Þ�� dl ¼

ZX

Re afð Þj jdl�ZX

afj jdl ¼ZX

aj j fj jdl

¼ZX

1 � fj jdl ¼ZX

fj jdl;

and hence ZX

f dl

��

ZX

fj jdl:

Thus, in all cases,RX f dl

�� RX fj jdl: ■

Theorem 1.136 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: For each n ¼ 1; 2; 3; . . .; let fn :X ! C be a measurable function. Let f : X ! C be a function. For every x 2 X; letlimn!1 fn xð Þ ¼ f xð Þ: (Clearly, f : X ! C is a measurable function.)

Let g : X ! 0;1½ Þ Cð Þ be a measurable function such that g 2 L1 lð Þ:Suppose that, for every x 2 X; and for every n ¼ 1; 2; 3; . . .; fn xð Þj j � g xð Þ: Then1. f 2 L1 lð Þ; 2. limn!1

RX fn � fj jdl� � ¼ 0; and 3. limn!1

RX fndl

� � ¼ RX f dl:Proof

1. We want to show that, f : X ! C is a measurable function. By Lemma 1.64, itsuffices to show that Re fð Þ : X ! R; Im fð Þ : X ! R are measurable functions.Since, for every x 2 X;


limn!1 fn xð Þ ¼ f xð Þ lim

n!1 Re fnð Þð Þ�

xð Þ�

þ i limn!1 Im fnð Þð Þ�

xð Þ�

¼ limn!1 Re fnð Þð Þ xð Þð Þ�

þ i limn!1 Im fnð Þð Þ xð Þð Þ�

¼ limn!1 Re fnð Þð Þ xð Þð Þþ i Im fnð Þð Þ xð Þð Þð Þ

¼ limn!1 fn xð Þ; f xð Þ

¼ Re fð Þð Þ xð Þð Þþ i Im fð Þð Þ xð Þð Þ; limn!1 Re fnð Þð Þ�

xð Þ;

limn!1 Im fnð Þð Þ�

xð Þ; Re fð Þð Þ xð Þ; Im fð Þð Þ xð Þ 2 R;

we have, for every x 2 X;

limn!1 Re fnð Þð Þ�

xð Þ ¼ Re fð Þð Þ xð Þ;

and

limn!1 Im fnð Þð Þ�

xð Þ ¼ Im fð Þð Þ xð Þ:

Since each fn : X ! C is a measurable function, by Lemma 1.65, each Re fnð Þ :X ! R is a measurable function, and each Im fnð Þ : X ! R is a measurablefunction. Since, each Re fnð Þ : X ! R is a measurable function, and, for everyx 2 X;

limn!1 Re fnð Þð Þ�

xð Þ ¼ Re fð Þð Þ xð Þ;

by Lemma 1.89, Re fð Þ : X ! R is a measurable function. Similarly, Im fð Þ :X ! R is a measurable function. Thus f : X ! C is a measurable function.Now, it remains to show that

RX fj jdl\1: Since, limn!1 fn ¼ f ; we have

limn!1 fnj j ¼ fj j: Since, for every n ¼ 1; 2; 3; . . .; fnj j � g; and limn!1 fnj j ¼fj j; we have fj j � g: Since f : X ! C is a measurable function, fj j : X !0;1½ Þ is a measurable function. Since fj j : X ! 0;1½ Þ and g : X ! 0;1½ Þ aremeasurable functions, and fj j � g; by Lemma 1.112, we haveRX fj jdl� R

X gdl: Since g : X ! 0;1½ Þ Cð Þ; we have g ¼ gj j: Sinceg 2 L1 lð Þ; Z

X

fj jdl�ZX

gdl ¼ZX

gj jdl\1;

and hence,RX fj jdl\1:


2. Since each fn : X ! C is a measurable function, by Lemma 1.65, each fnj j :X ! 0;1½ Þ is a measurable function. Since each fnj j : X ! 0;1½ Þ; and g :X ! 0;1½ Þ are measurable functions, and each fnj j � g; by Lemma 1.112, eachRX fnj jdl� R

X gdl: Since g : X ! 0;1½ Þ Cð Þ; we have g ¼ gj j: Sinceg 2 L1 lð Þ; Z

X

fnj jdl�ZX

gdl ¼ZX

gj jdl\1;

and hence,RX fnj jdl\1: It follows that each fn 2 L1 lð Þ: Since each fn 2 L1 lð Þ;

and f 2 L1 lð Þ; by Lemma 1.134, fn � f ¼ð Þ1fn þ �1ð Þf 2 L1 lð Þ; and hence, byLemma 1.135, we have fn � fj j 2 L1 lð Þ: It follows that eachRX fn � fj jdl 2 0;1½ Þ:Since, limn!1 fn ¼ f ; we have limn!1 fnj j ¼ fj j; and

lim supn!1

fn � fj j ¼ limn!1 fn � fj j ¼ 0:

Since limn!1 fnj j ¼ fj j; and each fnj j � g; we have fj j � g: Since

fn � fj j � fnj j þ �fj j � gþ �fj j ¼ gþ fj j ¼ gþ g ¼ 2g;

we have 0� 2g� fn � fj j: Since each fn � fj j 2 L1 lð Þ; and g 2 L1 lð Þ; byLemma 1.134(1), each 2g� fn � fj jð Þ 2 L1 lð Þ; and hence, each2g� fn � fj jð Þ : X ! 0;1½ Þ is a measurable function. Now, by Lemma 1.130,

2ZX

gdl ¼ZX

2g� 0ð Þdl ¼ZX

2g� limn!1 fn � fj j

� dl

¼ZX

2g� lim supn!1

fn � fj j� �

dl ¼ZX

2gþ � lim supn!1

fn � fj j� ��

dl

¼ZX

2gþ lim infn!1 � fn � fj jð Þ

� dl ¼

ZX

lim infn!1 2g� fn � fj jð Þ

� dl

� lim infn!1

ZX

2g� fn � fj jð Þdl0@

1A ¼ lim inf

n!1 2ZX

gdl�ZX

fn � fj jdl0@

1A

¼ 2ZX

gdlþ lim infn!1 �

ZX

fn � fj jdl0@

1A ¼ 2

ZX

gdlþ � lim supn!1

ZX

fn � fj jdl0@

1A

0@

1A:

It follows that

0�ð Þ lim supn!1

ZX

fn � fj jdl0@

1A� 0;


and hence

0� lim infn!1

ZX

fn � fj jdl0@

1A� lim sup

n!1

ZX

fn � fj jdl0@

1A ¼ 0:

Thus,

0 ¼ lim infn!1

ZX

fn � fj jdl0@

1A ¼ lim sup

n!1

ZX

fn � fj jdl0@

1A:

This shows that limn!1RX fn � fj jdl� �

exists, and limn!1RX fn � fj jdl� � ¼ 0:

3. We have to show that limn!1RXfndl

� �¼ RX f dl; that is,

limn!1RX fndl� RX f dl� � ¼ 0; that is, limn!1

RX fn � fð Þdl� � ¼ 0; that is,

limn!1RX fn � fð Þdl�� ¼ 0: Since, fn � fð Þ 2 L1 lð Þ; by Lemma 1.135, for each

n ¼ 1; 2; 3; . . .;

ZX

fn � fð Þdl��

��ZX

fn � fj jdl:

Since for each n ¼ 1; 2; 3; . . .; 0� RX fn � fð Þdl�� R

X fn � fj jdl; and, by 2,limn!1

RX fn � fj jdl� � ¼ 0; we have limn!1

RX fn � fð Þdl�� ¼ 0: ■

Theorem 1.136 is known as the Lebesgue’s dominated convergence theorem.

1.5 Sets of Measure Zero

The concept ‘a property holds almost everywhere on some set’ is related to the setsof measure zero. We shall exhibit here that many beautiful theorems remain valideven when condition holds only almost everywhere.

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let E 2 ℳ: Let P be any statement about x;where x 2 E: By ‘P holds almost everywhere on E’ we mean that there existsN 2 ℳ such that N E; l Nð Þ ¼ 0; and P is true at every point of E � Nð Þ:‘P holds almost everywhere on E’ is abbreviated as ‘P holds a.e. on E’.


Lemma 1.137 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! �1;1½ �; and g : X !�1;1½ � be measurable functions. Let l x : f xð Þ 6¼ g xð Þf gð Þ ¼ 0:Then f ¼ g holds a.e. on X:

Proof Let us take x : f xð Þ 6¼ g xð Þf g for N in the above definition. By Lemma1.124, x : f xð Þ ¼ g xð Þf g 2 ℳ: Since x : f xð Þ ¼ g xð Þf g 2 ℳ; and ℳ is a r-algebrain X;

x : f xð Þ 6¼ g xð Þf g ¼ð Þ x : f xð Þ ¼ g xð Þf gc2 ℳ;

and hence x : f xð Þ 6¼ g xð Þf g 2 ℳ: Also, x : f xð Þ 6¼ g xð Þf g X: It is given that

l x : f xð Þ 6¼ g xð Þf gð Þ ¼ 0:

For every y 2 X � x : f xð Þ 6¼ g xð Þf gð Þ; we have y 62 x : f xð Þ 6¼ g xð Þf g; andhence f yð Þ ¼ g yð Þ is true. Thus, f ¼ g holds a.e. on X: ■

Lemma 1.138 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: For every measurable function f :X ! �1;1½ �; and g : X ! �1;1½ �; by f � g; we shall mean that f ¼ g holds a.e. on X: Then � is an equivalence relation over the collection C of all measurablefunctions from X to �1;1½ �:Proof Reflexive: Let f : X ! �1;1½ � be a measurable function. We have toshow that f � f ; that is, f ¼ f holds a.e. on X: Since l x : f xð Þ 6¼ f xð Þf gð Þ ¼l ;ð Þ ¼ 0; we have, l x : f xð Þ 6¼ f xð Þf gð Þ ¼ 0; and hence, by Lemma 1.137, f ¼ fholds a.e. on X:

Symmetric: Let f : X ! �1;1½ �; g : X ! �1;1½ � be measurable functions.Suppose that f � g; that is, f ¼ g holds a.e. on X: We have to show that g� f ; thatis, g ¼ f holds a.e. on X: Since f ¼ g holds a.e. on X; there exists N 2 ℳ such thatN X; l Nð Þ ¼ 0; and

X � Nð Þ x : f xð Þ ¼ g xð Þf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ x : g xð Þ ¼ f xð Þf g:

It follows that g ¼ f holds a.e. on X:Transitive: Let f : X ! �1;1½ �; g : X ! �1;1½ �; and h : X ! �1;1½ � be

measurable functions. Suppose that f � g; and g� h: We have to show that f � h;that is, f ¼ h holds a.e. on X: Since f � g; f ¼ g holds a.e. on X; and hence thereexists N 2 ℳ such that N X; l Nð Þ ¼ 0; and X � Nð Þ x : f xð Þ ¼ g xð Þf g:Similarly, there exists N1 2 ℳ such that N1 X; l N1ð Þ ¼ 0; and

X � N1ð Þ x : g xð Þ ¼ h xð Þf g:

Since N;N1 2 ℳ; and ℳ is a r-algebra, N [N1ð Þ 2 ℳ: Clearly, N [N1ð Þ X:Since

1.5 Sets of Measure Zero 99

0� l N [N1ð Þ ¼ l N [ N1 � Nð Þð Þ¼ l Nð Þþ l N1 � Nð Þ� l Nð Þþ l N1ð Þ ¼ 0þ 0 ¼ 0ð Þ;

we have l N [N1ð Þ ¼ 0: It suffices to show that

X � Nð Þ \ X � N1ð Þ ¼ð Þ X � N [N1ð Þð Þ x : f xð Þ ¼ h xð Þf g;

that is

X � Nð Þ \ X � N1ð Þ x : f xð Þ ¼ h xð Þf g:

For this purpose, let us take any x 2 X � Nð Þ \ X � N1ð Þ: We have to show thatf xð Þ ¼ h xð Þ: Since

x 2 X � Nð Þ \ X � N1ð Þ X � Nð Þ y : f yð Þ ¼ g yð Þf gð Þ; f xð Þ ¼ g xð Þ:

Similarly, g xð Þ ¼ h xð Þ: Since f xð Þ ¼ g xð Þ, and g xð Þ ¼ h xð Þ; we havef xð Þ ¼ h xð Þ:

∎

Lemma 1.139 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! 0;1½ �; and g : X !0;1½ � be measurable functions. Let E 2 ℳ: Suppose that f ¼ g holds a:e: on E:Then Z

E

f dl ¼ZE

gdl:

Proof Since f ¼ g holds a:e: on E; there exists N 2 ℳ such that N E; l Nð Þ ¼0; and E � Nð Þ x : f xð Þ ¼ g xð Þf g: On using Lemma 1.131, we have

ZE

f dl ¼0@

1A Z

N [ E�Nð Þ

f dl ¼ZN

f dlþZ

E�Nð Þ

f dl;

and hence ZE

f dl ¼ZN

f dlþZ

E�Nð Þ

f dl:


Similarly, ZE

gdl ¼ZN

gdlþZ

E�Nð Þ

gdl:

It suffices to show that

1.R

E�Nð Þ f dl ¼ R E�Nð Þ gdl;

2.RN f dl ¼ RN gdl:

For 1: Since E � Nð Þ x : f xð Þ ¼ g xð Þf g; we have v E�Nð Þ � f�

¼v E�Nð Þ � g�

: By Lemma 1.121,

ZE�Nð Þ

f dl ¼ZX

v E�Nð Þ � f�

dl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZX

v E�Nð Þ � g�

dl ¼Z

E�Nð Þ

gdl;

so ZE�Nð Þ

f dl ¼Z

E�Nð Þ

gdl:

For 2: Since N 2 ℳ; and l Nð Þ ¼ 0; and f : X ! 0;1½ � is a measurable func-tion, by Lemma 1.117,

RN f dl ¼ 0: Similarly,

RN gdl ¼ 0: It follows thatR

N f dl ¼ RN gdl: ∎

Lemma 1.140 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: (In short, we say that the orderedtriplet X;ℳ; lð Þ is a measure space.) Put

ℳ� � E : E X; and there existsA;B 2 ℳ satisfyingA E B; and l B� Að Þ ¼ 0f g:

Then ℳ� is a r-algebra in X such that ℳ ℳ�:

Proof Let us take any A 2 ℳ: Since A A A; and l A� Að Þ ¼ l ;ð Þ ¼ 0; wehave, by the definition of ℳ�; A 2 ℳ�: Hence ℳ ℳ�:

1. Since ℳ ℳ�; and X 2 ℳ; we have X 2 ℳ�:2. Let E 2 ℳ�: We have to show that Ec 2 ℳ�: Since E 2 ℳ�; there exist A;B 2

ℳ satisfyingA E B; and l B� Að Þ ¼ 0:SinceA 2 ℳ;andℳ is ar-algebra,Ac 2 ℳ: Similarly,Bc 2 ℳ: Since A E B;we have Bc Ec Ac: Since


Ac � Bc ¼ Ac \ Bcð Þc¼ Ac \B ¼ B� A;

and l B� Að Þ ¼ 0; we have l Ac � Bcð Þ ¼ 0: Now, by the definition of ℳ�;Ec 2 ℳ�:

3. Let E1;E2; . . . be in ℳ�: We have to show that then E1 [E2 [ � � � is in ℳ�: Forevery i ¼ 1; 2; . . .; since Ei 2 ℳ�; there exists Ai;Bi 2 ℳ satisfyingAi Ei Bi; and l Bi � Aið Þ ¼ 0: It follows that

[1i¼1Ai; [1

i¼1Bi 2 ℳ; and [1i¼1Ai [1

i¼1Ei [1i¼1Bi:

Now, it suffices to show that l [1i¼1Bi

� �� [1i¼1Ai

� �� ¼ 0: Since

[1i¼1Bi

� �� [1i¼1Ai

� � ¼ [1i¼1Bi

� �\ [1i¼1Ai

� �c¼ [1i¼1Bi

� �\ \1i¼1 Aið Þcð Þ� �

¼ B1 \ \1i¼1 Aið Þcð Þ� �� [ B2 \ \1

i¼1 Aið Þcð Þ� �� [ � � � B1 \ A1ð Þcð Þð Þ [ B2 \ \1

i¼1 Aið Þcð Þ� �� [ � � � B1 \ A1ð Þcð Þð Þ [ B2 \ A2ð Þcð Þð Þ [ � � � B1 � A1ð Þ [ B2 � A2ð Þ [ � � � ;

we have

[1i¼1Bi

� �� [1i¼1Ai

� � B1 � A1ð Þ [ B2 � A2ð Þ [ � � � ;

and hence

0� l [1i¼1Bi

� �� [1i¼1Ai

� �� l B1 � A1ð Þ [ B2 � A2ð Þ [ � � �ð Þ� l B1 � A1ð Þþ l B2 � A2ð Þþ � � � ¼ 0þ 0þ � � � ¼ 0:

Thus, l [1i¼1Bi

� �� [1i¼1Ai

� �� ¼ 0: ■

Lemma 1.141 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Put

ℳ� � E : E X; and there existsA;B 2 ℳ satisfyingA E B; and l B� Að Þ ¼ 0f g:

By Lemma 1.140, ℳ� is a r-algebra in X containing ℳ: Take any E 2 ℳ�: Bythe definition of ℳ�; there exists A;B 2 ℳ satisfyingA E B; and l B� Að Þ ¼0: Also, let A1;B1 2 ℳ satisfyingA1 E B1; and l B1 � A1ð Þ ¼ 0: Thenl Að Þ ¼ l A1ð Þ:


Proof Here,

l Að Þ ¼ l A� A1ð Þ [ A\A1ð Þð Þ ¼ l A� A1ð Þþ l A\A1ð Þ;

so

l Að Þ ¼ l A� A1ð Þþ l A\A1ð Þ:

Similarly,

l A1ð Þ ¼ l A1 � Að Þþ l A\A1ð Þ:

Thus, it suffices to show that l A� A1ð Þ ¼ l A1 � Að Þ:Problem 1:142 A� A1ð Þ B1 � A1ð Þ:(Solution Take any x 2 A� A1ð Þ: It suffices to show that x 2 B1: If not, otherwise,let x 62 B1: We have to arrive at a contradiction. Since x 62 B1; and E B1; x 62 E:Since x 2 A� A1ð Þ; x 2 A Eð Þ; x 2 E: This is a contradiction. ■)

Since A� A1ð Þ B1 � A1ð Þ; and A� A1ð Þ; B1 � A1ð Þ 2 ℳ; we have0�ð Þl A� A1ð Þ� l B1 � A1ð Þ ¼ 0ð Þ; and therefore, l A� A1ð Þ ¼ 0: Similarly,l A1 � Að Þ ¼ 0: Thus, l A� A1ð Þ ¼ l A1 � Að Þ: ■

Lemma 1.143 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Put

ℳ� �E : E X; and there existA;B 2 ℳ satisfyingA E B; and l B� Að Þ ¼ 0f g

(By Lemma 1.140, ℳ� is a r-algebra in X containing ℳ:) Take any E 2 ℳ�:By the definition of ℳ�; there existsA;B 2 ℳ satisfyingA E B; and l B� Að Þ ¼ 0: Put

l1 Eð Þ � l Að Þ:

(By Lemma 1.141, l1 : ℳ� ! 0;1½ � is a well-defined function.) Then l1 is a

positive measure on ℳ�: Also,

a. l1 is an extension of l; in the sense that, if E 2 ℳ; then l1 Eð Þ ¼ l Eð Þ:b. if F 2 ℳ�; l1 Fð Þ ¼ 0; and E F; then E 2 ℳ�:

Proof

1. Countably additive: Let E1;E2;E3; . . .f g be a countable collection of membersin ℳ� such that i 6¼ j implies Ei \Ej ¼ ;: We have to show that

l1 E1 [E2 [E3 [ � � �ð Þ ¼ l1 E1ð Þþ l1 E2ð Þþ l1 E3ð Þþ � � � :


For every i ¼ 1; 2; . . .; since Ei 2 ℳ�; there existsAi;Bi 2 ℳ satisfyingAi Ei Bi; and l Bi � Aið Þ ¼ 0: It follows that [1

i¼1Ai; [1i¼1Bi 2 ℳ; and

[1i¼1Ai [1

i¼1Ei [1i¼1Bi:

Also, we have seen, in Lemma 1.140 (Proof 3), thatl [1

i¼1Bi� �� [1

i¼1Ai� �� ¼ 0: It follows, from the definition of l1; that

l1 [1i¼1Ei

� � ¼ l [1i¼1Ai

� �: Also, for every i ¼ 1; 2; . . .; l1 Eið Þ ¼ l Aið Þ: Since,

i 6¼ j ) Ei \Ej ¼ ;; and, for every n ¼ 1; 2; 3; . . .; An En; we find that i 6¼j ) Ai \Aj ¼ ;: It follows that

RHS ¼ l1 E1ð Þþ l1 E2ð Þþ l1 E3ð Þþ � � �¼ l A1ð Þþ l A2ð Þþ l A3ð Þþ � � � ¼ l [1

i¼1Ai� �

¼ l [1i¼1Ei

� � ¼ LHS:

2. Since l : ℳ ! 0;1½ � is a positive measure, there exists E 2 ℳ such thatl Eð Þ\1: Since E 2 ℳ ℳ�ð Þ; E 2 ℳ�: Since E E E; l E � Eð Þ ¼l ;ð Þ ¼ 0; by the definition of l1; l1 Eð Þ ¼ l Eð Þ \1ð Þ; and hence, l1 Eð Þ\1:Thus, l1 : ℳ

� ! 0;1½ � is a positive measure on ℳ�:

a. Let E 2 ℳ: We have to show that l1 Eð Þ ¼ l Eð Þ: Since E E E; andl E � Eð Þ ¼ l ;ð Þ ¼ 0; by the definition of l1; l1 Eð Þ ¼ l Eð Þ:

b. Let F 2 ℳ�; l1 Fð Þ ¼ 0; and E F: We have to show that E 2 ℳ�: SinceF 2 ℳ�; there existA;B 2 ℳ satisfyingA F B; and l B� Að Þ ¼ 0:Since E F; and F B; we have ; ð ÞE B: Since ℳ is a r-algebra, wehave ; 2 ℳ: Since ;;B 2 ℳ; and ; E B; it suffices to show thatl B� ;ð Þ ¼ 0:

LHS ¼ l B� ;ð Þ ¼ l Bð Þ¼ l A[ B� Að Þð Þ ¼ l Að Þþ l B� Að Þ¼ l Að Þþ 0 ¼ l Að Þ ¼ l1 Fð Þ ¼ 0 ¼ RHS:

∎

Definition Let X;ℳ; lð Þ and X;ℳ1; l1ð Þ be measure spaces. Suppose that ðF 2ℳ1; l1 Fð Þ ¼ 0; and E FÞ ) E 2 ℳ1: If ℳ ℳ1; and l1 is an extension of l;then we say that X;ℳ1;l1ð Þ is a completion of X;ℳ; lð Þ:

From Lemma 1.143, we find that, for every measure space X;ℳ; lð Þ; thereexists a measure space X;ℳ1; l1ð Þ such that X;ℳ1; l1ð Þ is a completion ofX;ℳ; lð Þ:Lemma 1.144 Let X;ℳ; lð Þ; X;ℳ1; l1ð Þ be measure spaces. Let X;ℳ1; l1ð Þ be acompletion of X;ℳ; lð Þ: Let f be a ‘complex-valued l-measurable functiondefined a.e. on X 0, in the sense that there exists S 2 ℳ such that


1. for every x 2 S; f xð Þ 2 C;2. l Scð Þ ¼ 0;3. for every open subset V of C; x : x 2 S and f xð Þ 2 Vf g 2 ℳ:

Suppose that f1 : X ! C is such that for every x 2 dom f ; f xð Þ ¼ f1 xð Þ: Then f1is a measurable function with respect to ℳ1:

Proof Let V be an open subset of C: We have to show that f1ð Þ�1 Vð Þ 2 ℳ1: Fromthe condition 1, S dom fð Þ; and hence X ¼ Sc [ dom fð Þ: It follows that

f1ð Þ�1 Vð Þ ¼ x : x 2 Sc and f1 xð Þ 2 Vf g[ x : x 2 S and f xð Þ ¼ð Þ f1 xð Þ 2 Vf g

and hence

f1ð Þ�1 Vð Þ ¼ x : x 2 Sc and f1 xð Þ 2 Vf g[ x : x 2 S and f xð Þ 2 Vf g:

By condition 3,

x : x 2 S and f xð Þ 2 Vf g 2 ℳ ℳ1ð Þ;

so,

x : x 2 S and f xð Þ 2 Vf g 2 ℳ1:

Since S 2 ℳ; and ℳ is a r-algebra, we have Sc 2 ℳ ℳ1ð Þ; and hence,Sc 2 ℳ1: Since Sc 2 ℳ1; l1 Scð Þ ¼ð Þl Scð Þ ¼ 0;

x : x 2 Sc and f1 xð Þ 2 Vf g Sc;

and X;ℳ1; l1ð Þ is a completion of X;ℳ; lð Þ; we have

x : x 2 Sc and f1 xð Þ 2 Vf g 2 ℳ1:

Since

x : x 2 Sc and f1 xð Þ 2 Vf g; x : x 2 S and f xð Þ 2 Vf g 2 ℳ1;

and ℳ1 is a r-algebra,

f1ð Þ�1 Vð Þ ¼ x : x 2 Sc and f1 xð Þ 2 Vf g[ x : x 2 S and f xð Þ 2 Vf g 2 ℳ1;

and hence, f1ð Þ�1 Vð Þ 2 ℳ1: ∎

Definition Let X;ℳ; lð Þ be a measure space. Let f 2 L1 lð Þ: Let E 2 ℳ: SinceE 2 ℳ; vE : X ! 0; 1f g Cð Þ is a measurable function. Since f 2 L1 lð Þ; f : X !C is a measurable function. Since vE and f are measurable functions, their productvE � fð Þ is a measurable function. Now, since vE � fð Þj j � fj j; we have


RX vE � fð Þj jdl� R

X fj jdl \1ð Þ; and hence,RX vE � fð Þj jdl\1: Thus vE � fð Þ 2

L1 lð Þ: Since vE � fð Þ 2 L1 lð Þ; we haveRX vE � fð Þdl 2 C:

ByRE f dl we mean

RX vE � fð Þdl:

Lemma 1.145 Let X;ℳ; lð Þ be a measure space. Let f ; g 2 L1 lð Þ: Let E 2 ℳ:

For every x 2 E; let f xð Þ ¼ g xð Þ: Then RE f dl ¼ RE gdl:Proof Since, for every x 2 E; f xð Þ ¼ g xð Þ; we have vE � fð Þ ¼ vE � gð Þ; and henceZ

E

f dl ¼ZX

vE � fð Þdl ¼ZX

vE � gð Þdl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ZE

gdl:

Thus,RE f dl ¼ RE gdl: ∎

Lemma 1.146 Let X;ℳ; lð Þ; X;ℳ1; l1ð Þ be measure spaces. Let X;ℳ1; l1ð Þ be acompletion of X;ℳ; lð Þ: Let f be a “complex-valued l-measurable functiondefined a.e. on X”. Let E 2 ℳ ℳ1ð Þ: Suppose that f1 : X ! C is such that forevery x 2 dom f ; f xð Þ ¼ f1 xð Þ: Suppose that f2 : X ! C is such that for everyx 2 dom f ; f xð Þ ¼ f2 xð Þ: By Lemma 1.144, f1; f2 are measurable functions withrespect to ℳ1: Let f1; f2 2 L1 l1ð Þ: Then

a.RE f1dl1 ¼

RE f2dl1; b.

RE f1j jdl1 ¼

RE f2j jdl1:

Proof Since f is a ‘complex-valued l-measurable function defined a.e. on X’, thereexists S 2 ℳ ℳ1ð Þ such that

1. for every x 2 S; f xð Þ 2 C;2. l Scð Þ ¼ 0;3. for every open subset V of C; x : x 2 S and f xð Þ 2 Vf g 2 ℳ:

a. Here, ZE

f1dl1 ¼ZX

vE � f1ð Þdl1 ¼Z

S[ Scð Þ

vE � f1ð Þdl1

¼ZS

vE � f1ð Þdl1 þZSc

vE � f1ð Þdl1

¼ZX

vS � vE � f1ð Þð Þdl1 þZSc

vE � f1ð Þdl1

¼ZX

vS\E � f1ð Þdl1 þZSc

vE � f1ð Þdl1

¼Z

S\E

f1dl1 þZSc

vE � f1ð Þdl1:


Since

ZSc

vE � f1ð Þdl1

�� ¼

ZX

v Scð Þ � vE � f1ð Þ�

dl1

��

ZX

v Scð Þ � vE � f1ð Þ� �� dl1

¼ZX

v Scð Þ�� vE � f1ð Þj j�

dl1 ¼ZX

v Scð Þ � vE � f1ð Þj j�

dl1

¼ZSc

vE � f1ð Þj jdl1;

we have

0�ZSc

vE � f1ð Þdl1

��

ZSc

vE � f1ð Þj jdl1:

Since l1 Scð Þ ¼ð Þl Scð Þ ¼ 0; we have

0�ZSc

vE � f1ð Þdl1

��

ZSc

vE � f1ð Þj jdl1 ¼ 0

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence,RSc vE � f1ð Þdl1

�� ¼ 0: SinceRSc vE � f1ð Þdl1

�� ¼ 0; we haveRSc vE � f1ð Þdl1 ¼ 0; and hence,

RE f1dl1 ¼

RS\E f1dl1: Similarly,

RE f2dl1 ¼R

S\E f2dl1: Thus, it suffices to show thatZS\E

f1dl1 ¼Z

S\E

f2dl1:

Since, for every x 2 dom fð Þ; f xð Þ ¼ f1 xð Þ; and f xð Þ ¼ f2 xð Þ; we have,for every x 2 dom fð Þ; f1 xð Þ ¼ f2 xð Þ: By 1, S domfð Þ: Thus, for every x 2S\E; f1 xð Þ ¼ f2 xð Þ: Now, by Lemma 1.145,Z

S\E

f1dl1 ¼Z

S\E

f2dl1:

b. Since for every x 2 domf ; f xð Þ ¼ f1 xð Þ; and f xð Þ ¼ f2 xð Þ; we have,for every x 2 domf ; f1 xð Þ ¼ f2 xð Þ: By 1, S domf : Thus, for every x 2S; f1 xð Þ ¼ f2 xð Þ; and hence, vS\E � f1j jð Þ ¼ vS\E � f2j jð Þ: Here,


ZE

f1j jdl1 ¼ZX

vE � f1j jð Þdl1 ¼Z

S[ Scð Þ

vE � f1j jð Þdl1

¼ZS

vE � f1j jð Þdl1 þZSc


¼ZX

vS � vE � f1j jð Þð Þdl1 þZSc


¼ZX

vS\E � f1j jð Þdl1 þZSc

vE � f1j jð Þdl1:

Since l1 Scð Þ ¼ð Þl Scð Þ ¼ 0; we haveRSc vE � f1j jð Þdl1 ¼ 0: Thus,Z

E

f1j jdl1 ¼ZX

vS\E � f1j jð Þdl1:

Similarly,RE f2j jdl1 ¼

RX vS\E � f2j jð Þdl1: Now, since vS\E � f1j jð Þ ¼

vS\E � f2j jð Þ; ZE

f1j jdl1 ¼ZE

f2j jdl1:

∎

Definition Let X;ℳ; lð Þ be a measure space. Let f be a “complex-valued mea-surable function defined a.e. on X”. Let X;ℳ1; l1ð Þ be the completion ofX;ℳ; lð Þ: Let E 2 ℳ ¼ ℳ1ð Þ: Suppose that f1 : X ! C is such that, for everyx 2 dom f ; f xð Þ ¼ f1 xð Þ: By Lemma 1.144, f1 is a measurable function with respectto ℳ1: Let f1 2 L1 l1ð Þ:

By Lemma 1.146, it is legitimate to defineRE f dl as

RE f1dl1; and

RE fj jdl asR

E f1j jdl1:Since f1 2 L1 l1ð Þ; Z

E

f dl ¼ZE

f1dl1 2 C

|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl};


and ZE

fj jdl ¼ZE

f1j jdl1 2 0;1½ Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

:

Thus,RE f dl 2 C; and

RE fj jdl 2 0;1½ Þ:

Lemma 1.147 Let X;ℳ; lð Þ be a measure space. For each n ¼ 1; 2; 3; . . .; letfn : X ! R be a measurable function. Then x : limn!1 fn xð Þ 2 Rf g 2 ℳ:

Proof

x : limn!1 fn xð Þ 2 R

n o¼ x : fn xð Þf g is a Cauchy sequencef g

x : for every n 2 N there exists n0 2 N such that m; pð Þ 2 N� n0 N� n0ð Þ implies fm � fp� �

xð Þ�� \ 1n

¼ \ n2N x : there exists n0 2 N such that m; pð Þ 2 N� n0 N� n0ð Þ implies fm � fp� �

xð Þ�� \ 1n

� �

¼ \ n2N [ n02N x : for every m; pð Þ 2 N� n0 N� n0ð Þ; fm � fp�� xð Þ\ 1

n

� ��

¼ \ n2N [ n02N \ m;pð Þ2 N� n0N� n0ð Þ x : fm � fp�� xð Þ\ 1

n

� ��

¼ \ n2N [ n02N \ m;pð Þ2 N� n0N� n0ð Þ fm � fp� ��1 � 1

n;1n

� �� 2 ℳ:

∎

Lemma 1.148 Let X;ℳ; lð Þ be a measure space. For each n ¼ 1; 2; 3; . . .; letfn : X ! R be a measurable function. Then x :

P1n¼1 fn xð Þ 2 R

� 2 ℳ:

Proof For every n ¼ 1; 2; . . .; and for every x 2 X; put sn xð Þ � f1 xð Þþ � � � þ fn xð Þ:Here, each sn : X ! R is a measurable function, and hence, by Lemma 1.147,

x :X1n¼1

fn xð Þ 2 R

( )¼ x : lim

n!1 sn xð Þ 2 R

n o2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

∎

Lemma 1.149 Let X;ℳ; lð Þ be a measure space. For each n ¼ 1; 2; 3; . . .; letfn : X ! C be a measurable function. Then

a. x :P1

n¼1 fn xð Þ 2 C� 2 ℳ; b. x :

P1n¼1 fnj j xð Þ 2 0;1½ Þ� 2 ℳ; and c.

x :P1

n¼1 fnj j xð Þ ¼ 1� 2 ℳ:


Proof

:

x :X1n¼1

fn xð Þ 2 C

( )¼ x :

X1n¼1

Re fnð Þð Þ xð Þ 2 R andX1n¼1

Im fnð Þð Þ xð Þ 2 R

( )

¼ x :X1n¼1

Re fnð Þð Þ xð Þ 2 R

( )\ x :

X1n¼1

Im fnð Þð Þ xð Þ 2 R

( )2 ℳ;

by Lemma 1.148.b. This is clear from Lemma 1.148.c. x :

P1n¼1 fnj j xð Þ ¼ 1� ¼ x :

P1n¼1 fnj j xð Þ 2 0;1½ Þ� c2 ℳ; by (b). ■

Lemma 1.150 Let X;ℳ; lð Þ be a measure space. For each n ¼ 1; 2; 3; . . .; let fn bea “complex-valued measurable function defined a.e. on X”. For every n ¼ 1; 2; . . .;suppose that

RX fnj jdl exists, and

RX fnj jdl 2 0;1½ Þ: LetZ

X

f1j jdlþZX

f2j jdlþZX

f3j jdlþ � � �\1:

Then f1 xð Þþ f2 xð Þþ f3 xð Þþ � � �ð Þ 2 C holds a.e. on X: (In other words, theseries f1 xð Þþ f2 xð Þþ f3 xð Þþ � � � converges almost for all x:)

Proof Let X;ℳ1; l1ð Þ be the completion of X;ℳ; lð Þ: For every n ¼ 1; 2; 3; . . .;since fn is a “complex-valued measurable function defined” a.e. on X; there existsSn 2 ℳ such that

1. for every x 2 Sn; fn xð Þ 2 C; 2. l Snð Þcð Þ ¼ 0; and 3. for every open subset V ofC; x : x 2 Sn and fn xð Þ 2 Vf g 2 ℳ:

From 1, for every n ¼ 1; 2; . . .; we have Sn domfnð Þ; and hence, domfnð ÞcSnð Þc: For every n ¼ 1; 2; . . .;

RX fnj jdl exists, so, for every n ¼ 1; 2; . . .; there exists

~fn : X ! C such that, for every x 2 domfn; we have fn xð Þ ¼ ~fn xð Þ; ~fn 2 L1 l1ð Þ;RX fndl ¼ RX ~fndl1; and RX fnj jdl ¼ RX ~fn

�� dl1: From 2, we have

0� l \1n¼1Sn

� �c� � ¼ l [1n¼1 Snð Þcð Þ� �� X1

n¼1

l Snð Þcð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl};

and hence, l \1n¼1Sn

� �c� � ¼ 0: From 1, for every x 2 \1n¼1Sn 2 ℳ ℳ1ð Þ; we

have

~f1 xð Þ ¼� �f1 xð Þ; ~f2 xð Þ ¼� �

f2 xð Þ; ~f3 xð Þ ¼� �f3 xð Þ; � � � 2 C:


Since

l1 \1n¼1Sn

� �c� � ¼ l \1n¼1Sn

� �c� � ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};we have l1 \1

n¼1Sn� �c� � ¼ 0: By Lemma 1.149,

x :X1n¼1

~fn xð Þ 2 C

( ); x :

X1n¼1

~fn�� xð Þ 2 0;1½ Þ

( ); x :

X1n¼1

~fn�� xð Þ ¼ 1

( )2 ℳ1:

Since

Zx:x2\1

n¼1Sn and~f1 xð Þj jþ ~f2 xð Þj jþ ~f3 xð Þj j þ ��¼1f g

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

�Z

x:x2\1n¼1Sn and

~f1 xð Þj jþ ~f2 xð Þj jþ ~f3 xð Þj j þ ��2 0;1½ Þf g~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

þZ

x:x2\1n¼1Sn and

~f1 xð Þj jþ ~f2 xð Þj jþ ~f3 xð Þj j þ ��¼1f g~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

¼Z

\1n¼1Snð Þ

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

¼Z

\1n¼1Snð Þ

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1 þ 0

¼Z

\1n¼1Snð Þ

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1 þZ

\1n¼1Snð Þcð Þ

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

¼Z

\1n¼1Snð Þ[ \1

n¼1Snð Þcð Þ~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

¼ZX

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1

¼ZX

~f1�� dl1 þ Z

X

~f2�� dl1 þ Z

X

~f3�� dl1 þ � � �

¼ZX

f1j jdlþZX

f2j jdlþZX

f3j jdlþ � � �\1;


we have Zx:x2\1

n¼1Sn and~f1 xð Þj j þ ~f2 xð Þj jþ ~f3 xð Þj jþ ��¼1f g

~f1�� þ ~f2

�� þ ~f3�� þ � � ��

dl1\1;

and hence

l1 x : x 2 \1n¼1Sn and f1 xð Þj j þ f2 xð Þ xð Þj j þ f3 xð Þj j þ � � � ¼ 1� � �

¼ l1 x : x 2 \1n¼1Sn and ~f1 xð Þ�� þ ~f2 xð Þ�� þ ~f3 xð Þ�� þ � � � ¼ 1� � � ¼ 0:

Since

l1 x : x 2 \1n¼1Sn and f1 xð Þj j þ f2 xð Þ xð Þj j þ f3 xð Þj j þ � � � ¼ 1� � � ¼ 0;

andl1 \1

n¼1Sn� �c� � ¼ 0;

we have

l1 x : f1 xð Þj j þ f2 xð Þ xð Þj j þ f3 xð Þj j þ � � � ¼ 1f g[ \1n¼1Sn

� �c� �¼ l1 x : x 2 \1

n¼1Sn and f1 xð Þj j þ f2 xð Þ xð Þj j þ f3 xð Þj j þ � � � ¼ 1� [ \1n¼1Sn

� �c� � ¼ 0;

and hence

l1 x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � �ð Þ ¼ 0:

It follows that

x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ domf1ð Þc [ domf2ð Þc [ � � � x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � � 2 ℳ1;

and

l1 x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � �ð Þ ¼ 0:

We have to find a set N 2 ℳ such that l Nð Þ ¼ 0; and

domf1ð Þc [ domf2ð Þc [ � � �ð Þ [ x : f1 xð Þþ f2 xð Þþ f3 xð Þþ � � � is not convergentf g N:

Since every absolutely convergent series is convergent, we have

x : f1 xð Þþ f2 xð Þþ f3 xð Þþ � � � is not convergentf g x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g:


Now, it suffices to show that there exists N 2 ℳ such that l Nð Þ ¼ 0; and

domf1ð Þc [ domf2ð Þc [ � � �ð Þ [ x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g N:

Since

x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � � 2 ℳ1;

and X;ℳ1; l1ð Þ is the completion of X;ℳ; lð Þ; there exist A;B 2 ℳ such that

A x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � � B;

and l B� Að Þ ¼ 0: It follows that

0 ¼ l1 x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ S1ð Þc [ S2ð Þc [ � � �ð Þ¼ l Að Þ ¼ 0þ l Að Þ ¼ l B� Að Þþ l Að Þ ¼ l Bð Þ:

On taking B for N; we get N 2 ℳ; l Nð Þ ¼ 0; and

x : f1j j xð Þþ f2j j xð Þþ f3j j xð Þþ � � � ¼ 1f g[ domf1ð Þc [ domf2ð Þc [ � � � N:

∎

Lemma 1.151 Let X;ℳ; lð Þ be a measure space. Let f : X ! 0;1½ � be a mea-surable function. Let E 2 ℳ: Let

RE f dl ¼ 0: Then f ¼ 0 a.e. on E.

Proof It suffices to show that l E \ x : 0\f xð Þf gð Þ ¼ 0: Observe that

limn!1 l E \ x :

1n� f xð Þ

� �� ¼ l [1

n¼1 E \ x :1n� f xð Þ

� ��

¼ l E \ [1n¼1 x :

1n� f xð Þ

� �� ¼ l E \ x : 0\f xð Þf gð Þ:

Since f : X ! 0;1½ � is a measurable function, for every positive integer n;x : 1n � f xð Þ� 2 ℳ: Since, for every positive integer n;

0� 1nl E \ x :

1n� f xð Þ

� ��

¼Z

E \ x:1n� f xð Þf g

1ndl

�Z

E \ x:1n� f xð Þf gf dl

�ZE

f dl ¼ 0;


we have, for every positive integer n;

1nl E \ x :

1n� f xð Þ

� �� ¼ 0;

and hence, for every positive integer n;

l E\ x :1n� f xð Þ

� �� ¼ 0:

It follows that

limn!1 l E \ x :

1n� f xð Þ

� �� ¼ 0:

∎

Lemma 1.152 Let X;ℳ; lð Þ be a measure space. Let f 2 L1 lð Þ: For every E 2 ℳ;

letRE f dl ¼ 0: Then f ¼ 0 a.e. on X.

Proof We have to show that Re fð Þð Þþ i Im fð Þð Þ ¼ð Þf ¼ 0 a.e. on X. It suffices toshow that Re fð Þ ¼ 0 a.e. on X; and Im fð Þ ¼ 0 a.e. on X: Here, f 2 L1 lð Þ; sof : X ! C is a measurable function, and hence Re fð Þ : X ! R is a measurablefunction. Since Re fð Þ : X ! R is a measurable function, x : 0� Re fð Þ xð Þf g 2 ℳ;and hence,

0 ¼Z

x:0� Re fð Þ xð Þf g

f dl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼

Zx:0� Re fð Þ xð Þf g

Re fð Þð Þdlþ iZ

x:0� Re fð Þ xð Þf g

Im fð Þð Þdl:

It follows that

0�ZX

Re fð Þð Þþ dl ¼Z

x:0� Refð Þ xð Þf g

Re fð Þð Þþ dl

¼Z


Re fð Þð Þþ dl� 0

¼Z


Re fð Þð Þþ dl�Z


Re fð Þð Þ�dl

¼Z


Re fð Þð Þdl ¼ 0;


and hence,RX Re fð Þð Þþ dl ¼ 0: It follows from Lemma 1.151, Re fð Þð Þþ¼ 0 a.e.

on X: Similarly, Re fð Þð Þ�¼ 0 a.e. on X: Since Re fð Þð Þþ¼ 0 a.e. on X; andRe fð Þð Þ�¼ 0 a.e. on X; Re fð Þ ¼ð Þ Re fð Þð Þþ� Re fð Þð Þ�¼ 0 a.e. on X; and hence,Re fð Þ ¼ 0 a.e. on X: Similarly, Im fð Þ ¼ 0 a.e. on X: ■

Lemma 1.153 Let X;ℳ; lð Þ be a measure space. Let f 2 L1 lð Þ: (By Lemma1.135, fj j 2 L1 lð Þ; and R

X f dl�� R

X fj jdl:) Let RX f dl�� ¼ RX fj jdl: Then thereexists a 2 C such that af ¼ fj j a.e. on X.

Proof Case I: whenRX f dl ¼ 0: It suffices to show that

l x : 0 6¼ fj j xð Þf gð Þ ¼ð Þl x : 0fð Þ xð Þ 6¼ fj j xð Þf gð Þ ¼ 0 SinceRX f dl

�� ¼ RX fj jdl;and

RX f dl ¼ 0;

RX fj jdl ¼ 0; and hence by Lemma 1.151 we have fj j ¼ 0 a.e. on

X: Since fj j ¼ 0 a.e. on X; l x : fj j xð Þ 6¼ 0f gð Þ ¼ 0:Case II: when

RX f dl 6¼ 0: It follows that

RX f dl

�� 2 C� 0f gð Þ: SinceRX f dl;

RX f dl

�� 2 C� 0f gð Þ; and C� 0f gð Þ is a multiplicative group, there existsa 2 C� 0f gð Þ such that

RX f dl

�� ¼ aRX f dl

� � ¼ RX afð Þdl; by Lemma 1:134 6ð Þ� �:

Since

ZX

f dl

�� ¼ a

ZX

f dl

0@

1A; and

ZX

f dl

�� 2 C� 0f gð Þ;

we have aj j ¼ 1: Since

ZX

f dl

�� ¼

ZX

afð Þdl ¼ZX



1A

þ iZX



1A;

and

ZX

f dl

��;

ZX



1A;

ZX



1A 2 R;

ZX



1A ¼ 0;


and henceZX

f dl

�� ¼

ZX



�ZX

Re afð Þð Þþ dlþZX


¼ZX

Re afð Þð Þþ þ Re afð Þð Þ�� dl ¼

ZX

Re afð Þj jdl

�ZX

afj jdl ¼ZX

aj j fj jdl ¼ZX

1 � fj jdl ¼ZX

fj jdl ¼ZX

f dl

��:

Hence,ZX

Re afð Þð Þdl ¼ZX


Re afð Þð Þ�dl ¼ZX

fj jdl:

SinceRX Re afð Þð Þdl ¼ RX fj jdl; it follows that RX fj j � Re afð Þð Þdl ¼ 0: Since

fj j ¼ 1 fj j ¼ aj j fj j ¼ afj j �Re afð Þ; 0� fj j � Re afð Þð Þ:

Since 0� fj j � Re afð Þð Þ; and RX fj j � Re afð Þð Þdl ¼ 0; by Lemma 1.151, fj j �Re afð Þ ¼ 0 a.e. on X; and hence, Re afð Þ ¼ fj j a.e. on X Thus,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Re afð Þð Þ2 þ Im afð Þð Þ2q

¼ afj j ¼ aj j fj j ¼ 1 fj j ¼ fj j ¼ Re afð Þ

a.e. on X: It follows that Im afð Þ ¼ 0 a.e. on X: Since Re afð Þ ¼ fj j a.e. on X; andIm afð Þ ¼ 0 a.e. on X;

af ¼ð ÞRe afð Þþ i Im afð Þð Þ ¼ fj j þ i0 ¼ fj jð Þ a:e: onX;

and hence af ¼ fj j a.e. on X: ■

Lemma 1.154 Let X;ℳ; lð Þ be a measure space. Let l Xð Þ\1: Let f 2 L1 lð Þ:Let S be a closed subset of C: (Since S is a closed subset of C; Sc is open in C:Since Sc is open in C; and f : X ! C is a measurable function,

x : f xð Þ 2 Sf gc¼ f�1 Sð Þ� �c¼ f�1 Scð Þ 2 ℳ|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl};and hence x : f xð Þ 2 Sf g 2 ℳ: Since X ð Þ x : f xð Þ 2 Sf gc2 ℳ; and l Xð Þ\1;we have l x : f xð Þ 2 Sf gcð Þ\1:Þ


Suppose that, for every E 2 ℳ satisfying l Eð Þ[ 0;

REfdl

l Eð Þ 2 S: Then f xð Þ 2 S a.

e. on X; in the sense l f�1 Scð Þð Þ ¼ 0:

Proof Since Sc is open in C; we can write

Sc ¼ D a1; r1½ � [D a2; r2½ � [D a3; r3½ � [ � � � ;

where each D ai; ri½ � is a closed disk with a center ai 2 Cð Þ and radius ri [ 0ð Þ: Since

0� l f�1 Scð Þ� � ¼ l f�1 D a1; r1½ � [D a2; r2½ � [D a3; r3½ � [ � � �ð Þ� �¼ l f�1 D a1; r1½ �ð Þ [ f�1 D a2; r2½ �ð Þ [ � � �� l f�1 D a1; r1½ �ð Þ� �þ l f�1 a2; r2½ �ð Þ� �þ � � � ;

it suffices to show that each l f�1 D an; rn½ �ð Þð Þ ¼ 0: For this purpose, let us fix anypositive integer n:We have to show that l f �1 D an; rn½ �ð Þð Þ ¼ 0: If not, otherwise, letl f�1 D an; rn½ �ð Þð Þ[ 0: We have to arrive at a contradiction. From the given con-dition, we have R

f�1 D an;rn½ �ð Þð Þ f dl

l f�1 D an; rn½ �ð Þð Þ 2 S:

Since

Rf�1 D an;rn½ �ð Þ f dl

l f�1 D an; rn½ �ð Þð Þ � an

�� ¼ 1

l f�1 D an; rn½ �ð Þð ÞZ

f�1 D an;rn½ �ð Þ

f dl� anl f�1 D an; rn½ �ð Þ� �0B@

1CA

��

¼ 1l f�1 D an; rn½ �ð Þð Þ

Zf�1 D an;rn½ �ð Þ

f dl�Z

f�1 D an;rn½ �ð Þ

andl

��

¼ 1l f�1 D an; rn½ �ð Þð Þ


f � anj jdl

0B@

1CA

� 1l f�1 D an; rn½ �ð Þð Þ


rndl

0B@

1CA

¼ 1l f�1 D an; rn½ �ð Þð Þ rnl f�1 D an; rn½ �ð Þ� �� ¼ rn;


we have Rf�1 D an;rn½ �ð Þ f dl

l f �1 D an; rn½ �ð Þð Þ � an

�� rn;

and hence Rf�1 D an;rn½ �ð Þ f dl

l f�1 D an; rn½ �ð Þð Þ 2 D an; rn½ � Scð Þ:

It follows that Rf�1 D an;rn½ �ð Þ f dl

l f�1 D an; rn½ �ð Þð Þ 62 S;

a contradiction. ■

Lemma 1.155 Let X;ℳ; lð Þ be a measure space. Let E1;E2; � � � 2 ℳ: Let eachl Enð Þ\1: Let l E1ð Þþ l E2ð Þþ � � �\1:

(Clearly,

x : n : x 2 Enf g is an infinite setf g ¼ x : vE1þ vE2

þ � � �� xð Þ ¼ 1�

:

Next,

x : n : x 2 Enf g is an infinite setf gc¼ x : n : x 2 Enf g is a finite setf g[ E1 [E2 [ � � �ð Þc

¼ [ k2N [ n1;��;nkð Þ2Nk En1 \ � � � \Enkð Þ� �

[ E1 [E2 [ � � �ð Þc2 ℳ;

and hence x : n : x 2 Enf g is an infinite setf g 2 ℳ:ÞThen, for almost all x in X; n : x 2 Enf g is a finite set, in the sense that

l x : n : x 2 Enf g is an infinite setf gð Þ ¼ 0; that is

l x : vE1þ vE2

þ � � �� xð Þ ¼ 1� � � ¼ 0:

Proof Since Zx: vE1 þ vE2 þ ��ð Þ xð Þ¼1f g

vE1þ vE2

þ � � �� dl

�ZX

vE1þ vE2

þ � � �� dl

¼ZX

vE1dlþ

ZX

vE2dlþ � � �

¼ l E1ð Þþ l E2ð Þþ � � �\1;


we have Zx: vE1 þ vE2 þ ��ð Þ xð Þ¼1f g

vE1þ vE2

þ � � �� dl\1;

and hence

l x : vE1þ vE2

þ � � �� xð Þ ¼ 1� � � ¼ 0:

∎

1.6 Preliminaries to Topology

The topological concept of support of a scalar-valued function is rarely usedelsewhere as vigorously as in Lebesgue integration. Urysohn’s lemma is provedhere in great detail, simply because of its many applications later on.

Lemma 1.156 Let X be a topological space. Let K be a compact subset of X: Let Fbe a closed subset of X: Let F K: Then F is compact.

Proof Let Gif gi2I be an open cover of F (that is, each Gi is open in X; andF [ i2IGið Þ). We have to find finite-many indices i1; . . .; in such that

F Gi1 [ � � � [Ginð Þ:

Since F is a closed subset of X; Fc is open in X: Since F [ i2IGið Þ; we have

[ i2IGið Þ [ Fcð Þð Þ ¼ X Kð Þ:

Since K [ i2IGið Þ [ Fcð Þð Þ; and each member of the family Gif gi2I [ Fcf g isopen, Gif gi2I [ Fcf g is an open cover of K: Now, since K is compact, by thedefinition of compact set, there exist finite-many indices i1; . . .; in such that

F ð ÞK Gi1 [ � � � [Gin [ Fcð Þð Þ:

It follows that

Gi1 [ � � � [Ginð Þc \F ¼ F \ Gi1 [ � � � [Ginð Þc \Fð Þ¼ F \ Gi1 [ � � � [Gin [ Fcð Þð Þc¼ ;;

and hence, Gi1 [ � � � [Ginð Þc \F ¼ ;: It follows that F Gi1 [ � � � [Ginð Þ: ■

Lemma 1.157 Let X be a topological space. Let B be a subset of X such that theclosure �B of B is a compact set. Let A B: Then �A is a compact set.


Proof Since A B; we have �A �B: Since �A is the closure of A; �A is a closed set.Since �B is a compact set, �A is a closed set, and �A �B; by Lemma 1.156, �A iscompact. ■

Lemma 1.158 Let X be a Hausdorff topological space (that is, X is a topologicalspace), and (x 6¼ y ) (there exist open neighborhood U of x; and open neighbor-hood V of y such that U \V ¼ ;)). Let K be a compact subset of X: Let p 62 K:Then there exist open sets U and V such that p 2 U; K V ; and U \V ¼ ;:Proof Let us take any q 2 K: Since q 2 K; and p 62 K; we have p 6¼ q: Now, sinceX is a Hausdorff topological space, there exists an open neighborhood Uq of p; andan open neighborhood Vq of q such that Uq \Vq ¼ ;: Since, for every q 2 K; Vq isan open neighborhood of q; Vq

� q2K is an open cover of K: Now, since K is

compact, there exist finite-many q1; . . .; qn such that K Vq1 [ � � � [Vqn

� �: Since,

for every i ¼ 1; . . .; n; Uqi is an open neighborhood of p; Uq1 \ � � � \Uqn is an openset such that p 2 Uq1 \ � � � \Uqn

� �: Put

U � Uq1 \ � � � \Uqn

� �; and V � Vq1 [ � � � [Vqn

� �:

Since U ¼ð Þ Uq1 \ � � � \Uqn

� �is an open set, U is an open set. Since each Vqi is

open, V ¼ð Þ Vq1 [ � � � [Vqn

� �is an open set, and hence V is an open set. Since

p 2 Uq1 \ � � � \Uqn

� � ¼ Uð Þ; p 2 U: Since K Vq1 [ � � � [Vqn

� � ¼ Vð Þ; K V :It remains to show that U \V ¼ ;: Since

; U \V ¼ Uq1 \ � � � \Uqn

� �\ Vq1 [ � � � [Vqn

� �¼ Uq1 \ � � � \Uqn

� �\Vq1

� �[ � � � [ Uq1 \ � � � \Uqn

� �\Vqn

� � Uq1 \Vq1

� �[ � � � [ Uqn \Vqn

� � ¼ ;[ � � � [ ; ¼ ;;

we have U \V ¼ ;:∎

Lemma 1.159 Let X be a Hausdorff topological space. Let K be a compact subsetof X: Then K is closed.

Proof It suffices to show that Kc is open. For this purpose, let us take any p 2 Kc:We have to find an open neighborhood U of p such that U Kc: Since p 2 Kc; wehave p 62 K; and hence, by Lemma 1.158, there exist open sets U and V such thatp 2 U; K V ; and U \V ¼ ;: It remains to show that U Kc: Since U \V ¼ ;;we have U Vc: Since K V ; we have U ð ÞVc Kc; and hence, U Kc: ■

Lemma 1.160 Let X be a Hausdorff topological space. Let K be a compact subsetof X: Let F be a closed subset of X: Then F \K is compact.

Proof Since K is a compact subset of X; by Lemma 1.159, K is closed. Since K isclosed, and F is closed, K ð ÞF \K is closed, and hence, F \K is a closed subsetof compact set K: Now, by Lemma 1.156, F \K is compact. ■


Lemma 1.161 Let X be a Hausdorff topological space. Let Kif gi2I be any family ofcompact subsets of X satisfying \ i2IKi ¼ ;: Then there exist finite-many indicesi1; . . .; in such that Ki1 \ � � � \Kinð Þ ¼ ;:Proof Let us fix any i0 2 I: Since \ i2IKi ¼ ;;

[ i2I Kið Þc¼ð Þ \ i2IKið Þc¼ ;c ¼ X Ki0ð Þ;

and hence, Ki0 [ i2I Kið Þcð Þ: It follows that Ki0 [ i2 I� i0f gð Þ Kið Þc� �: Since, each

Ki is compact, by Lemma 1.159, each Ki is closed, and hence, each Kið Þc is open.Since each Kið Þc is open, Ki0 [ i2 I� i0f gð Þ Kið Þc� �

; Kið Þcf g I� i0f gð Þ is an open cover

of Ki0 : Since Kið Þcf g I� i0f gð Þ is an open cover of Ki0 ; and Ki0 is a compact set, thereexist finite-many indices i1; . . .; in 2 I � i0f gð Þ such that

Ki0 Ki1ð Þc [ � � � [ Kinð Þcð Þ ¼ Ki1 \ � � � \Kinð Þcð Þ;

and hence

Ki0 \ Ki1 \ � � � \Kinð Þ ¼ð ÞKi0 \ Ki1 \ � � � \Kinð Þcð Þc¼ ;:

Thus, Ki0 \Ki1 \ � � � \Kin ¼ ;: ■

Definition Let X be a topological space. If, for every p 2 X; there exists an openneighborhood V of p such that the closure �V of V is compact, then we say that X isa locally compact space.

Problem 1.162 Every compact space is a locally compact space.

(Solution Let X be a compact space. We have to show that X is a locally compactspace. For this purpose, let us take any p 2 X: We have to find an open neigh-borhood V of p such that the closure �V of V is compact. Let us take X for V : Sincep 2 X; and X is open, X is an open neighborhood of p: Since �X ¼ X; and X iscompact, �X is compact. ■)

Lemma 1.163 Let X be a locally compact Hausdorff space. Let K be a compactsubset of X: Let U be an open subset of X: Let K U: Then there exists an open setV such that

1. K V �V|fflfflffl{zfflfflffl} U; and 2. �V is compact.

Proof Since X is a locally compact space, for every p 2 K; there exists an openneighborhood Vp of p such that the closure Vp

� �� of Vp is compact. Since, for everyp 2 K; Vp is an open neighborhood of p; the family Vp

� p2K is an open cover of K:

Now, since K is compact, there exist finite-many p1; . . .; pn 2 K such that

1.6 Preliminaries to Topology 121

K Vp1 [ � � � [Vpn

� � Vp1 [ � � � [Vpn

� �� :

Since each Vp is open, Vp1 [ � � � [Vpn

� �is an open set. Since, for every i ¼

1; . . .; n; Vpi is compact,

Vp1 [ � � � [Vpn

� ��¼� �Vp1

� �� [ � � � [ Vpn

� ��is compact, and hence, Vp1 [ � � � [Vpn

� ��is compact. Put

G � Vp1 [ � � � [Vpn

� �:

Here, we see that G is an open set, K G �G|fflfflffl{zfflfflffl}; and �G is compact.

Case I: when U ¼ X: Let us take G for V : The conclusion holds trivially.Case II: when U 6¼ X: Here, since U is open, Uc is a nonempty closed set. Since

K U; Uc Kc: Take any q 2 Uc Kcð Þ: It follows that q 62 K: By Lemma1.158, there exist open sets Wq and Vq such that q 2 Wq; K Vq; and Wq \ Vq ¼ ;:Since Wq is open, and q 2 Wq; Wq is an open neighborhood of q: Since Wq is anopen neighborhood of q; and Wq \ Vq ¼ ;;

q 62 Vq� �� \ r2Uc Vr

� �� :

It follows that

; \ r2Uc Ucð Þ \ �G\ Vr� ��

\ r2Uc Ucð Þ \ Vr� �� ¼ Ucð Þ \ \ r2Uc Vr

� �� ¼ ;;

and hence

\ r2Uc Ucð Þ \ �G\ Vr� �� ¼ ;:

Observe that, for every r 2 Uc; Uc \ �G\ Vr� �� Gð Þ is a closed set. Since, for

every r 2 Uc; Uc \ �G\ Vr� ��

is a closed subset of the compact set �G; by Lemma

1.156, Uc \ �G\ Vr� ��

is compact for every r 2 Uc: Thus, Uc \ �G\ Vr� ��

r2Uc is

a family of compact subsets of X; and \ r2Uc Uc \ �G\ Vr� �� ¼ ;: Now, by

Lemma 1.161, there exist finite-many q1; . . .; qn 2 Uc such that

Ucð Þ \ G\ Vq1 \ � � � \ Vqn

� ��¼ Ucð Þ \ �G\ Vq1

� �� \ � � � \ Vqn

� �� ¼ Uc \ �G\ Vq1

� �� \ � � � \ Uc \ �G\ Vqn

� �� ¼ ;:

Thus, Ucð Þ \ G\ Vq1 \ � � � \ Vqn

� ��¼ ;: It follows that


G\ Vq1 \ � � � \ Vqn

� � G\ Vq1 \ � � � \ Vqn

� �� U|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Let us take G\ Vq1 \ � � � \ Vqn for V : Since G; Vq1 ; . . .; Vqn are open sets,

G\ Vq1 \ � � � \ Vqn is an open set. Since

G\ Vq1 \ � � � \ Vqn

� �� G;

G\ Vq1 \ � � � \ Vqn

� ��is closed, and �G is compact, by Lemma 1.156,

G\ Vq1 \ � � � \ Vqn

� ��is compact. Now, it remains to show that K

G\ Vq1 \ � � � \ Vqn

� �: Since, K G; it suffices to show that K

Vq1 \ � � � \ Vqn

� �: Since each qi 2 Uc; each Vqi K; and hence

K Vq1 \ � � � \ Vqn

� �: ■

Definition Let X be a topological space. Let f : X ! �1;1½ � be a function.If, for every a 2 R; f�1 a;1ð �ð Þ is open in X; then we say that f is lower

semicontinuous. If, for every a 2 R; f�1 �1; a½ Þð Þ is open in X; then we say that fis upper semicontinuous.

Lemma 1.164 Let X be a topological space. Let f : X ! �1;1½ � be a continuousfunction. Then

1. f is lower semicontinuous, 2. f is upper semicontinuous.

Proof

1. Let a 2 R: We have to show that f�1 a;1ð �ð Þ is open in X: Here, a;1ð � is openin �1;1½ �: Since f : X ! �1;1½ � is a continuous function, and a;1ð � isopen in �1;1½ �; f �1 a;1ð �ð Þ is open in X:

2. Its proof is similar to Proof 1. ■

Lemma 1.165 Let X be a topological space. Let f : X ! �1;1½ � be a function.Let f be lower semicontinuous, and upper semicontinuous. Then f is continuous.

Proof Let G be a nonempty open subset of �1;1½ �:We have to show that f�1 Gð Þis open in X:

Case I: when1 62 G; and �1 62 G: Here, G is an open subset of R; so G can beexpressed as a countable union of open intervals, say

G � a1; b1ð Þ [ a2; b2ð Þ [ � � � ;

where, for every n ¼ 1; 2; . . .; an\bn: Since

f�1 Gð Þ ¼ f�1 a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ ¼ f�1 a1; b1ð Þð Þ [ f�1 a2; b2ð Þð Þ [ � � � ;

it suffices to show that each


f�1 an; bnð Þð Þ ¼ f�1 an;1 \� ½ �1; bnð Þð Þ ¼ f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ� �is open in X; that is, f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ is open in X: Since f is lowersemicontinuous, f�1 an;1ð �ð Þ is open in X: Since f is upper semicontinuous,f�1 �1; bnð Þð Þ is open in X: Since f�1 an;1ð Þð Þ is open in X; and f�1 �1; bnð Þð Þ isopen in X; f�1 an;1ð Þð Þ \ f�1 �1; bn½ Þð Þ is open in X:

Case II: when 1 2 G; and �1 62 G: Here, G is an open subset of �1;1½ �; soG can be expressed as:

G � a;1ð �[ a1; b1ð Þ [ a2; b2ð Þ [ � � � ;

where, for every n ¼ 1; 2; . . .; an\bn: Since f is lower semicontinuous,f�1 a;1ð �ð Þ is open in X: Since

f�1 Gð Þ ¼ f�1 a;1ð �[ a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ¼ f�1 a;1ð �ð Þ [ f�1 a1; b1ð Þð Þ [ f�1 a2; b2ð Þð Þ [ � � � ;

and f�1 a;1ð �ð Þ is open in X; it suffices to show that each

f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ ¼ f�1 an;1ð �\ �1; bn½ Þð Þ ¼� �f�1 an; bnð Þð Þ

is open in X; that is, f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ is open in X: Since f is lowersemicontinuous, f�1 an;1ð �ð Þ is open in X: Since f is upper semicontinuous,f�1 �1; bn½ Þð Þ is open in X: Since f�1 an;1ð �ð Þ is open in X; and f�1 �1; bn½ Þð Þ isopen in X; f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ is open in X:

Case III: when �1 2 G; and 1 62 G: This case is similar to case II.Case IV: when �1 2 G; and1 2 G: Here, G is an open subset of �1;1½ �; so

G can be expressed as:

G � a;1ð �[ �1; b½ Þ [ a1; b1ð Þ [ a2; b2ð Þ [ � � � ;

where, for every n ¼ 1; 2; . . .; an\bn: Since f is lower semicontinuous,f�1 a;1ð �ð Þ is open in X: Since f is upper semicontinuous, f�1 �1; b½ Þð Þ is open inX: Since

f�1 Gð Þ ¼ f �1 a;1ð �[ �1; b½ Þ [ a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ¼ f �1 a;1ð �ð Þ [ f�1 �1; b½ Þð Þ [ f�1 a1; b1ð Þð Þ [ f�1 a2; b2ð Þð Þ [ � � � ;

f�1 a;1ð �ð Þ; and f�1 �1; b½ Þð Þ are open in X; it suffices to show that each

f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ ¼ f�1 an;1ð �\ �1; bn½ Þð Þ ¼� �f�1 an; bnð Þð Þ


is open in X; that is, f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ is open in X: Since f is lowersemicontinuous, f�1 an;1ð �ð Þ is open in X: Since f is upper semicontinuous,f�1 �1; bn½ Þð Þ is open in X: Since f�1 an;1ð �ð Þ is open in X; and f�1 �1; bn½ Þð Þ isopen in X; f�1 an;1ð �ð Þ \ f�1 �1; bn½ Þð Þ is open in X: ■

Lemma 1.166 Let X be a topological space. Let G be an open set. Then vG :X ! 0; 1f g �1;1½ �ð Þ is lower semicontinuous.

Proof Let a 2 R: We have to show that vGð Þ�1 a;1ð �ð Þ is open in X: It is clear thateither

vGð Þ�1 a;1ð �ð Þ ¼ vGð Þ�1 0; 1f gð Þ ¼ Xð Þ or vGð Þ�1 a;1ð �ð Þ¼ vGð Þ�1 1f gð Þ ¼ Gð Þð Þ or vGð Þ�1 a;1ð �ð Þ ¼ vGð Þ�1 ;ð Þ ¼ ;ð Þ:

Now, since X; G; ; are open in X; vGð Þ�1 a;1ð �ð Þ is open in X: ■

Lemma 1.167 Let X be a topological space. Let F be a closed set. Then vF :X ! 0; 1f g �1;1½ �ð Þ is upper semicontinuous.

Proof Let a 2 R: We have to show that vFð Þ�1 �1; að �ð Þ is open in X: It is clearthat either

vFð Þ�1 �1; að �ð Þ ¼ vFð Þ�1 0; 1f gð Þ ¼ Xð Þ�

or

vFð Þ�1 �1; að �ð Þ ¼ vFð Þ�1 0f gð Þ ¼ Fcð Þ�

or

vFð Þ�1 �1; að �ð Þ ¼ vFð Þ�1 ;ð Þ ¼ ;ð Þ�

:

Since F is closed, Fc is open. Now, since X; Fc; ; are open in X;vFð Þ�1 �1; að �ð Þ is open in X: ■

Lemma 1.168 Let X be a topological space. For every index i 2 I; let fi : X !�1;1½ � be lower semicontinuous function. Then

supi2I

fi

� �: x 7! sup fi xð Þ : i 2 If g

is lower semicontinuous.

Proof Let a 2 R: Observe that

supi2I

fi

� ��1

a;1ð �ð Þ ¼ x : supi2I

fi

� �xð Þ 2 a;1ð �

� �¼ x : a\ sup

i2Ifi

� �xð Þ

� �¼ x : a\ sup fi xð Þ : i 2 If gf g ¼ x : there exists i 2 I such that a\fi xð Þf g¼ [ i2I x : a\fi xð Þf g ¼ [ i2I x : fi xð Þ 2 a;1ð �f g ¼ [ i2I fið Þ�1 a;1ð �ð Þ

� :


We have to show that supi2I fið Þ�1 a;1ð �ð Þ is open in X; that is,

[ i2I fið Þ�1 a;1ð �ð Þ�

is open in X: Since, each fi : X ! �1;1½ � is lower semi-

continuous, each fið Þ�1 a;1ð �ð Þ is open in X; and hence [ i2I fið Þ�1 a;1ð �ð Þ�

is

open in X: ■

Lemma 1.169 Let X be a topological space. For every index i 2 I; let fi : X !�1;1½ � be upper semicontinuous function. Then

infi2I

fi

� �: x 7! inf fi xð Þ : i 2 If g

is upper semicontinuous.


Definition Let X be a topological space. Let f : X ! C be any function. Theclosure f�1 C� 0f gð Þð Þ� of f�1 C� 0f gð Þ is called the support of f ; and is denotedby supp fð Þ:Lemma 1.170 Let X be a topological space. Let f : X ! C be a continuousfunction.

(Since 0f g is a closed subset of C; C� 0f g is open in C: Since C� 0f g is openin C; and f : X ! C is a continuous function, f�1 C� 0f gð Þ is open in X:)

Let the closed set supp fð Þ ¼ð Þ f�1 C� 0f gð Þð Þ� be compact. In, short, let f :X ! C be a continuous function with compact support. Let g : X ! C be a con-tinuous function with compact support.

It follows that f þ gð Þ : x 7! f xð Þþ g xð Þð Þ is a continuous function from X to C:ð Þ

Then supp f þ gð Þ is compact.

Proof Clearly,

f�1 0f gð Þ� �\ g�1 0f gð Þ� �� f þ gð Þ�1 0f gð Þ�

:

So,

f þ gð Þ�1C� 0f gð Þ ¼ f þ gð Þ�1 0f gð Þ

� c f�1 0f gð Þ� �\ g�1 0f gð Þ� �� c

¼ f�1 0f gð Þ� �c [ g�1 0f gð Þ� �c¼ f�1 C� 0f gð Þ� �[ g�1 C� 0f gð Þ� �and hence,


f þ gð Þ�1C� 0f gð Þ

� f�1 C� 0f gð Þ� �[ g�1 C� 0f gð Þ� ��

:

It follows that

supp f þ gð Þ ¼ð Þ f þ gð Þ�1C� 0f gð Þ

� � f�1 C� 0f gð Þ� �[ g�1 C� 0f gð Þ� ��

¼ f�1 C� 0f gð Þ� �� [ g�1 C� 0f gð Þ� ��¼ supp fð Þð Þ [ supp gð Þð Þ;

and hence

supp f þ gð Þð Þ supp fð Þð Þ [ supp gð Þð Þð Þ:

Since supp fð Þ; supp gð Þ are compact sets, supp fð Þð Þ [ supp gð Þð Þ is compact.Since supp f þ gð Þ is a closed set, supp fð Þð Þ [ supp gð Þð Þ is compact, andsupp f þ gð Þð Þ supp fð Þð Þ [ supp gð Þð Þð Þ; by Lemma 1.156, supp f þ gð Þ is com-pact. ■

Lemma 1.171 Let X be a topological space. Let f : X ! C be a continuousfunction with compact support. Let a 2 C:

It follows that afð Þ : x 7! a f xð Þð Þ is a continuous function from X to C:ð Þ

Then supp afð Þ is compact.

Proof Clearly,

f�1 0f gð Þ� � afð Þ�1 0f gð Þ�

:

So,

afð Þ�1C� 0f gð Þ ¼ afð Þ�1 0f gð Þ

� c f�1 0f gð Þ� �c|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f�1 C� 0f gð Þ� �

;

and hence

afð Þ�1C� 0f gð Þ f�1 C� 0f gð Þ� �

:

It follows that

supp afð Þ ¼ afð Þ�1C� 0f gð Þ

� � f �1 C� 0f gð Þ� ��|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ supp fð Þ;


and hence, supp afð Þð Þ supp fð Þð Þ: Since supp fð Þ is a compact set, supp afð Þ is aclosed set, and supp afð Þð Þ supp fð Þð Þ; by Lemma 1.156, supp afð Þ is compact. ■

Definition Let X be a topological space. The collection of all continuous functionsf : X ! C with compact support is denoted by Cc Xð Þ:

From Lemmas 1.170, and 1.171, it is clear that Cc Xð Þ is a complex linear space.

Lemma 1.172 Let X be a topological space. Let f : X ! C be a continuousfunction with compact support. Then f Xð Þ is a compact subset of C:

Proof Since

f�1 C� 0f gð Þ [ f�1 0f gð Þ ¼ f�1 C� 0f gð Þ [ 0f gð Þ ¼ f�1 Cð Þ ¼ X;

we have

f�1 C� 0f gð Þ [ f�1 0f gð Þ ¼ X:

Now, since

X f�1 C� 0f gð Þ� �� [ f�1 0f gð Þ;

we have

f Xð Þ f f�1 C� 0f gð Þ� �� [ f�1 0f gð Þ� � f f �1 C� 0f gð Þ [ f�1 0f gð Þ� � ¼ f Xð Þ;

and hence

f Xð Þ ¼ f f�1 C� 0f gð Þ� �� [ f�1 0f gð Þ� �¼ f sup p fð Þð Þ [ f�1 0f gð Þ� �¼ f sup p fð Þð Þ [ f f�1 0f gð Þ� �¼ f sup p fð Þð Þ [ 0f g

f sup p fð Þð Þ [ ;

�¼ f sup p fð Þð Þ [ 0f g

f sup p fð Þð Þ

�:

Thus,

f Xð Þ ¼ f supp fð Þð Þ [ 0f gf supp fð Þð Þ

�:

Since f : X ! C has compact support, supp fð Þ is a compact subset of X: Sincesupp fð Þ is a compact subset of X; and f : X ! C is continuous, f supp fð Þð Þ is


compact. Since f supp fð Þð Þ is compact, and 0f g is compact, f supp fð Þð Þ [ 0f g iscompact. Since f supp fð Þð Þ [ 0f g; f supp fð Þð Þ are compact, and

f Xð Þ ¼ f supp fð Þð Þ [ 0f gf supp fð Þð Þ

�;

f Xð Þ is compact. ■

Note 1.173 Let X be a locally compact Hausdorff space. Let K be a compact subsetof X: Let V be an open subset of X: Let K V :

Let us put r1 � 0; and r2 � 1: Since 0; 1ð Þ \Q Qð Þ; and the setQ of all rationalnumbers is countably infinite, 0; 1ð Þ \Q is countably infinite. Since, 0; 1ð Þ \Q iscountably infinite, we can arrange all the members of 0; 1ð Þ \Q in a sequence, say,r3; r4; r5; r6; . . .: For simplicity of argument, let us put r3 ¼ 0:8; r4 ¼ 0:3; r5 ¼0:6; r6 ¼ 0:55; r7 ¼ 0:17; r8 ¼ 0:01; r9 ¼ 0:88; r10 ¼ 0:56; r11 ¼ 0:35; � � � :

By Lemma 1.163, there exists an open set V0 ¼ Vr1ð Þ such that V0ð Þ� ¼ Vr1ð Þ�ð Þis compact, and K V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V : Again, by Lemma 1.163, there exists an

open set V1 ¼ Vr2ð Þ such that V1ð Þ� ¼ Vr2ð Þ�ð Þ is compact, and K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0: Thus,

K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V :

Observing r2 ¼ð Þ1[ 0:8 ¼ r3ð Þ[ 0 ¼ r1ð Þ; by Lemma 1.163, there exists anopen set V0:8 such that V0:8ð Þ� is compact, and V1ð Þ� V0:8 V0:8ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0: Thus,

K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0:8 V0:8ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V :

Observing r3 ¼ð Þ0:8[ 0:3 ¼ r4ð Þ[ 0 ¼ r1ð Þ; by Lemma 1.163, there exists anopen set V0:3 such that V0:3ð Þ� is compact, and V0:8ð Þ� V0:3 V0:3ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:

Thus,

K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0:8 V0:8ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:3 V0:3ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V :


Observing r3 ¼ð Þ0:8[ 0:6 ¼ r5ð Þ[ 0:3 ¼ r4ð Þ; by Lemma 1.163, there exists anopen set V0:6 such that V0:6ð Þ� is compact, and V0:8ð Þ� V0:6 V0:6ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:3:

Thus,

K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0:8 V0:8ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:6 V0:6ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:3 V0:3ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V :

Observing r5 ¼ð Þ0:6[ 0:55 ¼ r6ð Þ[ 0:3 ¼ r4ð Þ; by Lemma 1.163, there existsan open set V0:55 such that V0:55ð Þ� is compact, and V0:6ð Þ� V0:55 V0:55ð Þ�|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} V0:3: Thus,

K V1 V1ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V0:8 V0:8ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:6 V0:6ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0:55 V0:55ð Þ�|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} V0:3 V0:3ð Þ�|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} V0 V0ð Þ�|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} V ;

etc. Thus, we get a sequence of open sets

V0 ¼ Vr1ð Þ;V1 ¼ Vr2ð Þ;V0:8 ¼ Vr3ð Þ;V0:3 ¼ Vr4ð Þ;V0:6 ¼ Vr5ð Þ;V0:55 ¼ Vr6ð Þ; . . .

such that

V0ð Þ� ¼ Vr1ð Þ�ð Þ; V1ð Þ� ¼ Vr2ð Þ�ð Þ; V0:8ð Þ� ¼ Vr3ð Þ�ð Þ;V0:3ð Þ� ¼ Vr4ð Þ�ð Þ; V0:6ð Þ� ¼ Vr5ð Þ�ð Þ; V0:55ð Þ� ¼ Vr6ð Þ�ð Þ; . . .

are compact sets, and ri [ rj ) Vrið Þ� Vrj :

Let

f0 : X ! 0; 1½ � be the function defined by f0 xð Þ � 0 if x 2 V0 Kð Þ0 if x 2 V0ð Þc V0ð Þ�ð Þcð Þ;

�

f1 : X ! 0; 1½ � be the function defined by f1 xð Þ � 1 if x 2 V1 Kð Þ0 if x 2 V1ð Þc V0ð Þ�ð Þcð Þ;

�


f0:8 : X ! 0; 1½ � be the function defined by f0:8 xð Þ� 0:8 if x 2 V0:8 Kð Þ

0 if x 2 V0:8ð Þc V0ð Þ�ð Þcð Þ;�







etc. In short, for every r 2 0; 1½ � \Q; fr : X ! 0; 1½ � is the function defined by

fr xð Þ � r if x 2 Vr Kð Þ0 if x 2 Vrð Þc V0ð Þ�ð Þcð Þ:

�

Clearly, for every r 2 0; 1½ � \Q; fr is lower semicontinuous, and hence, byLemma 1.168,

supr2 0;1½ � \Q

fr

!: x 7! sup fr xð Þ : r 2 0; 1½ � \Qf g

is lower semicontinuous.For every x 2 K;

supr2 0;1½ � \Q

fr

!xð Þ ¼ sup fr xð Þ : r 2 0; 1½ � \Qf g ¼ sup r : r 2 0; 1½ � \Qf g ¼ 1:

Thus, for every x 2 K; supr2 0;1½ � \Q

fr

!xð Þ ¼ 1: It is clear that, for every

x 62 V0ð Þ�;

supr2 0;1½ � \Q

fr

!xð Þ ¼ sup fr xð Þ : r 2 0; 1½ � \Qf g ¼ sup 0 : r 2 0; 1½ � \Qf g ¼ 0:


Thus,

supr2 0;1½ � \Q

fr

!�1

C� 0f gð Þ V0ð Þ�:

It follows that

supp supr2 0;1½ � \Q

fr

!¼

!sup

r2 0;1½ � \Q

fr

!�1

C� 0f gð Þ0@

1A�

V0ð Þ�:

Since


fr

! V0ð Þ� Vð Þ;


fr

!is a closed set, and V0ð Þ� is a compact set, by Lemma 1.156,


fr

!is compact. Also, supp sup

r2 0;1½ � \Q

fr

! V :

Let

g0 : X ! 0; 1½ � be the function defined by g0 xð Þ� 1 if x 2 V0ð Þ� Kð Þ

0 if x 2 V0ð Þ�ð Þc V0ð Þ�ð Þcð Þ;�

g1 : X ! 0; 1½ � be the function defined by g1 xð Þ� 1 if x 2 V1ð Þ� Kð Þ

1 if x 2 V1ð Þ�ð Þc V0ð Þ�ð Þcð Þ;�

g0:8 : X ! 0; 1½ � be the function defined by g0:8 xð Þ� 1 if x 2 V0:8ð Þ� Kð Þ

0:8 if x 2 V0:8ð Þ�ð Þc V0ð Þ�ð Þcð Þ;�








etc. In short, for every r 2 0; 1½ � \Q; gr : X ! 0; 1½ � is the function defined by

gr xð Þ � 1 if x 2 Vrð Þ� Kð Þr if x 2 Vrð Þ�ð Þc V0ð Þ�ð Þcð Þ:

�

Clearly, for every r 2 0; 1½ � \Q; each gr is upper semicontinuous, and hence byLemma 1.169,

infr2 0;1½ � \Q

gr

� �: x 7! inf gr xð Þ : r 2 0; 1½ � \Qf g

is upper semicontinuous. Since, for every x 2 X; and for every r 2 0; 1½ � \Q;

fr xð Þ � r if x 2 Vr

0 if x 2 Vrð Þc�

and gr xð Þ � r if x 2 Vrð Þ�r0 if x 2 Vrð Þ�ð Þc ;

�

we have for every r 2 0; 1½ � \Q; fr � gr: Observe that if r; s 2 0; 1½ � \Q; and s\r;then Vr Vs; and hence by the definitions of fr and fs; fs � fr: Thus,

r; s 2 0; 1½ � \Q; and s\rð Þ ) fs � fr: � � � �ð Þ:Problem 1.174

gs xð Þ\fr xð Þ ) s\r; x 2 Vr; x 62 Vsð Þ�ð Þ: � � � ��ð Þ:(Solution Since gs xð Þ\fr xð Þ; and gs xð Þ; fr xð Þ 2 0; 1½ �; we have fr xð Þ 6¼ 0; andgs xð Þ 6¼ 1: Since fr xð Þ 6¼ 0; and

fr xð Þ ¼ r if x 2 Vr

0 if x 2 Vrð Þc�

;

we have x 2 Vr: Since gs xð Þ 6¼ 1; and

gs xð Þ ¼ 1 if x 2 Vsð Þ�s if x 2 Vsð Þ�ð Þc

�;

we have x 2 Vsð Þ�ð Þc; and hence, x 62 Vsð Þ�: Since x 62 Vsð Þ� Vsð Þ; we havex 62 Vs: Since x 62 Vs; and x 2 Vr; we have s 6¼ r: We claim that s\r: If not,otherwise, let r\s: We have to arrive at a contradiction.

Since r\s; we have Vsð Þ� Vr: Since


fs � gs; fs xð Þ� gs xð Þ \fr xð Þð Þ;

we have fs xð Þ\fr xð Þ: Since r\s; by �ð Þ; fr xð Þ� fs xð Þ: This is a contradiction. ■)Let us take any r; s 2 0; 1½ � \Q: Let us fix any x 2 X:

Problem 1.175 fr xð Þ� gs xð Þ:(Solution If not, otherwise, let gs xð Þ\fr xð Þ: We have to arrive at a contradiction.Since gs xð Þ\fr xð Þ, by ��ð Þ; s\r; x 2 Vr; x 62 Vsð Þ�: Since s\r; we havex 2 Vr ð Þ Vrð Þ� Vs Vsð Þ�ð Þ; and hence x 2 Vsð Þ�: This is a contradiction. ■)Since, for every r; s 2 0; 1½ � \Q; and, for every x 2 X; fr xð Þ� gs xð Þ; we have,

for every s 2 0; 1½ � \Q; and, for every x 2 X;

supr2 0;1½ � \Q

fr

!xð Þ ¼ sup fr xð Þ : r 2 0; 1½ � \Qf gð Þ� gs xð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence, for every s 2 0; 1½ � \Q; and, for every x 2 X;

supr2 0;1½ � \Q fr�

xð Þ� gs xð Þ: It follows that, for every x 2 X;

supr2 0;1½ � \Q

fr

!xð Þ� inf gs xð Þ : s 2 0; 1½ � \Qf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ inf

r2 0;1½ � \Qgr

� �xð Þ:

Thus,

supr2 0;1½ � \Q

fr

!� inf

r2 0;1½ � \Qgr

� �:

We claim that

supr2 0;1½ � \Q

fr

!¼ inf

r2 0;1½ � \Qgr

� �:

If not, otherwise, let there exist x 2 X such that

supr2 0;1½ � \Q

fr

!xð Þ

!\ inf

r2 0;1½ � \Qgr

� �xð Þ

� �:


We have to arrive at a contradiction. Here, there exist r1; r2 2 0; 1½ � \Q such that

sup fr xð Þ : r 2 0; 1½ � \Qf g ¼ supr2 0;1½ � \Q

fr

!xð Þ

!\r1\r2\ inf

r2 0;1½ � \Qgr

� �xð Þ

� �¼ inf gr xð Þ : r 2 0; 1½ � \Qf g:

Since

r1 if x 2 Vr10 if x 2 Vr1ð Þc

�¼ fr1 xð Þ� sup fr xð Þ : r 2 0; 1½ � \Qf gð Þ\r1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

it follows that

r1 if x 2 Vr10 if x 2 Vr1ð Þc

�6¼ r1;

and hence x 62 Vr1 : Since

r2\ inf gr xð Þ : r 2 0; 1½ � \Qf gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � gr2 xð Þ ¼ 1 if x 2 Vr2ð Þ�r2 if x 2 Vr2ð Þ�ð Þc

�;

we have

r2 6¼ 1 if x 2 Vr2ð Þ�r2 if x 2 Vr2ð Þ�ð Þc

�;

and hence x 2 Vr2ð Þ�: Since r1\r2; we have x 2ð Þ Vr2ð Þ� Vr1 ; and hence, x 2 Vr1 :This is a contradiction.

Since

supr2 0;1½ � \Q

fr

!¼ inf

r2 0;1½ � \Qgr

� �;

and infr2 0;1½ � \Q gr� �

is upper semicontinuous, supr2 0;1½ � \Q fr�

is upper semicon-

tinuous. Since supr2 0;1½ � \Q fr�

is upper semicontinuous, and supr2 0;1½ � \Q fr�

is

lower semicontinuous, by Lemma 1.165,

supr2 0;1½ � \Q

fr

!: X ! 0; 1½ � Cð Þ


is continuous. Since supr2 0;1½ � \Q fr�

: X ! C is continuous, and

supp supr2 0;1½ � \Q fr�

is compact, by the definition of Cc Xð Þ;supr2 0;1½ � \Q fr�

2 Cc Xð Þ:If we denote supr2 0;1½ � \Q fr

� by f ; we get the following

Conclusion 1.176 Let X be a locally compact Hausdorff space. Let K be a compactsubset of X: Let V be an open subset of X: Let K V : Then there exists f 2 Cc Xð Þsuch that

1. for all x 2 X; 0� f xð Þ� 1;2. for all x 2 K; f xð Þ ¼ 1;3. supp fð Þ V :

This result, known as the Urysohn’s lemma, is due to P. Urysohn (03.02.1898–17.08.1924, Soviet). He was of Jewish origin. He is best known for his contribu-tions to dimension theory and Urysohn’s lemma. All of them are fundamentalresults.

Note 1.177 Let X be a locally compact Hausdorff space. Let K be a compact subsetof X: Let V1; . . .;Vn be open subsets of X: Let K V1 [ � � � [Vn:

Let us take any x 2 K: Since x 2 K V1 [ � � � [Vnð Þ; there exists ix 21; . . .; nf g such that x 2 Vix ; and hence, xf g Vix : Since xf g Vix ; xf g is compact,

and Vix is an open set, by Lemma 1.163, there exists an open setWx such that Wxð Þ�is compact, and

xf g Wx Wxð Þ�|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} Vix :

Since xf g Wx; and Wx is open, Wx is an open neighborhood of x: It followsthat Wxf gx2K is an open cover of K: Now, since K is compact, there existfinite-many a1; . . .; am 2 K such that K Wa1 [ � � � [Wamð Þ: Observe that, foreach i ¼ 1; . . .;m; ai 2 Wai \Kð Þ; and hence, each Wai \K is nonempty. HereWaið Þ�: Waið Þ� V1f g is a finite collection of compact sets, so[ Waið Þ�: Waið Þ� V1f gð Þ V1ð Þ is a compact set. Since

[ Waið Þ�: Waið Þ� V1f gð Þ V1;

[ Waið Þ�: Waið Þ� V1f gð Þ is compact, and V1 is open, by Urysohn’s lemma,there exists g1 2 Cc Xð Þ such that

1. for all x 2 X; 0� g1 xð Þ� 1; 2. for all x 2 [ Waið Þ�: Waið Þ� V1f gð Þ;g1 xð Þ ¼ 1; and 3. supp g1ð Þ V1:


Similarly, there exists g2 2 Cc Xð Þ such that

1. for all x 2 X; 0� g2 xð Þ� 1;2. for all x 2 [ Waið Þ�: Waið Þ� V2f gð Þ; g2 xð Þ ¼ 1;3. supp g2ð Þ V2; etc.

Put

h1 � g1h2 � 1� g1ð Þg2

h3 � 1� g1ð Þ 1� g2ð Þg3...

hn � 1� g1ð Þ 1� g2ð Þ � � � 1� gn�1ð Þgn

9>>>>>=>>>>>;:

Since each gi 2 Cc Xð Þ; each gi : X ! 0; 1½ � is continuous, and hence, each hi :X ! 0; 1½ � is continuous. Since each gi 2 Cc Xð Þ; each supp gið Þ is compact. Itfollows that supp h1ð Þ ¼ð Þsupp g1ð Þ V1ð Þ is compact, and hence supp h1ð Þ iscompact. Also, supp h1ð Þ V1:

Since

h2ð Þ�1 0ð Þ ¼ 1� g1ð Þg2ð Þ�1 0ð Þ ¼ g1ð Þ�1 1ð Þ�

[ g2ð Þ�1 0ð Þ�

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} g2ð Þ�1 0ð Þ�

;

we have

h2ð Þ�1C� 0f gð Þ

� ¼ h2ð Þ�1 0ð Þ� c

g2ð Þ�1 0ð Þ� c

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ g2ð Þ�1C� 0f gð Þ

� ;

and hence


� � g2ð Þ�1

C� 0f gð Þ� �

¼ supp g2ð Þ:

Since


� � supp g2ð Þ;


� �is closed, and supp g2ð Þ is compact, by Lemma 1.156,

supp h2ð Þ ¼ð Þ h2ð Þ�1C� 0f gð Þ

� �is compact, and hence supp h2ð Þ is compact.

Also, supp h2ð Þ supp g2ð Þ V2ð Þ; so, supp h2ð Þ V2: Similarly, supp h3ð Þ iscompact, and supp h3ð Þ V3; etc.

Thus, for each i ¼ 1; . . .; n; supp hið Þ is compact, and supp hið Þ Vi:


Since each hi : X ! 0; 1½ � Cð Þ is continuous, and each supp hið Þ is compact, bythe definition of Cc Xð Þ; each hi 2 Cc Xð Þ:

Observe that

1� h1 þ h2ð Þ ¼ 1� g1 � 1� g1ð Þg2 ¼ 1� g1ð Þ 1� g2ð Þ;1� h1 þ h2 þ h3ð Þ ¼ 1� g1ð Þ 1� g2ð Þ � h3

¼ 1� g1ð Þ 1� g2ð Þ � 1� g1ð Þ 1� g2ð Þg3¼ 1� g1ð Þ 1� g2ð Þ 1� g3ð Þ; etc:

Thus,

1� h1 þ � � � þ hnð Þ ¼ 1� g1ð Þ � � � 1� gnð Þ:

Since each gi : X ! 0; 1½ �; each 1� gið Þ : X ! 0; 1½ �; and hence, the product1� g1ð Þ � � � 1� gnð Þð Þ : X ! 0; 1½ �: It follows that

h1 þ � � � þ hnð Þ ¼ð Þ 1� 1� g1ð Þ � � � 1� gnð Þð Þ : X ! 0; 1½ �:

Thus h1 þ � � � þ hnð Þ : X ! 0; 1½ �: Since each hi 2 Cc Xð Þ; we haveh1 þ � � � þ hnð Þ 2 Cc Xð Þ: Let us take any x 2 K: We shall show that

1� 1� g1 xð Þð Þ � � � 1� gn xð Þð Þ ¼ h1 xð Þþ � � � þ hn xð Þ ¼ 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is,

1� g1 xð Þð Þ � � � 1� gn xð Þð Þ ¼ 0;

that is, there exists k 2 1; . . .; nf g such that gk xð Þ ¼ 1: Since for all y 2[ Waið Þ�: Waið Þ� Vj� � �

; gj yð Þ ¼ 1; it suffices to show that there exists k 21; . . .; nf g such that x 2 [ Waið Þ�: Waið Þ� Vkf gð Þ:Since x 2ð ÞK Wa1 [ � � � [Wamð Þ; there exists j 2 1; . . .;mf g such that x 2

Waj : Here, aj 2 K; so

aj 2 Waj Waj

� ��|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} Vi ajð Þ ;

where i ajð Þ 2 1; . . .; nf g: Put k � i ajð Þ: Thus, k 2 1; . . .; nf g: Since

x 2 Waj Waj

� �� Vi ajð Þ ¼ Vk

� ;

we have x 2 Waj

� ��and Waj

� �� Vk; and hence,


x 2 [ Waið Þ�: Waið Þ� Vkf gð Þ:Conclusion 1.178 Let X be a locally compact Hausdorff space. Let K be a compactsubset of X: Let V1; . . .;Vn be open subsets of X: Let K V1 [ � � � [Vn: Then thereexist h1; . . .; hn 2 Cc Xð Þ such that

1. for all x 2 X; 0� h1 xð Þ� 1; . . .; 0� hn xð Þ� 1;2. for all x 2 K; h1 xð Þþ � � � þ hn xð Þ ¼ 1;3. supp h1ð Þ V1; . . .; supp hnð Þ Vn;4. h1 þ � � � þ hnð Þ : X ! 0; 1½ �:

Here, the collection h1; . . .; hnf g is called a partition of unity on K; subordinateto the cover V1; . . .;Vnf g:Notation Let X be a locally compact Hausdorff space. By K � f ; we shall meanthat

1. K is a compact subset of X;2. f 2 Cc Xð Þ;3. f : X ! 0; 1½ � Cð Þ;4. for every x 2 K; f xð Þ ¼ 1:

By f � V ; we shall mean that

1. V is an open subset of X;2. f 2 Cc Xð Þ;3. f : X ! 0; 1½ � Cð Þ;4. supp fð Þð Þ V :

By K � f � V ; we shall mean that K � f and f � V :In these notations, the above conclusion can be stated as follows.Let X be a locally compact Hausdorff space. Let K be a compact subset of X: Let

V1; . . .;Vn be open subsets of X: Let K V1 [ � � � [Vn: Then there existh1; . . .; hn 2 Cc Xð Þ such that 1. h1 � V1; . . .; hn � Vn; and 2. K � h1 þ � � � þ hnð Þ:

The Urysohn’s lemma can be restated as follows:Let X be a locally compact Hausdorff space. Let K be a compact subset of X: Let

V be an open subset of X: Let K V : Then there exists f 2 Cc Xð Þ such thatK � f � V :

1.7 Preliminaries to Riesz Representation Theorem

This theorem is regarded as one of the great advancement in mathematics. We shallsee later that an application of this theorem yields the concept of Lebesgue measureof sets in some large class of subsets of Euclidean space.


Definition Let X be a locally compact Hausdorff space. We know that Cc Xð Þ is acomplex linear space. Let K : Cc Xð Þ ! C be a function. By the linear functional Kon Cc Xð Þ; we mean that, for every a; b 2 C; and for every f ; g 2 Cc Xð Þ;

K af þ bgð Þ ¼ a K fð Þð Þþ b K gð Þð Þ:

By the positive linear functional K on Cc Xð Þ; we mean that

1. K is a linear functional on Cc Xð Þ;2. for every f 2 Cc Xð Þ satisfying f : X ! 0;1½ Þ Cð Þ; K fð Þ 2 0;1½ Þ:

Note 1.179 Let X be a locally compact Hausdorff space. Let K be a positive linearfunctional on Cc Xð Þ:

Now, let V be an open set in X:

Problem 1.180 0 � V :

(Solution Here, the constant function 0 : x 7! 0 is a continuous function on X.Since 0�1 C� 0f gð Þ ¼ ;; we have

supp 0ð Þ ¼ð Þ 0�1 C� 0f gð Þ� ��¼ ;� ¼ ;;which is a compact setð Þ;

and hence by the definition of Cc Xð Þ; 0 2 Cc Xð Þ: Clearly, 0 serves the purpose of‘zero vector’ in the complex linear space Cc Xð Þ; and 0 : X ! 0; 1½ �: Now, sincesupp 0ð Þ ¼ð Þ; V ; we have 0 � V : ∎)By Problem 1.180, 0 2 f : f � Vf g; and hence 0 ¼ð ÞK 0ð Þ 2 K fð Þ : f � Vf g:

Thus K fð Þ : f � Vf g is a nonempty set. If f � V ; then f 2 Cc Xð Þ; and f : X !0; 1½ � 0;1½ Þð Þ: Now, since K is a positive linear functional on Cc Xð Þ; f � V )K fð Þ 2 0;1½ Þ: Thus, K fð Þ : f � Vf g is a nonempty collection of nonnegative realnumbers, and hence sup K fð Þ : f � Vf gð Þ 2 0;1½ �:

We shall denote

sup K fð Þ : f � Vf gð Þ 2 0;1½ �ð Þ

by �l Vð Þ:Thus, for every open set V in X; �l Vð Þ 2 0;1½ �:

I. Problem 1.181

a. �l is monotone, in the sense that, if V1;V2 are open sets satisfying V1 V2; then�l V1ð Þ� �l V2ð Þ:

b. For every open set V in X; �l Vð Þ ¼ inf �l Wð Þ : V W ; and W is openf g:

(Solution a. Let V1;V2 be open sets satisfying V1 V2: We have to show that�l V1ð Þ� �l V2ð Þ: Since V1;V2 are open sets satisfying V1 V2;


f : f � V1f g f : f � V2f g;

and hence,

K fð Þ : f � V1f g ¼ K f : f � V1f gð Þ K f : f � V2f gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ K fð Þ : f � V2f g:

Thus

K fð Þ : f � V1f g K fð Þ : f � V2f g:

It follows that

�l V1ð Þ ¼ sup K fð Þ : f � V1f g� sup K fð Þ : f � V2f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ �l V2ð Þ;

and hence, �l V1ð Þ� �l V2ð Þ:b. Since V V ; and V is open; �l Vð Þ 2 �l Wð Þ : V W ; and W is openf g;

and hence,

inf �l Wð Þ : V W ; and W is openf g� �l Vð Þ:

It remains to show that �l Vð Þ� inf �l Wð Þ : V W ; and W is openf g:It suffices to show that �l Vð Þ is a lower bound of

�l Wð Þ : V W ; and W is openf g: For this purpose, let us take any W satisfyingV W ; and W is open: We have to show that �l Vð Þ� �l Wð Þ: Since, V W ; andV ;W are open sets, by (a), �l Vð Þ� �l Wð Þ: ∎)

II. Problem 1.182 �l ;ð Þ ¼ 0:

(Solution Since ; is an open set, �l ;ð Þ ¼ sup K fð Þ : f � ;f g: Let us take any fsatisfying f � ;: It follows that ; ð Þ supp fð Þð Þ ;; and hence,

; f �1 0ð Þ� �c¼ f�1 0f gð Þ� �c¼ f�1 C� 0f gð Þ f�1 C� 0f gð Þ� ��¼ supp fð Þð Þ ¼ ;|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} :

This shows that f�1 0ð Þð Þc¼ ;; and hence, f�1 0ð Þ ¼ X: Thus, f ¼ 0; and hence,K fð Þ ¼ K 0ð Þ ¼ 0ð Þ: This shows that K fð Þ : f � ;f g ¼ 0f g; and hence,

LHS ¼ �l ;ð Þ ¼ sup K fð Þ : f � ;f g ¼ sup 0f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0 ¼ RHS:

∎)

1.7 Preliminaries to Riesz Representation Theorem 141

For every subset E of X; X 2 V : E V ; and V is openf g; so for every subset Eof X; �l Vð Þ : E V ; and V is openf g is a nonempty set. Now, since for every openset V in X; �l Vð Þ 2 0;1½ �; �l Vð Þ : E V ; and V is openf g is a nonempty subset of0;1½ �; and hence inf �l Vð Þ : E V ; and V is openf gð Þ 2 0;1½ �:We shall denote

inf �l Vð Þ : E V ; and V is openf g 2 0;1½ �ð Þ

by l Eð Þ:Thus, for every subset E of X; l Eð Þ 2 0;1½ �: In other words, l : P Xð Þ !

0;1½ � is a function.

Problem 1.183 If V is an open set, then l Vð Þ ¼ �l Vð Þ:(Solution Let V be an open set. We have seen that �l Vð Þ ¼inf �l Wð Þ : V W ; and W is openf g ¼ l Vð Þð Þ: Thus, �l Vð Þ ¼ l Vð Þ: ∎)

By Problem 1.183, for every subset E of X;

l Eð Þ ¼ð Þinf �l Vð Þ : E V ; and V is openf g¼ inf l Vð Þ : E V ; and V is openf g:

Thus, for every subset E of X;

l Eð Þ ¼ inf l Vð Þ : E V ; and V is openf g:III. Problem 1.184 l is monotone, in the sense that, if E1;E2 are any subsets of Xsatisfying E1 E2; then l E1ð Þ� l E2ð Þ:(Solution Let E1;E2 be any subsets of X satisfying E1 E2: We have to show thatl E1ð Þ� l E2ð Þ: Since E1 E2;

V : E2 V ; and V is openf g V : E1 V ; and V is openf g;

and hence,

l Vð Þ : E2 V ; and V is openf g ¼ l V : E2 V ; and V is openf gð Þ l V : E1 V ; and V is openf gð Þ ¼ l Vð Þ : E1 V ; and V is openf g:

Thus,

l Vð Þ : E2 V ; and V is openf g l Vð Þ : E1 V ; and V is openf g:

It follows that

l E1ð Þ ¼ð Þinf l Vð Þ : E1 V ; and V is openf g� inf l Vð Þ : E2 V ; and V is openf g ¼ l E2ð Þð Þ:

Thus, l E1ð Þ� l E2ð Þ: ∎)


IV. Problem 1.185 l ;ð Þ ¼ 0:

(Solution Since ; is an open set, l ;ð Þ ¼ �l ;ð Þ: By II, l ;ð Þ ¼ð Þ�l ;ð Þ ¼ 0; sol ;ð Þ ¼ 0: ∎)

V. Problem 1.186 (i) If f � V1; and g � V2 then f � gð Þ � V1 \V2ð Þ; (ii) If K1 � f ;and K2 � g then K1 \K2ð Þ � f � gð Þ:(Solution (i) Let f � V1; and g � V2: We have to show that f � gð Þ � V1 \V2ð Þ;that is

a. V1 \V2 is an open subset of X;b. f � gð Þ 2 Cc Xð Þ;c. f � gð Þ : X ! 0; 1½ � Cð Þ;d. supp f � gð Þð Þ V1 \V2ð Þ:

For a: Since V1;V2 are open sets, V1 \V2 is an open set.For c, d, b: Since f � V1; f : X ! 0; 1½ �: Similarly, g : X ! 0; 1½ �: Since f :

X ! 0; 1½ �; and g : X ! 0; 1½ �; we have f � gð Þ : X ! 0; 1½ �: Since

f � gð Þ�1 0ð Þ ¼ f�1 0ð Þ [ g�1 0ð Þ;

we have

f � gð Þ�1C� 0f gð Þ ¼ f � gð Þ�1 0ð Þ

� c¼ f�1 0ð Þ [ g�1 0ð Þ� �c|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ f�1 0ð Þ� �c \ g�1 0ð Þ� �c¼ f�1 C� 0f gð Þ \ g�1 C� 0f gð Þ;

and hence

supp f � gð Þ ¼ f � gð Þ�1C� 0f gð Þ

� �¼ f�1 C� 0f gð Þ \ g�1 C� 0f gð Þ� ��|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

f�1 C� 0f gð Þ� �� \ g�1 C� 0f gð Þ� ��¼ supp fð Þ \ supp gð Þ V1 \ supp gð Þ V1 \V2:

Thus, supp f � gð Þ V1 \V2ð Þ: Since f � V1; we have f 2 Cc Xð Þ; and hence,supp fð Þ is compact. Similarly supp gð Þ is compact. Now, by Lemma 1.159, supp gð Þis closed. It follows, by Lemma 1.160, that supp fð Þ \ supp gð Þ is compact. Sincesupp f � gð Þ supp fð Þ \ supp gð Þ; supp f � gð Þ is closed, and supp fð Þ \ supp gð Þ iscompact, by Lemma 1.156, supp f � gð Þ is compact. Since f 2 Cc Xð Þ; f : X ! C iscontinuous. Similarly, g : X ! C is continuous. It follows that f � gð Þ : X ! C iscontinuous. Now, since supp f � gð Þ is compact, we have f � gð Þ 2 Cc Xð Þ:


(ii) Let K1 � f ; and K2 � g: We have to show that K1 \K2ð Þ � f � gð Þ; that is,a. K1 \K2 is a compact subset of X; b. f � gð Þ 2 Cc Xð Þ; c. f � gð Þ : X !

0; 1½ � Cð Þ; d. for every x 2 K1 \K2ð Þ; f � gð Þ xð Þ ¼ 1:For a: Since K1;K2 are compact, K1;K2 are closed sets, and hence K1 \K2 is a

closed set. Since K1 \K2 K1; K1 \K2 is closed, and K1 is compact, K1 \K2 iscompact.

c, b. Since K1 � f ; we have f : X ! 0; 1½ �: Similarly, g : X ! 0; 1½ �: Since f :X ! 0; 1½ �; and g : X ! 0; 1½ �; we have f � gð Þ : X ! 0; 1½ �: Since f � gð Þ�1 0ð Þ ¼f�1 0ð Þ [ g�1 0ð Þ; we have

f � gð Þ�1C� 0f gð Þ f � gð Þ�1 0ð Þ

� c¼ f�1 0ð Þ [ g�1 0ð Þ� �c|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ f�1 0ð Þ� �c \ g�1 0ð Þ� �c¼ f�1 C� 0f gð Þ \ g�1 C� 0f gð Þ;

and hence

supp f � gð Þ ¼ f � gð Þ�1C� 0f gð Þ

� �¼ f�1 C� 0f gð Þ \ g�1 C� 0f gð Þ� ��

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} f�1 C� 0f gð Þð Þ� \ g�1 C� 0f gð Þð Þ�

¼ supp fð Þ \ supp gð Þ:

Thus,

supp f � gð Þ supp fð Þ \ supp gð Þ:

Since K1 � f ; we have f 2 Cc Xð Þ; and hence supp fð Þ is compact. Similarly,supp gð Þ is compact. Now, by Lemma 1.159, supp gð Þ is closed. It follows, byLemma 1.160, that supp fð Þ \ supp gð Þ is compact. Since supp f � gð Þ supp fð Þ \ supp gð Þ; supp f � gð Þ is closed, and supp fð Þ \ supp gð Þ is compact, byLemma 1.156, supp f � gð Þ is compact. Since f 2 Cc Xð Þ; f : X ! C is continuous.Similarly, g : X ! C is continuous. It follows that f � gð Þ : X ! C is continuous.Now, since supp f � gð Þ is compact, f � gð Þ 2 Cc Xð Þ:

d. Let us take any x 2 K1 \K2ð Þ: We have to show that f � gð Þ xð Þ ¼ 1; that is,f xð Þ � g xð Þ ¼ 1: Since x 2 K1 \K2ð Þ; we have x 2 K1: Since K1 � f ; and x 2 K1;we have f xð Þ ¼ 1: Similarly, g xð Þ ¼ 1: Hence, f xð Þ � g xð Þ ¼ 1: ∎)

VI. Problem 1.187 Let V1;V2 be open sets in X: Then,l V1 [V2ð Þ� l V1ð Þþ l V2ð Þ:(Solution We have to show that l V1 [V2ð Þ� l V1ð Þþ l V2ð Þ: Observe that

sup K fð Þ : f � V1 [V2ð Þf g ¼ �l V1 [V2ð Þ ¼ l V1 [V2ð Þ;

and


l V1ð Þþ l V2ð Þ ¼ �l V1ð Þþ �l V2ð Þ ¼ sup K fð Þ : f � V1f gþ sup K fð Þ : f � V2f g

so, we have to show that

sup K fð Þ : f � V1 [V2ð Þf g� sup K fð Þ : f � V1f gþ sup K fð Þ : f � V2f g:

If not, otherwise, let

sup K fð Þ : f � V1f gþ sup K fð Þ : f � V2f g\sup K fð Þ : f � V1 [V2ð Þf g:

We have to arrive at a contradiction. There exists a real number a such that

sup K fð Þ : f � V1f gþ sup K fð Þ : f � V2f g\a\sup K fð Þ : f � V1 [V2ð Þf g:

Since

a\sup K fð Þ : f � V1 [V2ð Þf g;

there exists f0 2 Cc Xð Þ such that f0 � V1 [V2ð Þ; and a\K f0ð Þ: Since f0 �V1 [V2ð Þ; we have supp f0ð Þð Þ V1 [V2ð Þ: Since f0 2 Cc Xð Þ; supp f0ð Þ is compact.Since supp f0ð Þð Þ V1 [V2ð Þ; supp f0ð Þ is compact, and V1;V2 are open sets, byNote 1.177, there exist h1; h2 2 Cc Xð Þ such that h1 � V1; h2 � V2; and, for allx 2 supp f0ð Þ; h1 xð Þþ h2 xð Þ ¼ 1: Since f0 � V1 [V2ð Þ; and h1 � V1; by V,

f0 � h1ð Þ � V1 [V2ð Þ \V1 ¼ V1ð Þ:

Thus, f0 � h1ð Þ � V1; and hence K f0 � h1ð Þ� sup K fð Þ : f � V1f g: Similarly,K f0 � h2ð Þ� sup K fð Þ : f � V2f g: It follows that

K f0 � h1 þ h2ð Þð Þ ¼ K f0 � h1ð Þþ f0 � h2ð Þð Þ ¼ K f0 � h1ð ÞþK f0 � h2ð Þ� sup K fð Þ : f � V1f gþ sup K fð Þ : f � V2f g \að Þ:

Thus, K f0 � h1 þ h2ð Þð Þ\a: Since, for every x 2 X;

f0 � h1 þ h2ð Þð Þ xð Þ

¼ f0 xð Þ � h1 xð Þþ h2 xð Þð Þ if x 2 supp f0ð Þ f0ð Þ�1C� 0f gð Þ

� f0 xð Þ � h1 xð Þþ h2 xð Þð Þ if x 62 supp f0ð Þ

(

¼f0 xð Þ � h1 xð Þþ h2 xð Þð Þ if x 2 f0ð Þ�1

C� 0f gð Þf0 xð Þ � h1 xð Þþ h2 xð Þð Þ if x 2 supp f0ð Þ � f0ð Þ�1

C� 0f gð Þf0 xð Þ � h1 xð Þþ h2 xð Þð Þ if x 62 supp f0ð Þ

8><>:


¼f0 xð Þ � 1ð Þ if x 2 f0ð Þ�1

C� 0f gð Þ0 � h1 xð Þþ h2 xð Þð Þ if x 2 supp f0ð Þ � f0ð Þ�1

C� 0f gð Þ0 � h1 xð Þþ h2 xð Þð Þ if x 62 supp f0ð Þ

8><>:

¼f0 xð Þ if x 2 f0ð Þ�1

C� 0f gð Þ0 if x 2 supp f0ð Þ � f0ð Þ�1

C� 0f gð Þ0 if x 62 supp f0ð Þ

8><>:

¼ f0 xð Þ if x 2 f0ð Þ�1C� 0f gð Þ

0 if x 2 f0ð Þ�1 0ð Þ

(

¼ f0 xð Þ;

we have f0 � h1 þ h2ð Þ ¼ f0; and hence,

a[ð ÞK f0 � h1 þ h2ð Þð Þ ¼ K f0ð Þ:

Thus, K f0ð Þ\a: This a contradiction. ■)

VII. Problem 1.188 l is subadditive, in the sense that, if E1;E2; . . . are any subsetsof X; then

l E1 [E2 [ � � �ð Þ� l E1ð Þþ l E2ð Þþ � � � :(Solution Let E1;E2; . . . be any subsets of X: We have to show that

l E1 [E2 [ � � �ð Þ� l E1ð Þþ l E2ð Þþ � � � :

If any l Enð Þ ¼ 1; then the inequality is trivial. So, we consider the case wheneach l Enð Þ is finite. If not, otherwise, let l E1ð Þþ l E2ð Þþ � � �\l E1 [E2 [ � � �ð Þ;

that is,

inf l Vð Þ : E1 V ; and V is openf gþ inf l Vð Þ : E2 V ; and V is openf gþ � � �\ inf l Vð Þ : E1 [E2 [ � � �ð Þ V ; and V is openf g:

We have to arrive at a contradiction. Put

e � inf l Vð Þ : E1 [E2 [ � � �ð Þ V ; and V is openf g� inf l Vð Þ : E1 V ; and V is openf gþ inf l Vð Þ : E2 V ; and V is openf gþ � � �ð Þ[ 0ð Þ:There exists an open set V1 such that E1 V1; and

l V1ð Þ\inf l Vð Þ : E1 V ; and V is openf gþ e2:


There exists an open set V2 such that E2 V2; and

l V2ð Þ\inf l Vð Þ : E2 V ; and V is openf gþ e4; etc.

On adding these inequalities, we get

l V1ð Þþ l V2ð Þþ � � �\ inf l Vð Þ : E1 V ; and V is openf gþ inf l Vð Þ : E2 V ; and V is openf gþ � � �ð Þ þ e

¼ inf l Vð Þ : E1 [E2 [ � � �ð Þ V ; and V is openf g� l V1 [V2 [ � � �ð Þ ¼ �l V1 [V2 [ � � �ð Þ¼ sup K fð Þ : f � V1 [V2 [ � � �f g:

Thus,

l V1ð Þþ l V2ð Þþ � � �ð Þ\sup K fð Þ : f � V1 [V2 [ � � �f g:

It follows that there exists f0 2 Cc Xð Þ such that f0 � V1 [V2 [ � � � ; and

l V1ð Þþ l V2ð Þþ � � �ð Þ\K f0ð Þ:

Since f0 � V1 [V2 [ � � � ; we have supp f0ð Þ V1 [V2 [ � � � ; and supp f0ð Þ iscompact. It follows that there exists a positive integer n0 such that supp f0ð Þ V1 [ � � � [Vn0 : Since supp f0ð Þ V1 [ � � � [Vn0 ; and f0 � V1 [V2 [ � � � ; we havef0 � V1 [ � � � [Vn0 : This shows that

K f0ð Þ� sup K fð Þ : f � V1 [ � � � [Vn0f g¼ �l V1 [ � � � [Vn0ð Þ¼ l V1 [ � � � [Vn0ð Þ� l V1ð Þþ � � � þ l Vn0ð Þ� l V1ð Þþ l V2ð Þþ � � �\K f0ð Þ;

and hence, K f0ð Þ\K f0ð Þ: This is a contradiction. ■)

VIII. Problem 1.189 K : Cc Xð Þ ! C is monotone, in the sense that, if f1 : X !R Cð Þ; f2 : X ! R Cð Þ are members of Cc Xð Þ satisfying f1 � f2; thenK f1ð Þ;K f2ð Þ are real numbers, and K f1ð Þ�K f2ð Þ:(Solution Let f 2 Cc Xð Þ: Let f : X ! R: We first show that K fð Þ 2 R:

Problem 1:190 f þ 2 Cc Xð Þ:(Solution We must prove: a. f þ : X ! 0;1½ Þ is continuous, b. supp f þð Þ iscompact.

For a: Let us take any a 2 X: We have to show that f þ is continuous at a:Case I: when f að Þ[ 0: Let us take any e 2 0; f að Þð Þ: Since, f is continuous,

there exists an open neighborhood V of a such that f Vð Þ f að Þ � e; f að Þþ eð Þ:Since f Vð Þ f að Þ � e; f að Þþ eð Þ; and e 2 0; f að Þð Þ; we have


f þ að Þ � e; f þ að Þþ eð Þ ¼ f að Þ � e; f að Þþ eð Þ f Vð Þ ¼ f þ Vð Þ;

and hence, f þ Vð Þ f þ að Þ � e; f þ að Þþ eð Þ: Thus, f þ is continuous at a:Case II: when f að Þ ¼ 0: Let us take any real e[ 0: Since f is continuous, there

exists an open neighborhood V of a such that f Vð Þ f að Þ � e; f að Þþ eð Þ ¼ 0� e; 0þ eð Þ ¼ �e; eð Þð Þ: Since f Vð Þ �e; eð Þ; we have

f þ Vð Þ 0; e½ Þ �e; eð Þ ¼ f að Þ � e; f að Þþ eð Þ ¼ f þ að Þ � e; f þ að Þþ eð Þ;

and hence

f þ Vð Þ f þ að Þ � e; f þ að Þþ eð Þ:

Thus, f þ is continuous at a:Case III: when f að Þ\0: Let us take any e 2 0;�f að Þð Þ: Since, f is continuous,

there exists an open neighborhood V of a such that f Vð Þ f að Þ � e; f að Þþ eð Þ �1; 0ð Þð Þ: Since f Vð Þ �1; 0ð Þ; we have

f þ Vð Þ ¼ 0f g 0� e; 0þ eð Þ ¼ f þ að Þ � e; f þ að Þþ eð Þ;

and hence, f þ Vð Þ f þ að Þ � e; f þ að Þþ eð Þ: Thus, f þ is continuous at a:Thus, in all cases, f þ is continuous at a:For b: Observe that f þð Þ�1 0ð Þ ¼ f�1 �1; 0ð �ð Þ: It follows that

f þð Þ�1R� 0f gð Þ ¼ f þð Þ�1 0ð Þ

� c¼ f�1 �1; 0ð �ð Þ� �c¼ f�1 0;1ð Þ f�1 R� 0f gð Þ;

and hence

f þð Þ�1R� 0f gð Þ

� � f�1 R� 0f gð Þ� ��¼ supp fð Þ;

which is compact: It follows that

f þð Þ�1R� 0f gð Þ

� �¼ supp f þð Þð Þ

is compact. Thus supp f þð Þ is compact. ■)Similarly, f� 2 Cc Xð Þ: Since f þ : X ! 0;1½ Þ; and K is a positive linear

functional on Cc Xð Þ; we have K f þð Þ 2 0;1½ Þ: Similarly, K f�ð Þ 2 0;1½ Þ: It fol-lows that

K fð Þ ¼ K f þ � f�ð Þ ¼ð Þ K f þð Þ � K f�ð Þð Þ 2 R;

and hence, K fð Þ 2 R:


Now, let f1 : X ! R Cð Þ; f2 : X ! R Cð Þ be members of Cc Xð Þ satisfyingf1 � f2: It follows that f2 � f1ð Þ : X ! 0;1½ Þ Cð Þ is a member of Cc Xð Þ: Now,since K : Cc Xð Þ ! C is a positive linear functional on Cc Xð Þ; we haveK f2ð Þ � K f1ð Þ ¼ð ÞK f2 � f1ð Þ 2 0;1½ Þ: Since f1 : X ! R Cð Þ is a member ofCc Xð Þ; we have K f1ð Þ 2 R: Similarly; K f2ð Þ 2 R: Since K f1ð Þ;K f2ð Þ 2 R; andK f2ð Þ � K f1ð Þð Þ 2 0;1½ Þ; we have K f1ð Þ�K f2ð Þ: ∎)Let E be any subset of X:Since ; E; and ; is a compact set, ; 2 K : K E; and K is a compact setf g;

and hence

0 ¼ð Þl ;ð Þ 2 l Kð Þ : K E; and K is a compact setf g:

Thus, l Kð Þ : K E; and K is a compact setf g is a nonempty set. Now since,for every subset F of X; l Fð Þ 2 0;1½ �; l Kð Þ : K E; and K is a compact setf gis a nonempty subset of 0;1½ �; and hence

sup l Kð Þ : K E; and K is a compact setf gð Þ 2 0;1½ �:Problem 1.191 If K is a compact subset of X, then

l Kð Þ ¼ sup l K1ð Þ : K1 K; and K1 is a compact setf g:(Solution Let K be a compact subset of X. Since K K; and K is a compact set;we have l Kð Þ 2 l K1ð Þ : K1 K; and K1 is a compact setf g; and hencel Kð Þ� sup l K1ð Þ : K1 K; and K1 is a compact setf g: It remains to show thatsup l K1ð Þ : K1 K; and K1 is a compact setf g� l Kð Þ: It suffices to show thatl Kð Þ is an upper bound of l K1ð Þ : K1 K; and K1 is a compact setf g: For thispurpose, let us take any compact set K1 satisfying K1 K: We have to show thatl K1ð Þ� l Kð Þ: Since K1 K; by III, l K1ð Þ� l Kð Þ: ∎)

Let K be a compact subset of X:Since K X; K is compact and X is open, by Urysohn’s lemma, there exists

f 2 Cc Xð Þ such that K � f � X: It follows that f : K � ff g is nonempty, and henceK fð Þ : K � ff g is nonempty. If K � f ; then f 2 Cc Xð Þ; and

f : X ! 0; 1½ � 0;1½ Þð Þ. Since K is a positive linear functional on Cc Xð Þ; K � fimplies that K fð Þ is a nonnegative real number. Thus, K fð Þ : K � ff g is a none-mpty set of nonnegative real numbers, and hence, inf K fð Þ : K � ff g 2 0;1½ Þ:IX. Problem 1.192 Let K be a compact subset of X: Then l Kð Þ\1:

(Solution Since K X; K is compact, and X is open, by Urysohn’s lemma, thereexists f 2 Cc Xð Þ such that K � f � X: Since K � f ; f : X ! 0; 1½ � Cð Þ is con-tinuous, and, for every x 2 K; we have f xð Þ ¼ 1: It follows that

K f�1 1ð Þ � �f�1 1

2; 1

� ��

is an open set. Since K f�1 12 ; 1� ��

; we have


l Kð Þ� l f�1 12; 1

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ �l f�1 1

2; 1

� ��

¼ sup K gð Þ : g � f�1 12; 1

� �� :

Thus,

l Kð Þ� sup K gð Þ : g � f�1 12; 1

� �� :

Since f : X ! 0; 1½ �; f 2 Cc Xð Þ; and K is a positive functional on Cc Xð Þ; K fð Þ isa nonnegative real number. Since

l Kð Þ� sup K gð Þ : g � f�1 12; 1

� �� ;

and K fð Þ is a nonnegative real number, it suffices to show that the real number2 K fð Þð Þ is an upper bound of

K gð Þ : g � f�1 12; 1

� �� :

For this purpose, let us take any g 2 Cc Xð Þ satisfying g � f�1 12 ; 1� ��

: We haveto show that K gð Þ� 2 K fð Þð Þ:Problem 1:193 1

2 g� f :

(Solution Let us take any x 2 X: We have to show that 12 g xð Þ� f xð Þ:

Case I: when g xð Þ ¼ 0: In this case, 12 g xð Þ� f xð Þ is trivially true.

Case II: when g xð Þ 6¼ 0: In this case, x 2 supp gð Þ f�1 12 ; 1� ��

; and hence,f xð Þ 2 1

2 ; 1� �

: It follows that 12 g xð Þ� 1

2 � 1 ¼� �12\f xð Þ; and hence, 1

2 g xð Þ� f xð Þ:Thus, in all cases, 1

2 g xð Þ� f xð Þ: ∎)Since 1

2 g� f ; by VIII, 12 K gð Þð Þ ¼� �

K 12 g� ��K fð Þ; and hence, K gð Þ� 2K fð Þ: ∎)

We shall denote the collection

E : l Eð Þ\1; and l Eð Þ ¼ sup l Kð Þ : K E; and Kis a compact setf gf g

by ℳF :

X. Problem 1.194 Every compact subset of X is a member of ℳF :

(Solution Let K be a subset of X: We have seen that

l Kð Þ ¼ sup l K1ð Þ : K1 K; and K1 is a compact setf g:

Also, from IX, l Kð Þ\1: Now, by the definition of ℳF ; K 2 ℳF : ∎)


Let K be a compact subset of X:Since K X; K is compact and X is open, by Urysohn’s lemma, there exists

f0 2 Cc Xð Þ such that K � f0 � X: It follows that K fð Þ : K � ff g is nonempty. Forevery f satisfying K � f ; we have f 2 Cc Xð Þ; and f : X ! 0; 1½ �: Now, since K is apositive linear functional on Cc Xð Þ; for every f satisfying K � f ; K fð Þ is a non-negative real number. Thus, K fð Þ : K � ff g is a nonempty set of nonnegative realnumbers. It follows that inf K fð Þ : K � ff g 2 0;1½ Þ:XI. Problem 1.195 Let K be a compact subset of X: Thenl Kð Þ ¼ inf K fð Þ : K � ff g:(Solution We first try to show that l Kð Þ is a lower bound of K fð Þ : K � ff g: Forthis purpose, let us take any f satisfying K � f :We have to show that l Kð Þ�K fð Þ:

Let us take any a 2 0; 1ð Þ: Since K X; K is compact and X is open, byUrysohn’s lemma, there exists f 2 Cc Xð Þ such that K � f � X: Since K � f ; f :X ! 0; 1½ � Cð Þ is continuous, and for every x 2 K; f xð Þ ¼ 1: It follows that

K f�1 1ð Þ � �f�1 a; 1ð �ð Þ

is an open set. Since K f�1 a; 1ð �ð Þ; we have

l Kð Þ� l f�1 a; 1ð �ð Þ� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ �l f�1 a; 1ð �ð Þ� � ¼ sup K gð Þ : g � f�1 a; 1ð �ð Þ� :

Thus,

l Kð Þ� sup K gð Þ : g � f�1 a; 1ð �ð Þ� :

Since f : X ! 0; 1½ �; f 2 Cc Xð Þ; and K is a positive functional on Cc Xð Þ; K fð Þ isa nonnegative real number. Now, we shall try to show that the real number 1

a K fð Þð Þis an upper bound of K gð Þ : g � f�1 a; 1ð �ð Þ�

: For this purpose, let us take anyg 2 Cc Xð Þ satisfying g � f�1 a; 1ð �ð Þ: We have to show that K gð Þ� 1

a K fð Þð Þ:Problem 1:196 ag� f :

(Solution Let us take any x 2 X: We have to show that ag xð Þ� f xð Þ:Case I: when g xð Þ ¼ 0: In this case, ag xð Þ� f xð Þ is trivially true.Case II: when g xð Þ 6¼ 0: In this case, x 2 supp gð Þ f�1 a; 1ð �ð Þð Þ; and hence

f xð Þ 2 a; 1ð �: It follows that ag xð Þ� a � 1 ¼ð Þa\f xð Þ; and hence, ag xð Þ� f xð Þ:Thus, in all cases, ag xð Þ� f xð Þ: ∎)

Since ag� f ; by VIII, a K gð Þð Þ ¼ð ÞK agð Þ�K fð Þ; and hence K gð Þ� 1aK fð Þ:

We have seen that 1a K fð Þð Þ is an upper bound of K gð Þ : g � f�1 a; 1ð �ð Þ�

; andhence,


l Kð Þ�ð Þsup K gð Þ : g � f�1 a; 1ð �ð Þ� � 1a

K fð Þð Þ:

Since for every a 2 0; 1ð Þ; l Kð Þ� 1a K fð Þð Þ; on letting a ! 1; we get

l Kð Þ� K fð Þð Þ:Thus, l Kð Þ� inf K fð Þ : K � ff g: It remains to show that inf K fð Þ :f K �

f g� l Kð Þ: If not, otherwise, let l Kð Þ\inf K fð Þ : K � ff g: We have to arrive at acontradiction. Put e � inf K fð Þ : K � ff g � l Kð Þ [ 0ð Þ: Since

l Kð Þ ¼ inf l Vð Þ : K V ; and V is openf g;

and e[ 0; there exists an open set V0 such that K V0; and

sup K fð Þ : f � V0f g ¼ �l V0ð Þ ¼ l V0ð Þ\l Kð Þþ e|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ inf K fð Þ : K � ff g:

Thus,

sup K fð Þ : f � V0f g\inf K fð Þ : K � ff g:

Since K V0; by Urysohn’s lemma, there exists f0 2 Cc Xð Þ such that K �f0 � V0: It follows that K f0ð Þ� sup K fð Þ : f � V0f g; and inf K fð Þ :fK � f g�K f0ð Þ:

Thus, inf K fð Þ : K � ff g� sup K fð Þ : f � V0f g; which is a contradiction. ■)

XII. Problem 1.197 Let V be an open set in X: Then

l Vð Þ ¼ sup l Kð Þ : K V ; and K is a compact setf g:

Hence, if V is an open set satisfying l Vð Þ\1; then V 2 ℳF :

(Solution By the monotone property of l; l Vð Þ is an upper bound ofl Kð Þ : K V ; and K is a compact setf g; and hence,

sup l Kð Þ : K V ; and K is a compact setf g� l Vð Þ: It remains to show that

sup a : a is real; and a\l Vð Þf g ¼ l Vð Þ� sup l Kð Þ : K V ; and K is a compact setf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is,

sup a : a is real; and a\l Vð Þf g� sup l Kð Þ : K V ; and K is a compact setf g:

By IX, we observe that

a : a is real; and a\l Vð Þf g; and l Kð Þ : K V ; and K is a compact setf g

are nonempty sets of real numbers. It suffices to show that, for every real number asatisfying a\l Vð Þ; there exists a compact set K such that K V ; and a� l Kð Þ:For this purpose, let us take any real a satisfying


a\l Vð Þ|fflfflfflfflffl{zfflfflfflfflffl} ¼ �l Vð Þ ¼ sup K fð Þ : f � Vf g:

Since a\sup K fð Þ : f � Vf g; there exists f0 2 Cc Xð Þ such that f0 � V ; anda\K f0ð Þ: Since f0 � V ; supp f0ð Þ is a compact set, and supp f0ð Þ V : It suffices toshow that

a� l supp f0ð Þð Þ|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} ¼ inf l Wð Þ : supp f0ð Þ W ; and W is openf g;

that is, a� inf l Wð Þ : supp f0ð Þ W ; and W is openf g: It is enough to show thatthe real number a is a lower bound of l Wð Þ : supp f0ð Þ W ; and W is openf g: Forthis purpose, let us take any open set W such that supp f0ð Þ W : We have to showthat a� l Wð Þ:

Since f0 � V ; we have f0 2 Cc Xð Þ; and f0 : X ! 0; 1½ �: Since f0 2 Cc Xð Þ; f0 :X ! 0; 1½ �; W is open; and supp f0ð Þ W ; we have f0 � W ; and hence, K f0ð Þ 2K gð Þ : g � Wf g: It follows that

a\K f0ð Þ� sup K gð Þ : g � Wf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ �l Wð Þ ¼ l Wð Þ:

Hence a� l Wð Þ: ∎)

XIII. Problem 1.198 Let K1;K2 be compact subsets of X: Let K1 \K2 ¼ ;: Then,l K1 [K2ð Þ ¼ l K1ð Þþ l K2ð Þ:(Solution By VII,

l K1 [K2 [;[ ;[ � � �ð Þ� l K1ð Þþ l K2ð Þþ l ;ð Þþ l ;ð Þþ � � � :

Since

l K1 [K2ð Þ ¼ l K1 [K2 [;[ ;[ � � �ð Þ;

and

l K1ð Þþ l K2ð Þþ l ;ð Þþ l ;ð Þþ � � � ¼ l K1ð Þþ l K2ð Þþ 0þ 0þ � � �¼ l K1ð Þþ l K2ð Þ;

it remains to show that

l K1ð Þþ l K2ð Þ� l K1 [K2ð Þ:

If not, otherwise, let l K1 [K2ð Þ\l K1ð Þþ l K2ð Þ: We have to arrive at a con-tradiction. Since l K1 [K2ð Þ\l K1ð Þþ l K2ð Þ; by XI,


inf K fð Þ : K1 [K2 � ff g\inf K fð Þ : K1 � ff gþ inf K fð Þ : K2 � ff g:

Since K1 \K2 ¼ ;; we have K1 K2ð Þc: Since K2 is a compact subset of theHausdorff space X; by Lemma 1.159, K2 is a closed set. Since K2 is a closed set,K2ð Þc is an open set. Since K1 K2ð Þc; K1 is compact and K2ð Þc is open, byUrysohn’s lemma, there exists f0 2 Cc Xð Þ such that K1 � f0 � K2ð Þc: Since K1 �f0; for every x 2 K1; we have f0 xð Þ ¼ 1: Since f0 � K2ð Þc; we have

f0ð Þ�1 0ð Þ� c

¼ f0ð Þ�1C� 0f gð Þ f0ð Þ�1

C� 0f gð Þ� �

¼ supp f0ð Þ K2ð Þc;

and hence, K2 f0ð Þ�1 0ð Þ: It follows that, for every x 2 K2; we have f0 xð Þ ¼ 0:Since

inf K fð Þ : K1 [K2 � ff g\ inf K fð Þ : K1 � ff gþ inf K fð Þ : K2 � ff gð Þ;

there exists g 2 Cc Xð Þ such that K1 [K2 � g; and

K gð Þ\ inf K fð Þ : K1 � ff gþ inf K fð Þ : K2 � ff gð Þ:

Since K1 � f0; and K1 [K2 � g; by V, K1 ¼ð Þ K1 \ K1 [K2ð Þð Þ � f0 � gð Þ; andhence, K1 � f0 � gð Þ:Problem 1:199 K2 � 1� f0ð Þ � gð Þ:(Solution We must show

a. K2 is a compact subset of X;b. 1� f0ð Þ � gð Þ 2 Cc Xð Þ;c. 1� f0ð Þ � gð Þ : X ! 0; 1½ � Cð Þ;d. for every x 2 K2; 1� f0ð Þ � gð Þ xð Þ ¼ 1:

For a, c, d: These are trivially true.For b: Since f0 2 Cc Xð Þ; f0 is continuous. The constant function 1 is continuous.

It follows that 1� f0 is a continuous function. Since g 2 Cc Xð Þ; g is continuous.Since 1� f0ð Þ; g are continuous functions, their product 1� f0ð Þ � g is a continuousfunction. It remains to show that supp 1� f0ð Þ � gð Þ is compact. Since

1� f0ð Þ � gð Þ�1 0ð Þ ¼ 1� f0ð Þ�1 0ð Þ [ g�1 0ð Þ;

we have

1� f0ð Þ � gð Þ�1 0; 1ð �ð Þ ¼ 1� f0ð Þ � gð Þ�1 0ð Þ� c

¼ 1� f0ð Þ�1 0ð Þ� c

\ g�1 0ð Þ� �c|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1� f0ð Þ�1 0; 1ð �ð Þ \ g�1 0; 1ð �ð Þ;


and hence

supp 1� f0ð Þ � gð Þ ¼ 1� f0ð Þ � gð Þ�1 0; 1ð �ð Þ� �

¼ 1� f0ð Þ�1 0; 1ð �ð Þ \ g�1 0; 1ð �ð Þ� �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 1� f0ð Þ�1 0; 1ð �ð Þ� �

\ g�1 0; 1ð �ð Þ� ��¼ supp 1� f0ð Þð Þ \ supp gð Þð Þ:

Thus,

supp 1� f0ð Þ � gð Þ supp 1� f0ð Þð Þ \ supp gð Þð Þ:

Since g 2 Cc Xð Þ; supp gð Þ is compact. Since supp gð Þ is compact, andsupp 1� f0ð Þ is closed, supp 1� f0ð Þð Þ \ supp gð Þð Þ is compact. Since

supp 1� f0ð Þ � gð Þ supp 1� f0ð Þð Þ \ supp gð Þð Þ;

supp 1� f0ð Þ � gð Þ is a closed set, and supp 1� f0ð Þð Þ \ supp gð Þð Þ is a compactset, supp 1� f0ð Þ � gð Þ is a compact set. ■)

Since K1 � f0 � gð Þ; we have K f0 � gð Þ 2 K fð Þ : K1 � ff g; and hence,

l K1ð Þ ¼ inf K fð Þ : K1 � ff g�K f0 � gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, l K1ð Þ�K f0 � gð Þ: Since K2 � 1� f0ð Þ � gð Þ; we have

K 1� f0ð Þ � gð Þ 2 K fð Þ : K2 � ff g;and hence

l K2ð Þ ¼ inf K fð Þ : K2 � ff g�K 1� f0ð Þ � gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, l K2ð Þ�K 1� f0ð Þ � gð Þ: It follows that

l K1ð Þþ l K2ð Þ�K f0 � gð ÞþK 1� f0ð Þ � gð Þ ¼ K f0 � gþ 1� f0ð Þ � gð Þ¼ K gð Þ\inf K fð Þ : K1 � ff gþ inf K fð Þ : K2 � ff g ¼ l K1ð Þþ l K2ð Þ;

and hence

l K1ð Þþ l K2ð Þ\l K1ð Þþ l K2ð Þ:

This gives a contradiction. ■)

XIV. Problem 1.200 Let E1; . . .;En 2 ℳF : Suppose that E1; . . .;En are pairwisedisjoint. Then l E1 [ � � � [Enð Þ ¼ l E1ð Þþ � � � þ l Enð Þ:(Solution By VII,


l E1 [ � � � [Enð Þ ¼ l E1 [ � � � [En [;[ ;[ � � �ð Þ� l E1ð Þþ � � � þ l Enð Þþ l ;ð Þþ l ;ð Þþ � � �¼ l E1ð Þþ � � � þ l Enð Þþ 0þ 0þ � � �¼ l E1ð Þþ � � � þ l Enð Þ:

For every i ¼ 1; . . .; n; since Ei 2 ℳF ; by the definition of ℳF ; l Eið Þ\1; and

l Eið Þ ¼ sup l Kð Þ : K Ei; and K is a compact setf g:

It remains to show that

l E1ð Þþ � � � þ l Enð Þ� l E1 [ � � � [Enð Þ

that is,

sup l E1ð Þþ � � � þ l Enð Þ � e : e is a real number and e[ 0f g� l E1 [ � � � [Enð Þ:

Since l E1 [ � � � [Enð Þ� l E1ð Þþ � � � þ l Enð Þ; and, for each i ¼ 1; . . .; n;0� l Eið Þ\1; we have

0� l E1ð Þþ � � � þ l Enð Þ\1; and 0� l E1 [ � � � [Enð Þ\1:

Here, it suffices to show that l E1 [ � � � [Enð Þ is an upper bound of

l E1ð Þþ � � � þ l Enð Þ � e : e is a real number and e[ 0f g:

For this purpose, let us fix any real e[ 0: We have to show that

l E1ð Þþ � � � þ l Enð Þ � e� l E1 [ � � � [Enð Þ:

Observe that

sup l Kð Þ : K E1; and K is a compact setf gð Þ � en

� þ � � �

þ sup l Kð Þ : K En; and K is a compact setf g � en

� ¼ l E1ð Þþ � � � þ l Enð Þ � e:

Here e is a positive real number, and, for each i ¼ 1; . . .; n;

l Eið Þ ¼ sup l Kð Þ : K Ei; and K is a compact setf g;

it follows that, for each i ¼ 1; . . .; n; there exists a compact set Ki such that Ki Ei;and l Eið Þ � e

n\l Kið Þ: On adding these inequalities, we get


l E1ð Þþ � � � þ l Enð Þ � e\l K1ð Þþ � � � þ l Knð Þ:


l K1ð Þþ � � � þ l Knð Þ� l E1 [ � � � [Enð Þ:

Since E1; . . .;En are pairwise disjoint, and each Ki Ei; K1; . . .;Kn are pairwisedisjoint. Now, on using XIII,

l K1ð Þþ � � � þ l Knð Þ ¼ l K1 [ � � � [Knð Þ:

Thus, it is enough to show that

l K1 [ � � � [Knð Þ� l E1 [ � � � [Enð Þ:

Since each Ki Ei; we have

K1 [ � � � [Kn E1 [ � � � [En;

and hence

l K1 [ � � � [Knð Þ� l E1 [ � � � [Enð Þ:

∎)

XV. Problem 1.201 Let E1;E2; . . . be any members of ℳF : Suppose thatE1;E2; . . . are pairwise disjoint. Then,

l E1 [E2 [ � � �ð Þ ¼ l E1ð Þþ l E2ð Þþ � � � :(Solution Case I: when l Enð Þ ¼ 1 for some n: In this case,l E1ð Þþ l E2ð Þþ � � � ¼ 1: Since En E1 [E2 [ � � �ð Þ;

1 ¼ð Þl Enð Þ� l E1 [E2 [ � � �ð Þ;

we have l E1 [E2 [ � � �ð Þ ¼ 1: Hence,

l E1ð Þþ l E2ð Þþ � � � ¼ l E1 [E2 [ � � �ð Þ:

Case II: when l Enð Þ\1 for every positive integer n: By VII,

l E1 [E2 [ � � �ð Þ� l E1ð Þþ l E2ð Þþ � � � :



sup l E1ð Þþ � � � þ l Enð Þ : n 2 Nf g ¼ l E1ð Þþ l E2ð Þþ � � � � l E1 [E2 [ � � �ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :It suffices to show that l E1 [E2 [ � � �ð Þ is an upper bound of

l E1ð Þþ � � � þ l Enð Þ : n 2 Nf g:

For this purpose, let us fix any positive integer n: We have to show that

l E1ð Þþ � � � þ l Enð Þ� l E1 [E2 [ � � �ð Þ:

Since E1; . . .;En are pairwise disjoint members of ℳF ; by XIV,

l E1 [E2 [ � � �ð Þ� l E1 [ � � � [Enð Þ ¼ l E1ð Þþ � � � þ l Enð Þ;

and hence

l E1ð Þþ � � � þ l Enð Þ� l E1 [E2 [ � � �ð Þ:

∎)

XVI. Problem 1.202 Let E1;E2; . . . be any members of ℳF : Suppose thatE1;E2; . . . are pairwise disjoint. Let

l E1 [E2 [ � � �ð Þ\1:

Then, E1 [E2 [ � � � is a member of ℳF :

(Solution By the definition of ℳF ; it suffices to show that

l E1 [E2 [ � � �ð Þ� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g:

Since

sup l E1ð Þþ � � � þ l Enð Þ : n 2 Nf g ¼ l E1ð Þþ l E2ð Þþ � � � ¼ l E1 [E2 [ � � �ð Þ;

we have to show that

sup l E1ð Þþ � � � þ l Enð Þ : n 2 Nf g� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g:

Again, it suffices to show that, for every positive integer n; there exists acompact set K such that

K E1 [E2 [ � � �ð Þ; and l E1ð Þþ � � � þ l Enð Þ� l Kð Þ:


For this purpose, let us fix any positive integer n: Let us fix any real e[ 0: Sinceeach Ei 2 ℳF ; there exists a compact set Ki such that Ki Ei; and l Eið Þ �e2i \l Kið Þ: Here K1 [ � � � [Kn is a compact set such that

K1 [ � � � [Knð Þ E1 [E2 [ � � �ð Þ:


l E1ð Þþ � � � þ l Enð Þ� l K1 [ � � � [Knð Þ:

Since each Ki is a compact set, by X, each Ki 2 ℳF : Since, E1;E2; . . . arepairwise disjoint, and each Ki Ei; K1;K2; . . . are pairwise disjoint. SinceK1; . . .;Kn are pairwise disjoint members of ℳF ; on using XV, we get

l K1 [ � � � [Knð Þ ¼ l K1ð Þþ � � � þ l Knð Þ[ l E1ð Þ � e2

� þ � � �

þ l Enð Þ � e2n

� � l E1ð Þþ � � � þ l Enð Þð Þ � e;

and hence,

l E1ð Þþ � � � þ l Enð Þð Þ � e\l K1 [ � � � [Knð Þ:

On letting e ! 0; we get

l E1ð Þþ � � � þ l Enð Þð Þ� l K1 [ � � � [Knð Þ:

∎)

XVII. Problem 1.203 Let E 2 ℳF : Let e be any positive real. Then, there exists acompact set K; and an open set V such that K E V ; l V � Kð Þ\e; andV ;K; V � Kð Þ 2 ℳF :

(Solution Since E 2 ℳF ; we have l Eð Þ\1; and

l Eð Þ ¼ sup l Kð Þ : K E; and K is a compact setf g;

and hence, there exists a compact set K such that K E; and l Eð Þ � e2\l Kð Þ:

Since

l Eð Þ ¼ inf l Vð Þ : E V ; and V is openf g;

there exists an open set V such that K ð ÞE V ; and

l Vð Þ\l Eð Þþ e2|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}\ l Kð Þþ e

2

� þ e

2¼ l Kð Þþ e:


Hence, l Vð Þ\l Kð Þþ e: It follows that l Vð Þ is a nonnegative real number.Since V is open, by XII,

l Vð Þ ¼ sup l K1ð Þ : K1 V ; and K1 is a compact setf g:

Now, since l Vð Þ is a nonnegative real number, by the definition of ℳF ; V 2ℳF : Since K is compact, by X, K 2 ℳF : Since K is compact in the Hausdorffspace X; K is closed, and hence Kc is open. Since V is open, and Kc is open,V � Kð Þ ¼ð ÞV \ Kcð Þ is open. Since V � Kð Þ V ; we have

l V � Kð Þ� l Vð Þ \1ð Þ; and hence, l V � Kð Þ\1: Since V � Kð Þ is open, andl V � Kð Þ\1; on using XII, we get V � Kð Þ 2 ℳF : Since K V ; we have V ¼K [ V � Kð Þ; and K \ V � Kð Þ ¼ ;: Now, since K; V � Kð Þ 2 ℳF ; by XIV,l Vð Þ ¼ l Kð Þþ l V � Kð Þ: Now, since l Vð Þ; l V � Kð Þ; l Kð Þ are nonnegative realnumbers, l V � Kð Þ ¼ l Vð Þ � l Kð Þ \eð Þ: Thus, l V � Kð Þ\e; where K E V ;K is compact, and V is open. Also, V ;K; V � Kð Þ 2 ℳF : ■)

XVIII. Problem 1.204 Let E;F 2 ℳF : Then, E � Fð Þ 2 ℳF :

(Solution It is enough to show:

a. l E � Fð Þ\1;b. l E � Fð Þ� sup l Kð Þ : K E � Fð Þ; and K is a compact setf g:

For a: Since E 2 ℳF ; we have l Eð Þ\1: Since E � Fð Þ E; we havel E � Fð Þ� l Eð Þ \1ð Þ; and hence, l E � Fð Þ\1:

For b: Let us take any real e[ 0: Since E 2 ℳF ; by XVII, there exists acompact set K1; and an open set V1 such that K1 E V1; l V1 � K1ð Þ\ e

2 ; andV1;K1; V1 � K1ð Þ 2 ℳF : Similarly, there exists a compact set K2; and an open setV2 such that K2 E V2; l V2 � K2ð Þ\ e

2 ; and V2;K2; V2 � K2ð Þ 2 ℳF : Since,

E � Fð Þ V1 � Fð Þ V1 � K2ð Þ V1 � K1ð Þ [ K1 � V2ð Þ [ V2 � K2ð Þ;

by VII,

l E � Fð Þ� l V1 � K1ð Þþ l K1 � V2ð Þþ l V2 � K2ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\

e2þ l K1 � V2ð Þþ e

2¼ l K1 � V2ð Þþ e:

Since V2 is open, V2ð Þc is closed. Since V2ð Þc is closed and K1 is compact andK1 � V2 ¼ð ÞK1 \ V2ð Þcð Þ is compact, hence, K1 � V2 is compact. SinceK1 � V2ð Þ E � V2ð Þ E � Fð Þ; we have K1 � V2ð Þ E � Fð Þ: SinceK1 � V2ð Þ E � Fð Þ; and K1 � V2 is compact, we have

l K1 � V2ð Þ 2 l Kð Þ : K E � Fð Þ; and K is a compact setf g;

and hence


l E � Fð Þ � e\ð Þl K1 � V2ð Þ� sup l Kð Þ : K E � Fð Þ; and K is a compact setf g:

Since l E � Fð Þ � e\sup l Kð Þ : K E � Fð Þ; and K is a compact setf g; onletting e ! 0; we get l E � Fð Þ� sup l Kð Þ : K E � Fð Þ;fand K is a compact setg: ∎)

XIX. Problem 1.205 Let E;F 2 ℳF : Then, E \Fð Þ 2 ℳF :

(Solution By XVIII, E � Fð Þ 2 ℳF : Since E; E � Fð Þ 2 ℳF ; by XVIII,E \Fð Þ ¼ð Þ E � E � Fð Þð Þ 2 ℳF ; and hence, E \Fð Þ 2 ℳF : ■)

XX. Problem 1.206 Let E;F 2 ℳF : Let E \F ¼ ;: Then E[Fð Þ 2 ℳF :

(Solution It is enough to show:

a. l E [Fð Þ\1;b. l E [Fð Þ� sup l Kð Þ : K E [Fð Þ; and K is a compact setf g:

For a: Since E 2 ℳF ; l Eð Þ\1: Similarly, l Fð Þ\1: By XIV, l E[Fð Þ ¼l Eð Þþ l Fð Þ: Since l Eð Þ\1; and l Fð Þ\1; we havel E [Fð Þ ¼ð Þ l Eð Þþ l Fð Þð Þ\1; and hence, l E [Fð Þ\1:For b: Let us take any real e[ 0: By XIV, l E [Fð Þ ¼ l Eð Þþ l Fð Þ; so, we have

to show that

l Eð Þþ l Fð Þ� sup l Kð Þ : K E [Fð Þ; and K is a compact setf g:

Since E 2 ℳF ; we have l Eð Þ� sup l Kð Þ : K E; and K is a compact setf g;and hence, there exists a compact set K1 such that K1 E; and

sup l Kð Þ : K E; and K is a compact setf g � e2\l K1ð Þ:

Similarly, there exists a compact set K2 such that K2 F; and

sup l Kð Þ : K F; and K is a compact setf g � e2\l K2ð Þ:

Since K1 E; and K2 F; K1 [K2ð Þ E [Fð Þ: Since K1;K2 are compact,K1 [K2ð Þ is compact. Since K1 [K2ð Þ E [Fð Þ; and K1 [K2ð Þ is compact, wehave

l K1 [K2ð Þ 2 l Kð Þ : K E [Fð Þ; and K is a compact setf g;

and hence

l K1 [K2ð Þ� sup l Kð Þ : K E [Fð Þ; and K is a compact setf g:


Since K1 E; K2 F; and E \F ¼ ;; we have K1 \K2 ¼ ;: Since K1 is acompact set, by X, K1 2 ℳF : Similarly, K2 2 ℳF : Now, by XIV,

sup l Kð Þ : K E [Fð Þ; and K is a compact setf g� l K1 [K2ð Þ ¼ l K1ð Þþ l K2ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}[ sup l Kð Þ : K E; and K is a compact setf g � e

2

� þ sup l Kð Þ : K F; and K is a compact setf g � e

2

� ¼ l Eð Þ � e

2

� þ l Fð Þ � e

2

� ¼ l Eð Þþ l Fð Þð Þ � e;

and hence

l Eð Þþ l Fð Þð Þ � e\sup l Kð Þ : K E [Fð Þ; and K is a compact setf g:

On letting e ! 0; we get l Eð Þþ l Fð Þð Þ� sup l Kð Þ : K E [Fð Þ;fand K is a compact setg: ∎)

XXI. Problem 1.207 Let E;F 2 ℳF : Then, E [Fð Þ 2 ℳF :

(Solution By XVIII, E � Fð Þ 2 ℳF ; and F � Eð Þ 2 ℳF : Now, sinceE � Fð Þ \ F � Eð Þ ¼ ;; by XX, E � Fð Þ [ F � Eð Þð Þ 2 ℳF : By XIX, E \Fð Þ 2ℳF : Now, since

E \Fð Þ \ E � Fð Þ [ F � Eð Þð Þ ¼ ;; and

E \Fð Þ [ E � Fð Þ [ F � Eð Þð Þ ¼ E [Fð Þ;

by XX, E [Fð Þ 2 ℳF : ∎)

XXII. Problem 1.208 Let E 2 ℳF : Then, for every compact set K; E \K 2 ℳF :

(Solution Let us take any compact set K: We have to show that E \K 2 ℳF :Since K is a compact set, by X, K 2 ℳF : Now, by XIX, E \K 2 ℳF : ■)

Notation We shall denote the collection

E : for every compact set K; E \Kð Þ 2 ℳFf g

by ℳ: From XXII, ℳF ℳ:

XXIII. Problem 1.209 Let V be an open set in X: Then, V 2 ℳ:

(Solution Let us take any compact set K: We have to show that

K � K \ Vcð Þð Þ ¼ K � K � Vð Þ ¼ V \K 2 ℳF|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} :


Since V is open, Vc is closed. Since Vc is closed and K is compact, by Lemma1.160, K \ Vcð Þð Þ is compact. Now, by X, K \ Vcð Þð Þ 2 ℳF : Since K is compact,by X, K 2 ℳF : Since K; K \ Vcð Þð Þ 2 ℳF ; by XVIII,

V \K ¼ K � K \ Vcð Þð Þ 2 ℳF|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence, V \K 2 ℳF : ∎)

XXIV. Problem 1.210 Let E;F 2 ℳ: Then, E [Fð Þ; E � Fð Þ 2 ℳ:

(Solution Let us take any compact set K: We have to show that

E\Kð Þ [ F \Kð Þ ¼ð Þ E [Fð Þ \K; E \Kð Þ � F \Kð Þ ¼ð Þ E � Fð Þ \K 2 ℳF :

Since E 2 ℳ; and K is compact, we have E \K 2 ℳF : Similarly, F \K 2 ℳF :Now, by XXI, E [Fð Þ \K ¼ð Þ E \Kð Þ [ F \Kð Þ 2 ℳF ; and hence, E[Fð Þ \K 2ℳF : Next, by XVIII,

E � Fð Þ \K ¼ E \Kð Þ � F \Kð Þ 2 ℳF|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence, E � Fð Þ \K 2 ℳF : ■)

XXV. Problem 1.211 Let E;F 2 ℳ: Then, E \Fð Þ; ; 2 ℳ:

(Solution Since E;F 2 ℳ; by XXIV, E [Fð Þ; E � Fð Þ; F � Eð Þ 2 ℳ: SinceE; E � Fð Þ 2 ℳ; by XXIV, E \F ¼ð ÞE � E � Fð Þ 2 ℳ; and hence, E \Fð Þ 2ℳ: Since, E 2 ℳ; by XXIV, ; ¼ð Þ E � Eð Þ 2 ℳ; and hence, ; 2 ℳ: ■)

XXVI. Problem 1.212 Let E1;E2;E3; � � � 2 ℳ: Let E1;E2;E3; . . . be pairwisedisjoint. Then, E1 [E2 [E3 [ � � �ð Þ 2 ℳ:

(Solution Let us take a compact set K: We have to show that

E1 \Kð Þ [ E2 \Kð Þ [ E3 \Kð Þ [ � � � ¼ E1 [E2 [E3 [ � � �ð Þ \K 2 ℳF|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is, E1 \Kð Þ [ E2 \Kð Þ [ E3 \Kð Þ [ � � � 2 ℳF : Since each En 2 ℳ; and K iscompact, each En \Kð Þ 2 ℳF : Since K is compact, by IX, l Kð Þ\1: SinceE1 [E2 [E3 [ � � �ð Þ \K K; we have

l E1 [E2 [E3 [ � � �ð Þ \Kð Þ� l Kð Þ \1ð Þ;

and hence


l E1 \Kð Þ [ E2 \Kð Þ [ E3 \Kð Þ [ � � �ð Þ ¼ l E1 [E2 [E3 [ � � �ð Þ \Kð Þ\1:

It follows that

l E1 \Kð Þ [ E2 \Kð Þ [ E3 \Kð Þ [ � � �ð Þ\1:

Since E1;E2;E3; . . . are pairwise disjoint sets, E1 \Kð Þ; E2 \Kð Þ;E3 \Kð Þ; . . . are pairwise disjoint sets: Since E1;E2;E3; � � � 2 ℳ; and K is com-pact, we have E1 \Kð Þ; E2 \Kð Þ; E3 \Kð Þ; � � � 2 ℳF : Now, by XVI,

E1 \Kð Þ [ E2 \Kð Þ [ E3 \Kð Þ [ � � � 2 ℳF :

∎)

XXVII. Problem 1.213 Let E1;E2;E3; � � � 2 ℳ: Then, E1 [E2 [E3 [ � � �ð Þ 2 ℳ:

(Solution We can write:

E1 [E2 [E3 [ � � �ð Þ¼ E1 [ E2 � E1ð Þ [ E3 � E1 [E2ð Þð Þ [ E4 � E1 [E2 [E3ð Þð Þ [ � � � :

Observe that E1; E2 � E1ð Þ; E3 � E1 [E2ð Þð Þ; E4 � E1 [E2 [E3ð Þð Þ; . . . arepairwise disjoint sets. By XXIV,E1; E2 � E1ð Þ; E3 � E1 [E2ð Þð Þ; E4 � E1 [E2 [E3ð Þð Þ; � � � 2 ℳ: Now, by XXVI,

E1 [E2 [E3 [ � � �ð Þ¼ E1 [ E2 � E1ð Þ [ E3 � E1 [E2ð Þð Þ [ E4 � E1 [E2 [E3ð Þð Þ [ � � �ð Þ 2 ℳ;

and hence E1 [E2 [E3 [ � � �ð Þ 2 ℳ: ∎)

XXVIII. Problem 1.214 ℳ is a r-algebra containing all Borel sets.

(Solution In view of XXIV, XXV and XXVII, ℳ is a r-algebra. By XXIII, ℳcontains all open sets. Since ℳ is a r-algebra containing all open sets, the‘smallest’ r-algebra containing all open sets is a subset of ℳ: Now, since themembers of the smallest r-algebra containing all open sets are called Borel sets, allBorel sets are members of ℳ: ∎)

XXIX. Problem 1.215 Let E 2 ℳ: Let l Eð Þ\1: Then, E 2 ℳF :

(Solution It suffices to show that l Eð Þ� sup l Kð Þ : K E; and K is a compact setf g:Let us take a real e[ 0: Since

1[ð Þl Eð Þ ¼ inf l Vð Þ : E V ; and V is openf g;

there exists an open set V0 such that E V0; and l V0ð Þ\l Eð Þþ e: It follows thatl V0ð Þ\1: Since V0 is an open set satisfying l V0ð Þ\1; by XII, V0 2 ℳF : Since


V0 2 ℳF ; and e is a positive real, by XVII, there exists a compact set K1; and anopen set V1 such that K1 V0 V1; l V1 � K1ð Þ\e; and V1;K1; V1 � K1ð Þ 2 ℳF :Since E 2 ℳ; and K1 is compact, E \K1 2 ℳF ; and hence

l E \K1ð Þ ¼ sup l Kð Þ : K E \K1ð Þ; and K is a compact setf g:

It follows that there exists a compact set K2 such that K2 E \K1ð Þ Eð Þ; andl E \K1ð Þ � e\l K2ð Þ: Since K2 E; and K2 is a compact set;

l K2ð Þ 2 l Kð Þ : K E; and K is a compact setf g;

and hence

l K2ð Þ� sup l Kð Þ : K E; and K is a compact setf g:

Since E V0 V1; we have E E \K1ð Þ [ V1 � K1ð Þ; and hence

l Eð Þ� l E \K1ð Þþ l V1 � K1ð Þ\l E \K1ð Þþ e\ l K2ð Þþ eð Þþ e

¼ l K2ð Þþ 2e� sup l Kð Þ : K E; and K is a compact setf gþ 2e:

Thus, l Eð Þ\sup l Kð Þ : K E; and K is a compact setf gþ 2e: On letting e !0; we get

l Eð Þ� sup l Kð Þ : K E; and K is a compact setf g:

∎)

XXX. Problem 1.216 ℳF ¼ E : E 2 ℳ; and l Eð Þ\1f g:(Solution Let E 2 ℳF : Now, by the definition of ℳF ; l Eð Þ\1: Since E 2 ℳF ;and ℳF ℳ; we have E 2 ℳ: Thus, LHS RHS: Next, let E 2 RHS: Hence,E 2 ℳ; and l Eð Þ\1: Now, by XXIX, E 2 ℳF : Thus, RHS LHS: ∎)

XXXI. Problem 1.217 l is a positive measure on ℳ:

(Solution By XXVIII, ℳ is a r-algebra, so ; 2 ℳ: By IV, l ;ð Þ ¼ 0 \1ð Þ:Countably additive: Let E1;E2;E3; . . .f g be a countable collection of members

in ℳ such that i 6¼ j ) Ei \Ej ¼ ;: We have to show that

l E1 [E2 [E3 [ � � �ð Þ ¼ l E1ð Þþ l E2ð Þþ l E3ð Þþ � � � :

Case I: when there exists a positive integer n such that l Enð Þ ¼ 1: In this case,l E1ð Þþ l E2ð Þþ l E3ð Þþ � � � ¼ 1: Since En E1 [E2 [E3 [ � � � ; we have

1 ¼ð Þl Enð Þ� l E1 [E2 [E3 [ � � �ð Þ;

and hence, l E1 [E2 [E3 [ � � �ð Þ ¼ 1: Thus,



Case II: when each l Enð Þ\1: Since l E1ð Þ\1; and E1 2 ℳ; by XXX, E1 2ℳF : Similarly, E2;E3; � � � 2 ℳF : Since each Ei 2 ℳF ; and E1;E2;E3; . . . arepairwise disjoint, by XV,


Thus, l is a positive measure on ℳ: ∎)

XXXII. Problem 1.218 Let E 2 ℳ satisfying l Eð Þ ¼ 0: Let A E: Then,A 2 ℳ:

(Solution Let us fix any compact set K: We must prove A\K 2 ℳF ; that is,l A\Kð Þ\1; and

l A\Kð Þ ¼ sup l K1ð Þ : K1 A\Kð Þ and K1 is a compact setf g:

Here, it suffices to show

a. l A\Kð Þ ¼ 0;b. 0 ¼ sup l K1ð Þ : K1 A\Kð Þ and K1is a compact setf g:

For a: Since A\K A E; we have A\K E; and hence,0� l A\Kð Þ� l Eð Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0: It follows that l A\Kð Þ ¼ 0:

For b: Let us take any compact set K1 such that K1 A\Kð Þ: It follows that0� l K1ð Þ� l A\Kð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0; and hence l K1ð Þ ¼ 0: Thus,

l K1ð Þ : K1 A\Kð Þ and K1 is a compact setf g ¼ 0f g;

and hence

sup l K1ð Þ : K1 A\Kð Þ and K1 is a compact setf g ¼ sup 0f g ¼ 0ð Þ:

Thus

sup l K1ð Þ : K1 A\Kð Þ and K1 is a compact setf g ¼ 0:

∎)

Conclusion 1.219 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Then there exists a r-algebra ℳ in X that contains allBorel sets in X; (see XXVIII) and there exists a positive measure l on ℳ (seeXXXI) satisfying the following conditions:


1. for every compact subset K of X; K 2 ℳ; and l Kð Þ\1 (see IX, X, and XXII),2. for every E 2 ℳ; l Eð Þ ¼ inf l Vð Þ : E V ; and V is openf g (from the defini-

tion of l Eð ÞÞ,3. for every open set V in X; V 2 ℳ; and l Vð Þ ¼ sup l Kð Þ : K V ;f

and K is a compact setg (see XII and XXIII),4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E; andf

K is a compact setg (see XXIX),5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ (see XXXII),6. for every compact subset K of X; l Kð Þ ¼ inf K fð Þ : K � ff g (see XI),7. for every open set V in X; l Vð Þ ¼ sup K fð Þ : f � Vf g (see definition),8. if f1 : X ! R; f2 : X ! R are members of Cc Xð Þ satisfying f1 � f2; then

K f1ð Þ;K f2ð Þ are real numbers, and

K f1ð Þ�K f2ð Þ see VIIIð Þ:

1.8 Riesz Representation Theorem

Using the results of previous sections, we are now able to prove the Riesz repre-sentation theorem in its generality enough for our purposes.

Note 1.220 Let X be a locally compact Hausdorff space. Let K be a positive linearfunctional on Cc Xð Þ: By the Conclusion 1.219, there exists a r-algebraℳ in X thatcontains all Borel sets in X; and there exists a positive measure l on ℳ satisfyingthe following conditions:

1. for every compact subset K of X; K 2 ℳ; and l Kð Þ\1;2. for every E 2 ℳ; l Eð Þ ¼ inf l Vð Þ : E V ; andV is openf g;3. for every open set V in X; V 2 ℳ; and l Vð Þ ¼ sup l Kð Þ : K V ; andf

K is a compact setg;4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E; andK isf

a compact setg;5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;6. for every compact subset K of X; l Kð Þ ¼ inf K fð Þ : K � ff g;7. for every open set V in X; l Vð Þ ¼ sup K fð Þ : f � Vf g;8. if f1 : X ! R; f2 : X ! R are members of Cc Xð Þ satisfying f1 � f2; then

K f1ð Þ;K f2ð Þ are real numbers, and K f1ð Þ�K f2ð Þ:Let f 2 Cc Xð Þ: Let f : X ! R:Since f 2 Cc Xð Þ; f is continuous. Since ℳ is a r-algebra containing all Borel

sets, and f is continuous, f is ℳ-measurable, and hence,RX f dl is meaningful.

We want to show:


K fð Þ�ZX

f dl

Since f 2 Cc Xð Þ; by Lemma 1.172, f Xð Þ is a compact subset of real numbers.Now, by the Heine-Borel theorem, f Xð Þ is bounded, so there exist real numbers a; bsuch that a\b; and f Xð Þ a; b½ �:

Let us take any e 2 0; b�a2

� �:

We can find real numbers y0; y1; . . .; yn such that I. y0\a\y1\ � � �\yn ¼ b; II.y1 � y0\e; y2 � y1\e; . . .; yn � yn�1\e:

Since f is continuous, each f�1 y0; y1 þ 1n

� �� is an open set, and hence, each

f�1 y0; y1 þ 1n

� �� 2 ℳ: Now,

f�1 y0; y1ð �Þð Þ ¼ f�1 \1n¼1 y0; y1 þ 1

n

� �� ¼ \1

n¼1 f�1 y0; y1 þ 1n

� �� 2 ℳ:

Thus, f�1 y0; y1ð �Þð Þ 2 ℳ: Since supp fð Þ is compact, by 1, supp fð Þ 2 ℳ: Sincef�1 y0; y1ð �Þð Þ; supp fð Þ 2 ℳ; we have f�1 y0; y1ð �Þð Þ \ supp fð Þ 2 ℳ: Similarly,

f�1 y1; y2ð �Þð Þ \ supp fð Þ 2 ℳ; f�1 y2; y3ð �Þð Þ \ supp fð Þ 2 ℳ; etc:

Since

f�1 y0; y1ð �Þð Þ; f�1 y1; y2ð �Þð Þ; f�1 y2; y3ð �Þð Þ; . . .

are pairwise disjoint,

f�1 y0; y1ð �Þð Þ \ supp fð Þ; f�1 y1; y2ð �Þð Þ \ supp fð Þ; f�1 y2; y3ð �Þð Þ \ supp fð Þ; . . .

are pairwise disjoint sets. Since,

supp fð Þ X f�1 a; b½ �ð Þ f�1 y0; y1ð � [ y1; y2ð � [ � � � [ yn�1; ynð �ð Þ¼ f�1 y0; y1ð �Þð Þ [ f�1 y1; y2ð �Þð Þ [ f�1 y2; y3ð �Þð Þ [ � � � ;

the union of

f�1 y0; y1ð �Þð Þ \ supp fð Þ; f�1 y1; y2ð �Þð Þ \ supp fð Þ; f�1 y2; y3ð �Þð Þ \ supp fð Þ; . . .

is supp fð Þ: In short, supp fð Þ is partitioned into

f�1 y0; y1ð �Þð Þ \ supp fð Þ; f�1 y1; y2ð �Þð Þ \ supp fð Þ; f�1 y2; y3ð �Þð Þ \ supp fð Þ; . . .:


Here, each f�1 yi�1; yið �Þð Þ \ supp fð Þ 2 ℳ: By 2,

l f�1 y0; y1ð �Þð Þ \ supp fð Þ� � ¼ inf l Vð Þ : f�1 y0; y1ð �Þð Þ \ supp fð Þ V ; andV is open�

;

so there exists an open set V1 such that

f�1 y0; y1ð �Þð Þ \ supp fð Þ V1; and l V1� �

\l f�1 y0; y1ð �Þð Þ \ supp fð Þ� �þ en:

Similarly, there exists an open set V2 such that

f�1 y1; y2ð �Þð Þ \ supp fð Þ V2; and l V2� �

\l f�1 y1; y2ð �Þð Þ \ supp fð Þ� �þ en; etc:

Since, V1 is open, and f�1 �1; y1 þ eð Þð Þ is open, V1 \ f�1 �1; y1 þ eð Þð Þ is anopen set. Similarly, V2 \ f�1 �1; y2 þ eð Þð Þ is an open set, etc. In short,

V1 \ f�1 �1; y1 þ eð Þð Þ; V2 \ f�1 �1; y2 þ eð Þð Þ; . . .; Vn \ f�1 1; yn þ eð Þð Þ

are open sets. Here,

l Vi \ f�1 �1; yi þ eð Þð Þ� �� l Vi� �

\l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �þ en

� ;

so

l Vi \ f�1 �1; yi þ eð Þð Þ� �\l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �þ e

n:

Observe that

V1 \ f�1 �1; y1 þ eð Þð Þ� �[ V2 \ f �1 �1; y2 þ eð Þð Þ� �[ � � � f�1 y0; y1ð �Þð Þ \ supp fð Þ� �\ f�1 �1; y1 þ eð Þð Þ� �[ f�1 y1; y2ð �Þð Þ \ supp fð Þ� �\ f�1 �1; y2 þ eð Þð Þ� �[ � � �¼ f�1 y0; y1ð �Þð Þ \ supp fð Þ� �� [ f�1 y1; y2ð �Þð Þ \ supp fð Þ� �� [ � � �¼ f�1 y0; y1ð �Þð Þ [ f�1 y1; y2ð �Þð Þ [ � � �� \ supp fð Þ¼ f�1 y0; y1ð � [ y1; y2ð � [ � � �Þð Þ� �\ supp fð Þ¼ f�1 y0; ynð �Þð Þ \ supp fð Þ ¼ X \ supp fð Þ ¼ supp fð Þ

so,

supp fð Þ V1 \ f�1 �1; y1 þ eð Þð Þ� �[ V2 \ f�1 �1; y2 þ eð Þð Þ� �[ � � � :

Now, since supp fð Þ is compact, and each Vi \ f�1 �1; yi þ eð Þð Þ is open, byConclusion 1.178, there exist h1; . . .; hn 2 Cc Xð Þ such that

1.8 Riesz Representation Theorem 169

a. h1 � V1 \ f�1 �1; y1 þ eð Þð Þ; . . .; hn � Vn \ f�1 �1; yn þ eð Þð Þ; b.supp fð Þ � h1 þ � � � þ hnð Þ:

From b,

f ¼ f � h1 þ � � � þ hnð Þ ¼ f � h1 þ � � � þ f � hnð Þ:

Since

supp fð Þ � h1 þ � � � þ hnð Þ;

and supp fð Þ is compact, by 6,

l supp fð Þð Þ�K h1 þ � � � þ hnð Þ ¼ K h1ð Þþ � � � þK hnð Þð Þ;

and hence,

l supp fð Þð Þ�K h1ð Þþ � � � þK hnð Þ:Problem 1.221 For every i ¼ 1; . . .; n;

f � hi � yi þ eð Þ � hi:(Solution Let us fix any x 2 X; and let us fix any i 2 1; . . .; nf g: We have to showthat

f xð Þ � hi xð Þ� yi þ eð Þ � hi xð Þ:

Case I: when x 62 Vi \ f�1 �1; yi þ eð Þð Þ: Since hi � Vi \ f�1 �1; yi þ eð Þð Þ; wehave

supp hið Þ Vi \ f�1 �1; yi þ eð Þð Þ:

Since x 62 Vi \ f �1 �1; yi þ eð Þð Þ supp hið Þð Þ; we have x 62 supp hið Þ; andhence, hi xð Þ ¼ 0: It follows that yi þ eð Þ � hi xð Þ ¼ 0 ¼ f xð Þ � hi xð Þ; and hence,f xð Þ � hi xð Þ� yi þ eð Þ � hi xð Þ:

Case II: when x 2 Vi \ f�1 �1; yi þ eð Þð Þ: It follows that x 2 f�1 �1; yi þ eð Þð Þ;and hence f xð Þ\ yi þ eð Þ: Now, since 0� hi xð Þ� 1; we havef xð Þ � hi xð Þ� yi þ eð Þ � hi xð Þ:

Thus, in all cases, f xð Þ � hi xð Þ� yi þ eð Þ � hi xð Þ: ■)Since, for every i ¼ 1; . . .; n; f � hi � yi þ eð Þ � hi; we have

f ¼ f � h1 þ � � � þ hnð Þ ¼Xni¼1

f � hið Þ�Xni¼1

yi þ eð Þ � hið Þ¼ y1 þ eð Þ � h1 þ � � � þ yn þ eð Þ � hn;

and hence,


f � y1 þ eð Þ � h1 þ � � � þ yn þ eð Þ � hn:

Problem 1.222 For every i ¼ 1; . . .; n;

0\ aj j þ yi:

(Solution Case I: when a� 0: In this case aj j ¼ a: For every i ¼1; . . .; n; 0�ð Þa\yi; so, for every i ¼ 1; . . .; n; 0\aþ yi ¼ aj j þ yið Þ:

Case II: when a\0: In this case, aj j ¼ �a: For every i ¼ 1; . . .; n; a\yi; so forevery i ¼ 1; . . .; n; 0\ �að Þþ yi ¼ aj j þ yið Þ: ■)

Problem 1.223 For every i ¼ 1; . . .; n; and for every x 2 f�1 yi�1; yið �Þð Þ \ supp fð Þ;yi � eð Þ� f xð Þ:(Solution Let us fix any i 2 1; . . .; nf g; and fix any x 2 f�1 yi�1; yið �Þð Þ \ supp fð Þ:It follows that yi � e\ð Þyi�1\f xð Þ� yi; and hence, yi � eð Þ� f xð Þ: ■)

Since f � y1 þ eð Þ � h1 þ � � � þ yn þ eð Þ � hn; f : X ! R is a member of Cc Xð Þ;and y1 þ eð Þ � h1 þ � � � þ yn þ eð Þ � hnð Þ : X ! R is a member of Cc Xð Þ; by 8,

K fð Þ�K y1 þ eð Þ � h1 þ � � � þ yn þ eð Þ � hnð Þ

¼Xni¼1

yi þ eð Þ K hið Þð Þ

¼Xni¼1

aj j þ yi þ eð Þ K hið Þð Þ � aj jXni¼1

K hið Þð Þ

¼Xni¼1

aj j þ yi þ eð Þ K hið Þð Þ � aj j K h1 þ � � � þ hnð Þð Þ;

and hence,

K fð Þ�Xni¼1

aj j þ yi þ eð Þ K hið Þð Þ � aj j K h1 þ � � � þ hnð Þð Þ:

Since, supp fð Þ � h1 þ � � � þ hnð Þ; we have

K h1 þ � � � þ hnð Þ 2 K fð Þ : supp fð Þð Þ � ff g:

Since supp fð Þ is compact, by 6,

l supp fð Þð Þ ¼ inf K fð Þ : supp fð Þð Þ � ff g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} �K h1 þ � � � þ hnð Þ:


Thus,

� K h1 þ � � � þ hnð Þð Þð Þ� �l supp fð Þð Þð Þ:

On using this inequality, we get

K fð Þ�Xni¼1

aj j þ yi þ eð Þ K hið Þð Þ � aj j l supp fð Þð Þð Þ:

Since, for every i ¼ 1; . . .; n; hi � Vi \ f�1 �1; yi þ eð Þð Þ� �we have, for every

i ¼ 1; . . .; n;

K hið Þ 2 K fð Þ : f � Vi \ f�1 �1; yi þ eð Þð Þ� �� :

Now, since Vi \ f�1 �1; yi þ eð Þð Þ is open, by 7, for every i ¼ 1; . . .; n;

K hið Þ� l Vi \ f�1 �1; yi þ eð Þð Þ� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �þ en:

We have seen that, for every i ¼ 1; . . .; n; 0\ aj j þ yi \ aj j þ yi þ eð Þ: On usingthese inequalities, we get

K fð Þ�Xni¼1

aj j þ eð Þþ yið Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �þ en

� � aj j l supp fð Þð Þð Þ

¼ aj j þ eð ÞXni¼1

l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ Xni¼1

yi l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e

n

Xni¼1

aj j þ eð Þþ yið Þ � aj j l supp fð Þð Þð Þ:

Since supp fð Þ is partitioned into f�1 y0; y1ð �Þð Þ \ supp fð Þ;f�1 y1; y2ð �Þð Þ \ supp fð Þ; f�1 y2; y3ð �Þð Þ \ supp fð Þ; . . .; each f�1 yi�1; yið �Þð Þ \supp fð Þ 2 ℳ; and l is a positive measure, we get

l supp fð Þð Þ ¼Xni¼1

l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� :

On using this equality, we get


K fð Þ� aj j þ eð Þ l supp fð Þð Þð ÞþXni¼1

yi l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e

n

Xni¼1

aj j þ eð Þþ yið Þ � aj j l supp fð Þð Þð Þ

¼ e l supp fð Þð Þð ÞþXni¼1

yi l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ en

Xni¼1

aj j þ eð Þþ yið Þ



n aj j þ eð ÞþXni¼1

yi

!

� e l supp fð Þð Þð ÞþXni¼1


n aj j þ eð Þþ nbð Þ


yi l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e aj j þ eþ bð Þ

¼Xni¼1

yi l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e l supp fð Þð Þþ aj j þ eþ bð Þ

¼Xni¼1

yi � eð Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ eXni¼1

l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �þ e l supp fð Þð Þþ aj j þ eþ bð Þ

¼Xni¼1

yi � eð Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e l supp fð Þð Þð Þþ e l supp fð Þð Þþ aj j þ eþ bð Þ

¼Xni¼1

yi � eð Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e 2l supp fð Þð Þþ aj j þ eþ bð Þ:

Thus,

K fð Þ�Xni¼1

yi � eð Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� þ e 2l supp fð Þð Þþ aj j þ eþ bð Þ:

Since for every i ¼ 1; . . .; n; and, for every x 2 f �1 yi�1; yið �Þð Þ \ supp fð Þ;yi � eð Þ� f xð Þ;

f�1 y0; y1ð �Þð Þ \ supp fð Þ; f�1 y1; y2ð �Þð Þ \ supp fð Þ; f�1 y2; y3ð �Þð Þ \ supp fð Þ; . . .constitute a partition of supp fð Þ; and f ¼ 0 on supp fð Þð Þc; we have

Xni¼1

yi � eð Þ l f�1 yi�1; yið �Þð Þ \ supp fð Þ� �� ZX

f dl:

On using this inequality, we get


K fð Þ�ZX

f dlþ e 2l supp fð Þð Þþ aj j þ eþ bð Þ:

On letting e ! 0; we get K fð Þ� RX f dl.

Conclusion 1.224 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Then there exists a r-algebra ℳ in X which containsall Borel sets in X; and there exists a positive measure l on ℳ satisfying thefollowing conditions:

1. for every compact subset K of X; K 2 ℳ; and l Kð Þ\1;2. for every E 2 ℳ; l Eð Þ ¼ inf l Vð Þ : E V ; andV is openf g;3. for every open set V in X; V 2 ℳ; and l Vð Þ ¼ sup l Kð Þ : K V ;f

and K is a compact setg;4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E; andK isf

a compact setg;5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;6. for every f 2 Cc Xð Þ satisfying f : X ! R; K fð Þ� R

X f dl:

Theorem 1.225 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Then there exists a r-algebra ℳ in X that contains allBorel sets in X; and there exists a unique positive measure l on ℳ satisfying thefollowing conditions:

1. for every compact subset K of X; K 2 ℳ; and l Kð Þ\1;2. for every E 2 ℳ; l Eð Þ ¼ inf l Vð Þ : E V ; andV is openf g;3. for every open set V in X; V 2 ℳ; and l Vð Þ ¼ sup l Kð Þ : K V ; andf

K is a compact setg;4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E; andf

K is a compact setg;5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;6. for every f 2 Cc Xð Þ; K fð Þ ¼ RX f dl:Proof On applying Conclusion 1.224, we see that only (6) remains to be proved.

For 6: Let f 2 Cc Xð Þ:Case I: when f : X ! R: In this case, by Conclusion 1.224, we have

K fð Þ� RX f dl: Since f : X ! R; �fð Þ : X ! R: Since f 2 Cc Xð Þ; �fð Þ 2 Cc Xð Þ:

Now, by Conclusion 1.224, we have

� K fð Þð Þ ¼ K �fð Þ�ZX

�fð Þdl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ �ZX

f dl;


and hence,RX f dl�K fð Þ: Since

RX f dl�K fð Þ; and K fð Þ� R

X f dl; we haveK fð Þ ¼ RX f dl:

Case II: when f : X ! C: Here Re fð Þ : X ! R; and Im fð Þ : X ! R: Since f 2Cc Xð Þ; f is continuous, and hence Re fð Þ : X ! R is continuous. Since

f�1 0ð Þ ¼ Re fð Þð Þ�1 0ð Þ�

\ Im fð Þð Þ�1 0ð Þ�

;

we have

f�1 C� 0f gð Þ ¼ f�1 0ð Þ� �c¼ Re fð Þð Þ�1 0ð Þ�

\ Im fð Þð Þ�1 0ð Þ� � c

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ Re fð Þð Þ�1 0ð Þ� c

[ Im fð Þð Þ�1 0ð Þ� c

¼ Re fð Þð Þ�1R� 0f gð Þ

� [ Im fð Þð Þ�1

R� 0f gð Þ�

;

and hence

supp fð Þ ¼ f�1 C� 0f gð Þ� ��¼ Re fð Þð Þ�1R� 0f gð Þ

� [ Im fð Þð Þ�1

R� 0f gð Þ� � �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ Re fð Þð Þ�1

R� 0f gð Þ� �

[ Im fð Þð Þ�1R� 0f gð Þ

� �¼ supp Re fð Þð Þð Þ [ supp Im fð Þð Þð Þ supp Re fð Þð Þ:

Thus, supp Re fð Þð Þ supp fð Þ: Since f 2 Cc Xð Þ; supp fð Þ is compact. Sincesupp Re fð Þð Þ supp fð Þ; supp Re fð Þð Þ is closed and supp fð Þ is compact,supp Re fð Þð Þ is compact. Since supp Re fð Þð Þ is compact, and Re fð Þ : X ! R iscontinuous, Re fð Þ 2 Cc Xð Þ: Similarly, Im fð Þ 2 Cc Xð Þ; and Im fð Þ : X ! R: SinceRe fð Þ 2 Cc Xð Þ; and Re fð Þ : X ! R; by case I, K Re fð Þð Þ ¼ RX Re fð Þdl:Similarly, K Im fð Þð Þ ¼ RX Im fð Þdl: Now,

LHS ¼ K fð Þ ¼ K Re fð Þð Þþ i Im fð Þð Þð Þ ¼ K Re fð Þð Þþ i K Im fð Þð Þð Þ¼ZX

Re fð Þdlþ iZX

Im fð Þdl ¼ZX

f dl ¼ RHS:

Uniqueness: Let l1 : ℳ ! 0;1½ � and l2 : ℳ ! 0;1½ � be two positive mea-sures on ℳ satisfying the following conditions:

1. for every compact subset K of X; K 2 ℳ; l1 Kð Þ\1; and l2 Kð Þ\1;2. for every E 2 ℳ; l1 Eð Þ ¼ inf l1 Vð Þ : E V ; and V is openf g; and

l2 Eð Þ ¼ inf l2 Vð Þ : E V ; and V is openf g;3. for every open set V in X; V 2 ℳ; l1 Vð Þ ¼ sup l1 Kð Þ : K V ; andf

K is a compact setg; and l2 Vð Þ ¼ sup l2 Kð Þ : K V ; and K is a compact setfg;


4.

(i) for every E 2 ℳ satisfying l1 Eð Þ\1; l1 Eð Þ ¼ sup l1 Kð Þ : K E;fand K is a compact setg;

(ii) for every E 2 ℳ satisfying l2 Eð Þ\1; l2 Eð Þ ¼ sup l2 Kð Þ : K E;fand K is a compact setg;

5.

(i) if E 2 ℳ; l1 Eð Þ ¼ 0; and A E; then A 2 ℳ;(ii) if E 2 ℳ; l2 Eð Þ ¼ 0; and A E; then A 2 ℳ;

6. for every f 2 Cc Xð Þ; K fð Þ ¼ RX f dl1; and K fð Þ ¼ RX f dl2:We have to show that l1 ¼ l2: For this purpose, let us take any E 2 ℳ: We

have to show that l1 Eð Þ ¼ l2 Eð Þ:Case I: when E is a compact set. Let us take a real e[ 0: Since E is a compact

set, by 1, l1 Eð Þ\1; and l2 Eð Þ\1: Also, by 2, there exists an open set V suchthat E V ; and l2 Vð Þ\l2 Eð Þþ e: Since E V ; E is compact, and V is open, byUrysohn’s lemma, there exists f 2 Cc Xð Þ such that E � f � V : Since E � f ; wehave vE � f ; and hence

l1 Eð Þ ¼ZX

vEdl1 �ZX

f dl1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:

Thus, l1 Eð Þ� RX f dl1: Since f 2 Cc Xð Þ; by 6, l1 Eð Þ�

ZXf dl1 ¼

ZXf dl2|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence, l1 Eð Þ� RX f dl2: Since f � V ; we have supp fð Þ V ; and 0� f � 1;

and hence f � vV : It follows that

l1 Eð Þ�ZX

f dl2 �ZX

vV

|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}dl2 ¼ l2 Vð Þ\l2 Eð Þþ e;

and hence l1 Eð Þ\l2 Eð Þþ e: Now, on letting e ! 0; we get l1 Eð Þ� l2 Eð Þ:Similarly, l2 Eð Þ � l1 Eð Þ: It follows that l1 Eð Þ ¼ l2 Eð Þ:

Case II: when E is an open set. By 3, l1 Eð Þ ¼ sup l1 Kð Þ : K E;fand K is a compact setg; and l2 Eð Þ ¼ sup l2 Kð Þ : K E; and K is afcompact setg: By Case I,

l1 Kð Þ : K E; and K is a compact setf g¼ l2 Kð Þ : K E; and K is a compact setf g;

and hence


l1 Eð Þ ¼ð Þ sup l1 Kð Þ : K E; and K is a compact setf g¼ sup l2 Kð Þ : K E; and K is a compact setf g ¼ l2 Eð Þð Þ:

Thus, l1 Eð Þ ¼ l2 Eð Þ:Case III: when E 2 ℳ: By 2,

l1 Eð Þ ¼ inf l1 Vð Þ : E V ; and V is openf g; andl2 Eð Þ ¼ inf l2 Vð Þ : E V ; and V is openf g:

By Case II,

l1 Vð Þ : E V ; andV is openf g ¼ l2 Vð Þ : E V ; andV is openf g;

and hence

l1 Eð Þ ¼ð Þ inf l1 Vð Þ : E V ; and V is openf g¼ inf l2 Vð Þ : E V ; and V is openf g ¼ l2 Eð Þð Þ:

Thus,

l1 Eð Þ ¼ l2 Eð Þ:

∎The Theorem 1.225, known as the Riesz representation theorem, is due to F.

Riesz (20.01.1880–28.02.1956, Hungarian). He made fundamental contributions tofunctional analysis. His work also has many applications in physics.

1.9 Borel Measure

It is true that Borel measure is not as general as Lebesgue measure, but for mostpurposes Borel measure is good enough to yield results. The situation is similar tothat of rational numbers and real numbers.

Note 1.226

Definition Let X be a locally compact Hausdorff space. Let B be the r-algebra ofall Borel sets in X: Let l : B ! 0;1½ � be a mapping. If l is a measure, then we saythat l is a Borel measure on X:

In Theorem 1.225, B ℳ; and the restriction of l to B is a Borel measure on X:

Definition Let X be a locally compact Hausdorff space. Let ℳ be a r-algebra in X;which contains all Borel sets in X: Let l : ℳ ! 0;1½ � be a measure on X: Let E bea Borel set in X:


1. If

l Eð Þ ¼ inf l Vð Þ : E V ; andV is openf g;

then we say that E is outer regular.

2. If

l Eð Þ ¼ sup l Kð Þ : K V ; and K is a compact setf g;

then we say that E is inner regular.In Theorem 1.225, every Borel set is outer regular, and every open set is inner

regular.

Problem 1.227 Every compact subset of X is inner regular.

(Solution Let K be a compact subset of X:We have to show that K is inner regular.Since K is a compact subset of Hausdorff space X; K is a closed set, and hence

Kc is open in X: It follows that Kc is a Borel set in X: Since the collection of allBorel sets in X is a r-algebra, and Kc is a Borel set in X; K ¼ð Þ Kcð Þc is a Borel setin X; and hence, K is a Borel set in X: It remains to show that


Since, K is a compact subset of X; by Problem 1.194, K is a member ofℳF ; andhence,


■)

Definition Let X be a topological space. Let E X: If there exists a countablecollection K1;K2; . . .f g of compact subsets of X such that E ¼ K1 [K2 [ � � � ; thenwe say that E is r-compact.

It is known that every open set in Rk is r-compact.

Problem 1.228 In Theorem 1.225, every r-compact set is a Borel set.

(Solution Let E be a r-compact set. We have to show that E is a Borel set in X:From definition, there exists a countable collection K1;K2; . . .f g of compact subsetsof X such that E ¼ K1 [K2 [ � � � : Since each Ki is compact, eachKi is a Borel setin X: Now, since the collection of all Borel sets in X is a r-algebra in X;E ¼ð Þ K1 [K2 [ � � �ð Þ is a Borel set in X; and hence E is a Borel set in X: ■)

Definition Let X be a nonempty set, ℳ be a r-algebra in X; and l : ℳ ! 0;1½ �be a positive measure on X: Let E X: If there exists a countable collectionE1;E2; . . .f g of members in ℳ such that E ¼ E1 [E2 [ � � � ; and each l Eið Þ\1;

then we say that E has r-finite measure.


Problem 1.229 If E has r-finite measure, then there exists a countable collectionE1;E2; . . .f g of members in ℳ such that E ¼ E1 [E2 [ � � � ; each l Eið Þ\1; and

E1;E2; . . . are pairwise disjoint.

(Solution Suppose that E has r-finite measure. Then there exists a countablecollection F1;F2; . . .f g of members in ℳ such that E ¼ F1 [F2 [ � � � ; and eachl Fið Þ\1: Put

E1 � F1; E2 � F2 � F1; E3 � F3 � F1 [F2ð Þ; E4 � F4 � F1 [F2 [F3ð Þ; etc:

Now, since each Fi 2 ℳ; and ℳ is a r-algebra in X; each Ei 2 ℳ: Also,E1;E2; . . . are pairwise disjoint sets, and E ¼ E1 [E2 [ � � � : It suffices to show thateach l Eið Þ\1: Since, for every i ¼ 1; 2; . . .; Ei Fi; by Lemma 1.99,l Eið Þ� l Fið Þ \1ð Þ; and hence each l Eið Þ\1: ■)

Problem 1.230 In Theorem 1.225, let E 2 ℳ: Suppose that E has r-finite mea-sure. Then

l Eð Þ ¼ sup l Kð Þ : K E; and K is a compact setf g:(Solution Let us take any e[ 0: Since E has r-finite measure, there exists acountable collection E1;E2; . . .f g of members in ℳ such that E ¼ E1 [E2 [ � � � ;each l Eið Þ\1; and E1;E2; . . . are pairwise disjoint. Since E1 2 ℳ; andl E1ð Þ\1; we have

l E1ð Þ ¼ sup l Kð Þ : K E1; andK is a compact setf g;

and hence, there exists a compact set K1 such that K1 E1; and l E1ð Þ �e21 \l K1ð Þ: Similarly, there exists a compact set K2 such that K2 E2; and l E2ð Þ �e22 \l K2ð Þ; etc. It suffices to show that

sup l E1ð Þþ � � � þ l Enð Þ : n ¼ 1; 2; � � �f g ¼ l E1ð Þþ l E2ð Þþ � � �¼ l E1 [E2 [ � � �ð Þ� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is,

sup l E1ð Þþ � � � þ l Enð Þ : n ¼ 1; 2; � � �f g� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g:

1.9 Borel Measure 179

For this purpose, let us fix any positive integer n: It suffices to show that

l E1ð Þþ � � � þ l Enð Þ� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g:

Here,

l E1ð Þþ � � � þ l Enð Þ\ l K1ð Þþ e21

� þ � � � þ l Knð Þþ e

2n

� \ l K1ð Þþ � � � þ l Knð Þð Þþ e ¼ l K1 [ � � � [Knð Þþ e;

so,

l E1ð Þþ � � � þ l Enð Þ\l K1 [ � � � [Knð Þþ e:

Since K1; . . .;Kn are compact sets, K1 [ � � � [Kn E1 [ � � � [En E1 [E2 [ � � �ð Þis compact, and hence

l E1ð Þþ � � � þ l Enð Þ � eð Þ\ l K1 [ � � � [Knð Þ 2 l Kð Þ : K E1 [E2 [ � � �ð Þ; andK is a compact setf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

It follows that

l E1ð Þþ � � � þ l Enð Þð Þ� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf gþ e:

On letting e ! 0; we get

l E1ð Þþ � � � þ l Enð Þ� sup l Kð Þ : K E1 [E2 [ � � �ð Þ; and K is a compact setf g:�Þ

Definition Let X be a locally compact Hausdorff space. Let ℳ be a r-algebra in X;which contains all Borel sets in X: Let l : ℳ ! 0;1½ � be a measure on X: If everyBorel set in X is both outer regular and inner regular, then we say that l is regular.

Definition Let X be a topological space. Let E X:

1. If there exist closed sets F1;F2; . . . such that E ¼ F1 [F2 [ � � � ; then we saythat E is an Fr;

2. If there exist open sets G1;G2; . . . such that E ¼ G1 \G2 \ � � � ; then we say thatE is a Gd:

Lemma 1.231 Let X be a locally compact Hausdorff space. Let X be r-compact.Let K be a positive linear functional on Cc Xð Þ: Then there exists a r-algebra ℳ inX that contains all Borel sets in X; and there exists a positive measure l on ℳsatisfying the following conditions:



and K is a compact setg;4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E;f

and K is a compact setg;5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;6. for every f 2 Cc Xð Þ; K fð Þ ¼ RX f dl;7. for every E 2 ℳ; and for every e[ 0; there exist a closed set F; and an open set

V such that F E V ; and l V � Fð Þ\e;8. l is a regular Borel measure;9. for every E 2 ℳ; there exist sets A and B such that A is an Fr; B is a Gd;

A E B; and l B� Að Þ ¼ 0:

Proof By Theorem 1.225, there exists a r-algebra ℳ in X that contains all Borelsets in X; and there exists a positive measure l on ℳ satisfying the followingconditions:


and K is a compact setg;4. for every E 2 ℳ satisfying l Eð Þ\1; l Eð Þ ¼ sup l Kð Þ : K E;f

and K is a compact setg;5. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;6. for every f 2 Cc Xð Þ; K fð Þ ¼ RX f dl:

It remains to prove 7, 8, 9.For 7: Take any E 2 ℳ; and e[ 0: Since X is r-compact, there exists a

countable collection K1;K2; . . .f g of compact subsets of X such that X ¼K1 [K2 [ � � � : It follows that

E ¼ E \K1ð Þ [ E \K2ð Þ [ � � � :

Since K1 is a compact subset of X; by 1, K1 2 ℳ; and l K1ð Þ\1: SinceK1 2 ℳ; E 2 ℳ; and ℳ is a r-algebra, E \K1 2 ℳ: Since K1 ð ÞE \K1 2 ℳ;we have l E \K1ð Þ� l K1ð Þ \1ð Þ; and hence l E \K1ð Þ\1: By 2,

l E \K1ð Þ ¼ inf l Vð Þ : E \K1ð Þ V ; andV is openf g:

It follows that there exists an open set V1 such that E \K1ð Þ V1; and

l E \K1ð Þþ l V1 � E \K1ð Þð Þ ¼ l E \K1ð Þ [ V1 � E \K1ð Þð Þð Þ¼ l V1ð Þ\l E \K1ð Þþ e

22;


and hence l V1 � E \K1ð Þð Þ\ e22 : Similarly, there exists an open set V2 such that

l V2 � E\K2ð Þð Þ\ e23 ; etc. Since, for every i ¼ 1; 2; . . .; E \Kið Þ Vi; we have

E ¼ð Þ E \K1ð Þ [ E \K2ð Þ [ � � �ð Þ V1 [V2 [ � � �ð Þ:

Thus, E V1 [V2 [ � � �ð Þ: Since, each Vi is open, V1 [V2 [ � � �ð Þ is open. SinceE 2 ℳ; and ℳ is a r-algebra, we have Ec 2 ℳ: Now, as above, for every n ¼1; 2; . . .; there exists open sets Vn such that Ec \Knð Þ Vn; andl Vn � Ec \Knð Þ� �

\ e2nþ 1 : Since, for every n ¼ 1; 2; . . .; Ec \Knð Þ Vn; we have

Ec ¼ð Þ Ec \K1ð Þ [ Ec \K2ð Þ [ � � �ð Þ V1 [ V2 [ � � �� ;

and hence, V1 [ V2 [ � � �� c E: Since each Vn is open, V1 [ V2 [ � � �� is open,

and hence, V1 [ V2 [ � � �� cis closed. It suffices to show that

l V1 [V2 [ � � �ð Þ � V1 [ V2 [ � � �� c� \e:

Here,

l V1 [V2 [ � � �ð Þ � V1 [ V2 [ � � �� c� ¼ l V1 [V2 [ � � �ð Þ � Eð Þ [ E � V1 [ V2 [ � � �� c� � ¼ l V1 [V2 [ � � �ð Þ � Eð Þþ l E � V1 [ V2 [ � � �� c� ¼ l V1 � Eð Þ [ V2 � Eð Þ [ � � �ð Þ þ l E \ V1 [ V2 [ � � �� ¼ l V1 � Eð Þ [ V2 � Eð Þ [ � � �ð Þ þ l V1 � Ec

� �[ V2 � Ec� �[ � � ��

� l V1 � E \K1ð Þð Þ [ V2 � E \K2ð Þð Þ [ � � �ð Þ þ l V1 � Ec \K1ð Þ� �[ V2 � Ec \K2ð Þ� �[ � � �� l V1 � E \K1ð Þð Þþ l V2 � E\K2ð Þð Þþ � � �ð Þþ l V1 � Ec \K1ð Þ� �þ l V2 � Ec \K2ð Þ� �þ � � �� \

e22

þ e23

þ � � ��

þ e22

þ e23

þ � � ��

¼ e:

For 8: Since l : ℳ ! 0;1½ � is a positive measure, and ℳ contains the r-algebra, say B; of all Borel sets in X; the restriction of l to B is a Borel measure.Now, it remains to show that l is regular. For this purpose, let us take any Borel setE in X:We have to show that E is both outer regular and inner regular. From 2, E isouter regular. It remains to show that E is inner regular, that is,

l Eð Þ ¼ sup l Kð Þ : K E; and K is a compact setf g:

Since X is r-compact, there exists a countable collection K1;K2; . . .f g ofcompact subsets of X such that X ¼ K1 [K2 [ � � � : It follows that E ¼K1 \Eð Þ [ K2 \Eð Þ [ � � � : Since K1 is a compact subset of the Hausdorff space X;K1 is closed, and hence K1 2 B: Since K1;E 2 B; and B is a r-algebra, K1 \Eð Þ 2


B ℳð Þ: Since K1 is a compact subset of X; by 1, l K1 \Eð Þ�ð Þl K1ð Þ\1; andhence, l K1 \Eð Þ\1: Similarly, l K2 \Eð Þ\1; etc. Since

E ¼ K1 \Eð Þ [ K2 \Eð Þ [ � � � ;

each Ki \Eð Þ 2 ℳ; and each l Ki \Eð Þ\1; E has r-finite measure, and hence, byProblem 1.230,

l Eð Þ ¼ sup l Kð Þ : K E; and K is a compact setf g:

For 9: By 7, for every positive integer n; there exist a closed set Fn; and an openset Vn such that Fn E Vn; and l Vn � Fnð Þ\ 1

n : It follows that

[1n¼1Fn

� � E \1n¼1Vn

� �:

Clearly, [1n¼1Fn is an Fr; and \1

n¼1Vn is a Gd: It suffices to show that

l \1n¼1Vn

� �� [1n¼1Fn

� �� ¼ 0:

If not, otherwise, let l \1n¼1Vn

� �� [1n¼1Fn

� �� 6¼ 0: We have to arrive at acontradiction. Since,

l \1n¼1Vn

� �� [1n¼1Fn

� �� 6¼ 0;

there exists a positive integer n0 such that

1n0

\l \1n¼1Vn

� �� [1n¼1Fn

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � l Vn0 � [1

n¼1Fn� �� l Vn0 � Fn0ð Þ\ 1

n0:

Thus, we get a contradiction. ∎

Note 1.232 Let X be a locally compact Hausdorff space. Let k be any positiveBorel measure on X: Suppose that

I. every open set in X is r-compact;II. for every compact set K; k Kð Þ\1:

Let f 2 Cc Xð Þ:It follows that supp fð Þ is a compact subset of X; and hence, by II,

k supp fð Þð Þ\1: Since f 2 Cc Xð Þ; f : X ! C is continuous, and hence, f : X ! C

is measurable. It follows that fj j : X ! C is measurable, and hence,


ZX

fj jdk ¼Z

supp fð Þð Þ [ supp fð Þð Þcð Þ

fj jdk

¼Z

supp fð Þ

fj jdkþZ

supp fð Þð Þcð Þ

fj jdk

¼Z

supp fð Þ

fj jdkþZ

supp fð Þð Þcð Þ

0j jdk

¼Z

supp fð Þ

fj jdkþ 0

¼Z

supp fð Þ

fj jdk:

Thus, ZX

fj jdk ¼Z

supp fð Þ

fj jdk:

Since f : X ! C is continuous, fj j : X ! 0;1½ Þ is continuous. Since fj j : X !0;1½ Þ is continuous, and supp fð Þ is a compact subset of X; fj j supp fð Þð Þ is acompact set of real numbers, and hence, fj j supp fð Þð Þ is bounded. Sincefj j supp fð Þð Þ is bounded, there exists a positive real number a such that fj j � a onsupp fð Þ: It follows thatZ

X

fj jdk ¼Z

supp fð Þ

fj jdk� a k supp fð Þð Þð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\1;

and hence,RX fj jdk\1: Thus f 2 L1 lð Þ; and hence,

RX f dk

� � 2 C:

Now, we can define a function K : Cc Xð Þ ! C as follows: for every f 2 Cc Xð Þ;

K fð Þ �ZXf dk:

From the properties of Lebesgue integral, K is a positive linear functional onCc Xð Þ: Now, by Lemma 1.231, there exists a r-algebra ℳ in X that contains allBorel sets in X; and there exists a positive measure l on ℳ satisfying the followingconditions:


1. for every f 2 Cc Xð Þ; RX f dk ¼� �K fð Þ ¼ RX f dl;

2. for every E 2 ℳ; and for every e[ 0; there exist a closed set F; and an open setV such that F E V ; and l V � Fð Þ\e;

3. l is a regular Borel measure.

We shall try to show: for every Borel set E in X; l Eð Þ ¼ k Eð Þ:For this purpose, let us take any Borel set E:We have to show that l Eð Þ ¼ k Eð Þ:Case I: when E is an open set.Since E is an open set in X; by I, E is r-compact, and hence there exists a

countable collection K1;K2; . . .f g of compact subsets of X such that E ¼K1 [K2 [ � � � : Here, K1 E; K1 is a compact subset of X; and E is an open set inX; by Urysohn’s lemma, there exists f1 2 Cc Xð Þ such that K1 � f1 � E: Similarly,there exists f2 2 Cc Xð Þ such that K2 � f2 � E; etc. For every positive integer n; put

gn � max f1; . . .; fnf g:

Since each fi 2 Cc Xð Þ; each fi is continuous, and hence each fi is Borel mea-surable. Since each fi is continuous, and ℳ contains all Borel sets in X; each fi isℳ-measurable. Now, by Lemma 1.90, each gn ¼ð Þmax f1; . . .; fnf g is both ℳ-measurable and Borel measurable. Thus, for every n ¼ 1; 2; . . .; gn : X ! 0; 1½ � isan ℳ-measurable function, and Borel measurable function. Since each fi 2 Cc Xð Þ;it is easy to see that each

gn ¼ð Þmax f1; . . .; fnf g 2 Cc Xð Þ;

and hence, each gi 2 Cc Xð Þ: Since, for every n ¼ 1; 2; . . .;max f1; . . .; fnf g�max f1; . . .; fn; fnþ 1f g; we have, for every x 2 X;

g1 xð Þ� g2 xð Þ� � � � :Problem 1.233 For every x 2 X;

limn!1 gn xð Þ ¼ vE xð Þ:

(Solution Case I: when x 62 E: Since f1 � E; we have supp f1ð Þ E 63xð Þ; and hencex 62 supp f1ð Þ: It follows that f1 xð Þ ¼ 0: Similarly, f2 xð Þ ¼ 0; f3 xð Þ ¼ 0; etc. Hence,each gn xð Þ ¼ 0: Thus limn!1 gn xð Þ ¼ 0 ¼ vE xð Þð Þ:

Case II: when x 2 E: Since x 2ð ÞE ¼ K1 [K2 [ � � � ; there exists a positiveinteger n0 such that x 2 Kn0 : Since x 2 Kn0 ; and Kn0 � fn0 ; we have fn0 xð Þ ¼ 1:Since, for each n; 0� fn � 1; and fn0 xð Þ ¼ 1; we have

gn0 xð Þ ¼ 1; gn0 þ 1 xð Þ ¼ 1; gn0 þ 2 xð Þ ¼ 1; etc.,

and hence limn!1 gn xð Þ ¼ 1 ¼ vE xð Þð Þ:Thus, in all cases, limn!1 gn xð Þ ¼ vE xð Þ: ■)


Now, by Theorem 1.125,

limn!1

ZX

gndk

0@

1A ¼

ZX

vEdk

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ k Eð Þ:

Thus, k Eð Þ ¼ limn!1RX gndk

� �: Similarly, l Eð Þ ¼ limn!1

RX gndl

� �: Since

each gn 2 Cc Xð Þ; by 1,RX gndk ¼ RX gndl; and hence

k Eð Þ ¼ limn!1

ZXgndk

� �¼ lim

n!1

ZXgndl

� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ l Eð Þ:

Thus, k Eð Þ ¼ l Eð Þ:Case II: when E is not an open set. We have to show that l Eð Þ ¼ k Eð Þ: If not,

otherwise, let l Eð Þ 6¼ k Eð Þ: We have to arrive at a contradiction. Pute � l Eð Þ � k Eð Þj j [ 0ð Þ:

By 2, there exist a closed set F; and an open set V such that F E V ; andl V � Fð Þ\e: Since V is an open set in X; by Case I, l Vð Þ ¼ k Vð Þ: Since V is anopen set in X; and F is a closed set, V � Fð Þ is an open set, and hence, by Case I,k V � Fð Þ ¼ l V � Fð Þ: Since, l Vð Þ ¼ k Vð Þ; we have

k Eð Þ � l Eð Þ� k Vð Þ � l Eð Þ ¼ l Vð Þ � l Eð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ l V � Eð Þ� l V � Fð Þ\e;

and hence k Eð Þ � l Eð Þ\e: Since, k V � Fð Þ ¼ l V � Fð Þ; we have

� k Eð Þ � l Eð Þð Þ ¼ l Eð Þ � k Eð Þ� l Vð Þ � k Eð Þ ¼ k Vð Þ � k Eð Þ¼ k V � Eð Þ� k V � Fð Þ ¼ l V � Fð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\e;

and hence � k Eð Þ � l Eð Þð Þ\e: Since � k Eð Þ � l Eð Þð Þ\e; and k Eð Þ � l Eð Þ\e;we have

l Eð Þ � k Eð Þj j\e ¼ l Eð Þ � k Eð Þj jð Þ:

This gives a contradiction.Thus, l ¼ k on the collection of all Borel sets in X: It follows, from 3, that k is a

regular measure.

Conclusion 1.234 Let X be a locally compact Hausdorff space. Let k be anypositive Borel measure on X: Suppose that

1. every open set in X is r-compact;2. for every compact set K; k Kð Þ\1:

Then k is regular.


1.10 Lebesgue Measure

The discovery of Lebesgue measure via Riesz representation theorem is a won-derful topic in mathematics. We shall investigate this patiently below.Note 1.235

Definition Let a � a1; . . .; akð Þ 2 Rk: Let d be a positive real. The set

a1; a1 þ d½ Þ � � � ak; ak þ d½ Þ Rk� �

is denoted by Q a; dð Þ; and is called the d-box with corner at a:

Notation j1 12 ; . . .; jk 1

2

� �: j1; . . .; jk 2 Z

� Rk� �

is denoted by P1;

j1 14; . . .; jk 1

4

� �: j1; . . .; jk 2 Z

� � Rk� �

is denoted by P2;

j1 18; . . .; jk 1

8

� �: j1; . . .; jk 2 Z

� � Rk� �

is denoted by P3; etc. In short, for every positive integer n;

Pn � j1 12n

; . . .; jk 12n

� �: j1; . . .; jk 2 Z

� �:

Clearly, P1 P2 P3 � � � :The collection

Q a;12

� �: a 2 P1

� �

is denoted by X1:

In the case of R2;

X1 ¼ Qm2;n2

� ;12

� �: m; n 2 Z

� �

¼ m2;m2þ 1

2

� � n

2;n2þ 1

2

� �: m; n 2 Z

� �

¼ m2;mþ 12

� � n

2;nþ 12

� �: m; n 2 Z

� �:

1.10 Lebesgue Measure 187

Clearly,

X1 ¼ð Þ m2;mþ 12

� � n

2;nþ 12

� �: m; n 2 Z

� �

is a partition of R2; and hence, X1 is a partition of R2: Similarly, in the case of Rk;

X1 is a partition of Rk:

The collection Q a; 14� �

: a 2 P2�

is denoted by X2: As above, X2 is a partitionof Rk: The collection Q a; 18

� �: a 2 P3

� is denoted by X3; etc. In short, for every

positive integer n;

Xn � Q a;12n

� �: a 2 Pn

� �:

Clearly, each Q a; 12n

� � Rk� �

is a convex set.I. As above, Xn is a partition of Rk: In the case of R2; we have seen that

X1 ¼ m2;mþ 12

� � n

2;nþ 12

� �: m; n 2 Z

� �:

Similarly,

X2 ¼ m4;mþ 14

� � n

4;nþ 14

� �: m; n 2 Z

� �;

X3 ¼ m8;mþ 18

� � n

8;nþ 18

� �: m; n 2 Z

� �; etc:

Observe that one of the partition of

m2;mþ 12

� � n

2;nþ 12

� �2 X1ð Þ

is

2m4

;2mþ 1

4

� � 2n

4;2nþ 1

4

� �;2mþ 1

4;2mþ 2

4

� � 2n

4;2nþ 1

4

� �;

�2m4

;2mþ 1

4

� � 2nþ 1

4;2nþ 2

4

� �;2mþ 1

4;2mþ 2

4

� � 2nþ 1

4;2nþ 2

4

� �� X2ð Þ:

It follows that if Q0 2 X1; and Q00 2 X2; then Q00 Q0 or Q00 \Q0 ¼ ;ð Þ:II. As above, in Rk; if n\r; Q0 2 Xn; and Q00 2 Xr; then

Q00 Q0 or Q00 \Q0 ¼ ;ð Þ:


In the case of R2; let us observe that all the points of P3 that lie in

m2;mþ 12

� � n

2;nþ 12

� �

are

4m8

;4n8

� �;

4mþ 18

;4n8

� �;

4mþ 28

;4n8

� �;

4mþ 38

;4n8

� �;

4m8

;4nþ 1

8

� �;

4mþ 18

;4nþ 1

8

� �;

4mþ 28

;4nþ 1

8

� �;

4mþ 38

;4nþ 1

8

� �;

4m8

;4nþ 2

8

� �;

4mþ 18

;4nþ 2

8

� �;

4mþ 28

;4nþ 2

8

� �;

4mþ 38

;4nþ 2

8

� �;

4m8

;4nþ 3

8

� �;

4mþ 18

;4nþ 3

8

� �;

4mþ 28

;4nþ 3

8

� �;

4mþ 38

;4nþ 3

8

� �:

So, in the case of R2; the number of points of P3 that lie in

m2;mþ 12

� � n

2;nþ 12

� �2 X1ð Þ

is 24 ¼ 2 3�1ð Þ� �2� :

III. As above, in Rk; if n\r; and Q 2 Xn; then the number of points of Pr that lie

in Q is 2 r�nð Þ� �k ¼ 2k r�nð Þ� �:

IV. Problem 1.236 Every nonempty open set in Rk can be expressed as a disjointunion of countable-many sets in X1 [X2 [X3 [ � � � :(Solution Let V be a nonempty open set. Take any x 2 V : Since V is open, thereexists a positive integer n such that the open sphere S x; 1

2n� � V : There exist a

positive integer N; and a 2 PN such that x 2 Q a; 12N

� � S x; 12n

� � Vð Þ: Since a 2PN ; we have

x 2 Q a;12N

� �2 XN|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} X1 [X2 [X3 [ � � �ð Þ:

It follows that V can be expressed as a union of sets in X1 [X2 [X3 [ � � � :Since each Xn is countable, X1 [X2 [X3 [ � � � is a countable collection, and henceV can be expressed as a union of countable-many sets in X1 [X2 [X3 [ � � � : So,we can suppose that


V ¼ Q1;1 [Q1;2 [ � � �� [ Q2;1 [Q2;2 [ � � �� [ Q3;1 [Q3;2 [ � � �� [ � � � ;

where each Qi;j is a member of Xi; and Qi;j are distinct.First, choose Q1;1 [Q1;2 [ � � ��

: Now, by I, Q1;1 [Q1;2 [ � � � is a disjoint union.By II, either Q2;1 is contained in some Q1;j or Q2;1 is disjoint from each Q1;j: If Q2;1

is contained in some Q1;j; then remove it from

V ¼ Q1;1 [Q1;2 [ � � �� [ Q2;1 [Q2;2 [ � � �� [ Q3;1 [Q3;2 [ � � �� [ � � � :

If Q2;1 is disjoint from each Q1;j; then retain it in

V ¼ Q1;1 [Q1;2 [ � � �� [ Q2;1 [Q2;2 [ � � �� [ Q3;1 [Q3;2 [ � � �� [ � � � :

Similarly, if Q2;2 is contained in some Q1;j; then remove it from

V ¼ Q1;1 [Q1;2 [ � � �� [ Q2;1 [Q2;2 [ � � �� [ Q3;1 [Q3;2 [ � � �� [ � � � :

If Q2;2 is disjoint from each Q1;j; then retain it in

V ¼ Q1;1 [Q1;2 [ � � �� [ Q2;1 [Q2;2 [ � � �� [ Q3;1 [Q3;2 [ � � �� [ � � � ; etc:

In this way, V is expressed as a disjoint union of countable-many sets inX1 [X2 [X3 [ � � � : ■)

Definition Let W R: If there exist real numbers a; b such that a\b; and

W ¼ a; bð Þ or a; b½ Þ or a; bð � or a; b½ �;

then we say that W is a 1-cell. If I1; I2 are 1-cells, then I1 I2 is called a 2-cell.Similar definitions can be supplied for 3-cell, 4-cell etc. A 2-cell a1; b1ð Þ a2; b2ð Þis called an open 2-cell, and a 2-cell a1; b1½ � a2; b2½ � is called a closed 2-cell, etc.

Notation Let f : Rk ! C: Let supp fð Þ be compact. Let n be a positive integer.Since supp fð Þ is a compact subset of Rk; by Heine-Borel theorem, supp fð Þ is a

bounded subset of Rk; and hence supp fð Þ contains only finite-many points of Pn: Itfollows that

Px2Pn \ supp fð Þð Þ f xð Þ is a finite sum of complex numbers. If x 2

Pn \ supp fð Þð Þcð Þ; then f xð Þ ¼ 0: That is whyP

x2Pn \ supp fð Þð Þ f xð Þ is also denotedbyP

x2Pnf xð Þ: By Kn fð Þ, we shall mean

12nkXx2Pn

f xð Þ:

Let f 2 Cc Rk� �

: Let f : Rk ! R: Let f 6¼ 0: Let e be a positive real.


Since f 6¼ 0; supp fð Þ is nonempty. Since f 2 Cc Rk� �

; f : Rk ! R is continuous.So, for every p 2 Rk; there exists a real dp [ 0 such that x� pj j\dp impliesf xð Þ � f pð Þj j\ e

2 :

Thus S p; 12 dp� �

: p 2 supp fð Þ� is an open cover of the compact set supp fð Þ: It

follows that there exists p1; . . .; pn 2 supp fð Þ such that

supp fð Þ S p1;12dp1

� �[ � � � [ S pn;

12dpn

� �:

Put

d � min12dp1 ; . . .;

12dpn

� �[ 0ð Þ:

Put

G � S p1;12dp1

� �[ � � � [ S pn;

12dpn

� � supp fð Þð Þ:

Here, G is a bounded open set containing supp fð Þ: Let x; y 2 G satisfyingx� yj j\d: We shall show that f xð Þ � f yð Þj j\e: Since

x 2 G ¼ S p1;12dp1

� �[ � � � [ S pn;

12dpn

� �� ;

there exists l 2 1; . . .; nf g such that x 2 S pl; 12 dpl� �

; and hence

x� plj j\ 12dpl \dpl� �

:

Next,

y� plj j � y� xj j þ x� plj j\ y� xj j þ 12dpl\dþ 1

2dpl

¼ min12dp1 ; � � � ;

12dpn

� �þ 1

2dpl �

12dpl þ

12dpl ¼ dpl ;

so y� plj j\dpl ; and hence, f yð Þ � f plð Þj j\ e2 : Since x� plj j\dpl ; we have

f xð Þ � f plð Þj j\ e2 : Since f xð Þ � f plð Þj j\ e

2 ; and f yð Þ � f plð Þj j\ e2 ; we have

f xð Þ � f yð Þj j\e:

Conclusion 1.237 Let f 2 Cc Rk� �

: Let f : Rk ! R: Let f 6¼ 0: Let e be a positivereal. Then there exist a bounded open set G; and a positive real number d such that


a. supp fð Þ G;b. for every x; y 2 G satisfying x� yj j\d; f xð Þ � f yð Þj j\e:

Note 1.238 Let f 2 Cc Rk� �

: Let f : Rk ! R: Let f 6¼ 0: Let e be a positive real.Let a1;b1ð Þ � � � ak; bkð Þ be an open k-cell containing the compact subsetsupp fð Þ of Rk:

Since each projection map pi from Rk to R is continuous, and supp fð Þ is acompact subset of Rk; for each i ¼ 1; . . .; k; min pi supp fð Þð Þð Þ andmax pi supp fð Þð Þð Þ exist. It follows that, for each i ¼ 1; . . .; k; there exist

ai; bi 2 supp fð Þ a1;b1ð Þ � � � ak; bkð Þð Þ

such that

min pi supp fð Þð Þð Þ ¼ pi aið Þ; and max pi supp fð Þð Þð Þ ¼ pi bið Þ:

Since

a1 2 a1; b1ð Þ � � � ak; bkð Þ;

we have p1 a1ð Þ 2 a1; b1ð Þ: Similarly, p1 b1ð Þ 2 a1; b1ð Þ: Thus,a1\p1 a1ð Þ� p1 b1ð Þ\b1: Similarly, a2\p2 a2ð Þ� p2 b2ð Þ\b2; etc. Thus,

supp fð Þ a1 þ p1 a1ð Þ2

;p1 b1ð Þþ b1

2

� � � � � ak þ pk akð Þ

2;pk bkð Þþ bk

2

� � a1;b1ð Þ � � � ak; bkð Þ:

By Conclusion 1.237, there exists a bounded open set G; and a positive realnumber d such that

a. supp fð Þ G;b. for every x; y 2 G satisfying x� yj j\d; f xð Þ � f yð Þj j\ e

4 :

There exists a positive integer N such that, for every Q 2 XN ; and, for everyx; y 2 �Q; we have x� yj j\d; and

12N

\minp1 a1ð Þ � a1

2;b1 � p1 b1ð Þ

2; . . .;

pk akð Þ � ak2

;bk � pk bkð Þ

2

� �:

Let us take any Q 2 XN :

Case I: when Q supp fð Þ: It follows that �Q supp fð Þ; and �Q is compact.Since �Q is a nonempty compact set, and f is continuous, f �Qð Þ assumes its mini-mum. Now, there exists aQ 2 �Q such that f aQð Þ ¼ min f �Qð Þð Þ: Put, for every x 2 Q;


g xð Þ � f aQð Þ:Problem 1.239 g is constant on Q; and g� f on Q: Also, f � gð Þ\ e

2 on Q:

(Solution Let x 2 Q: We have to show that f xð Þ � g xð Þð Þ\ e2 : Since x 2 Q �Qð Þ;

we have x 2 �Q: Since x; aQ 2 �Q; we have x� aQ�� \d; and hence

f xð Þ � g xð Þ ¼ f xð Þ � g xð Þj j ¼ f xð Þ � f aQð Þ�� \ e4|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus, f xð Þ � g xð Þð Þ\ e2 : ■)

Since �Q is a nonempty compact set, and f is continuous, f �Qð Þ assumes itsmaximum. Now, there exists bQ 2 �Q such that f bQð Þ ¼ max f �Qð Þð Þ: Put, for everyx 2 Q;

h xð Þ � f bQð Þ:

Problem 1.240 h is constant on Q; and f � h on Q: Also, h� fð Þ\ e2 on Q:

(Solution Let x 2 Q: We have to show that h xð Þ � f xð Þð Þ\ e2 : Since x 2 Q �Qð Þ;

we have x 2 �Q: Since x; bQ 2 �Q; we have x� bQ�� \d; and hence,

h xð Þ � f xð Þ ¼ f xð Þ � h xð Þj j ¼ f xð Þ � f bQð Þ�� \ e4|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus, h xð Þ � f xð Þð Þ\ e2 : ■)

Since f � gð Þ\ e2 on Q; and h� fð Þ\ e

2 on Q; we have h� gð Þ\e on Q:Case II: when Q supp fð Þð Þc: Put, for every x 2 Q; g xð Þ � 0: Clearly, g is

constant on Q: Here, for every x 2 Q supp fð Þð Þcð Þ; we have f xð Þ ¼ 0: It followsthat g� f on Q: Put, for every x 2 Q; h xð Þ � 0: Clearly, h is constant on Q: Here,for every x 2 Q supp fð Þð Þcð Þ; we have f xð Þ ¼ 0: It follows that f � h on Q:Clearly, h� gð Þ\e on Q:

Case III: when Q 6 supp fð Þð Þc; and Q 6 supp fð Þ: There exist a 2 Q\ supp fð Þ;and b 2 Q\ supp fð Þð Þc: Thus, b 62 supp fð Þ; and a; b 2 Q: Since a; b 2 Q; and Q isconvex, for every t 2 0; 1½ �; we have 1� tð Þaþ tb 2 Q: Clearly, a 6¼ b:

Since the mapping u : t 7! 1� tð Þaþ tbð Þ from compact set 0; 1½ � onto compactset 1� tð Þaþ tb : t 2 0; 1½ �f g is 1-1 and continuous, the mapping u�1 from com-pact set 1� tð Þaþ tb : t 2 0; 1½ �f g to 0; 1½ � is continuous, and hence

max u�1 1� tð Þaþ tb : t 2 0; 1½ �f g \ supp fð Þð Þ� � 0; 1½ �ð Þ

exists. Now, since


u�1 1� tð Þaþ tb : t 2 0; 1½ �f g \ supp fð Þð Þ¼ u�1 1� tð Þaþ tb : t 2 0; 1½ �f gð Þ \u�1 supp fð Þð Þ¼ u�1 1� tð Þaþ tbð Þ : t 2 0; 1½ �� \u�1 supp fð Þð Þ¼ t : t 2 0; 1½ �f g \u�1 supp fð Þð Þ¼ 0; 1½ � \u�1 supp fð Þð Þ ¼ 0; 1½ � \ t : u tð Þ 2 supp fð Þf g¼ t : t 2 0; 1½ �; and 1� tð Þaþ tb 2 supp fð Þf g;

max t : t 2 0; 1½ �; and 1� tð Þaþ tb 2 supp fð Þf g exists. Suppose that

max t : t 2 0; 1½ �; and 1� tð Þaþ tb 2 supp fð Þf g ¼ t0:

Clearly, t0 2 0; 1½ Þ: Since 1� t0ð Þaþ t0b 2 supp fð Þ Gð Þ; t0 2 0; 1½ Þ; and G isopen, there exists t1 2 0; 1½ Þ such that t0\t1; and 1� t1ð Þaþ t1b 2 G: Clearly,1� t1ð Þaþ t1b 2 Q: Since

max t : t 2 0; 1½ �; and 1� tð Þaþ tb 2 supp fð Þf g ¼ t0\t1|fflffl{zfflffl};we have 1� t1ð Þaþ t1b 62 supp fð Þ; and hence f 1� t1ð Þaþ t1bð Þ ¼ 0:

Let us take any x 2 Q\ supp fð Þ Q\G Gð Þ: Since 1� t1ð Þaþ t1b 2Q �Qð Þ; x� 1� t1ð Þaþ t1bð Þj j\d: Since x; 1� t1ð Þaþ t1b 2 G; andx� 1� t1ð Þaþ t1bð Þj j\d; we have

f xð Þj j ¼ f xð Þ � 0j j ¼ f xð Þ � f 1� t1ð Þaþ t1bð Þj j\ e4;

and hence � e4\f xð Þ\ e

4 : Put, for every x 2 Q; g xð Þ � � e4 : Clearly, g is constant

on Q; and g� f on Q: Put, for every x 2 Q; h xð Þ � e4 : Clearly, h is constant on Q;

and f � h on Q: Also, h� gð Þ\e on Q:Since supp fð Þ is a bounded set, Q : Q 2 XN ; and Q\ supp fð Þ 6¼ ;f g is a finite

set. Now, since

12N

\minp1 a1ð Þ � a1

2;b1 � p1 b1ð Þ

2; . . .;

pk akð Þ � ak2

;bk � pk bkð Þ

2

� �;

we have

[ �Q : Q 2 XN ; and Q\ supp fð Þ 6¼ ;f g a1 þp1 a1ð Þ

2 ; p1 b1ð Þþ b12

h i � � � ak þ pk akð Þ

2 ; pk bkð Þþbk2

h i a1; b1ð Þ � � � ak; bkð Þ:


From the construction of g in all cases,

x : g xð Þ 6¼ 0f g [ Q : Q 2 XN ; and Q\ supp fð Þ 6¼ ;f g;

and hence,

supp gð Þ ¼ x : g xð Þ 6¼ 0f g� [ Q : Q 2 XN ; and Q\ supp fð Þ 6¼ ;f gð Þ�|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ [ �Q : Q 2 XN ; and Q\ supp fð Þ 6¼ ;f g a1; b1ð Þ � � � ak; bkð Þ:

Thus, supp gð Þ a1; b1ð Þ � � � ak; bkð Þ: Similarly, supp hð Þ a1; b1ð Þ � � � ak; bkð Þ: Now, since XN is a partition of Rk; we have g : Rk ! R; andh : Rk ! R:

Conclusion 1.241 Let f 2 Cc Rk� �

: Let f : Rk ! R: Let f 6¼ 0: Let e be a positivereal. Let W be an open k-cell containing the compact subset supp fð Þ of Rk: Thenthere exists a positive integer N; a function g : Rk ! R; and a function h : Rk ! R

such that

1. g is constant on each Q in XN ; and h is constant on each Q in XN ;2. g� f � h;3. h� g\e;4. supp gð Þ W ; and supp hð Þ W :

Note 1.242 Let f 2 Cc Rk� �

: Let f : Rk ! R: Let f 6¼ 0: Let e be a positive real.Let a1;b1ð Þ � � � ak; bkð Þ be an open k-cell containing the compact subsetsupp fð Þ of Rk: By Conclusion 1.241, there exists a positive integer N; a functiong : Rk ! R; and a function h : Rk ! R such that

1. g is constant on each Q in XN ; and h is constant on each Q in XN ;2. g� f � h;3. h� g\e;4. supp gð Þ a1; b1ð Þ � � � ak; bkð Þ; and supp hð Þ a1; b1ð Þ � � � ak; bkð Þ:

Since, supp gð Þ a1; b1ð Þ � � � ak; bkð Þ; supp gð Þ is a bounded subset of Rk:Now, since supp gð Þ is a closed set, by Heine-Borel theorem, supp gð Þ is a compactset.

Problem 1.243 KN gð Þ ¼ KNþ 1 gð Þ:(Solution


LHS ¼ KN gð Þ ¼ 12Nk

Xx2PN

g xð Þ ¼ 12Nk

XQ x; 1

2Nð Þ2XN

g xð Þ

¼ 12Nk

XQ y; 1

2N þ 1

� �2XN þ 1

g yð Þ2k

¼ 12Nk

12k

XQ y; 1

2Nþ 1

� �2XN þ 1

g yð Þ

¼ 12 Nþ 1ð Þk

XQ y; 1

2N þ 1

� �2XN þ 1

g yð Þ ¼ 12 Nþ 1ð Þk

Xy2PNþ 1

g yð Þ ¼ KNþ 1 gð Þ ¼ RHS:

∎)Similarly, KNþ 1 gð Þ ¼ KNþ 2 gð Þ; etc. Also, KN hð Þ ¼ KNþ 1 hð Þ ¼ KNþ 2 hð Þ ¼

� � � : For every m[N;

KN gð Þ ¼ Km gð Þ ¼ 12mk

Xx2Pm

g xð Þ� 12mk

Xx2Pm

f xð Þ

¼ Km fð Þ� 12mk

Xx2Pm

h xð Þ ¼ Km hð Þ ¼ KN hð Þ;

and hence, for every m; n[N;

KN gð Þ�Km fð Þ�KN hð ÞKN gð Þ�Kn fð Þ�KN hð Þ

�:

It follows that, for every m; n[N;

Km fð Þ � Kn fð Þj j �KN hð Þ � KN gð Þ ¼ 12Nk

Xx2PN

h xð Þ � 12Nk

Xx2PN

g xð Þ

¼ 12Nk

Xx2PN

h xð Þ � g xð Þð Þ ¼ 12Nk

Xx2PN \ a1;b1ð Þ�� ak ;bkð Þð Þ

h xð Þ � g xð Þð Þ

\12Nk

Xx2PN \ a1;b1ð Þ�� ak ;bkð Þð Þ

eð Þ ¼ eX

x2PN \ a1;b1ð Þ�� ak ;bkð Þð Þ

12N

� � � 12N|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

k factors

0BB@

1CCA

� e b1 � a1ð Þ � � � bk � akð Þð Þ:

Thus, for every m; n[N; Km fð Þ � Kn fð Þj j\e b1 � a1ð Þ � � � bk � akð Þð Þ:It follows that the sequence Kn fð Þf g is a Cauchy sequence of real numbers.

Now, since R is complete, Kn fð Þf g is a convergent sequence. It follows that thereexists a real number K fð Þ such that limn!1 Kn fð Þ ¼ K fð Þ:

Thus, for every f 2 Cc Rk� �

satisfying f : Rk ! R; and f 6¼ 0; we haveK fð Þ ¼ limn!1 Kn fð Þ:


For 0 2 Cc Rk� �

; put K 0ð Þ � 0: Let f 2 Cc Rk� �

satisfying f : Rk ! R; anda 2 R:

Problem 1.244 K afð Þ ¼ a K fð Þð Þ:(Solution Case I: when f ¼ 0:LHS ¼ K afð Þ ¼ K a0ð Þ ¼ K 0ð Þ ¼ 0 ¼ a 0ð Þ ¼ a K 0ð Þð Þ ¼ a K fð Þð Þ ¼ RHS:

Case II: when a ¼ 0:LHS ¼ K 0fð Þ ¼ K 0ð Þ ¼ 0 ¼ 0 K fð Þð Þ ¼ a K fð Þð Þ ¼ RHS:

Case III: when a 6¼ 0; and f 6¼ 0: It follows that af 6¼ 0:

LHS ¼ K afð Þ ¼ limn!1Kn afð Þ ¼ lim

n!112nkXx2Pn

afð Þ xð Þ ¼ limn!1

12nkXx2Pn

a f xð Þð Þ

¼ a limn!1

12nkXx2Pn

f xð Þ ¼ a limn!1Kn fð Þ�

¼ a K fð Þð Þ ¼ RHS:

∎)Let f ; g 2 Cc Rk

� �satisfying f : Rk ! R; and g : Rk ! R:

Problem 1.245 K f þ gð Þ ¼ K fð ÞþK gð Þ:(Solution Case I: when f ¼ 0:LHS ¼ K 0þ gð Þ ¼ K gð Þ ¼ 0þK gð Þ ¼ K 0ð ÞþK gð Þ ¼ K fð ÞþK gð Þ ¼ RHS:

Case II: when g ¼ 0: This case is similar to the Case I.Case III: when f 6¼ 0; g 6¼ 0; and f þ g ¼ 0: It follows that g ¼ �f ; and hence,

K gð Þ ¼ K �fð Þ ¼ K �1ð Þfð Þ ¼ �1ð Þ K fð Þð Þ ¼ � K fð Þð Þ:LHS ¼ K f þ gð Þ ¼ K 0ð Þ ¼ 0 ¼ K fð Þþ � K fð Þð Þð Þ ¼ K fð ÞþK gð Þ ¼ RHS:

Case IV: when f 6¼ 0; g 6¼ 0; and f þ g 6¼ 0:

LHS ¼ K f þ gð Þ ¼ limn!1Kn f þ gð Þ ¼ lim

n!112nkXx2Pn

f þ gð Þ xð Þ

¼ limn!1

12nkXx2Pn

f xð Þþ g xð Þð Þ ¼ limn!1

12nkXx2Pn

f xð Þþ 12nkXx2Pn

g xð Þ !

¼ limn!1

12nkXx2Pn

f xð Þþ limn!1

12nkXx2Pn

g xð Þ

¼ limn!1Kn fð Þþ lim

n!1Kn gð Þ ¼ K fð ÞþK gð Þ ¼ RHS:

∎)Now, for every f 2 Cc Rk

� �; we define


K fð Þ � K Re fð Þð Þþ i K Im fð Þð Þð Þ:

Let a 2 C; and f 2 Cc Rk� �

:

Problem 1.246 K afð Þ ¼ a K fð Þð Þ:Solution

ðLHS ¼ K afð Þ ¼ K Re að Þð Þ Re fð Þð Þ � Im að Þð Þ Im fð Þð Þð Þðþ i Re að Þð Þ Im fð Þð Þþ Im að Þð Þ Re fð Þð Þð ÞÞ

¼ K Re að Þð Þ Re fð Þð Þ � Im að Þð Þ Im fð Þð Þð Þþ i K Re að Þð Þ Im fð Þð Þþ Im að Þð Þ Re fð Þð Þð Þð Þ

¼ Re að Þð Þ K Re fð Þð Þð Þ � Im að Þð Þ K Im fð Þð Þð Þð Þþ i Re að Þð Þ K Im fð Þð Þð Þþ Im að Þð Þ K Re fð Þð Þð Þð Þ

¼ Re að Þþ iIm að Þð Þ K Re fð Þð Þþ iK Im fð Þð Þð Þ¼ a K Re fð Þð Þþ iK Im fð Þð Þð Þ ¼ a K fð Þð Þ ¼ RHS:

∎)Let f ; g 2 Cc Rk

� �:

Problem 1.247 K f þ gð Þ ¼ K fð ÞþK gð Þ:(Solution

LHS ¼ K f þ gð Þ ¼ K Re fð ÞþRe gð Þð Þþ i Im fð Þþ Im gð Þð Þð Þ¼ K Re fð ÞþRe gð Þð Þþ i K Im fð Þþ Im gð Þð Þð Þ¼ K Re fð Þð ÞþK Re gð Þð Þð Þþ i K Im fð Þð ÞþK Im gð Þð Þð Þ¼ K Re fð Þð Þþ i K Im fð Þð Þð Þð Þþ K Re gð Þð Þþ i K Im gð Þð Þð Þð Þ¼ K fð ÞþK gð Þ ¼ RHS:

∎)

Problem 1.248 Let f 2 Cc Rk� �

: Let f : Rk ! 0;1½ Þ: Then, K fð Þ 2 0;1½ Þ:(Solution Case I: when f ¼ 0: Here K fð Þ ¼ K 0ð Þ ¼ 0 2 0;1½ Þ:

Case II: when f 6¼ 0: Since f : Rk ! 0;1½ Þ; we have

K fð Þ ¼ limn!1Kn fð Þ ¼ lim

n!112nkXx2Pn

f xð Þ !

2 0;1½ Þ:

∎)Thus, K : Cc Rk

� �! C is a positive linear functional on Cc Rk� �

: Further, it isknown that Rk; with the usual topology, is a locally compact Hausdorff space. Also,


Rk is r-compact. By Lemma 1.231, there exists a r-algebraℳ in Rk that containsall Borel sets in Rk; and there exists a positive measure m on ℳ satisfying thefollowing conditions:

1. for every compact subset K of Rk; K 2 ℳ; and m Kð Þ\1;2. for every E 2 ℳ; m Eð Þ ¼ inf m Vð Þ : E V ; andV is openf g;3. for every open set V in Rk; V 2 ℳ; and m Vð Þ ¼ sup m Kð Þ : K V ;f

and K is a compact setg;4. for every E 2 ℳ satisfying m Eð Þ\1; m Eð Þ ¼ sup m Kð Þ : K E;f

and K is a compact setg;5. if E 2 ℳ; m Eð Þ ¼ 0; and A E; then A 2 ℳ; that is, m is complete;6. for every f 2 Cc Rk

� �;

K fð Þ ¼ZRk

f dm;

7. for every E 2 ℳ; and for every e[ 0; there exists a closed set F; and an openset V such that F E V ; and m V � Fð Þ\e;

8. m is a regular Borel measure;9. for every E 2 ℳ; there exist sets A and B such that A is an Fr; B is a Gd;

A E B; and m B� Að Þ ¼ 0:

I. Problem 1.249ℳ ¼ E : there exist sets A andB such that A is anFr;B is aGd;A E B;fand m B� Að Þ ¼ 0g:

(Solution From 9,

ℳ E : there exist sets A and B such that A is an Fr;B is a Gd;A E B;fand m B� Að Þ ¼ 0g:It remains to show that

E : there exist sets A and B such that A is anFr;B is a Gd;A E B;fandm B� Að Þ ¼ 0g ℳ:

For this purpose, let E be a subset of Rk such that there exist setsA and B satisfying A is an Fr; B is a Gd;A E B; and m B� Að Þ ¼ 0: Wehave to show that A[ E � Að Þ ¼ E 2 ℳ|fflfflffl{zfflfflffl} :

Since A is an Fr; there exist closed sets F1;F2; . . . such that

A ¼ F1 [F2 [ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ F1ð Þc \ F2ð Þc \ � � �ð Þc2 ℳ;


and hence A 2 ℳ: Similarly, B 2 ℳ: It follows that B� Að Þ 2 ℳ: Now, it sufficesto show that E � Að Þ 2 ℳ: Since B� Að Þ 2 ℳ;m B� Að Þ ¼ 0; and E � Að Þ B� Að Þ; by 5, we have E � Að Þ 2 ℳ: ■)Let a1; b1ð Þ � � � ak; bkð Þ be an open k-cell of Rk:

Since a1; b1ð Þ � � � ak; bkð Þ is open in Rk; a1; b1ð Þ � � � ak; bkð Þ is a Borelset in Rk; and hence a1; b1ð Þ � � � ak; bkð Þ 2 ℳ: There exist a positive integerN; and a 2 PN such that

Q a;12N

� �� a1; b1ð Þ � � � ak; bkð Þ:

Since a 2 PN ; there exist integers j1; . . .; jk such that a ¼ j12N ; . . .;

jk2N

� �: Here

Q a;12N

� �¼ Q

j12N

; . . .;jk2N

� �;12N

� �

¼ j12N

;j12N

þ 12N

� � � � � jk

2N;jk2N

þ 12N

� �

¼ j12N

;j1 þ 12N

� � � � � jk

2N;jk þ 12N

� �;

and hence

Q a;12N

� �� ¼ j1

2N;j1 þ 12N

� � � � � jk

2N;jk þ 12N

� ��

¼ j12N

;j1 þ 12N

� � � � � jk

2N;jk þ 12N

� �:

Thus, Q a; 12N

� �� is a compact subset of Rk: Now, observe that

x : x 2 PN ; and Q x;12N

� �� a1; b1ð Þ � � � ak; bkð Þ

� �

is a nonempty finite subset of PN : It follows that

[ x2PN ; and Q x; 12Nð Þð Þ� a1;b1ð Þ�� ak ;bkð ÞQ x;

12N

� ��

¼ [ x2PN ; and Q x; 12Nð Þð Þ� a1;b1ð Þ�� ak ;bkð Þ Q x;

12N

� ��

a1; b1ð Þ � � � ak; bkð Þð Þ:


Hence,


12N

� ��

is a compact subset of Rk; and


12N

� ��

is contained in the open subset a1; b1ð Þ � � � ak; bkð Þ of Rk: Put W � a1; b1ð Þ � � � ak; bkð Þ; and, for every non-negative integer r; put

Er � [ x2PN þ r ; and Q x; 12Nþ rð Þð Þ� a1;b1ð Þ�� ak ;bkð ÞQ x;

12Nþ r

� �:

Clearly, E0 E1 E2 � � � ; and Erð Þ� W for every nonnegative integer r:Now, since Rk is a locally compact Hausdorff space, by Urysohn’s lemma, thereexists f0 2 Cc Rk

� �� 0f g� �such that

E0ð Þ�� f0 � W :

Similarly, there exists f1 2 Cc Rk� �� 0f g� �

such that

E1ð Þ�� f1 � W :

Also, there exists f2 2 Cc Rk� �� 0f g� �

such that

E2ð Þ�� f2 � W ;

etc. For every nonnegative integer n; put

gn � max f0; f1; . . .; fnf g:

Clearly, g0 � g1 � g2 � � � � : Since each fi is continuous, eachgn ¼ð Þmax f0; f1; . . .; fnf g is continuous, and hence each gi is a measurable function.Since supp max f0; f1; . . .; fnf gð Þ ¼ supp f0ð Þ [ � � � [ supp fnð Þ; and each supp fið Þ iscompact, supp max f0; f1; . . .; fnf gð Þ is compact. Thus, each gn 2 Cc Rk

� �:

Since each fi is nonzero, and 0� fi � 1, each gn ¼ð Þmax f0; f1; . . .; fnf g is non-zero. Clearly, for each nonnegative integer n; 0� fn � gn � 1: Since each supp fið Þ W ; we have, for every x 2 Wc; each fi xð Þ ¼ 0; and hence, for every x 2 Wc; eachgi xð Þ ¼ 0: Clearly, for every x 2 W ;

1� limn!1 gn xð Þ� lim

n!1 fn xð Þ ¼ 1|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl};


so, for every x 2 W ;

limn!1 gn xð Þ ¼ 1|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ vW xð Þ:

For every x 2 Wc;

limn!1 gn xð Þ ¼ lim

n!1max f0 xð Þ; f1 xð Þ; . . .; fn xð Þf g ¼ limn!1max 0f g ¼ 0 ¼ vW xð Þ:

Thus, for every x 2 Rk; limn!1 gn xð Þ ¼ vW xð Þ: Since each gi is a measurablefunction, g0 � g1 � g2 � � � � ; and limn!1 gn xð Þ ¼ vW xð Þ; by Theorem 1.125, wehave

limn!1

ZRk

gndm

0B@

1CA ¼

ZRk

vWdm

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ m Wð Þ:

Since each gn 2 Cc Rk� �

; by 6, for each n; K gnð Þ ¼ RRk gndm; and hence,

limn!1K gnð Þ ¼ lim

n!1

ZRk

gndm

0B@

1CA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ m Wð Þ ¼ m a1; b1ð Þ � � � ak; bkð Þð Þ:

Thus,

limn!1K gnð Þ ¼ m a1; b1ð Þ � � � ak; bkð Þð Þ:

Let r; n be positive integers satisfying n[ r[N: Here,

Kn frð Þ ¼ 12nkXx2Pn

fr xð Þ� 12nkXx2Pn

gr xð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ Kn grð Þ;

so, Kn frð Þ�Kn grð Þ; and hence

K frð Þ ¼ limn!1Kn frð Þ� lim

n!1Kn grð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ K grð Þ:

Thus, K frð Þ�K grð Þ: Since


Kn frð Þ ¼ 12nkXx2Pn

fr xð Þ

¼X

x2Pn \En

fr xð Þ 12n

� � � 12n|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

k factors

0BB@

1CCAþ

Xx2Pn \ Enð Þc

fr xð Þ 12n


k factors

0BB@

1CCA

�X

x2Pn \En

fr xð Þ 12n


k factors

0BB@

1CCA ¼

Xx2Pn \En

112n


k factors

0BB@

1CCA ¼ vol Enð Þ;

we have

vol Wð Þ ¼ limn!1 vol Enð Þ� lim

n!1Kn frð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ K frð Þ:

Thus, vol Wð Þ�K frð Þ: Since

Kn grð Þ ¼ 12nkXx2Pn

gr xð Þ ¼X

x2Pn \W

gr xð Þ 12n


k factors

0BB@

1CCAþ

Xx2Pn \ Wð Þc

gr xð Þ 12n


k factors

0BB@

1CCA

¼X

x2Pn \W

gr xð Þ 12n


k factors

0BB@

1CCAþ

Xx2Pn \ Wð Þc

0 � 12n


k factors

0BB@

1CCA

¼X

x2Pn \W

gr xð Þ 12n


k factors

0BB@

1CCA�

Xx2Pn \W

112n


k factors

0BB@

1CCA;

we have

K grð Þ ¼ limn!1Kn grð Þ� lim

n!1

Xx2Pn \W

12n


k factors

0BB@

1CCA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ vol Wð Þ:

Thus, K grð Þ� vol Wð Þ: Since


K grð Þ� vol Wð Þ; vol Wð Þ�K frð Þ; and K frð Þ�K grð Þ;

we have

K grð Þ ¼ vol Wð Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} ¼ vol a1; b1ð Þ � � � ak; bkð Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ:

Thus, for every r[N;

K grð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ:

It follows that

m a1; b1ð Þ � � � ak; bkð Þð Þ ¼ limn!1K gnð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

m a1; b1ð Þ � � � ak; bkð Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ:

Thus, we get the following result:II. m a1; b1ð Þ � � � ak; bkð Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ:

III. Problem 1.250 m a1; b1½ Þ � � � ak; bk½ Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ:(Solution

LHS ¼ m a1; b1½ Þ � � � ak; bk½ Þð Þ

¼ m \1n¼1 a1 � 1

n; b1

� � � � � ak � 1

n; bk

� ��

¼ limn!1m a1 � 1

n; b1

� � � � � ak � 1

n; bk

� ��

¼ limn!1 b1 � a1 � 1

n

� �� bk � ak � 1

n

� �� ¼ b1 � a1ð Þ � � � bk � akð Þ ¼ RHS:

∎)

IV. Problem 1.251 Let k be a positive Borel measure. Suppose that, for every boxQ in X1 [X2 [ � � � ; k Qð Þ ¼ m Qð Þ: Then, for all Borel sets E in Rk; k Eð Þ ¼ m Eð Þ:(Solution Case I: when E is a nonempty open set in Rk: By Note 1.235(IV), thereexist boxes Q1;Q2; . . . in X1 [X2 [ � � � such that Q1;Q2; . . . are pairwise disjoint,and E ¼ Q1 [Q2 [ � � � : Now,


LHS ¼ k Eð Þ ¼ k Q1 [Q2 [ � � �ð Þ¼ k Q1ð Þþ k Q2ð Þþ � � � ¼ m Q1ð Þþm Q2ð Þþ � � �¼ m Q1 [Q2 [ � � �ð Þ ¼ m Eð Þ ¼ RHS:

Case II: when E is any Borel set in Rk: We know that Rk is a locally compactHausdorff space, and every open set in Rk is r-compact. Next, let K ba a compactsubset of Rk: We shall try to show that k Kð Þ\1: By 1, m Kð Þ\1; and, by 2, wehave

1[ m Kð Þ ¼ inf m Vð Þ : K V ; andV is openf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ inf k Vð Þ : K V ; andV is openf g� k Kð Þ;

so, k Kð Þ\1: Now, by Conclusion 1.234, k is regular. It follows that

k Eð Þ ¼ inf k Vð Þ : E V ; and V is openf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ inf m Vð Þ : E V ; andV is openf g:

By 8, m is a regular Borel measure, and hence,

m Eð Þ ¼ inf m Vð Þ : E V ; andV is openf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ k Eð Þ:

Thus, k Eð Þ ¼ m Eð Þ: ■)

V. Problem 1.252 Let E 2 ℳ; and a 2 Rk: Then xþ a : x 2 Ef g �ð ÞEþ að Þ 2 ℳ:

(Solution Since, E 2 ℳ; by (I), there exist sets A andB such that A is an Fr;B is a Gd;A E B; and m B� Að Þ ¼ 0: SinceA is an Fr; there exist closed sets F1;F2; . . . such that A ¼ F1 [F2 [ � � � : SinceB is a Gd; there exist open sets G1;G2; . . . such that B ¼ G1 \G2 \ � � � : SinceA ¼ F1 [F2 [ � � � ; we have Aþ að Þ ¼ F1 þ að Þ [ F2 þ að Þ [ � � � : Since each Fi isclosed, each Fi þ að Þ is closed. Since each Fi þ að Þ is closed, and Aþ að Þ ¼F1 þ að Þ [ F2 þ að Þ [ � � � ; Aþ að Þ is an Fr: Similarly, Bþ að Þ is a Gd: Since A E B; we have Aþ að Þ Eþ að Þ Bþ að Þ: Next

m Bþ að Þ � Aþ að Þð Þ ¼ m B� Að Þ ¼ 0:

Since Aþ að Þ Eþ að Þ Bþ að Þ; Aþ að Þ is an Fr; Bþ að Þ is a Gd; andm Bþ að Þ � Aþ að Þð Þ ¼ 0; by (I), we have Eþ að Þ 2 ℳ: ■)

VI. Problem 1.253 Let E 2 ℳ; and a � a1; . . .; akð Þ 2 Rk: Then,m Eþ að Þ ¼ m Eð Þ:


In short, we say that m is translational invariant.

(Solution For every A 2 ℳ; put

k Að Þ � m Aþ að Þ:

Thus, k : ℳ ! 0;1½ �: We shall try to show that k is a measure. For thispurpose, let us take any countable collection A1;A2;A3; . . .f g of members in ℳsuch that i 6¼ j ) Ai \Aj ¼ ;: We have to show that

k A1 [A2 [A3 [ � � �ð Þ ¼ k A1ð Þþ k A2ð Þþ k A3ð Þþ � � � ;

that is,

m A1 [A2 [A3 [ � � �ð Þþ að Þ ¼ m A1 þ að Þþm A2 þ að Þþm A3 þ að Þþ � � � :

Clearly,

A1 [A2 [A3 [ � � �ð Þþ a ¼ A1 þ að Þ [ A2 þ að Þ [ A3 þ að Þþ � � � :

Also, if i 6¼ j; then Ai þ að Þ \ Aj þ a� � ¼ Ai \Aj ¼ ;: Thus

LHS ¼ m A1 [A2 [A3 [ � � �ð Þþ að Þ ¼ m A1 þ að Þ [ A2 þ að Þ [ A3 þ að Þþ � � �ð Þ¼ m A1 þ að Þþm A2 þ að Þþm A3 þ að Þþ � � � ¼ RHS:

Next k ;ð Þ ¼ m ;þ að Þ ¼ m ;ð Þ ¼ 0\1: Thus, k is a positive Borel measure.Let Q b; 1

2n� �

be any box where

b � j12n

; . . .;jk2n

� �2 Pnð Þ; and j1; . . .; jk 2 Z:

Thus,

Q b;12n

� �¼ j1

2n;j12n

þ 12n

� � � � � jk

2n;jk2n

þ 12n

� �;

and hence, by III,

m Q b;12n

� ��

¼ mj12n

;j12n

þ 12n

� � � � � jk

2n;jk2n

þ 12n

� �� ¼ j1

2nþ 1

2n

� �� j12n

� � � � � jk

2nþ 1

2n

� �� jk2n

� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ 12n


k factors

¼ 12nk

:


Next,

k Q b;12n

� �� ¼ m Q b;

12n

� �þ a

� �¼ m Q bþ a;

12n

� ��

¼ mj12n

þ a1;j12n

þ a1 þ 12n

� � � � � jk

2nþ ak;

jk2n

þ ak þ 12n

� ��

¼ j12n

þ a1 þ 12n

� �� j1

2nþ a1

� �� jk

2nþ ak þ 1

2n

� �� jk

2nþ ak

� ��

¼ 12nk

¼ m Q b;12n

� �� :

Thus

m Q b;12n

� �� ¼ k Q b;

12n

� �� :

Here, E 2 ℳ: So, by I, there exist sets A and B such that A is an Fr;B is a Gd;A E B; and m B� Að Þ ¼ 0: Since A is an Fr; A is a Borel set, and hence, byIV, k Að Þ ¼ m Að Þ: Similarly, B is a Borel set, and k Bð Þ ¼ m Bð Þ: Since A;B areBorel sets, B� A is a Borel set, and hence 0 ¼ð Þm B� Að Þ ¼ k B� Að Þ: Now, since

k Eð Þ ¼ k A[ E � Að Þð Þ ¼ k Að Þþ k E � Að Þ ¼ m Að Þþ k E � Að Þ;

and

m Eð Þ ¼ m A[ E � Að Þð Þ ¼ m Að Þþm E � Að Þ;

it suffices to show that k E � Að Þ ¼ m E � Að Þ: Since A E B; we haveE � Að Þ B� Að Þ; and hence

0� k E � Að Þ� k B� Að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0; and 0� m E � Að Þ�m B� Að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0:

It follows that k E � Að Þ ¼ 0; and m E � Að Þ ¼ 0; and thereforek E � Að Þ ¼ m E � Að Þ:

∎)

VII. Problem 1.254 Let l be a positive Borel measure on Rk. Let l be transla-tional invariant. Then there exists a nonnegative real number c such that, for everyBorel set E;

l Eð Þ ¼ c m Eð Þð Þ:(Solution Put c � l Q 0; 1ð Þð Þ � 0ð Þ:


Problem 1:255 If Q a; 12n

� �;Q b; 1

2n� � 2 Xn; where a; b 2 Pn; then

l Q a; 12n

� �� ¼ l Q b; 12n

� �� :

(Solution Since Q a; 12n

� � ¼ Q 0; 12n

� �þ a; and l is translational invariant,

l Q a;12n

� �� ¼ l Q 0;

12n

� �� :

Similarly,

l Q b;12n

� �� ¼ l Q 0;

12n

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ l Q a;

12n

� �� ;

and hence,

l Q a;12n

� �� ¼ l Q b;

12n

� �� :

∎)Now since, for every positive integer n; Q 0; 1ð Þ is the disjoint union of 2nk-many

sets of the form Q a; 12n

� �in Xn; we have

l Q 0; 1ð Þð Þ ¼ l Q 0;12n

� �� þ � � � þ l Q 0;

12n

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

2nk terms|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 2nk l Q 0;

12n

� �� ;

and hence,

l Q 0; 1ð Þð Þ ¼ 2nk l Q 0;12n

� �� :

Since, by VI, m is a translational invariant positive Borel measure on Rk; asabove, we get

m Q 0; 1ð Þð Þ ¼ 2nk m Q 0;12n

� �� :

Also, by III,

2nk m Q 0;12n

� �� ¼ m Q 0; 1ð Þð Þ ¼ 0þ 1ð Þ � 0ð Þ � � � 0þ 1ð Þ � 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

k factors|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1;


and hence

c m Q 0;12n

� �� ¼ l Q 0; 1ð Þð Þð Þ m Q 0;

12n

� ��

¼ 2nk l Q 0;12n

� �� m Q 0;

12n

� ��

¼ 2nk m Q 0;12n

� �� l Q 0;

12n

� ��

¼ 1 � l Q 0;12n

� �� :

Thus, for every positive integer n;

l Q 0;12n

� �� ¼ c m Q 0;

12n

� �� :

It follows that, for every box Q in X1 [X2 [ � � � ; l Qð Þ ¼ c m Qð Þð Þ:Case I: when c 6¼ 0: In this case, for every box Q in X1 [X2 [ � � � ; we have

1c l� �

Qð Þ ¼ m Qð Þ: Since c[ 0; and l is a positive Borel measure on Rk, 1c l is a

positive Borel measure on Rk: Since l is translational invariant, 1c l is translationalinvariant. Now, by IV, for all Borel sets E in Rk; 1

c l� �

Eð Þ ¼ m Eð Þ; and hence, forevery Borel set E;

l Eð Þ ¼ c m Eð Þð Þ:

Case II: when c ¼ 0: Here, it suffices to show that, for every Borel set E;l Eð Þ ¼ 0:

Situation I: when E is in X1 [X2 [ � � �. Here, there exists a positive integer n;and a 2 Pn such that Q a; 1

2n� � ¼ E: Now,

l Eð Þ ¼ l Q a;12n

� �� ¼ l Q 0;

12n

� �þ a

� �¼ l Q 0;

12n

� �� :

Since

0 ¼ c ¼ l Q 0; 1ð Þð Þ ¼ l Q 0;12n

� �� þ � � � þ l Q 0;

12n

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

2nk terms

¼ 2nk l Q 0;12n

� ��

¼ 2nk l Q 0;12n

� �þ a

� �� ¼ 2nk l Q a;

12n

� �� ¼ 2nk l Eð Þð Þ;

we have l Eð Þ ¼ 0:


Situation II: when E is a nonempty open set in Rk: By Note 1.235(IV), thereexist boxes Q1;Q2; . . . in X1 [X2 [ � � � such that Q1;Q2; . . . are pairwise disjoint,and E ¼ Q1 [Q2 [ � � � : It follows, from Situation I, that

LHS ¼ l Eð Þ ¼ l Q1 [Q2 [ � � �ð Þ¼ l Q1ð Þþ l Q2ð Þþ � � � ¼ 0þ 0þ � � �¼ 0 ¼ RHS:

Situation III: when E is any Borel set in Rk: Since E ð ÞRk is a nonempty openset, by Situation II, 0� l Eð Þ� l Rk

� � ¼ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}; and hence, l Eð Þ ¼ 0:

So, in all situations, for every Borel set E; l Eð Þ ¼ 0: ∎)

VIII. Problem 1.256 Let T : Rk ! Rk be a linear transformation. Then, thereexists a nonnegative real number D Tð Þ such that, for every E 2 ℳ;

a. T Eð Þ 2 ℳ;b. m T Eð Þð Þ ¼ D Tð Þð Þ m Eð Þð Þ:

(Solution We first show that, for every Borel set E; T Eð Þ 2 ℳ; andm T Eð Þð Þ ¼ D Tð Þð Þ m Eð Þð Þ:

Case I: when T Rk� � 6¼ Rk: Take D Tð Þ ¼ 0: Since T : Rk ! Rk is a linear

transformation, T Rk� �

is a linear subspace of Rk: Now, since T Rk� � 6¼ Rk; we have

dim T Rk� ��

\k; and hence, m T Rk� �� ¼ 0: Since m T Rk

� �� ¼ 0; T Eð Þ T Rk� �

; by 5, T Eð Þ 2 ℳ; and m T Eð Þð Þ ¼ 0: Now,

LHS ¼ m T Eð Þð Þ ¼ 0 ¼ 0ð Þ m Eð Þð Þ ¼ D Tð Þð Þ m Eð Þð Þ ¼ RHS:

Case II: when T Rk� � ¼ Rk: In this case, the linear transformation T : Rk ! Rk

is 1-1, and onto. Thus T : Rk ! Rk is a linear isomorphism. Since T : Rk ! Rk is alinear transformation, T is continuous. Similarly, T�1 : Rk ! Rk exists, and iscontinuous. Thus T : Rk ! Rk is a homeomorphism. Since T : Rk ! Rk is 1-1,and onto, T as a set function from the power set P Rk

� �to P Rk

� �is 1-1, and onto.

Since T : Rk ! Rk is a homeomorphism, we have T Oð Þ ¼ O; and T�1 Oð Þ ¼ O:

Since T : Rk ! Rk is 1-1, and onto, we have, for every r-algebra B; T Bð Þ is a r-algebra and T�1 Bð Þ is a r-algebra. Clearly,

T Að Þ : T Að Þ is a r-algebra; and O T Að Þf g¼ B : B is a r-algebra; and O Bf g:

Let E be a Borel set. We want to show that T Eð Þ is a Borel set. Since E is a Borelset, we have


E 2 \ A : A is a r-algebra; and O Af g:

We have to prove:

T Eð Þ 2 \ B : B is a r-algebra; and O Bf g:

Since

E 2 \ A : A is a r-algebra; and O Af g;T Eð Þ 2 T \ A : A is a r-algebra; and O Af gð Þ¼ \ T Að Þ : A is a r-algebra; and O Af g¼ \ B : B is a r-algebra; and O Bf g¼ \ T Að Þ : T Að Þ is a r-algebra; and T Oð Þ T Að Þf g¼ \ T Að Þ : T Að Þ is a r-algebra; and O T Að Þf g¼ \ B : B is a r-algebra; and O Bf g:

Thus,

T Eð Þ 2 \ B : B is a r-algebra; and O Bf g:

For every Borel set E; put l Eð Þ � m T Eð Þð Þ: Since, T is 1-1, and onto, and m is apositive measure, l is a positive measure. Since m is translational invariant, and T isa linear transformation, l is translational invariant. Now, by VII, there exists anonnegative real number c such that, for every Borel set E; m T Eð Þð Þ ¼

l Eð Þ ¼ c m Eð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}; and hence, for every Borel set E; m T Eð Þð Þ ¼ c m Eð Þð Þ: Thus, for

every Borel set E;

T Eð Þ 2 ℳ; and m T Eð Þð Þ ¼ D Tð Þð Þ m Eð Þð Þ �ð Þ:

Now, we consider the general case when E 2 ℳ: By 9, there exist sets A and Bsuch that A is an Fr; B is a Gd; A E B; and m B� Að Þ ¼ 0: Since A is an Fr; Ais a Borel set. Similarly, B is a Borel set. Now, from (�),

T Að Þ; T Bð Þ 2 ℳ; m T Að Þð Þ ¼ D Tð Þð Þ m Að Þð Þ; and m T Bð Þð Þ ¼ D Tð Þð Þ m Bð Þð Þ:

Here,

T Að Þ T Eð Þ T Bð Þ; and ℳ 3ð ÞT Bð Þ � T Að Þ ¼ T B� Að Þ:

Since A;B are Borel sets, B� A is a Borel set, and hence, by (�), T B� Að Þ 2ℳ; and


m T Bð Þ � T Að Þð Þ ¼ m T B� Að Þð Þ ¼ D Tð Þð Þ m B� Að Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ D Tð Þð Þ 0ð Þ ¼ 0:

Since

m T Bð Þ � T Að Þð Þ ¼ 0; and T Eð Þ � T Að Þð Þ T Bð Þ � T Að Þð Þ;

by 5, we have T Eð Þ � T Að Þð Þ 2 ℳ; and m T Eð Þ � T Að Þð Þ ¼ 0: SinceT Eð Þ � T Að Þð Þ; T Að Þ 2 ℳ; we have T Eð Þ ¼ð Þ T Eð Þ � T Að Þð Þ [T Að Þ 2 ℳ; andhence T Eð Þ 2 ℳ: Since m B� Að Þ ¼ 0; E � Að Þ is a Borel set, and E � Að Þ B� Að Þ; we have m E � Að Þ ¼ 0: Now,

LHS ¼ m T Eð Þð Þ ¼ m T Eð Þ � T Að Þð Þ [ T Að Þð Þ¼ m T Eð Þ � T Að Þð Þþm T Að Þð Þ ¼ 0þm T Að Þð Þ ¼ m T Að Þð Þ¼ D Tð Þð Þ m Að Þð Þ ¼ D Tð Þð Þ m Eð Þ � m E � Að Þð Þ¼ D Tð Þð Þ m Eð Þ � 0ð Þ ¼ D Tð Þð Þ m Eð Þð Þ ¼ RHS:

∎)

IX. Problem 1.257 Let T : Rk ! Rk be a linear transformation. Then D Tð Þ ¼det Tð Þj j; where D Tð Þ is a nonnegative real number as described in VIII.

(Solution Case I: when T 1; 0; . . .; 0ð Þð Þ; T 0; 1; 0; . . .; 0ð Þ; . . .; T 0; . . .; 0; 1ð Þð Þð Þð Þ isa transposition of 1; 0; . . .; 0ð Þ; 0; 1; 0; . . .; 0ð Þ; . . .; 0; . . .; 0; 1ð Þð Þ: In this case,det Tð Þ ¼ �1; and hence det Tð Þj j ¼ 1: It remains to show that D Tð Þ ¼ 1: By VIII,

m T Q 0; 1ð Þð Þð Þ ¼ D Tð Þð Þ m Q 0; 1ð Þð Þð Þ¼ D Tð Þð Þ m 0; 0þ 1½ Þ � � � 0; 0þ 1½ Þð Þð Þ¼ D Tð Þð Þ m 0; 1½ Þ � � � 0; 1½ Þð Þð Þ¼ D Tð Þð Þ 1� 0ð Þ � � � 1� 0ð Þð Þ ¼ D Tð Þ;

and hence,

m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ m T Q 0; 1ð Þð Þð Þ ¼ D Tð Þ:

Thus, it remains to show that m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ 1: Since

T 1; 0; . . .; 0ð Þð Þ; T 0; 1; 0; . . .; 0ð Þ; . . .; T 0; . . .; 0; 1ð Þð Þð Þð Þ

is a transposition of

1; 0; . . .; 0ð Þ; 0; 1; 0; . . .; 0ð Þ; . . .; 0; . . .; 0; 1ð Þð Þ;


and T : Rk ! Rk is a linear transformation,

T 0; 1½ Þ � � � 0; 1½ Þð Þ ¼ 0; 1½ Þ � � � 0; 1½ Þ:

It follows that

m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ m 0; 1½ Þ � � � 0; 1½ Þð Þ ¼ 1ð Þ;

and hence m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ 1Case II: when T 1; 0; . . .; 0ð Þð Þ ¼ a 1; 0; . . .; 0ð Þ for some real number a; and

T 0; 1; 0; . . .; 0ð Þð Þ ¼ 0; 1; 0; . . .; 0ð Þ; . . .; T 0; . . .; 0; 1ð Þð Þ ¼ 0; . . .; 0; 1ð Þ:In this case, det Tð Þ ¼ a; and hence det Tð Þj j ¼ aj j: It suffices to show that

D Tð Þ ¼ aj j: By VIII,


and hence

det Tð Þj j ¼ aj j ¼ m 0; a½ Þ � � � 0; 1½ Þð Þ or m a; 0½ Þ � � � 0; 1½ Þð Þð Þ¼ m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ m T Q 0; 1ð Þð Þð Þ ¼ D Tð Þ:

Thus, D Tð Þ ¼ aj j:Case III: when T 1; 0; . . .; 0ð Þð Þ ¼ 1; 1; 0; . . .; 0ð Þ; and T 0; 1; 0; . . .; 0ð Þð Þ ¼

0; 1; 0; . . .; 0ð Þ; . . .; T 0; . . .; 0; 1ð Þð Þ ¼ 0; . . .; 0; 1ð Þ: In this case, det Tð Þ ¼ 1; andhence det Tð Þj j ¼ 1: It suffices to show that, D Tð Þ ¼ 1: By VIII,


and hence,

D Tð Þ ¼ m T Q 0; 1ð Þð Þð Þ ¼ m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ:

Thus, it remains to show that m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ 1: Let

S � 0; 1½ Þ � � � 0; 1½ Þð Þ \ T 0; 1½ Þ � � � 0; 1½ Þð Þ:


Now, since

T 1; 0; . . .; 0ð Þð Þ ¼ 1; 1; 0; . . .; 0ð Þ; and T 0; 1; 0; . . .; 0ð Þð Þ¼ 0; 1; 0; . . .; 0ð Þ; . . .; T 0; . . .; 0; 1ð Þð Þ ¼ 0; . . .; 0; 1ð Þ;

we find that 0; 1½ Þ � � � 0; 1½ Þð Þ is the disjoint union of S and

T 0; 1½ Þ � � � 0; 1½ Þð Þ � Sð Þ � 0; 1; 0; . . .; 0ð Þ:

It follows that

1 ¼ð Þm 0; 1½ Þ � � � 0; 1½ Þð Þ ¼ m Sð Þþm T 0; 1½ Þ � � � 0; 1½ Þð Þ � Sð Þ � 0; 1; 0; . . .; 0ð Þð Þð Þ¼ m Sð Þþm T 0; 1½ Þ � � � 0; 1½ Þð Þ � Sð Þ ¼ m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ:

Thus, m T 0; 1½ Þ � � � 0; 1½ Þð Þð Þ ¼ 1:Case IV: when T is any linear transformation. In this case, by linear algebra, we

can write T as a product, say T1T2 � � � Tn of linear transformations Tis, such thateach Ti is either of the type in Case I, or, of the type in Case II, or of the type inCase III. We have to show that D Tð Þ ¼ det Tð Þj j: Since D T1T2 � � � Tnð Þ ¼ D Tð Þ; and

det Tð Þj j ¼ det T1T2 � � � Tnð Þj j ¼ det T1ð Þ � � � det Tnð Þj j¼ det T1ð Þj j � � � det Tnð Þj j;


D Tð Þ ¼ D T1T2 � � � Tnð Þ ¼ det T1ð Þj j � � � det Tnð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is,

D Tð Þ ¼ det T1ð Þj j � � � det Tnð Þj j:

By VIII,

m T2 Q 0; 1ð Þð Þð Þ ¼ D T2ð Þð Þ m Q 0; 1ð Þð Þð Þ ¼ D T2ð Þð Þ 1ð Þ ¼ D T2ð Þ:

By VIII,

m T1T2ð Þ Q 0; 1ð Þð Þð Þ ¼ m T1 T2 Q 0; 1ð Þð Þð Þð Þ ¼ D T1ð Þð Þ m T2 Q 0; 1ð Þð Þð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ D T1ð Þð Þ D T2ð Þð Þ:


Thus,

m T1T2ð Þ Q 0; 1ð Þð Þð Þ ¼ D T1ð Þð Þ D T2ð Þð Þ:

Similarly,

m T1T2 � � � Tnð Þ Q 0; 1ð Þð Þð Þ ¼ D T1ð Þð Þ D T2ð Þð Þ � � � D Tnð Þð Þ:

Since each Ti is either of the type in Case I, or, of the type in Case II, or of thetype in Case III, by Cases I, II, III, for every i ¼ 1; . . .; n; we have D Tið Þ ¼det Tið Þj j: It follows that

D T1T2 � � �Tnð Þ ¼ D T1T2 � � � Tnð Þð Þ 1ð Þ ¼ D T1T2 � � � Tnð Þð Þ m Q 0; 1ð Þð Þð Þ¼ m T1T2 � � � Tnð Þ Q 0; 1ð Þð Þð Þ ¼ D T1ð Þð Þ D T2ð Þð Þ � � � D Tnð Þð Þ¼ det T1ð Þj j � � � det Tnð Þj j

and hence, D T1T2 � � � Tnð Þ ¼ det T1ð Þj j � � � det Tnð Þj j:∎)

Conclusion 1.258 There exists a r-algebra ℳ in Rk that contains all Borel setsin Rk; and there exists a positive measure m on ℳ satisfying the followingconditions:

1. for every compact subset K of Rk; K 2 ℳ; and m Kð Þ\1;2. for every E 2 ℳ; m Eð Þ ¼ inf m Vð Þ : E V ; andV is openf g,3. for every open set V in Rk; V 2 ℳ; and m Vð Þ ¼ sup m Kð Þ : K V ;f

and K is a compact setg,4. for every E 2 ℳ satisfying m Eð Þ\1; m Eð Þ ¼ sup m Kð Þ : K E;f

andK is a compact setg;5. if E 2 ℳ; m Eð Þ ¼ 0; and A E; then A 2 ℳ; that is, m is complete,6. for every f 2 Cc Rk

� �; K fð Þ ¼ R

Rk f dm,7. for every E 2 ℳ; and for every e[ 0; there exist a closed set F; and an open

set V such that F E V ; and m V � Fð Þ\e,8. m is a regular Borel measure,9. for every E 2 ℳ; there exist sets A and B such that A is an Fr; B is a Gd;

A E B; and m B� Að Þ ¼ 0,10. ℳ ¼ E : there exist sets A and B such that A is an Fr; B isf a Gd; A E

B; and m B� Að Þ ¼ 0g; and m is regular,11.

a. m a1; b1ð Þ � � � ak; bkð Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ;b. m a1; b1½ Þ � � � ak; bk½ Þð Þ ¼ b1 � a1ð Þ � � � bk � akð Þ; etc.,

12. if E 2 ℳ; and a 2 Rk; then Eþ að Þ 2 ℳ; and m Eþ að Þ ¼ m Eð Þ; that is, m istranslational invariant,


13. if l is a positive Borel measure on Rk , which is translational invariant, thenthere exists a nonnegative real number c such that, for every Borel set E;l Eð Þ ¼ c m Eð Þð Þ,

14. if T : Rk ! Rk is a linear transformation, then, for every E 2 ℳ; T Eð Þ 2 ℳ;and

m T Eð Þð Þ ¼ det Tð Þj jð Þ m Eð Þð Þ:

Here, members ofℳ are called Lebesgue measurable sets in Rk , and m is calledthe Lebesgue measure on Rk: A subset of Rk; which is not a member ofℳ is calleda non-Lebesgue measurable set in Rk:

1.11 Existence of Non-Lebesgue Measurable Sets

The proof of existence of a non-Lebesgue measurable subset of R is considered as aremarkable advancement in Lebesgue integration theory. In short, we shall provethat, for k ¼ 1; ℳ(P Rk

� �: This establishes that the Lebesgue integral is intrin-

sically different from the Riemann integral. In this section, another beautiful the-orem is Luzin’s theorem.

Note 1.259 We know that R is an additive group, and Q is a subgroup of R: So, bygroup theory (cf. NJ[1], p. 39), Qf g[ Qþ n : n 62 Qf g is a partition of R: By theaxiom of choice, there exists a set E consisting of exactly one element of eachmember of Qf g[ Qþ n : n 62 Qf g:I. Problem 1.260 If r; s are distinct members of Q; then Eþ rð Þ \ Eþ sð Þ ¼ ;:(Solution Let r; s be distinct members of Q: We claim that Eþ rð Þ \ Eþ sð Þ ¼ ;:If not, otherwise, let Eþ rð Þ \ Eþ sð Þ 6¼ ;: We have to arrive at a contradiction.

Since Eþ rð Þ \ Eþ sð Þ 6¼ ;; there exists g 2 Eþ rð Þ \ Eþ sð Þ: It follows thatthere exist f1; f2 2 E such that g ¼ f1 þ r; and g ¼ f2 þ s: Thus, f1 þ r ¼ f2 þ s:Since f1 þ r ¼ f2 þ s; and r 6¼ s; f1 6¼ f2: Since f1 6¼ f2; and f1; f2 2 E; by thedefinition of E; there exist n1; n2 2 R such that f1 2 Qþ n1ð Þ; f2 2 Qþ n2ð Þ; ands� r ¼ð Þ f1 � f2ð Þ 62 Q; and hence, s� rð Þ 62 Q: Since r; s are members of Q;s� rð Þ 2 Q: This is a contradiction. ■)

II. Problem 1.261 [ Eþ rð Þ : r 2 Qf g ¼ R:

(Solution It is clear that [ Eþ rð Þ : r 2 Qf g R: It remains to show that R [ Eþ rð Þ : r 2 Qf g: For this purpose, let us take any n 2 R: We have to show thatn 2 [ Eþ rð Þ : r 2 Qf g:

Since n 2 R; and Qf g[ Qþ n : n 62 Qf g is a partition of R; either n 2 Q orn 2 [ Qþ g : g 62 Qf g:


Case I: when n 2 Q: By the definition of E; E contains exactly one element ofQ; say r0: Thus,

n ¼ r0 þ n� r0ð Þ 2 Eþ n� r0ð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} [ Eþ rð Þ : r 2 Qf g:

It follows that n 2 [ Eþ rð Þ : r 2 Qf g:Case II: when n 2 [ Qþ g : g 62 Qf g: It follows that there exists g0 62 Q such

that n 2 Qþ g0ð Þ: Since n 2 Qþ g0ð Þ; and g0 62 Q; n 62 Q: Since n 2 Qþ g0ð Þ; wehave Qþ g0ð Þ ¼ Qþ nð Þ: By the definition of E; there exists exactly one element,say r0 þ g0ð Þ of Qþ g0; where r0 2 Q: Since n 2 Qþ g0ð Þ; there exists s0 2 Q

such that n ¼ s0 þ g0ð Þ: Here r0 þ n� s0ð Þ ¼ r0 þ g0ð Þ 2 E|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}; so

n 2 Eþ s0 � r0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} [ Eþ rð Þ : r 2 Qf g; and hence, n 2 [ Eþ rð Þ : r 2 Qf g: ■)

From I, and II, Eþ rð Þ : r 2 Qf g is a partition of R:Let A be a Lebesgue measurable subset of R: Suppose that every subset of A is

Lebesgue measurable, and m Að Þ\1: We shall try to show that m Að Þ ¼ 0:Since for every r 2 Q; Eþ rð Þ \Að Þ A; by the assumption on A; for every

r 2 Q; Eþ rð Þ \A is a Lebesgue measurable set. Since Eþ rð Þ : r 2 Qf g is apartition of R; Eþ rð Þ \A : r 2 Qf g is a partition of A: Since

Eþ rð Þ \A : r 2 Qf g

is a partition of A; each Eþ rð Þ \A is Lebesgue measurable, and Q is countable, wehave m Að Þ ¼Pr2Q m Eþ rð Þ \Að Þ: Now, it suffices to show that, for each r 2 Q;

m Eþ rð Þ \Að Þ ¼ 0: For this purpose, let us fix any r0 2 Q: We have to show that

sup m Kð Þ : K Eþ r0ð Þ \A; and K is a compact setf g¼ m Eþ r0ð Þ \Að Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is,

sup m Kð Þ : K Eþ r0ð Þ \A; andK is a compact setf g ¼ 0:

For this purpose, let us take any compact set K satisfying K Eþ r0ð Þ \A|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Eþ r0ð Þ: It suffices to show that m Kð Þ ¼ 0: Since K is a compact subset of R; K isbounded, and hence

[ K þ t : t 2 0; 1½ � \Qf g

is bounded. It follows that

1.11 Existence of Non-Lebesgue Measurable Sets 217

m [ Kþ t : t 2 0; 1½ � \Qf gð Þ\1:

Problem 1.262 Kþ t : t 2 0; 1½ � \Qf g is a countable collection of Lebesguemeasurable subsets of R: Also, they are pairwise disjoint.

(Solution Let s; t 2 0; 1½ � \Q: Let s 6¼ t:We have to show that Kþ sð Þ \ Kþ tð Þ ¼;: Since K Eþ r0ð Þ; we have

Kþ sð Þ Eþ r0 þ sð Þð Þð Þ; and K þ tð Þ Eþ r0 þ tð Þð Þð Þ:

Since s 6¼ t; we have r0 þ sð Þ 6¼ r0 þ tð Þ; and hence, by I,

; Kþ sð Þ \ K þ tð Þ Eþ r0 þ sð Þð Þ \ Eþ r0 þ tð Þð Þ ¼ ;:

Thus, Kþ sð Þ \ Kþ tð Þ ¼ ;: ■)It follows that

1[ð Þm [ Kþ t : t 2 0; 1½ � \Qf gð Þ ¼X

t2 0;1½ � \Q

m K þ tð Þ ¼X

t2 0;1½ � \Q

m Kð Þ;

and hence,P

t2 0;1½ � \Q m Kð Þ\1: This shows that m Kð Þ ¼ 0:

Conclusion 1.263 Let A be a Lebesgue measurable subset of R: Suppose that everysubset of A is Lebesgue measurable, and m Að Þ\1: Then m Að Þ ¼ 0:

Corollary 1.264 Let A be a Lebesgue measurable subset of R such that0\m Að Þ\1: Then there exists a subset B of A such that B is not Lebesguemeasurable set. Hence there exists a non-Lebesgue measurable subset of R:

Proof If not, otherwise, suppose that every subset of A is Lebesgue measurable set.We have to arrive at a contradiction. By the above conclusion, m Að Þ ¼ 0: This is acontradiction. ■

Note 1.265 Let X be a locally compact Hausdorff space. Let K be a positive linearfunctional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179. Letf : X ! 0; 1½ Þ be any measurable function. Let A be a compact subset of X:Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real.

Since A is compact, by Note 1.179(IX), l Að Þ\1:Let us define a function u3 : 0;1½ Þ ! 0;1½ Þ as follows: For every t 2 0;1½ Þ;

u3 tð Þ �

0 if t 2 0; 1 123

� �1 1

23 if t 2 1 123 ; 2 1

23� �

2 123 if t 2 2 1

23 ; 3 123

� �...

26 � 1� � 1

23 if t 2 26 � 1� � 1

23 ; 26 1

23� �

26 123 if t 2 26 1

23 ;1� �

:

8>>>>>>><>>>>>>>:


Since

0; 1 123

� �; 1 1

23; 2 1

23

� �; 2 1

23; 3 1

23

� �; . . .; 26 � 1

� � 123

; 26 123

� �; 26 1

23;1

� �

are Borel sets of real numbers, and

u3 0;1½ Þð Þ ¼ 0; 1 123

; 2 123

; . . .; 26 � 1� � 1

23; 26 1

23

� ��

has only finite-many elements, u3 is a simple measurable function on 0;1½ Þ. If wedraw the graph of u3; the straight line t 7! t; and the straight line t 7! t � 1

23 ; it iseasy to observe that, for every t in 0; 23½ �; t � 1

23� �

\u3 tð Þ� t:Let us define a function u4 : 0;1½ Þ ! 0;1½ Þ as follows: For every t 2 0;1½ Þ;

u4 tð Þ �

0 if t 2 0; 1 124

� �1 1

24 if t 2 1 124 ; 2 1

24� �

2 124 if t 2 2 1

24 ; 3 124

� �...

28 � 1ð Þ 124 if t 2 28 � 1ð Þ 1

24 ; 28 1

24� �

28 124 if t 2 28 1

24 ;1� �

:

8>>>>>>><>>>>>>>:

As above, u4 is a simple measurable function on 0;1½ Þ, and, for every t in0; 24½ �; t � 1

24� �

\u4 tð Þ� t: Similar definitions can be supplied for u1;u2;u5;u6;

etc. If we draw the graphs of u1;u2;u3;u4; . . .; it is easy to observe that for every tin 0;1½ Þ; 0�u1 tð Þ�u2 tð Þ�u3 tð Þ�u4 tð Þ� � � � : Since for every t in 0; 2n½ �;

t � 12n

� �\un tð Þ� t and lim

n!1 t � 12n

� �¼ t;

we have, for every t in 0;1½ Þ; limn!1 un tð Þ ¼ t: Since f : X ! 0; 1½ � is a mea-surable function, and un : 0;1½ Þ ! 0;1½ Þ is a simple measurable mapping, byLemma 1.84, each composite un � fð Þ : X ! 0;1½ Þ is a simple measurable func-tion. Since for every t in 0;1½ Þ;

0�u1 tð Þ�u2 tð Þ�u3 tð Þ�u4 tð Þ� � � � ;

it follows that, for every x in X;

0�u1 f xð Þð Þ�u2 f xð Þð Þ�u3 f xð Þð Þ�u4 f xð Þð Þ� � � � ;

and hence for every x in X;


0� u1 � fð Þ xð Þ� u2 � fð Þ xð Þ� u3 � fð Þ xð Þ� u4 � fð Þ xð Þ� � � � :

Since, for every t in 0;1½ �; limn!1 un tð Þ ¼ t; we have, for every x in X;limn!1 un f xð Þð Þ ¼ f xð Þ: Thus, for every x in X; limn!1 un � fð Þ xð Þ ¼ f xð Þ: Since

u1 � f þ u2 � u1ð Þ � f þ u3 � u2ð Þ � f þ u4 � u3ð Þ � f þ � � �¼ u1 � f þ u2 � f � u1 � fð Þþ u3 � f � u2 � fð Þþ u4 � f � u3 � fð Þþ � � �¼ lim

n!1 u1 � f þ u2 � f � u1 � fð Þþ u3 � f � u2 � fð Þþ � � � þ un � f � un�1 � fð Þð Þ¼ lim

n!1 un � fð Þ ¼ f ;

we have

f ¼ u1 � f þ u2 � u1ð Þ � f þ u3 � u2ð Þ � f þ u4 � u3ð Þ � f þ � � � :

Let us observe that, for every t 2 0;1½ Þ;

22u2

� �tð Þ �

0 if t 2 0; 1 122

� �1 if t 2 1 1

22 ; 2 122

� �2 if t 2 2 1

22 ; 3 122

� �3 if t 2 3 1

22 ; 4 122

� �4 if t 2 4 1

22 ; 5 122

� �...

15 if t 2 15 122 ; 16 1

22� �

16 if t 2 16 122 ;1

� �;

8>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>:

and

23u3

� �tð Þ �

0 if t 2 0; 1 123

� �1 if t 2 1 1

23 ; 2 123

� �2 if t 2 2 1

23 ; 3 123

� �3 if t 2 3 1

23 ; 4 123

� �4 if t 2 4 1

23 ; 5 123

� �5 if t 2 5 1

23 ; 6 123

� �6 if t 2 6 1

23 ; 7 123

� �7 if t 2 7 1

23 ; 8 123

� �8 if t 2 8 1

23 ; 9 123

� �...

8>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>:


Now, since f : X ! 0; 1½ Þ; for every x 2 0;1½ Þ;

22u2

� �f xð Þð Þ �

0 if f xð Þ 2 0; 1 122

� �1 if f xð Þ 2 1 1

22 ; 2 122

� �2 if f xð Þ 2 2 1

22 ; 3 122

� �3 if f xð Þ 2 3 1

22 ; 4 122

� �;

8>>><>>>: and

23u3

� �f xð Þð Þ �

0 if f xð Þ 2 0; 1 123

� �1 if f xð Þ 2 1 1

23 ; 2 123

� �2 if f xð Þ 2 2 1

23 ; 3 123

� �3 if f xð Þ 2 3 1

23 ; 4 123

� �4 if f xð Þ 2 4 1

23 ; 5 123

� �5 if f xð Þ 2 5 1

23 ; 6 123

� �6 if f xð Þ 2 6 1

23 ; 7 123

� �7 if f xð Þ 2 7 1

23 ; 8 123

� �

8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:

and hence,

23u3

� �f xð Þð Þ � 23u2

� �f xð Þð Þ ¼

0� 2 0 if f xð Þ 2 0; 1 123

� �1� 2 0 if f xð Þ 2 1 1

23 ; 2 123

� �2� 2 1 if f xð Þ 2 2 1

23 ; 3 123

� �3� 2 1 if f xð Þ 2 3 1

23 ; 4 123

� �4� 2 2 if f xð Þ 2 4 1

23 ; 5 123

� �5� 2 2 if f xð Þ 2 5 1

23 ; 6 123

� �6� 2 3 if f xð Þ 2 6 1

23 ; 7 123

� �7� 2 3 if f xð Þ 2 7 1

23 ; 8 123

� �

8>>>>>>>>>><>>>>>>>>>>:

¼ 1 if f xð Þ 2 1 123 ; 2 1

23� �[ 3 1

23 ; 4 123

� �[ 5 123 ; 6 1

23� �[ 7 1

23 ; 8 123

� �0 if f xð Þ 2 0; 1 1

23� �[ 2 1

23 ; 3 123

� �[ 4 123 ; 5 1

23� �[ 6 1

23 ; 7 123

� �(

¼ 1 if x 2 f�1 1 123 ; 2 1

23� �[ 3 1

23 ; 4 123

� �[ 5 123 ; 6 1

23� �[ 7 1

23 ; 8 123

� �� 0 if x 2 f �1 1 1

23 ; 2 123

� �[ 3 123 ; 4 1

23� �[ 5 1

23 ; 6 123

� �[ 7 123 ; 8 1

23� �� c

:

(

Thus,

23 u3 � u2ð Þ � f ¼ vf�1 1 1

23;2 1

23

� �[ 3 1

23;4 1

23

� �[ 5 1

23;6 1

23

� �[ 7 1

23;8 1

23

� �� :Similarly, for every n ¼ 2; 3; 4; . . .;

2n un � un�1ð Þ � f ¼ vf�1 1 12n;2 1

2n½ Þ [ 3 12n;4 1

2n½ Þ[ �� [ 2n�1ð Þ 12n;2

n 12n½ Þð Þ:


Since

21u1

� �f xð Þð Þ ¼ 0 if f xð Þ 2 0; 1 1

21� �

1 if f xð Þ 2 1 121 ; 2 1

21� �

(

¼ 1 if x 2 f�1 1 121 ; 2 1

21� ��

0 if x 2 f�1 1 121 ; 2 1

21� �� c

(

¼ vf�1 1 1

21;2 1

21

� �� xð Þ;

we have

21 u1 � fð Þ ¼ vf�1 1 1

21;2 1

21

� �� :Since A is a compact subset of X; and X is open, by Lemma 1.163, there exists

an open set V such that A V �V|fflfflffl{zfflfflffl} X; and �V is compact. Since f : X ! 0; 1½ Þ is

a measurable function, and, for every n ¼ 1; 2; 3; 4; . . .;

1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �

is a Borel subset of 0; 1½ Þ;

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� 2 ℳ:

Since, for every x 2 Ac; f xð Þ ¼ 0; we have Ac f�1 0ð Þ; and hence

f�1 1 12n ; 2 1

2n� �[ 3 1

2n ; 4 12n

� �[ � � � [ 2n � 1ð Þ 12n ; 2

n 12n

� �� f�1 0f gcð Þ ¼ f�1 0ð Þð Þc A:

Thus,

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� A:

It follows that

l f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� l Að Þ \1ð Þ;


and hence

l f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� \1:

Since, for every n ¼ 1; 2; 3; 4; . . .;

l f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� \1;

and

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� 2 ℳ;

by XXX and XVII of Note 1.179, for every n ¼ 1; 2; 3; 4; . . .; there exist a compactset Kn; and an open set ~Vn such that

Kn f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� ~Vn;

and l ~Vn � Kn� �

\ e2n : Since V ; ~Vn are open sets, V \ ~Vn is an open set. Put Vn �

V \ ~Vn: Since

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� A V ;

and

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� ~Vn;

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� V \ ~Vn� � ¼ Vnð Þ:

Thus, for every n ¼ 1; 2; 3; 4; . . .;

Kn f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� Vn:


Since

l Vn � Knð Þ ¼ l V \ ~Vn� �� Kn� �� l ~Vn � Kn

� �\

e2n|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl};

we have l Vn � Knð Þ\ e2n : Since, for every n ¼ 1; 2; 3; 4; . . .; Kn Vn; Kn is

compact, and Vn is open, by Urysohn’s lemma, for every n ¼ 1; 2; 3; 4; . . .; thereexists hn 2 Cc Xð Þ such that Kn � hn � Vn: Since hn � Vn; hn : X ! 0; 1½ � is con-tinuous, and supp hnð Þ Vn: Since hn : X ! 0; 1½ �; by Weierstrass M-test,

g : x 7! 12

h1 xð Þð Þþ 14

h2 xð Þð Þþ 18

h3 xð Þð Þþ � � � � 12þ 1

4þ 1

8þ � � � ¼ 1\1

� �

converges uniformly on X: Since

g : x 7! 12

h1 xð Þð Þþ 14

h2 xð Þð Þþ 18

h3 xð Þð Þþ � � ��

converges uniformly on X; and each hn is continuous, g : X ! R is continuous.

Problem 1.266 supp gð Þ �V :

(Solution Since each hn : X ! 0; 1½ �; for every x 2 X satisfying

12

h1 xð Þð Þþ 14

h2 xð Þð Þþ 18

h3 xð Þð Þþ � � � ¼ g xð Þ 6¼ 0|fflfflfflfflffl{zfflfflfflfflffl};there exists a positive integer n0 such that hn0 xð Þ 6¼ 0; and hence

x 2 supp hn0ð Þ|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Vn0 ¼ V \ ~Vn0

� � V :

Thus, g�1 R� 0f gð Þð Þ V ; and hence supp gð Þ ¼ g�1 R� 0f gð Þð Þ� �� V|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} : ■)

Since, �V is compact, supp gð Þ is closed, supp gð Þ �V ; and X is Hausdorff, wefind that supp gð Þ is compact. Since supp gð Þ is compact, and g : X ! R is contin-uous, we have g 2 Cc Xð Þ: Since g : X ! R is continuous, g is a measurablefunction. Since g is a measurable function, and f is a measurable function, f � gð Þis a measurable function, and hence x : f xð Þ 6¼ g xð Þf g ¼

f � gð Þ�1R� 0f gð Þ 2 ℳ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} : Thus, x : f xð Þ 6¼ g xð Þf g 2 ℳ:


Problem 1.267 l x : f xð Þ 6¼ g xð Þf gð Þ\e:

(Solution Let us fix any positive integer n: Take any x 2 Kn: Since

Kn f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� Vn;

and

2n un � un�1ð Þ � f ¼ vf�1 1 12n;2 1

2n½ Þ [ 3 12n;4 1

2n½ Þ[ �� [ 2n�1ð Þ 12n;2

n 12n½ Þð Þ;

we have 2n un � un�1ð Þ � fð Þ xð Þ ¼ 1: Since, Kn � hn; and x 2 Kn; we havehn xð Þ ¼ 1|fflfflfflfflfflffl{zfflfflfflfflfflffl} ¼ 2n un � un�1ð Þ � fð Þ xð Þ: Thus, 2n un � un�1ð Þ � fð Þ ¼ hn on Kn:

Next, let us take any y 2 Vnð Þc: Since y 2 Vnð Þc; and

f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� Vn;

we have

y 62f�1 1 12n

; 2 12n

� �[ 3 1

2n; 4 1

2n

� �[ � � � [ 2n � 1ð Þ 1

2n; 2n 1

2n

� �� ;

and hence

2n un � un�1ð Þ � fð Þ yð Þ ¼ð Þvf�1 1 12n;2 1

2n½ Þ[ 3 12n;4 1

2n½ Þ[ �� [ 2n�1ð Þ 12n;2

n 12n½ Þð Þ yð Þ ¼ 0:

Thus, 2n un � un�1ð Þ � fð Þ yð Þ ¼ 0: Since hn � Vn; and y 2 Vnð Þc; we havehn yð Þ ¼ 0|fflfflfflfflfflffl{zfflfflfflfflfflffl} ¼ 2n un � un�1ð Þ � fð Þ yð Þ: Thus, 2n un � un�1ð Þ � fð Þ ¼ hn on Vnð Þc:

We have seen that, for every n ¼ 2; 3; 4; . . .; un � un�1ð Þ � f ¼ 12n hn on

Vn � Knð Þc: Clearly, u1 � f ¼ 12 h1 on V1 � K1ð Þc: It follows that, for every

n ¼ 2; 3; 4; . . .;

x : un � un�1ð Þ � fð Þ xð Þ 6¼ 12n

hn xð Þð Þ� �

Vn � Knð Þ;

and

x : u1 � fð Þ xð Þ 6¼ 12h1 xð Þ

� � V1 � K1ð Þ:


Hence,

x : f xð Þ 6¼ g xð Þf g x : f xð Þ 6¼ 12

h1 xð Þð Þþ 14

h2 xð Þð Þþ 18

h3 xð Þð Þþ � � ��

¼ x : u1 � f þ u2 � u1ð Þ � f þ u3 � u2ð Þ � f þ u4 � u3ð Þ � f þ � � �ð Þ xð Þf

6¼ 12

h1 xð Þð Þþ 14

h2 xð Þð Þþ 18

h3 xð Þð Þþ � � ��

[1n¼1 Vn � Knð Þ:

Thus,

x : f xð Þ 6¼ g xð Þf g [1n¼1 Vn � Knð Þ:

It follows that

l x : f xð Þ 6¼ g xð Þf gð Þ� l [1n¼1 Vn � Knð Þ� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} �

X1n¼1

l Vn � Knð Þ\X1n¼1

e2n

¼ e:

Thus,

l x : f xð Þ 6¼ g xð Þf gð Þ\e:

∎)

Conclusion 1.268 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179.Let f : X ! 0; 1½ Þ be any measurable function. Let A be a compact subset of X:Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real number. Thenthere exists g 2 Cc Xð Þ such that


Theorem 1.269 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179.Let f : X ! C be any measurable bounded function. Let A be a compact subset ofX: Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real number.Then there exists g 2 Cc Xð Þ such that


Proof Case I: when f : X ! 0;1½ Þ: Since f is bounded, there exists a positive realnumber c\1 such that cf : X ! 0; 1½ Þ: Since f is a measurable function, cf is ameasurable function. Since for every x 2 Ac; f xð Þ ¼ 0; for every x 2 Ac; cfð Þ xð Þ ¼0: Now, by Conclusion 1.268, there exists g 2 Cc Xð Þ such thatl x : cfð Þ xð Þ 6¼ g xð Þf gð Þ\e: Since g 2 Cc Xð Þ; and c is a positive real number, wehave 1

c g 2 Cc Xð Þ: Since


x : cfð Þ xð Þ 6¼ g xð Þf g ¼ x : f xð Þ 6¼ 1cg

� �xð Þ

� �;

we have

e[ l x : cfð Þ xð Þ 6¼ g xð Þf gð Þ ¼ l x : f xð Þ 6¼ 1cg

� �xð Þ

� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence,

l x : f xð Þ 6¼ 1cg

� �xð Þ

� �� \e:

Case II: when f : X ! R: Since f is a measurable function, f þ : X ! 0;1½ Þ;and f� : X ! 0;1½ Þ are measurable functions. Also, f ¼ f þ � f�: Since for everyx 2 Ac; f xð Þ ¼ 0; we have, for every x 2 Ac; f þ xð Þ ¼ max f xð Þ; 0f g ¼

max f xð Þf g ¼ 0|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}; and hence for every x 2 Ac; f þ xð Þ ¼ 0: Since f is bounded, f þ is

bounded. It follows, from Case I, that there exists g 2 Cc Xð Þ such thatl x : f þ xð Þ 6¼ g xð Þf gð Þ\ e

2 : Similarly, there exists h 2 Cc Xð Þ such thatl x : f� xð Þ 6¼ h xð Þf gð Þ\ e

2 : Since g; h 2 Cc Xð Þ; we have g� hð Þ 2 Cc Xð Þ: Since

x : f xð Þ 6¼ g� hð Þ xð Þf g ¼ x : f þ xð Þ � f� xð Þ 6¼ g xð Þ � h xð Þf g x : f þ xð Þ 6¼ g xð Þf g [ x : f� xð Þ 6¼ h xð Þf g;

we have

l x : f xð Þ 6¼ g� hð Þ xð Þf gð Þ� l x : f þ xð Þ 6¼ g xð Þf g[ x : f� xð Þ 6¼ h xð Þf gð Þ� l x : f þ xð Þ 6¼ g xð Þf gð Þþ l x : f� xð Þ 6¼ h xð Þf gð Þ\ e

2þ e

2¼ e:

Thus,

l x : f xð Þ 6¼ g� hð Þ xð Þf gð Þ\e:

Case III: when f : X ! C: Since f is bounded, Re fð Þ : X ! R; and Im fð Þ :X ! R are bounded functions. Since for every x 2 Ac;


Re fð Þð Þ xð Þð Þþ i Im fð Þð Þ xð Þð Þ ¼ f xð Þ ¼ 0|fflfflfflfflffl{zfflfflfflfflffl};we have, for every x 2 Ac; Re fð Þð Þ xð Þ ¼ 0; and Im fð Þð Þ xð Þ ¼ 0: Now, by Case II,there exists g 2 Cc Xð Þ such that l x : Re fð Þð Þ xð Þ 6¼ g xð Þf gð Þ\ e

2 : Similarly, thereexists h 2 Cc Xð Þ such that l x : Im fð Þð Þ xð Þ 6¼ h xð Þf gð Þ\ e

2 : Since g; h 2 Cc Xð Þ; wehave gþ ihð Þ 2 Cc Xð Þ: Since

x : f xð Þ 6¼ gþ ihð Þ xð Þf g¼ x : Re fð Þð Þ xð Þð Þþ i Im fð Þð Þ xð Þð Þ 6¼ g xð Þð Þþ i h xð Þð Þf g x : Re fð Þð Þ xð Þ 6¼ g xð Þf g[ x : Im fð Þð Þ xð Þ 6¼ h xð Þf g;

we have

l x : f xð Þ 6¼ gþ ihð Þ xð Þf gð Þ� l x : Re fð Þð Þ xð Þ 6¼ g xð Þf g[ x : Im fð Þð Þ xð Þ 6¼ h xð Þf gð Þ� l x : Re fð Þð Þ xð Þ 6¼ g xð Þf gð Þþ l x : Im fð Þð Þ xð Þ 6¼ h xð Þf gð Þ\

e2þ e

2¼ e:

Thus,

l x : f xð Þ 6¼ gþ ihð Þ xð Þf gð Þ\e:

Thus, in all cases, there exists g 2 Cc Xð Þ such that


∎

Theorem 1.270 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179.Let f : X ! C be any bounded measurable function. Let A 2 ℳ satisfyingl Að Þ\1: Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real.Then there exists g 2 Cc Xð Þ such that


Proof Since A 2 ℳ satisfying l Að Þ\1; by Problems 1.216 and 1.203, there exista compact set K; and an open set V such that K A V ; and

l A� Kð Þ� l V � Kð Þ\ e2|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} : Let us define a function f : X ! C as follows: For

every x 2 X;


f xð Þ � f xð Þ if x 2 K0 if x 2 Kc:

�

Problem 1:271 f : X ! C is a measurable function.

(Solution Let G be a nonempty open set in C: We have to show that f�1 Gð Þ 2 ℳ:Here,

f�1 Gð Þ ¼ f��K

� �1Gð Þ

� �[ f

��Kc

� �1Gð Þ

� �¼ f jK

� ��1Gð Þ

� [ 0jKc

� ��1Gð Þ

� ¼ f�1 Gð Þ� �\K� �[ 0jKc

� ��1Gð Þ

� ¼ f�1 Gð Þ� �\K� �[; or f�1 Gð Þ� �\K

� �[ Kcð Þ:

Thus,

f�1 Gð Þ ¼ f�1 Gð Þ� �\K� �

or f�1 Gð Þ� �[ Kcð Þ� �:

It follows that f�1 Gð Þ 2 ℳ: ■)Since f : X ! C is a bounded function, f : X ! C is a bounded function. Now,

by Theorem 1.269, there exists g 2 Cc Xð Þ such that l x : f xð Þ 6¼ g xð Þ� � �\ e

2 :

Clearly, f ¼ f on A� Kð Þc; and hence

x : f xð Þ 6¼ g xð Þf g x : f xð Þ 6¼ g xð Þ� [ A� Kð Þ:

It follows that

l x : f xð Þ 6¼ g xð Þf gð Þ\l x : f xð Þ 6¼ g xð Þ� [ A� Kð Þ� �� l x : f xð Þ 6¼ g xð Þ� � �þ l A� Kð Þ\

e2þ e

2¼ e;

and hence


∎

Theorem 1.272 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179.Let f : X ! C be any measurable function. Let A 2 ℳ satisfying l Að Þ\1:


Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real number. Thenthere exists g 2 Cc Xð Þ such that


Proof Since f : X ! C is a measurable function, fj j : X ! 0;1½ Þ is a measurablefunction, and hence

x : 0\ fj j xð Þf g; x : 1\ fj j xð Þf g; x : 2\ fj j xð Þf g; x : 3\ fj j xð Þf g; . . .

are in ℳ: Also,

x : 0\ fj j xð Þf g x : 1\ fj j xð Þf g x : 2\ fj j xð Þf g x : 3\ fj j xð Þf g � � � :

Since for every x 2 Ac; f xð Þ ¼ 0; we have Ac f�1 0ð Þ; and hence

x : 0\ fj j xð Þf g f�1 C� 0f gð Þ ¼ f�1 0ð Þ� �c A|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} :It follows that

l x : 0\ fj j xð Þf gð Þ� l Að Þ \1ð Þ:

Thus, l x : 0\ fj j xð Þf gð Þ\1: Now, by Lemma 1.99,

limn!1 l x : n\ fj j xð Þf gð Þ ¼ l x : 0\ fj j xð Þf g\ x : 1\ fj j xð Þf g\ x : 2\ fj j xð Þf g\ � � �ð Þ ¼ l ;ð Þ ¼ 0:

Thus, limn!1 l x : n\ fj j xð Þf gð Þ ¼ 0: There exists a positive integer N such

that l x : N\ fj j xð Þf gð Þ\ e2 : Clearly, 1� v x:N\ fj j xð Þf g

� � f is a bounded measur-

able function. Since f ¼ 0 on Ac; 1� v x:N\ fj j xð Þf g�

� f ¼ 0 on Ac: Now, by

Theorem 1.270, there exists g 2 Cc Xð Þ such that

l x : 1� v x:N\ fj j xð Þf g�

� f�

xð Þ 6¼ g xð Þn o�

\e2:

Since

x : f xð Þ 6¼ g xð Þf g x : 1� v x:N\ fj j xð Þf g�

� f�

xð Þ 6¼ g xð Þn o

[ x : N\ fj j xð Þf g;


we have

l x : f xð Þ 6¼ g xð Þf gð Þ� l x : 1� v x:N\ fj j xð Þf g�

� f�

xð Þ 6¼ g xð Þn o

[ x : N\ fj j xð Þf g�

� l x : 1� v x:N\ fj j xð Þf g�

� f�

xð Þ 6¼ g xð Þn o�

þ l x : N\ fj j xð Þf gð Þ\ e2þ e

2¼ e;

and hence l x : f xð Þ 6¼ g xð Þf gð Þ\e: ■

Theorem 1.273 Let X be a locally compact Hausdorff space. Let K be a positivelinear functional on Cc Xð Þ: Let ℳ and l be the same as constructed in Note 1.179.Let f : X ! C be any measurable function. Let A 2 ℳ satisfying l Að Þ\1:Suppose that, for every x 2 Ac; f xð Þ ¼ 0: Let e be any positive real number. Thenthere exists g 2 Cc Xð Þ such that

1. sup g xð Þj j : x 2 Xf g� sup f xð Þj j : x 2 Xf g; and 2. l x : f xð Þ 6¼ g xð Þf gð Þ\e:

Proof By Theorem 1.272, there exists g1 2 Cc Xð Þ such thatl x : f xð Þ 6¼ g1 xð Þf gð Þ\e: If sup f xð Þj j : x 2 Xf g ¼ 1; then the theorem is triviallytrue. If sup f xð Þj j : x 2 Xf g ¼ 0; then f ¼ 0; and hence, 0 serves the purpose of g:So, it remains to consider the case when 0\sup f yð Þj j : x 2 Xf g\1:

Let us define g : X ! C as follows: For every x 2 X;

g xð Þ � g1 xð Þ if g1 xð Þ 2 D 0; sup f yð Þj j : y 2 Xf g½ �sup f yð Þj j : y 2 Xf gð Þ g1 xð Þ

g1 xð Þj j if g1 xð Þ 62 D 0; sup f yð Þj j : y 2 Xf gð Þ:

(

Since g1 2 Cc Xð Þ; g1 : X ! C is continuous, and hence, by the definition of g;g : X ! C is continuous. Also, by the definition of g; g1ð Þ�1 0ð Þ g�1 0ð Þ: It fol-lows that g�1 0ð Þð Þc g1ð Þ�1 0ð Þ

� c; and hence

supp gð Þ ¼ g�1 0ð Þ� �c� �� g1ð Þ�1 0ð Þ� c� �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ supp g1ð Þ:

Since g1 2 Cc Xð Þ; supp g1ð Þ is compact. Since supp g1ð Þ is compact, supp gð Þ supp g1ð Þ; and supp gð Þ is closed, supp gð Þ is compact. Since supp gð Þ is compact, andg : X ! C is continuous, g 2 Cc Xð Þ:

For 1: Let us take any x 2 X: It suffices to show thatg xð Þj j � sup f yð Þj j : y 2 Xf g: By the definition of g; it is clear that, in all cases,g xð Þj j � sup f yð Þj j : y 2 Xf gð Þ:For 2: Clearly, ran fð Þ D 0; sup f yð Þj j : y 2 Xf g½ �: It follows that, if

sup f yð Þj j : y 2 Xf g\ g1 xð Þj j; then g1 xð Þ 62 D 0; sup f yð Þj j : y 2 Xf g½ � ran fð Þð Þ;and hence g1 xð Þ 62 ran fð Þ: Thus, i sup f yð Þj j : y 2 Xf g\ g1 xð Þj j; then f xð Þ 6¼ g1 xð Þ:It follows that


x : f xð Þ 6¼ g xð Þf g¼ x : g1 xð Þ 2 D 0; sup f yð Þj j : y 2 Xf g½ �; and f xð Þ 6¼ g xð Þf g[ x : g1 xð Þ 62 D 0; sup f yð Þj j : y 2 Xf g½ �; and f xð Þ 6¼ g xð Þf g x : f xð Þ 6¼ g1 xð Þf g[ x : g1 xð Þ 62 D 0; sup f yð Þj j : y 2 Xf g½ �; and f xð Þ 6¼ g xð Þf g x : f xð Þ 6¼ g1 xð Þf g[ x : g1 xð Þ 62 D 0; sup f yð Þj j : y 2 Xf g½ �f g x : f xð Þ 6¼ g1 xð Þf g[ x : f xð Þ 6¼ g1 xð Þf g ¼ x : f xð Þ 6¼ g1 xð Þf g;

and hence

l x : f xð Þ 6¼ g xð Þf gð Þ� l x : f xð Þ 6¼ g1 xð Þf gð Þ \eð Þ:

Thus, l x : f xð Þ 6¼ g xð Þf gð Þ\e: ■Theorem 1.273, known as Luzin’s theorem, is due to N. N. Luzin (09.12.1883 –

25.02.1950, Soviet). He is known for his work in descriptive set theory. He alsofound a relationship between point-set topology and mathematical analysis.

Note 1.274

Problem 1.275 Let X be a non-empty set. Let ℳ be a r-algebra in X: Let ℳ be aninfinite set. Show that

a. X is an infinite set,b. there exists a sequence B1;B2; . . .f g of members in ℳ such that

;( � � �(B2(B1(X and, for each n; A\Bnf gA2ℳ is infinite. (Here, ( stands for‘is a proper subset of’.)

Problem 1:276 a. X is an infinite set.

(Solution If not, otherwise, let X be a finite set. We have to arrive at a contra-diction. Since X is a finite set, the largest r-algebra in X (that is, P Xð Þ) is finite, andhence all r-algebras in X are finite. This contradicts the assumption. ■)

Since ℳ is an infinite set, ℳ� ;;Xf g is an infinite set. Now, let us fix anyC 2 ℳ� ;;Xf g: It follows that Cc 2 ℳ� ;;Xf g:Problem 1:277 A\Cf gA2ℳ is an infinite collection or A\Ccf gA2ℳ is an infinitecollection.

(Solution If not, otherwise, let A\Cf gA2ℳ and A\Ccf gA2ℳ be finite collections.We have to arrive at a contradiction. It follows that A\Cð Þ [ A\Ccð Þf gA2ℳ is afinite collection. Now since,

ℳ ¼ Af gA2ℳ¼ A\Xf gA2ℳ¼ A\ C [Ccð Þf gA2ℳ¼ A\Cð Þ [ A\Ccð Þf gA2ℳℳ is finite, a contradiction. ■)Put B0 � X: Since ℳ is an infinite set, and


A\B0f gA2ℳ¼ A\Xf gA2ℳ¼ Af gA2ℳ¼ ℳ;

A\B0f gA2ℳ is an infinite set.Case I: when A\Cf gA2ℳ is an infinite set: Put B1 � C: Since B1 ¼

C 2 ℳ� ;;Xf g|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}; we have ;(B1(X|fflfflfflfflffl{zfflfflfflfflffl} ¼ B0: Thus, ;(B1(B0: Since

A\Cf gA2ℳ is infinite; and B1 ¼ C; A\B1f gA2ℳ is infinite:Case II: when A\Ccf gA2ℳ is infinite: Put B1 � Cc: Since B1 ¼

Cc 2 ℳ� ;;Xf g|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}; we have ;(B1(X|fflfflfflfflffl{zfflfflfflfflffl} ¼ B0: Thus, ;(B1(B0: Since

A\Ccf gA2ℳ is infinite; and B1 ¼ Cc; A\B1f gA2ℳ is infinite:Thus, in all cases, B1 2 ℳ� ;;Xf g; ;(B1(B0; and A\B1f gA2ℳ is infinite:Since A\B1f gA2ℳ is infinite, A\B1f gA2ℳ� ;;X;B1; B1ð Þcf g is an infinite set.

It follows that there exists D 2 ℳ such that D\B1 62 ;;X;B1; B1ð Þcf g: It followsthat ;(D\B1(B1; and D\B1ð Þc2 ℳ� ;;X;B1; B1ð Þcf g:Problem 1:278 A\ D\B1ð Þf gA2ℳ is infinite or A\ Dc \B1ð Þf gA2ℳ is infinite.

(Solution If not, otherwise, let A\ D\B1ð Þf gA2ℳ and A\ Dc \B1ð Þf gA2ℳ befinite. We have to arrive at a contradiction. It follows thatA\ D\B1ð Þð Þ [ A\ Dc \B1ð Þð Þf gA2ℳ is finite. Now since,

A\B1f gA2ℳ¼ A\B1ð Þ \Xf gA2ℳ¼ A\B1ð Þ \ D[Dcð Þf gA2ℳ¼ A\ D\B1ð Þð Þ [ A\ Dc \B1ð Þð Þf gA2ℳ;

A\B1f gA2ℳ is finite. This is a contradiction. ■)Case I: when A\ D\B1ð Þf gA2ℳ is infinite. Put B2 � D\B1: Now, since

D\B1 62 ;;X;B1; B1ð Þcf g; we have ;(D\B1(B1; and hence, ;(B2(B1: SinceA\B2f gA2ℳ¼� �

A\ D\B1ð Þf gA2ℳ is infinite, A\B2f gA2ℳ is infinite. Since;(B2(B1(X; we have B2 2 ℳ� ;;Xf g:

Case II: when A\ Dc \B1ð Þf gA2ℳ is infinite. Put B2 � Dc \B1: Now, sinceD\B1 62 ;;X;B1; B1ð Þcf g; ;(D\B1(B1; and hence ;( Dc \B1ð Þ(B1: Thisshows that ;(B2(B1(X: Since A\B2f gA2ℳ¼� �

A\ Dc \B1ð Þf gA2ℳ is infinite,A\B2f gA2ℳ is infinite. Since ;(B2(X; we have B2 2 ℳ� ;;Xf g:Thus, in all cases, B2 2 ℳ� ;;Xf g; ;(B2(B1(B0; and

A\B2f gA2ℳ is infinite:Continuing as above, we get a sequence B1;B2; . . .f g of members in ℳ such

that ;( � � �(B2(B1(X and, for each positive integer n; A\Bnf gA2ℳ is infinite.■)


Problem 1.279 Does there exist an infinite r-algebra that has only countable-manyelements?

(Solution No.

Explanation Let X be a non-empty set. Let ℳ be a r-algebra in X: Let ℳ be aninfinite set. It suffices to show that ℳ is uncountable. By the Problem 1.275, thereexists a sequence B1;B2; . . .f g of members in ℳ such that ;( � � �(B2(B1(X: Itfollows that B1 � B2;B2 � B3;B3 � B4; . . .f g. is a sequence of nonempty membersin ℳ such that each pair of members in B1 � B2;B2 � B3;B3 � B4; . . .f g aredisjoint.

Put Cn � Bn � Bnþ 1 n ¼ 1; 2; . . .ð Þ: Thus C1;C2;C3; . . .f g is a sequence ofmembers in ℳ such that each Cn 6¼ ;; and i 6¼ j ) Ci \Cj ¼ ;� �

:

Problem 1:280 The collection C of all sequences n1; n2; n3; . . .f g of positiveintegers satisfying n1\n2\n3\ � � � is uncountable.(Solution If not, otherwise, let C be countable. We have to arrive at a contradiction.Since C is countable, all the elements of C can be arranged in a sequence, say,

n11; n12; n13; . . .f g;n21; n22; n23; . . .f g;n31; n32; n33; . . .f g;

..

.

It is clear that

n12;max n12; n23f gþ 1;max n12; n23; n34f gþ 2; . . .f g 2 C:

so,

n12;max n12; n23f gþ 1;max n12; n23; n34f gþ 2; � � �f g 6¼ n11; n12; n13; . . .f g;n12;max n12; n23f gþ 1;max n12; n23; n34f gþ 2; � � �f g 6¼ n21; n22; n23; . . .f g;n12;max n12; n23f gþ 1;max n12; n23; n34f gþ 2; � � �f g 6¼ n31; n32; n33; . . .f g; etc:

This contradicts the fact that

C ¼ n11; n12; n13; . . .f g; n21; n22; n23; . . .f g; n31; n32; n33; . . .f g; � � �f g:

∎)Let us consider the map

f : n1; n2; n3; . . .f g 7!Cn1 [Cn2 [Cn3 [ � � � 2 ℳð Þ

from C to ℳ:


Problem 1:281 f is 1-1.

(Solution Let n1; n2; n3; . . .f g and m1;m2;m3; . . .f g be distinct members of C: Wehave to show that

Cn1 [Cn2 [Cn3 [ � � � 6¼ Cm1 [Cm2 [Cm3 [ � � � :

Since n1; n2; n3; . . .f g and m1;m2;m3; . . .f g are distinct, there exists a smallestpositive integer j such that nj 6¼ mj: For simplicity, let j ¼ 2: It follows that n1 ¼m1: Since n2 6¼ m2; we have n2\m2 or m2\n2:

Case I: when n2\m2: Since, n2 6¼ m2; we have Cn2 \Cm2 ¼ ;; Cn2 6¼ ;; andCm2 6¼ ;: Since Cn2 6¼ ;; there exists a 2 Cn2 : Since m1 ¼ n1\n2\m2\m3\ � � � ;we have n2 6¼ mj j ¼ 1; 2; 3; . . .ð Þ; and hence Cn2 \Cmj ¼ ; j ¼ 1; 2; 3; . . .ð Þ: SinceCn2 \Cmj ¼ ; j ¼ 1; 2; 3; . . .ð Þ; and

a 2 Cn2 Cn1 [Cn2 [Cn3 [ � � �ð Þ;

we have a 62 Cm1 [Cm2 [Cm3 [ � � � ; and hence

Cn1 [Cn2 [Cn3 [ � � � 6¼ Cm1 [Cm2 [Cm3 [ � � � :

Case II: when m2\n2: Similar to Case I,

Cn1 [Cn2 [Cn3 [ � � � 6¼ Cm1 [Cm2 [Cm3 [ � � � :

Thus, in all cases,

Cn1 [Cn2 [Cn3 [ � � � 6¼ Cm1 [Cm2 [Cm3 [ � � � :

∎Since the map f : C ! ℳ is 1-1, and C is uncountable, ℳ is uncountable. ■)

Conclusion 1.282 There does not exist an infinite r-algebra that hascountable-many elements.

Exercises

1:1 Show that, for each positive integer p, the sum of the seriesP1

n¼01n!ð Þp is

irrational.1:2 Suppose that ℳ is an infinite collection of sets of real numbers. Let ℳ be

countable. Show that ℳ is not a r-algebra.1:3 Suppose that A is a set of real numbers such that every subset of A is Lebesgue

measurable. Show that, if the Lebesgue measure of A is finite, then theLebesgue measure of A is zero.

1:4 Let X be a locally compact Hausdorff space. Let K be a positive linearfunctional on Cc Xð Þ: Show that there exists a r-algebraℳ in X that containsall Borel sets in X; and there exists a positive measure l on ℳ such that


a. if E 2 ℳ; l Eð Þ ¼ 0; and A E; then A 2 ℳ;

b. for every f 2 Cc Xð Þ satisfying f : X ! R; K fð Þ� RX f dl:

1:5 Let f 2 L1 �1; 1½ �ð Þ such thatR 1�1 f tð Þdt

�� ¼ R 1�1 f tð Þj jdt: Show that there exists

a complex number a such that af ¼ fj j a.e. on �1; 1½ �.1:6 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let f : X ! �1;1½ �;

and g : X ! �1;1½ � be measurable functions. Show that

x : f xð Þþ g xð Þ� 1f g 2 ℳ:

1:7 Let X be any nonempty set. Letℳ be a r-algebra in X: For every n ¼ 1; 2; . . .;let fn : X ! �1;1½ � be a measurable function. Show thatlim infn!1 fnð Þ lim supn!1 fnð Þ : X ! �1;1½ � is a measurable function.

1:8 Let X be a nonempty set, ℳ be an algebra, and l : ℳ ! 0;1½ � be a finitelyadditive set function. The function

l� : A 7! infX1n¼1

l Anð Þ : eachAn 2 S; andA [1n¼1An

( )

from power set P Xð Þ to 0;1½ � is called the outer measure of l: Let A be thecollection of all sets A such that, for every subset B of X;

l� Bð Þ ¼ l� B\Að Þþ l� B\ Acð Þð Þ:

Show that A is a r-algebra.1:9 For every a; bð Þ; c; dð Þ 2 R2; define

q a; bð Þ; c; dð Þð Þ � a� cj j if b ¼ da� cj j þ 2 if b 6¼ d:

�

Show that R2; q� �

is a metric space. Further, the topology induced by themetric q makes R2 a locally compact space.

1:10 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ ! 0;1½ �be a positive measure on ℳ: For each n ¼ 1; 2; 3; . . .; let fn : X ! 0;1½ � be ameasurable function. Let E be a member of ℳ: Suppose that, for everyn ¼ 1; 2; 3; . . .; f2n�1 � 1� vE and f2n � vE: Show that

ZX

lim infn!1 fn

� dl\ lim inf

n!1

ZX

fndl

0@

1A:


Chapter 2Lp-Spaces

The notion of convex function is more stringent than that of continuous function. Itis the source of many interesting inequalities in real analysis. Here we shall studyLp-space as an example of Banach space, and L2-space as an example of Hilbertspace. The theories of Banach spaces and Hilbert spaces pervade in several areas ofmathematics. In a sense, convexity, subspace, orthogonality and completeness aregeometric properties of Hilbert space. Further, many analytic properties oftrigonometric series are more natural in the framework of Banach space. In thischapter, first of all, we introduce convex functions. Next we prove the Riesz-Fischertheorem, and that Lp-space is a Banach space. Finally, we derive some properties ofBanach algebra.

2.1 Convex Functions

Many important inequalities have their origin in convex functions.

Note 2.1

Definition Let a; b 2 �1;1½ �: Let a\b: Let u : a; bð Þ ! R: If, for every x; y 2a; bð Þ; and, for every k 2 0; 1½ �;

u 1� kð Þxþ kyð Þ� 1� kð Þ u xð Þð Þþ k u yð Þð Þ;

then we say that u is convex.Let a; b 2 �1;1½ �: Let a\b: Let u : a; bð Þ ! R be convex. Let s; t; u 2

a; bð Þ: Let s\t\u:

Problem 2.2 u tð Þ�u sð Þt�s � u uð Þ�u tð Þ

u�t :


237




(Solution Since s\ t\ u; we have 0\ t � sð Þ\ u� sð Þ; and hence 0\ t�sð Þu�sð Þ\1:

Now, since u : a; bð Þ ! R is convex, we have

u tð Þ ¼ ut u� sð Þu� s

� �¼ u

u� tð Þsþ t � sð Þuu� s

� �¼ u 1� t � sð Þ

u� sð Þ� �

sþ t � sð Þu� sð Þ u

� �� 1� t � sð Þ

u� sð Þ� �

u sð Þð Þþ t � sð Þu� sð Þ u uð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ u� tð Þu� sð Þ

� �u sð Þð Þþ t � sð Þ

u� sð Þ u uð Þð Þ ¼ u� tð Þ u sð Þð Þþ t � sð Þ u uð Þð Þu� sð Þ ;

and hence

u� tð Þ u tð Þð Þþ t � sð Þ u tð Þð Þ ¼ u� sð Þ u tð Þð Þ� u� tð Þ u sð Þð Þþ t � sð Þ u uð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :It follows that

u� tð Þ u tð Þ � u sð Þð Þ� t � sð Þ u uð Þ � u tð Þð Þ;

and hence

u tð Þ � u sð Þt � s

� u uð Þ � u tð Þu� t

: ∎)

Let a; b 2 �1;1½ �: Let a\b: Let u : ða; bÞ ! R be a function. Suppose that,for every s; t; u 2 a; bð Þ satisfying s\t\u;



:

Problem 2.3 u is convex.

(Solution For this purpose, let us take any x; y 2 a; bð Þ; and k 2 0; 1½ �: We have toshow that

u 1� kð Þxþ kyð Þ� 1� kð Þ u xð Þð Þþ k u yð Þð Þ:

Case I: when k ¼ 0: In this case,

u 1� kð Þxþ kyð Þ ¼ u 1� 0ð Þxþ 0yð Þ ¼ u xð Þ� 1� 0ð Þ u xð Þð Þþ 0 u yð Þð Þ¼ 1� kð Þ u xð Þð Þþ k u yð Þð Þ;

238 2 Lp-Spaces

and hence,


Case II: when k ¼ 1: This case is similar to the case I.Case III: when 0\k\1:Situation I: when x ¼ y: Here

u 1� kð Þxþ kyð Þ ¼ u 1� kð Þxþ kxð Þ¼ u xð Þ� 1� kð Þ u xð Þð Þþ k u xð Þð Þ¼ 1� kð Þ u xð Þð Þþ k u yð Þð Þ;

so,


Situation II: when x\y: Here,

1� kð Þxþ kyð Þ � x ¼ k y� xð Þ:

Since 0\k\1; and x\y; we have

1� kð Þxþ kyð Þ � x ¼ k y� xð Þ[ 0|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl};and hence, x\ 1� kð Þxþ ky: Similarly, 1� kð Þxþ ky\y: Thus, 1� kð Þxþ kyð Þ 2x; yð Þ: Since x; y 2 a; bð Þ; and x\y; we have

1� kð Þxþ kyð Þ 2ð Þ x; yð Þ � a; bð Þ;

and hence 1� kð Þxþ kyð Þ 2 a; bð Þ: Since x; 1� kð Þxþ kyð Þ; y 2 a; bð Þ andx\ 1� kð Þxþ kyð Þ\y; by the given condition, we have

u 1� kð Þxþ kyð Þ � u xð Þk y� xð Þ ¼ u 1� kð Þxþ kyð Þ � u xð Þ

1� kð Þxþ kyð Þ � x� u yð Þ � u 1� kð Þxþ kyð Þ

y� 1� kð Þxþ kyð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ u yð Þ � u 1� kð Þxþ kyð Þ

1� kð Þ y� xð Þ ;

and hence

1� kð Þ u 1� kð Þxþ kyð Þ � u xð Þð Þ� k u yð Þ � u 1� kð Þxþ kyð Þð Þ:

2.1 Convex Functions 239

It follows that


Situation III: when y\x: This situation is similar to situation II.Thus, in all cases

u 1� kð Þxþ kyð Þ� 1� kð Þ u xð Þð Þþ k u yð Þð Þ: ∎)

Let a; b 2 �1;1½ �: Let a\b: Let u : a; bð Þ ! R be convex. Let u be differ-entiable. Let s; t 2 a; bð Þ: Let s\t:

Problem 2.4 u0 sð Þ�u0 tð Þ:(Solution Let us take any u; v 2 a; bð Þ such that s\u\v\t: Since u : a; bð Þ ! Ris convex, we have

u uð Þ � u sð Þu� s

� u vð Þ � u uð Þv� u

� u tð Þ � u vð Þt � v

;

and hence

u0 sð Þ ¼ limu!s

u uð Þ � u sð Þu� s

� u tð Þ � u vð Þt � v|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

It follows that

u0 sð Þ� limv!t

u tð Þ � u vð Þt � v|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ u0 tð Þ:

and hence u0 sð Þ�u0 tð Þ: ∎)Let a; b 2 �1;1½ �: Let a\b: Let u : a; bð Þ ! R be differentiable. Suppose

that, for every s; t 2 a; bð Þ satisfying s\t; u0 sð Þ�u0 tð Þ:Problem 2.5 u is convex.

(Solution For this purpose, let us take any s; t; u 2 a; bð Þ such that s\t\u: Itsuffices to show that



:

By the mean value theorem, there exists v 2 s; tð Þ such that

u0 vð Þ ¼ u tð Þ � u sð Þt � s

:

240 2 Lp-Spaces

Similarly, there exists w 2 t; uð Þ such that

u0 wð Þ ¼ u uð Þ � u tð Þu� t

:

Since v 2 s; tð Þ; w 2 t; uð Þ and s\t\u; v\w; and hence by assumption


¼ u0 vð Þ�u0 wð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ u uð Þ � u tð Þu� t

;

It follows that



:

∎)

Problem 2.6 The exponential function t 7! et from �1;1ð Þ to 0;1ð Þ is a dif-ferentiable convex function.

(Solution SincedðetÞdt

¼ et; and t 7! et is an increasing function, by Problem 2.5,

t 7! et is a convex function. ∎)Let a; b 2 �1;1½ �: Let a\b: Let u : ða; bÞ ! R be convex.

Problem 2.7 u is continuous.

(Solution Let us fix any c 2 a; bð Þ: Let us take any u; v 2 a; bð Þ such that u\c\v:It suffices to show that

limu!c

u uð Þ ¼ u cð Þ ¼ limv!c

u vð Þ:

Here, u : ða; bÞ ! R is convex, u; c; v 2 a; bð Þ; and u\c\v; so

u uð Þ ¼ u cð Þ � u vð Þ � u cð Þv� c

� �c� uð Þ;

and hence

limu!c

u uð Þ ¼ limu!c

u cð Þ � u vð Þ � u cð Þv� c

� �c� uð Þ

� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ u cð Þ � u vð Þ � u cð Þ

v� c

� �c� cð Þ ¼ u cð Þ:

Thus, limu!c u uð Þ ¼ u cð Þ: Similarly, limv!c u vð Þ ¼ u cð Þ: ∎)


Let X be any nonempty set. Letℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be apositive measure on ℳ: Let l Xð Þ ¼ 1: Let a; b 2 �1;1ð Þ satisfying a\b: Letf : X ! a; bð Þ be a measurable function. Let f 2 L1 lð Þ: Let c; d be real numberssuch that c\d; and a; b½ � � c; dð Þ: Let u : c; dð Þ ! R be a convex function.

Since u : c; dð Þ ! R is convex, u : c; dð Þ ! R is continuous: Now, sincea; b½ � � c; dð Þ; u a; b½ �ð Þ is compact,and hence, u a; b½ �ð Þ is bounded. Since u a; b½ �ð Þis bounded, and f : X ! a; bð Þ; u � f is bounded, and hence, u � fð Þþ ; u � fð Þ� arebounded nonnegative functions. Since a; b½ � � c; dð Þ; u : c; dð Þ ! R is continuous,and f : X ! a; bð Þ is a measurable function, by Lemma 1.61, u � f : X ! R is ameasurable function and hence u � fð Þþ ; u � fð Þ� are measurable functions. Sinceu � fð Þþ is a nonnegative bounded measurable function on X, and l Xð Þ ¼ 1;RX u � fð Þþ dl exists, and is a real number. Similarly,

RX u � fð Þ�dl exists, and is a

real number. Thus,RX u � fð Þdl exists and is a real number. Since f 2 L1 lð Þ; we

haveRX fj jdl 2 0;1½ Þ; and RX f dl 2 R:

Since for every x 2 X; a\ f xð Þ\ b; we have

a ¼ a � 1 ¼ a l Xð Þð Þ�ZX

f dl� b l Xð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ b � 1 ¼ b;

and hence ZX

f dl 2 a; b½ �:

Problem 2.8RX f dl 6¼ a:

(Solution If not, otherwise, letZX

f dl ¼ a

|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}¼ a � 1 ¼ a � l Xð Þð Þ ¼

ZX

a dl:

We have to arrive at a contradiction. SinceRX f dl ¼ RX a dl; we haveZ

X

f � að Þdl ¼ 0:

Now, by Lemma 1.151,

f � að Þ ¼ 0 a:e:;

242 2 Lp-Spaces

and hence l x : f xð Þ 6¼ af g ¼ 0ð Þ: Since for every x 2 X; a\f xð Þ; we havex : f xð Þ ¼ af g ¼ ;; and hence

1 ¼ l Xð Þ ¼ l Xð Þ � 0 ¼ l Xð Þ � l x : f xð Þ 6¼ af gð Þ¼ l X� x : f xð Þ 6¼ af gð Þ ¼ l x : f xð Þ ¼ af gð Þ ¼ l ;ð Þ ¼ 0:

Thus, we get a contradiction. ∎)Similarly,

RX f dl 6¼ b: Since,

RX f dl 2 a; b½ �; RX f dl 6¼ a; and

RX f dl 6¼ b; we

haveRX f dl 2 a; bð Þ: Since,RX f dl 2 a; bð Þ � c; dð Þð Þ; and u : c; dð Þ ! R is con-

vex, uRX f dl

� �is a real number. Thus, u

RX f dl

� �;RX u � fð Þdl are real numbers.

Problem 2.9 uRX f dl

� �� RX u � fð Þdl:

(Solution Put t0 �RX f dl: Since

RX f dl 2 a; bð Þ; we have t0 2 a; bð Þ: We have to

show that

u t0ð Þ�ZX

u � fð Þdl:

Since t0 2 a; bð Þ; and u : a; bð Þ ! R is convex, for every s 2 a; t0ð Þ; we have

u t0ð Þ � u sð Þt0 � s

� u t0 þ b2

� �� u t0ð Þt0 þ b2 � t0

;

and hence,

u t0 þ b2

� �� u t0ð Þt0 þ b2 � t0

is an upper bound of


: s 2 a; t0ð Þ� �

:

It follows that

supu t0ð Þ � u sð Þ

t0 � s: s 2 a; t0ð Þ

� �exists, and is a real number, say b: Thus, for every s 2 a; t0ð Þ;


� b:


It follows that for every s 2 a; t0ð Þ;

u t0ð Þ � u sð Þ� b � t0 � sð Þ:

Let us take any s 2 a; t0ð Þ and u 2 t0; bð Þ: Now, since u : ða; bÞ ! R is convex,we have


� u uð Þ � u t0ð Þu� t0

;

and hence for every u 2 t0; bð Þ;u uð Þ � u t0ð Þ

u� t0

is an upper bound of


: s 2 a; t0ð Þ� �

:

It follows that for every u 2 t0; bð Þ;

b ¼ð Þsup u t0ð Þ � u sð Þt0 � s

: s 2 a; t0ð Þ� �

� u uð Þ � u t0ð Þu� t0

;

and hence for every u 2 t0; bð Þ; b � u� t0ð Þ�u uð Þ � u t0ð Þ: Thus, for everyu 2 t0; bð Þ;

u t0ð Þ � u uð Þ� b � t0 � uð Þ:

Since for every s 2 t0; bð Þ;u t0ð Þ � u sð Þ� b � t0 � sð Þ;

and, for every s 2 a; t0ð Þ;u t0ð Þ � u sð Þ� b � t0 � sð Þ;

it follows that for every s 2 a; bð Þ;u t0ð Þ � u sð Þ� b � t0 � sð Þ:

Let us take any x 2 X: Since f : X ! ða; bÞ we have f xð Þ 2 a; bð Þ; and hence

u t0ð Þ � u f xð Þð Þ� b � t0 � f xð Þð Þ:

244 2 Lp-Spaces

Thus, for every x 2 X;

u t0ð Þ � b � t0ð Þþ bfð Þ xð Þ ¼ u t0ð Þ � b � t0 � f xð Þð Þ�u f xð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ u � fð Þ xð Þ;

and hence

0�u � f � u t0ð Þþ b � t0 � bf :

Now, by Lemma 1.112,

0 ¼ZX

0 dl�ZX

u � f � u t0ð Þþ b � t0 � bfð Þdl

¼ZX

u � fð Þdlþ �u t0ð Þþ b � t0ð Þ l Xð Þð Þ � bZX

f dl

¼ZX

u � fð Þdlþ �u t0ð Þþ b � t0ð Þ 1ð Þ � bZX

f dl

¼ZX

u � fð Þdlþ �u t0ð Þþ b � t0ð Þ � bt0 ¼ZX

u � fð Þdl� u t0ð Þ;

so,

0�ZX

u � fð Þdl� u t0ð Þ:

Thus, u t0ð Þ� RX u � fð Þdl: ∎)

Conclusion 2.10 Let X be any nonempty set. Let ℳ be a r-algebra in X. Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let l Xð Þ ¼ 1: Let a; b 2 �1;1ð Þsatisfying a\b: Let f : X ! a; bð Þ be a measurable function. Let f 2 L1 lð Þ: Letc; d be real numbers such that c\d and a; b½ � � c; dð Þ: Let u : ðc; dÞ ! R be aconvex function. Then

uZX

f dl

0@ 1A�ZX

u � fð Þdl:

This conclusion, known as the Jensen’s inequality, is due to J. L. Jensen(08.05.1859–05.03.1925, Danish). He is mostly renowned for his famous Jensen’sinequality. He also proved Jensen’s formula in complex analysis.


a. If we take the exponential function t 7! et for u; the Jensen’s inequality takes theform

eðRX f dlÞ �

ZX

e f dl:

b. Let us take any finite set p1; p2; . . .; pnf g for X; and let us define l :

p1; p2; . . .; pnf g ! 0; 1½ � as l pkf gð Þ � 1n for every k ¼ 1; . . .; n: Let a1; . . .; an be

any real numbers. Now, let us define f : p1; p2; . . .; pnf g ! R as f pkð Þ � ak forevery k ¼ 1; . . .; n: The Jensen’s inequality in (a) takes the form

e a1ð Þ1nþ a2ð Þ1nþ �� þ anð Þ1nð Þ � ea1ð Þ 1nþ ea2ð Þ 1

nþ � � � þ eanð Þ 1

n;

that is

ea1 þ �� þ an

nð Þ � ea1 þ � � � þ ean

n:

Here, let us put xk � eak [ 0ð Þ for every k ¼ 1; . . .; n: We get

x1 � � � xnð Þ1n ¼ eln x1��xnð Þ1n� �

¼ e1n ln x1��xnð Þð Þ

¼ eln x1 þ �� þ ln xn

nð Þ � x1 þ � � � þ xnn

:

Thus, for every x1; . . .; xn 2 0;1ð Þ;

x1 � � � xnð Þ1n � x1 þ � � � þ xnn

:

In short, for positive real numbers, G.M.�A.M.

c. Let us take any finite set p1; p2; . . .; pnf g for X. Let m1; . . .; mn be any non-negative real numbers such that

m1 þ � � � þ mn ¼ 1:

Let us define l : fp1; p2; . . .; png ! ½0; 1� as l pkf gð Þ � mk for every k ¼1; . . .; n: Let a1; . . .; an be any real numbers. Now, let us define f :p1; p2; . . .; pnf g ! R as f pkð Þ � ak for every k ¼ 1; . . .; n: The Jensen’s inequality

in (a) takes the form

246 2 Lp-Spaces

e a1ð Þm1 þ a2ð Þm2 þ �� þ anð Þmnð Þ � ea1ð Þm1 þ ea2ð Þm2 þ � � � þ eanð Þmn;

that is

e a1m1 þ �� þ anmnð Þ � ea1ð Þm1 þ ea2ð Þm2 þ � � � þ eanð Þmn:

Here, let us put xk � eak [ 0ð Þ for every k ¼ 1; . . .; n: We get

x1ð Þm1 � � � xnð Þmn ¼ e ln x1ð Þm1�� xnð Þmnð Þð Þ

¼ e ln x1ð Þm1 þ �� þ ln xnð Þmnð Þ � x1m1 þ � � � þ xnmn|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, for every x1; . . .; xn 2 0;1ð Þ; and for every nonnegative real numbers

m1; . . .; mn satisfying

m1 þ � � � þ mn ¼ 1;

we have

x1ð Þm1 � � � xnð Þmn � x1m1 þ � � � þ xnmn:

Note 2.11

Definition Let p; q 2 0;1ð Þ: Let pq ¼ pþ q that is; 1p þ 1q ¼ 1

: Clearly, p; q 2

1;1ð Þ: Here, we say that p and q are pair of conjugate exponents. Clearly, 2 and 2are pair of conjugate exponents. Also, 3 and 3

2 are pair of conjugate exponents.Suppose that p and q are pair of conjugate exponents. Let X be any nonempty

set. Let ℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be a positive measure on ℳ:Let f : X ! 0;1½ �; and g : X ! 0;1½ � be measurable functions.

Since p and q are pair of conjugate exponents, we have 1\p; 1\q; and pq ¼pþ q: Since f : X ! 0;1½ � is a measurable function, and t 7! tp from 0;1½ � to0;1½ � is continuous, their composite f p : x 7! f xð Þð Þp from X to 0;1½ � is a mea-surable function. It follows that

RX f

pdl exists, andRX f

pdl� � 2 0;1½ �: Now, since

t 7! t1p is a function from 0;1½ � to 0;1½ �;

ZX

f pdl

0@ 1A1p

2 0;1½ �:

Similarly, ZX

gqdl

0@ 1A1q

2 0;1½ �:


It follows that the product

ZX

f pdl

0@ 1A1p Z

X

gqdl

0@ 1A1q

2 0;1½ �:

Since f : X ! 0;1½ �; and g : X ! 0;1½ � are measurable functions, their pro-duct f � gð Þ : X ! 0;1½ � is a measurable function, and hence

RX f � gð Þdl exists andR

X f � gð Þdl� � 2 0;1½ �: Thus,

ZX

f � gð Þdl0@ 1A;

ZX

f pdl

0@ 1A1p Z

X

gqdl

0@ 1A1q

2 0;1½ �:

Problem 2.12RX f � gð Þdl� �� R

X fpdl

� �1pRX g

qdl� �1

q ð Þ(Solution If either

RX f

pdl ¼ 1; orRX g

qdl ¼ 1; then ð Þ is trivially true.IfRX f

pdl ¼ 0; then, by Lemma 1.151, f p ¼ 0 a.e. on X: It follows thatl x : f pð Þ xð Þ 6¼ 0f gð Þ ¼ 0: Since

x : f pð Þ xð Þ 6¼ 0f g ¼ x : f xð Þð Þp 6¼ 0f g ¼ x : f xð Þ 6¼ 0f g � x : f xð Þð Þ g xð Þð Þ 6¼ 0f g¼ x : f � gð Þ xð Þ 6¼ 0f g;

we have

0�ð Þl x : f � gð Þ xð Þ 6¼ 0f gð Þ� l x : f pð Þ xð Þ 6¼ 0f gð Þ ¼ 0ð Þ;

and hence

l x : f � gð Þ xð Þ 6¼ 0f gð Þ ¼ 0:

It follows that f � g ¼ 0 a.e. on X; and hence,RX f � gð Þdl ¼ 0: SinceR

X f � gð Þdl ¼ 0; ð Þ holds. Similarly, ifRX g

qdl ¼ 0; then ð Þ holds.The case when

RX f

pdl;RX g

qdl 2 0;1ð Þ; is the only case that remains to beverified. Here, let us put

A �ZX

f pdl

0@ 1A1p

; and B �ZX

gqdl

0@ 1A1q

:

248 2 Lp-Spaces

Now, since

Ap ¼ZX

f pdl 2 0;1ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

;

and p 2 1;1ð Þ; we have A 2 0;1ð Þ: Similarly, B 2 0;1ð Þ: We have to show that

ZX

f � gð Þdl0@ 1A�AB;

that is ZX

1Af

� �� 1

Bg

� �� dl ¼ 1

AB

ZX

f � gð Þdl� 1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is ZX

1Af

� �� 1

Bg

� �� dl� 1:

Since f : X ! 0;1½ �; and A 2 0;1ð Þ; we have, for every x 2 X;0� 1

A f xð Þ�1: Similarly, for every x 2 X; 0� 1B g xð Þ�1:

Let us fix any x 2 X such that 0\ 1A f xð Þ\1; and 0\ 1

B g xð Þ\1: Since0\ 1

A f xð Þ\1; and t 7! et is a mapping from �1;1ð Þ onto 0;1ð Þ; there existss1 2 �1;1ð Þ such that es1 ¼ 1

A f xð Þ: Since s1 2 �1;1ð Þ; and p 2 1;1ð Þ; wehave ps1 2 �1;1ð Þ: Put s � ps1: Thus, s 2 �1;1ð Þ: Also, es

1p ¼ 1

A f xð Þ:Similarly, there exists t 2 �1;1ð Þ such that et

1q ¼ 1

B g xð Þ: Since 1p þ 1

q ¼ 1; and

exponential is a convex function, we have

1Af xð Þ � 1

Bg xð Þ ¼ es

1p � et1q

¼ e s1pþ t1qð Þ � esð Þ 1pþ etð Þ 1

q|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1

Af xð Þ

� �p1pþ 1

Bg xð Þ

� �q1q


Hence, for every x 2 X satisfying 0\ 1A f xð Þ\1 and 0\ 1

B g xð Þ\1; we have

1Af xð Þ � 1

Bg xð Þ� 1

Af xð Þ

� �p1pþ 1

Bg xð Þ

� �q1q:

It follows that, for every x 2 X;

1AB

f � gð Þ� �

xð Þ ¼ 1Af xð Þ � 1

Bg xð Þ� 1

Af xð Þ

� �p1pþ 1

Bg xð Þ

� �q1q|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ 1p1Ap f

p þ 1q1Bq g

q

� �xð Þ;

and hence,

1AB

f � gð Þ� 1p1Ap

f p þ 1q1Bq

gq� �

:

Now, since A;B 2 0;1ð Þ; we have

0�ð Þf � g� 1p

BAp�1 f

p þ 1q

ABq�1 g

q:

On integration, we getZX

f � gð Þdl�ZX

1p

BAp�1 f

p þ 1q

ABq�1 g

q

� �dl

¼ 1p

BAp�1

ZX

f pdlþ 1q

ABq�1

ZX

gqdl

¼ 1p

BAp�1 A

p þ 1q

ABq�1 B

q

¼ AB1pþ 1

q

� �¼ AB 1ð Þ ¼ AB:

Thus, ZX

f � gð Þdl�AB:∎)

250 2 Lp-Spaces

Conclusion 2.13 Suppose that p and q are pair of conjugate exponents. Let X beany nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be a positivemeasure on ℳ: Let f : X ! 0;1½ �; and g : X ! 0;1½ � be measurable functions.Then Z

X

f � gð Þdl0@ 1A�

ZX

f pdl

0@ 1A1p Z

X

gqdl

0@ 1A1q

:

This conclusion, known as the Holder’s inequality, is due to L. O. Holder(22.12.1859–29.08.1937, German). He worked on the convergence of Fourierseries. Later he became interested in group theory, and proved the uniqueness offactor group in a composition series.

Suppose that p 2 1;1ð Þ: Let X be any nonempty set. Letℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! 0;1½ �; and g :X ! 0;1½ � be measurable functions.

Put q � pp�1 : Thus, p and q are a pair of conjugate exponents, and 1\q: Since

f : X ! 0;1½ � is a measurable function, and t 7! tp from 0;1½ � to 0;1½ � is con-tinuous, their composite f p : x 7! f xð Þð Þp from X to 0;1½ � is a measurable function.

It follows thatRX f

pdl exists, andRX f

pdl� � 2 0;1½ �: Now, since t 7! t

1p is a

function from 0;1½ � to 0;1½ �; we have

ZX

f pdl

0@ 1A1p

2 0;1½ �:

Similarly,

ZX

gpdl

0@ 1A1p

2 0;1½ �:

It follows that

ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

0B@1CA 2 0;1½ �:

Since f : X ! 0;1½ �; and g : X ! 0;1½ � are measurable functions, f þ gð Þ :X ! 0;1½ � is a measurable function. Since f þ gð Þ : X ! 0;1½ � is a measurablefunction, and t 7! tp from 0;1½ � to 0;1½ � is continuous, their composite


f þ gð Þp: x 7! f xð Þþ g xð Þð Þp

from X to 0;1½ � is a measurable function. It follows thatRX f þ gð Þpdl exists, andR

X f þ gð Þpdl� � 2 0;1½ �: Thus,

ZX

f þ gð Þpdl0@ 1A;

ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

0B@1CA 2 0;1½ �:

Problem 2.14RX f þ gð Þpdl� �1

p � RX f

pdl� �1

p þ RX g

pdl� �1

p ð Þ(Solution If either

RX f

pdl ¼ 1; orRX g

qdl ¼ 1; then ð Þ is trivially true. IfRX f þ gð Þpdl ¼ 0; then ð Þ is trivially true. When

RX f

pdl;RX g

qdl 2 0;1½ Þ; andRX f þ gð Þpdl 2 0;1ð � is the only case that remained to verify.Since t 7! tp is a differentiable function from 0;1ð Þ to 0;1ð Þ; and its derivative

t 7! ptp�1 is an increasing function, t 7! tp is a convex function. It follows that

12p

f þ gð Þp¼ 1� 12

� �f þ 1

2g

� �p

� 1� 12

� �f pð Þþ 1

2gpð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼

12

f p þ gpð Þ;

and hence

f þ gð Þp � 2p�1 f p þ gpð Þ:

On integration, we get

0\ð ÞZX

f þ gð Þpdl� 2p�1ZX

f pdlþZX

gpdl

0@ 1A \1ð Þ:

Thus, ZX

f þ gð Þpdl 2 0;1ð Þ;

and hence

ZX

f þ gð Þpdl0@ 1A1

q

2 0;1ð Þ:

252 2 Lp-Spaces

Since,

f þ gð Þp¼ f þ gð Þ � f þ gð Þp�1¼ f � f þ gð Þp�1 þ g � f þ gð Þp�1;

we haveZX

f þ gð Þpdl ¼ZX

f � f þ gð Þp�1 þ g � f þ gð Þp�1

dl

¼ZX

f � f þ gð Þp�1

dlþZX

g � f þ gð Þp�1

dl

�ZX

f pdl

0@ 1A1p Z

X

f þ gð Þp�1 q

dl

0@ 1A1q

þZX


dl

¼ZX

f pdl

0@ 1A1p Z

X

f þ gð Þpq�qdl

0@ 1A1q

þZX


dl

¼ZX

f pdl

0@ 1A1p Z

X


q

þZX


dl

�ZX

f pdl

0@ 1A1p Z

X


q

þZX

gpdl

0@ 1A1p Z

X

f þ gð Þp�1 q

dl

0@ 1A1q

¼ZX

f pdl

0@ 1A1p Z

X


q

þZX

gpdl

0@ 1A1p Z

X

f þ gð Þpq�qdl

0@ 1A1q

¼ZX

f pdl

0@ 1A1p Z

X


q

þZX

gpdl

0@ 1A1p Z

X


q

¼ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

0B@1CA Z

X


q

;

and hence

ZX

f þ gð Þpdl�ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

0B@1CA Z

X


q

:


Now, since

ZX


q

2 0;1ð Þ;

we have

ZX


p

¼ZX

f þ gð Þpdl0@ 1A1�1

q

�ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

ZX


p

�ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

:

∎)

Conclusion 2.15 Suppose that p 2 1;1ð Þ: Let X be any nonempty set. Let ℳ be ar-algebra in X: Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X !0;1½ �; and g : X ! 0;1½ � be measurable functions. Then

ZX


p

�ZX

f pdl

0@ 1A1p

þZX

gpdl

0@ 1A1p

:

This conclusion, known as theMinkowski’s inequality, is due to H. Minkowski(22.06.1864–12.01.1909, German). He was Einstein’s former mathematics profes-sor. He made numerous contributions in number theory and in the theory of relativity.

Definition Let p 2 0;1ð Þ: Let X be any nonempty set. Let ℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure onℳ: Let f : X ! C be any measurablefunction.

It follows that fj j : x 7! f xð Þj j from X to 0;1½ Þ is a measurable function. Sincefj j : X ! 0;1½ Þ is a measurable function, and t 7! tp from 0;1½ Þ to 0;1½ Þ is acontinuous function, their composite fj jp: x 7! f xð Þj jp from X to 0;1½ Þ is a mea-surable function, and hence

RX fj jpdl exists and

RX fj jpdl� � 2 0;1½ �: Now, since

t 7! t1p is a function from 0;1½ � to 0;1½ �; we have

ZX

fj jpdl0@ 1A1

p

2 0;1½ �:

254 2 Lp-Spaces

Notation The collection of all measurable functions f : X ! C for which

ZX

fj jpdl0@ 1A1

p

6¼ 1

is denoted by Lp lð Þ: Thus, Lp lð Þ is the collection of all measurable functionsf : X ! C for which

0�ZX

fj jpdl0@ 1A1

p

\1:

Definition If f 2 Lp lð Þ; then the nonnegative real numberRX fj jpdl� �1

p is denotedby kf kp; and is called the Lp-norm of f :

Definition Let E be any subset of X: If

l Eð Þ �1 if E is an infinite set0 if E ¼ ;n if E is a finite set, and n is the number of elements in E;

8<:then l : P Xð Þ ! 0;1½ � is a measure on X: This l is called the counting measureon X: If l is a counting measure on X; then Lp lð Þ is denoted by ‘p Xð Þ: If X iscountable, then ‘p Xð Þ is denoted by ‘p:

An element of ‘p may be regarded as a sequence of complex numbers. Thus, ‘p

is the collection of those sequences nnf g of complex numbers for which

n1j jp�1þ n2j jp�1þ � � �ð Þ\1:

Also,

nnf gk kp¼ n1j jp þ n2j jp þ � � �ð Þ1p:

Notation If l is the Lebesgue measure on Rk; then Lp lð Þ is denoted by Lp Rk� �

:

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure onℳ: Let g : X ! 0;1½ � be any measurable function.Since g : X ! 0;1½ � is a measurable function, and for every real a� 0; a;1ð �

is open in 0;1½ �; we have, for every real a� 0;

g�1 a;1ð �ð Þ 2 ℳ:


Observe that, if

a : a 2 0;1½ Þ�; and l g�1 a;1ð �ð Þ� � ¼ 0� � 6¼ ;;

then 0 is a lower bound of

a : a 2 0;1½ Þ�; and l g�1 a;1ð �ð Þ� � ¼ 0� �

;

and hence

inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

exists and

0� inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

\1:

Suppose that

a : a 2 0;1½ Þ�; and l g�1 a;1ð �ð Þ� � ¼ 0� � 6¼ ;:

Put

b � inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

:

Problem 2.16

b 2 a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

:

(Solution Since

0� inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

\1;

and

b ¼ inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

;

we have 0� b\1; and hence b 2 0;1½ Þ: It remains to show thatl g�1 b;1ð �ð Þð Þ ¼ 0: Since

b ¼ inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

;

for every positive integer n; there exists an 2 0;1½ Þ such that l g�1 an;1ð �ð Þð Þ ¼ 0;and an\bþ 1

n : It follows that, for every positive integer n;

l g�1 bþ 1n;1

� � �� ¼ 0;

256 2 Lp-Spaces

and hence

l g�1 b;1ð �ð Þ� � ¼ l g�1 [1n¼1 bþ 1

n;1

� � �� ¼ l [1

n¼1g�1 bþ 1

n;1

� � �� ¼ lim

n!1 l g�1 bþ 1n;1

� � �� ¼ 0:

Thus, l g�1 b;1ð �ð Þð Þ ¼ 0: ∎)By Problem 2.16, it follows that, if

a : a 2 0;1½ Þ�; and l g�1 a;1ð �ð Þ� � ¼ 0� � 6¼ ;;

then

inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

¼ min a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þ� � ¼ 0� �

:

We define

b � inf a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� �

if a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� � 6¼ ;

1 if a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� � ¼ ;:

�

We can also write

b � min a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� �

if a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� � 6¼ ;

1 if a : a 2 0;1½ Þ; and l g�1 a;1ð �ð Þð Þ ¼ 0� � ¼ ;:

(

Thus, b 2 0;1½ �: Here, b is called the essential supremum of g:

Definition Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let f : X ! C be any measurable function. Itfollows that fj j : x 7! f xð Þj j from X to 0;1½ Þ is a measurable function. Thus, theessential supremum of fj j is in 0;1½ �:

The collection of all measurable functions f : X ! C for which the essentialsupremum of fj j is different from 1 is denoted by L1 lð Þ: If f 2 L1 lð Þ; then theessential supremum of fj j is denoted by fk k1; and is called the L1-norm of f :

Thus, L1 lð Þ is the collection of all measurable functions f : X ! C for which0� fk k1\1:

Let X be any nonempty set. Letℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be apositive measure onℳ: Let f : X ! C be any measurable function. Let f 2 L1 lð Þ:Let k 2 0;1½ Þ: Let fk k1 � k:


Since f : X ! C is a measurable function, fj j : x 7! f xð Þj j from X to 0;1½ Þ is ameasurable function. It follows that fj j�1 k;1ð �ð Þ 2 ℳ:

Problem 2.17 lðfx : f xð Þj j�kgÞ

(Solution Here, fk k1 � k; and k 2 0;1½ Þ; so fk k16¼ 1; and hence

fa : a 2 0;1½ Þ; and lð fj j�1 a;1ð �ð ÞÞ ¼ 0g 6¼ ;;

and

k�ð Þ fk k1¼ minfa : a 2 0;1½ Þ; and lð fj j�1 a;1ð �ð ÞÞ ¼ 0g:

It follows that

l x : f xð Þj j�k

� �� ¼ l x : k\ f xð Þj jf gð Þ

¼ l fj j�1 k;1ð �ð Þ

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence lðfx : f xð Þj j�kgÞ ¼ 0: ∎)

Conclusion 2.18 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! C be any measurablefunction. Let f 2 L1 lð Þ: Let k 2 0;1½ Þ: Let fk k1 � k: Then f xð Þj j � k holds a.e.on X: Also, f xð Þj j � fk k1 holds a.e. on X:

Let X be any nonempty set. Letℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be apositive measure onℳ: Let f : X ! C be any measurable function. Let k 2 0;1½ Þ:Let f xð Þj j � k holds a.e. on X; that is,

l fj j�1 k;1ð �ð Þ

¼ l x : k\ f xð Þj jf gð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Problem 2.19 fk k1 � k:

(Solution Since l fj j�1 k;1ð �ð Þ

¼ 0; and k 2 0;1½ Þ;

k 2 a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0n o

;

and hence

a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0n o

6¼ ;:

258 2 Lp-Spaces

It follows that f 2 L1 lð Þ; and

fk k1¼ min a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0n o

� kð Þ;

and hence

fk k1 � k: ∎)

Conclusion 2.20 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! C be any measurablefunction. Let k 2 0;1½ Þ: Let f xð Þj j � k holds a.e. on X: Then f 2 L1 lð Þ; andfk k1 � k:

Definition The members of L1 lð Þ are called essentially bounded measurablefunctions on X:

Notation If l is the Lebesgue measure on Rk; then L1 lð Þ is denoted by L1 Rk� �

:

If l : P Xð Þ ! 0;1½ � is a counting measure on X; then L1 lð Þ is denoted by‘1 Xð Þ: If X is countable, then ‘1 Xð Þ is denoted by ‘1: An element of ‘1 may beregarded as a sequence of complex numbers.

Thus, ‘1 Xð Þ is the collection of all functions f : X ! C for which there existsa 2 0;1½ Þ such that

countingmeasure of x : a\ f xð Þj jf gð Þ ¼ l fj j�1 a;1ð �ð Þ

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is ‘1 Xð Þ is the collection of all functions f : X ! C for which there existsa 2 0;1½ Þ such that x : a\ f xð Þj jf g ¼ ;; that is ‘1 Xð Þ is the collection of allfunctions f : X ! C for which there exists a 2 0;1½ Þ such that for all x 2 X;f ðxÞj j � a. In short, ‘1 Xð Þ is the collection of all measurable bounded functionsf : X ! C: Also,

fk k1¼ min a : a 2 0;1½ Þ; and for all x 2 X; f xð Þj j � af g:

Further, ‘1 is the collection of those sequences nnf g of complex numbersfor which there exists a 2 0;1½ Þ such that for all positive integers n; nnj j � a:In short, ‘1 is the collection of all bounded sequences of complex numbers.Also,

nnf gk k1¼ min a : a 2 0;1½ Þ; and for all positive integer n; nnj j � af g:


2.2 The Lp-Spaces

Here we shall assume that X is any measure space, and l is a positive measure.

Lemma 2.21 Suppose that p and q are pair of conjugate exponents. Thus, p; q 21;1ð Þ: Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ !0;1½ � be a positive measure on ℳ: Let f : X ! C; and g : X ! C be measurablefunctions. Let f 2 Lp lð Þ; and g 2 Lq lð Þ: Then,1. f � gð Þ 2 L1 lð Þ;2. f � gk k1 � fk kp gk kq:Proof

Problem 2:22 f � gð Þ 2 L1 lð Þ:(Solution Since f 2 Lp lð Þ; f : X ! C is a measurable function for which

0�ZX

fj jpdl0@ 1A1

p

\1:

Similarly, g : X ! C is a measurable function for which

0�ZX

gj jqdl0@ 1A1

q

\1:

Since

ZX

fj jpdl0@ 1A1

p

;

ZX

gj jqdl0@ 1A1

q

2 0;1½ Þ;

we have

ZX

fj jpdl0@ 1A1

p

�ZX

gj jqdl0@ 1A1

q

0B@1CA 2 0;1½ Þ:

Since f : X ! C and g : X ! C are measurable functions, their product f � gð Þ :X ! C is a measurable function. It suffices to show thatZ

X

f :gj jdl\1:

260 2 Lp-Spaces

Since f : X ! C is a measurable function, fj j : X ! 0;1½ Þ is a measurablefunction. Similarly, gj j : X ! 0;1½ Þ is a measurable function. Now, by Holder’sinequality,

ZX

f � gj jdl ¼ZX

fj j � gj jð Þdl0@ 1A�

ZX

fj jpdl0@ 1A1

p ZX

gj jqdl0@ 1A1

q

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\1;

and hence ZX

f � gj jdl\1∎)

1: By Problem 2.22, f � gð Þ 2 L1 lð Þ:2: We have seen above that

f � gk k1¼ZX

f � gj jdl�ZX

fj jpdl0@ 1A1

p ZX

gj jqdl0@ 1A1

q

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ fk kp gk kq;

so f � gk k1 � fk kp gk kq: ∎

Lemma 2.23 Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l :ℳ ! 0;1½ � be a positive measure on ℳ: Let f : X ! C; and g : X ! C bemeasurable functions. Let f 2 L1 lð Þ; and g 2 L1 lð Þ: Then,1. f � gð Þ 2 L1 lð Þ;2. f � gk k1 � fk k1 gk k1:

Proof Since f : X ! C is a measurable function, fj j : X ! 0;1½ Þ is a measurablefunction. Similarly, gj j : X ! 0;1½ Þ is a measurable function. Since fj j : X !0;1½ Þ; and gj j : X ! 0;1½ Þ are measurable functions, their productf � gj j ¼ð Þ fj j gj j : X ! 0;1½ Þ is a measurable function, and hence f � gj j : X !0;1½ Þ is a measurable function.Since f 2 L1 lð Þ; we have

RX fj jdl\1: Since, g 2 L1 lð Þ; we have

0� gk k1\1: Since gk k1\1; we have

a : a 2 0;1½ Þ; and l gj j�1 a;1ð �ð Þ

¼ 0n o

6¼ ;

2.2 The Lp-Spaces 261

and

1[ð Þ gk k1¼ min a : a 2 0;1½ Þ; and l gj j�1 a;1ð �ð Þ

¼ 0n o

:

Thus, there exists a 2 0;1½ Þ such that

l x : a\ g xð Þj jf gð Þ ¼ 0; and gk k1 � a:

Since gk k1 � a; and a 2 0;1½ Þ; by Conclusion 2.18, g xð Þj j � a holds a.e. on X;and hence

f � gð Þ xð Þj j � f xð Þj ja holds a.e onX:

It follows that ZX

f � gj jdl�ZX

fj jadl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ aZX

fj jdl\1;

and hence ZX

f � gj jdl\1:

Now, it suffices to show that

ZX

f � gj jdl�ZX

fj jdl0@ 1A min b : b 2 0;1½ Þ; and l gj j�1 b;1ð �ð Þ

¼ 0

n o :

For this purpose, let us fix any a 2 0;1½ Þ satisfying l gj j�1 a;1ð �ð Þ

¼ 0: It is

enough to show that ZX

f � gj jdl� aZX

fj jdl0@ 1A:

Since a 2 0;1½ Þ satisfying l gj j�1 a;1ð �ð Þ

¼ 0; we have gk k1 � a: Since

gk k1 � a; and a 2 0;1½ Þ; by Conclusion 2.18, g xð Þj j � a holds a.e. on X; andhence f � gð Þ xð Þj j � f xð Þj ja holds a.e. on X: It follows thatZ

X

f � gj jdl�ZX

fj jadl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ aZX

fj jdl;

and henceRX f � gj jdl� a

RX fj jdl: ∎

262 2 Lp-Spaces

Definition By the conjugate exponent of 1 we shall mean 1: By the conjugateexponent of 1 we shall mean 1.

From the above discussion, we get the following

Lemma 2.24 Let p; q 2 1;1½ �: Suppose that p and q are pair of conjugate expo-nents. Let X be any nonempty set. Let ℳ be a r-algebra in X: Let l : ℳ ! 0;1½ �be a positive measure on ℳ: Let f 2 Lp lð Þ; and g 2 Lq lð Þ: Then1. f � gð Þ 2 L1 lð Þ;2. f � gk k1 � fk kp gk kq:

Lemma 2.25 Let p 2 1;1½ �: Let X be any nonempty set. Let ℳ be a r-algebra inX: Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let f ; g 2 Lp lð Þ: Then1. f þ gð Þ 2 Lp lð Þ;2. f þ gk kp � fk kp þ gk kp:

Proof Case I: when p 2 1;1ð Þ: Since f 2 Lp lð Þ; fj j : X ! 0;1½ � is a measurablefunction and

fk kp¼ZX

fj jpdl0@ 1A1

p

2 0;1½ Þð Þ:

Similarly, gj j : X ! 0;1½ � is a measurable function and

gk kp¼ZX

gj jpdl0@ 1A1

p

2 0;1½ Þð Þ:

By the Minkowski’s inequality,

0�ZX

f þ gj jpdl0@ 1A1

p

�ZX

fj j þ gj jð Þpdl0@ 1A1

p

�ZX

fj jpdl0@ 1A1

p

þZX

gj jpdl0@ 1A1

p

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}2 0;1½ Þ;

so ZX

f þ gj jpdl0@ 1A1

p

2 0;1½ Þ:


It follows that f þ gð Þ 2 Lp lð Þ: We have seen that

f þ gk kp¼ZX

f þ gj jpdl0@ 1A1

p

�ZX

fj jpdl0@ 1A1

p

þZX

gj jpdl0@ 1A1

p

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ fk kp þ gk kp:

Thus,

f þ gk kp � fk kp þ gk kp:

Case II: when p ¼ 1: Since f 2 L1 lð Þ; fj j : X ! 0;1½ � is a measurable functionand

fk k1¼ZX

fj jdl 2 0;1½ Þð Þ:


gk k1¼ZX

gj jdl 2 0;1½ Þð Þ:

Since f þ gj j � fj j þ gj j; we have

0�ZX

f þ gj jdl�ZX

fj j þ gj jð Þdl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ZX

fj jdl þZX

gj jdl0@ 1A 2 0;1½ Þ;

and henceRX f þ gj jdl 2 0;1½ Þ: Thus, f þ gð Þ 2 L1 lð Þ: We have seen that

f þ gk k1¼ZX

f þ gj jdl�ZX

fj jdlþZX

gj jdl|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ fk k1 þ gk k1;

so

f þ gk k1 � fk k1 þ gk k1:

Case III: when p ¼ 1: Since f 2 L1 lð Þ; fj j : X ! 0;1½ � is a measurablefunction and

fk k1¼ min a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0n o

2 0;1½ Þð Þ:

264 2 Lp-Spaces


gk k1¼ min a : a 2 0;1½ Þ; and l gj j�1 a;1ð �ð Þ

¼ 0n o

2 0;1½ Þð Þ:

It follows that

l x : fk k1\ f xð Þj j� �� ¼ 0;

and

l x : gk k1\ g xð Þj j� �� ¼ 0:

Since

x : fk k1 þ gk k1\ f þ gð Þ xð Þj j� � ¼ x : fk k1 þ gk k1\ f xð Þþ g xð Þj j� �� x : fk k1\ f xð Þj j� �[ x : gk k1\ g xð Þj j� �

;

we have

0� l x : fk k1 þ gk k1\ f þ gð Þ xð Þj j� �� ¼ l x : fk k1 þ gk k1\ f xð Þþ g xð Þj j� �� l x : fk k1\ f xð Þj j� �[ x : gk k1\ g xð Þj j� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}� l x : fk k1\ f xð Þj j� �� þ l x : gk k1\ g xð Þj j� �� ¼ 0þ 0 ¼ 0;

and hence

l x : fk k1 þ gk k1\ f þ gð Þ xð Þj j� �� ¼ 0:

This shows that

f þ gð Þ 2 L1 lð Þ; and f þ gk k1 � fk k1 þ gk k1: ∎

Lemma 2.26 Let p 2 1;1½ �: Let X be any nonempty set. Let ℳ be a r-algebra inX: Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let f 2 Lp lð Þ: Let a 2 C:Then

1. afð Þ 2 Lp lð Þ;2. afk kp¼ aj j fk kp:

Proof If a ¼ 0; then 1, and 2 are trivially true. So, we consider the case whena 6¼ 0:


Case I: when p 2 1;1½ Þ: Since f 2 Lp lð Þ; fj j : X ! 0;1½ � is a measurablefunction and

fk kp¼ZX

fj jpdl0@ 1A1

p

2 0;1½ Þð Þ:

Since

ZX

afj jpdl0@ 1A1

p

¼ZX

aj j fj jð Þpdl0@ 1A1

p

¼ZX

aj jp fj jpdl0@ 1A1

p

¼ aj jpZX

fj jpdl0@ 1A1

p

¼ aj jpð Þ1pZX

fj jpdl0@ 1A1

p

¼ aj jZX

fj jpdl0@ 1A1

p

2 0;1½ Þð Þ;

we have

ZX

afj jpdl0@ 1A1

p

2 0;1½ Þ;

and hence afð Þ 2 Lp lð Þ: We have seen that

afk kp¼ZX

afj jpdl0@ 1A1

p

¼ aj jZX

fj jpdl0@ 1A1

p

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ aj j fk kp;

so afk kp¼ aj j fk kp:Case II: when p ¼ 1: Since f 2 L1 lð Þ; we have fj j : X ! 0;1½ � is a

measurable function and

fk k1¼ min b : b 2 0;1½ Þ; and l fj j�1 b;1ð �ð Þ

¼ 0n o

2 0;1½ Þð Þ;

and hence

fk k12 0;1½ Þ; and l fj j�1 fk k1;1� �� ¼ 0:

266 2 Lp-Spaces

Since fk k12 0;1½ Þ; and a 2 C; we have aj j fk k1� � 2 0;1½ Þ: Since

afj j�1 aj j fk k1;1� �� ¼ x : aj j fk k1\ afð Þ xð Þj j� �¼ x : aj j fk k1\ aj j f xð Þj j� �¼ x : fk k1\ f xð Þj j� �¼ fj j�1 fk k1;1� ��

;

we have

0�ð Þl afj j�1 aj j fk k1;1� �� ¼ l fj j�1 fk k1;1� ��

¼ 0ð Þ;

and hence

l afj j�1 aj j fk k1;1� �� ¼ 0:

It follows that afð Þ 2 L1 lð Þ and afk k1 � aj j fk k1: Since

fk k1¼ 1a

afð Þ��

1� 1

a

�� afk k1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼1aj j afk k1;

we have aj j fk k1 � afk k1: Since aj j fk k1 � afk k1; and afk k1 � aj j fk k1; wehave afk k1¼ aj j fk k1: ∎)

In view of Lemmas 2.25 and 2.26, Lp lð Þ is a complex linear space.

Lemma 2.27 Let p 2 1;1½ �: Let X be any nonempty set. Letℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure onℳ: For every f ; g 2 Lp lð Þ; let us write:

f g if and only if f � gk kp¼ 0:

Then

a. is an equivalence relation over Lp lð Þ;b. if f ; g; f1; g1 2 Lp lð Þ; f f1 and g g1 then f1 � g1k kp¼ f � gk kp;c. if f ; g; f1; g1 2 Lp lð Þ; f f1; and g g1; then f þ gð Þ f1 þ g1ð Þ;d. if f ; f1 2 Lp lð Þ; f f1; and a 2 C; then afð Þ af1ð Þ:

Proof

a. We must prove:

1. for every f 2 Lp lð Þ; f f ;2. for every f ; g 2 Lp lð Þ; if f g then g f ;3. for every f ; g; h 2 Lp lð Þ; if f g; and g h then f h:


For 1: Let us take any f 2 Lp lð Þ: We have to show that f f ; that is, f � fk kp¼ 0;that is, 0k kp¼ 0:

Case I: when p 2 1;1½ Þ: Here

0k kp¼ZX

0j jpdl0@ 1A1

p

¼ZX

0 dl

0@ 1A1p

¼ 0ð Þ1p¼ 0:

Case II: when p ¼ 1: We have to show that 0k k1¼ 0:

LHS ¼ 0k kp¼ 0 ¼ min b : b 2 0;1½ Þ; and l 0j j�1 b;1ð �ð Þ

¼ 0n o

¼ min b : b 2 0;1½ Þ; and l x : b\ 0 xð Þj jf gð Þ ¼ 0f g¼ min b : b 2 0;1½ Þf g ¼ 0 ¼ RHS:

For 2: Let us take any f ; g 2 Lp lð Þ: Let f g: We have to show that g f ; thatis, g� fk kp¼ 0: Since f g; we have

g� fk kp¼ 1 � g� fk kp¼ �1j j g� fk kp¼ �1ð Þ g� fð Þk kp¼ f � gk kp¼ 0|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};and hence g� fk kp¼ 0:

For 3: Let us take any f ; g; h 2 Lp lð Þ: Let f g; and g h:We have to show thatf h; that is f � hk kp¼ 0: Since f g; we have f � gk kp¼ 0: Since g h; we haveg� hk kp¼ 0: On using Lemma 2.6,

LHS ¼ f � hk kp¼ f � gð Þþ g� hð Þk kp� f � gk kp þ g� hk kp¼ 0þ 0 ¼ 0 ¼ RHS

:

b. Let f ; g; f1; g1 2 Lp lð Þ; f f1; and g g1: It follows that f � f1k kp¼ 0; andg� g1k kp¼ 0: Since

f � gk kp ¼ f � f1ð Þþ f1 � g1ð Þþ g1 � gð Þk kp� f � f1k kp þ f1 � g1k kp þ g1 � gk kp¼ 0þ f1 � g1k kp þ g1 � gk kp¼ f1 � g1k kp þ g� g1k kp¼ f1 � g1k kp þ 0 ¼ f1 � g1k kp;

we have f � gk kp � f1 � g1k kp: Similarly, f1 � g1k kp � f � gk kp: It follows thatf � gk kp¼ f1 � g1k kp:

268 2 Lp-Spaces

c. Let f ; g; f1; g1 2 Lp lð Þ satisfying f f1; and g g1: We have to show thatf þ gð Þ f1 þ g1ð Þ; that is, f þ gð Þ � f1 þ g1ð Þk kp¼ 0: Since f f1; we havef � f1k kp¼ 0: Similarly, g� g1k kp¼ 0: Since

0� f þ gð Þ � f1 þ g1ð Þk kp¼ f � f1ð Þþ g� g1ð Þk kp � f � f1k kp þ g� g1k kp¼ 0þ 0 ¼ 0;

we have

f þ gð Þ � f1 þ g1ð Þk kp¼ 0:

d. Let f ; f1 2 Lp lð Þ; f f1; and a 2 C: We have to show that afð Þ af1ð Þ; that is,af � af1k kp¼ 0: Since f f1; we have f � f1k kp¼ 0 and hence

af � af1k kp¼ a f � f1ð Þk kp¼ aj j f � f1k kp¼ aj j � 0 ¼ 0:

Thus, af � af1k kp¼ 0: ∎

By Lemma 2.27(a), we find that Lp lð Þ is partitioned into equivalent classes.Lemma 2.8(b) shows that

d : f½ �; g½ �ð Þ 7! f � gk kp

is a well-defined function from Lp lð Þ= ð Þ Lp lð Þ= ð Þ to 0;1½ Þ:Problem 2.28 d : f½ �; g½ �ð Þ 7! f � gk kp is a metric over Lp lð Þ= ð Þ:(Solution We must show:

1. for every f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ; d f½ �; g½ �ð Þ � 0;2. for every f½ � 2 Lp lð Þ= ð Þ; where f 2 Lp lð Þ; d f½ �; f½ �ð Þ ¼ 0;3. for every f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ; if d f½ �; g½ �ð Þ ¼ 0; then

f½ � ¼ g½ �;4. for every f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ; d f½ �; g½ �ð Þ ¼ d g½ �; f½ �ð Þ;5. for every f½ �; g½ �; h½ � 2 Lp lð Þ= ð Þ; where f ; g; h 2 Lp lð Þ; d f½ �; g½ �ð Þ�

d f½ �; h½ �ð Þþ d h½ �; g½ �ð Þ:For 1: Let us take any f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ: We have to

show that d f½ �; g½ �ð Þ� 0; that is f � gk kp � 0: This is clearly true from thedefinition of � � �k kp:

For 2: Let us take any f½ � 2 Lp lð Þ= ð Þ; where f 2 Lp lð Þ:We have to show thatd f½ �; f½ �ð Þ ¼ 0; that is, f � fk kp¼ 0; that is, 0k kp¼ 0: This is clearly true from thedefinition of � � �k kp:


For 3: Let us take any f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ: Letd f½ �; g½ �ð Þ ¼ 0: We have to show that f½ � ¼ g½ �; that is, f g; that is, f � gk kp¼ 0:Since f � gk kp¼ d f½ �; g½ �ð Þ ¼ 0|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}; we have f � gk kp¼ 0:

For 4: Let us take any f½ �; g½ � 2 Lp lð Þ= ð Þ; where f ; g 2 Lp lð Þ: We have toshow that d f½ �; g½ �ð Þ ¼ d g½ �; f½ �ð Þ:

LHS ¼ d f½ �; g½ �ð Þ ¼ f � gk kp¼ �1ð Þ g� fð Þk kp¼ �1j j g� fk kp¼ g� fk kp¼ d g½ �; f½ �ð Þ ¼ RHS:

For 5: Let us take any f½ �; g½ �; h½ � 2 Lp lð Þ= ð Þ; where f ; g; h 2 Lp lð Þ: We haveto show that d f½ �; g½ �ð Þ� d f½ �; h½ �ð Þþ d h½ �; g½ �ð Þ: Here

d f½ �; g½ �ð Þ ¼ f � gk kp¼ f � hð Þþ h� gð Þk kp � f � hk kp þ h� gk kp¼ d f½ �; h½ �ð Þþ d h½ �; g½ �ð Þ;

so

d f½ �; g½ �ð Þ� d f½ �; h½ �ð Þþ d h½ �; g½ �ð Þ:

Thus, Lp lð Þ= ð Þ; dð Þ is a metric space. ∎)From Lemma 2.27(c), we find that þ : f½ �; g½ �ð Þ 7! f þ g½ � is a well-defined

mapping from

Lp lð Þ= ð Þ Lp lð Þ= ð Þ to Lp lð Þ= ð Þ:

Thus, þ is a binary operation over Lp lð Þ= ð Þ. From Lemma 2.27(d), we findthat a; f½ �ð Þ 7! af½ � is a well-defined mapping from

C Lp lð Þ= ð Þ to Lp lð Þ= ð Þ:

In this sense, it is easy to verify that Lp lð Þ= ð Þ is a complex linear space.

Notation For the sake of simplicity, Lp lð Þ= ð Þ is denoted by Lp lð Þ; and theequivalence class f½ � by f :

Thus, Lp lð Þ is a complex linear space as well as a metric space.

Note 2.29 Let p 2 1;1½ �: Let X be any nonempty set. Let ℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let f 2 Lp lð Þ:Problem 2.30

1. fj j 2 Lp lð Þ;2. fj jð Þk kp¼ fk kp:

270 2 Lp-Spaces

(Solution Since f 2 Lp lð Þ; f : X ! C. is measurable, and hence fj j : X ! 0;1½ Þis measurable.

Case I: when p 2 1;1½ Þ: Since f 2 Lp lð Þ; we have

0�ZX

fj jpdl0@ 1A1

p

\1:

Since fj jð Þj j ¼ fj j; we have

ZX

fj jð Þj jpdl0@ 1A1

p

¼ZX

fj jpdl0@ 1A1

p

2 0;1½ Þð Þ;

and hence

ZX


p

2 0;1½ Þ:

It follows that fj j 2 Lp lð Þ: Since

fj jð Þk kp¼ZX


p

¼ZX

fj jpdl0@ 1A1

p

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ fk kp;

we have fj jð Þk kp¼ fk kp:Case II: when p ¼ 1: Since f 2 L1 lð Þ; we have

0� l x : fk k1\ fj j xð Þð Þj j� �� ¼ l x : fk k1\ f xð Þj j� �� ¼ l fj j�1 fk k1;1� ��

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence,

l x : fk k1\ fj j xð Þð Þj j� �� ¼ 0:

It follows that fj j 2 Lp lð Þ: Since, for every a 2 0;1½ Þ;

fj j�1 a;1ð �ð Þ ¼ fj jð Þj j�1 a;1ð �ð Þ;


we have

a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0n o

¼ a : a 2 0;1½ Þ; and l fj jð Þj j�1 a;1ð �ð Þ

¼ 0n o

;

and hence

fk k1¼� �min a : a 2 0;1½ Þ; and l fj j�1 a;1ð �ð Þ

¼ 0

n o¼ min a : a 2 0;1½ Þ; and l fj jð Þj j�1 a;1ð �ð Þ

¼ 0

n o¼ fj jð Þ1� �

:

Thus, fj jð Þk k1¼ fk k1: ∎)Let p 2 1;1½ Þ: Let fnf g be a Cauchy sequence in the metric space Lp lð Þ; dð Þ:There exists a positive integer n1 such that m; n� n1 implies

fm � fnk kp¼ d fm; fnð Þ\12|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} :

There exists a positive integer n2 [ n1 such that m; n� n2 implies

fm � fnk kp¼ d fm; fnð Þ\ 122|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} :

There exists a positive integer n3 [ n2 such that m; n� n3 implies

fm � fnk kp¼ d fm; fnð Þ\ 123|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} :

There exists a positive integer n4 [ n3 such that m; n� n4 implies fm � fnk kp¼d fm; fnð Þ\ 1

24|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}; etc.It follows that

fn2 � fn1k kp\121

; fn3 � fn2k kp\122

; fn4 � fn3k kp\123

; fn5 � fn4k kp\124

; etc:

Also,

n1\n2\n3\n4\ � � � :

Thus, for each k ¼ 1; 2; 3; . . .; fnkþ 1 � fnk��

p\12k : Since each fnk 2 Lp lð Þ; and

Lp lð Þ is a complex linear space, each fnkþ 1 � fnk� � 2 Lp lð Þ; and hence each

fnkþ 1 � fnk�� 2 Lp lð Þ; and

fnkþ 1 � fnk��

p¼ fnkþ 1 � fnk��

p|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\ 12k

:

272 2 Lp-Spaces

It follows that, for each k ¼ 1; 2; 3; . . .;

Xki¼1

fniþ 1 � fni�� !��

��p

�Xki¼1

fniþ 1 � fni��

p\Xki¼1

12i\1 \1ð Þ;

and hence for each k ¼ 1; 2; 3; . . .;

Xki¼1

fniþ 1 � fni�� 2 Lp lð Þ;

and

ZX

Xki¼1

fniþ 1 � fni�� !p

dl

0@ 1A1p

¼ZX

Xki¼1


��p

dl

0@ 1A1p

¼Xki¼1

fniþ 1 � fni�� !��

��p

\1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus, for each k ¼ 1; 2; 3; . . .;

0�ZX

Xki¼1


dl\1;

and hence by Lemma 1.130,

ZX

X1i¼1


dl ¼ZX

limk!1

Xki¼1

fniþ 1 � fni�� ! !p

dl

¼ZX

limk!1

Xki¼1

fniþ 1 � fni�� !p ! !

dl

¼ZX

lim infk!1

Xki¼1

fniþ 1 � fni�� !p ! !

dl� lim infk!1

ZX

Xki¼1


dl

0@ 1A|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

� 1:

It follows that

ZX

X1i¼1


dl

0@ 1A1p

� 1 \1ð Þ;


and henceP1

i¼1 fniþ 1 � fni�� 2 Lp lð Þ: Also, since

X1i¼1


��p

¼ZX

X1i¼1


dl

0@ 1A1p

� 1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

we have

X1i¼1


��p

� 1:

This shows that

X1i¼1

fniþ 1 xð Þ � fni xð Þ�� \1 holds a:e: on X;

and hence there exists a set S 2 ℳ such thatP1

i¼1 fniþ 1 xð Þ � fni xð Þ� �converges

absolutely for all x 2 Sc; and l Sð Þ ¼ 0: Let us define a function f : X ! C asfollows: For every x 2 X;

f xð Þ � fn1 xð Þþ P1i¼1

fniþ 1 xð Þ � fni xð Þ� �if x 2 Sc

0 if x 2 S;

8<:that is,

f xð Þ � limk!1

fnk xð Þ if x 2 Sc

0 if x 2 S;

�that is, f ¼ limk!1 fnk a.e. on X:

Problem 2.31 f 2 Lp lð Þ:(Solution Since fnf g is a Cauchy sequence in the metric space Lp lð Þ; dð Þ; thereexists a positive integer N such that m; n�N implies

ZX

fm � fnj jpdl0@ 1A1

p

¼ fm � fnk kp¼ d fm; fnð Þ\1|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};

274 2 Lp-Spaces

and hence, m; n�N implies ZX

fm � fnj jpdl\1:

It follows that

ZX

lim infi!1

fni � fNj jpð Þ� �

dl� lim infi!1

ZX

fni � fNj jpdl0@ 1A� 1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:

Now, sinceZX

f � fNj jpð Þdl ¼ZX

limi!1

fni

� �� fN

�� p� �dl

¼ZX

limi!1


dl ¼ZX

lim infi!1


dl;

we have

ZX

f � fNj jpð Þdl0@ 1A1

p

� 1 \1ð Þ:

This shows that f � fNð Þ 2 Lp lð Þ: Since f � fNð Þ; fN 2 Lp lð Þ; and Lp lð Þ is acomplex linear space, we have

f ¼ð Þ f � fNð Þþ fN 2 Lp lð Þ;

and hence f 2 Lp lð Þ: ∎)

Problem 2.32 limm!1 f � fmk kp¼ 0:

(Solution Let us take any e[ 0: Since fnf g is a Cauchy sequence in the metricspace Lp lð Þ; dð Þ; there exists a positive integer N such that m; n�N implies

ZX

fm � fnj jpdl0@ 1A1

p

¼ fm � fnk kp¼ d fm; fnð Þ\ e2|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};


and hence m; n�N implies ZX

fm � fnj jpdl\ ep

2p:

It follows that, for every m0 �N; we haveZX

lim infi!1

fni � fm0j jpð Þ� �

dl� lim infi!1

ZX

fni � fm0j jpdl0@ 1A� ep

2p|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:

Now, since m0 �N impliesZX

f � fm0j jpð Þdl ¼ZX

limi!1

fni

� �� fm0

�� p� �dl

¼ZX

limi!1


dl ¼ZX

lim infi!1


dl;

we have m0 �N implies

f � fm0k kp¼ZX

f � fm0j jpð Þdl0@ 1A1

p

� e2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\e:

This proves that limm!1 f � fmk kp¼ 0: ∎)

Conclusion 2.33

1. For p 2 1;1½ Þ; the metric space Lp lð Þ; dð Þ is complete.2. If fnf g is a Cauchy sequence in the metric space Lp lð Þ; dð Þ; which converges to

f 2 Lp lð Þ; then there exists a subsequence fnkf g of fnf g such that limk!1 fnk ¼ fa.e. on X:

Let fnf g be a Cauchy sequence in the metric space L1 lð Þ; dð Þ:Since for every n ¼ 1; 2; . . .; fn 2 L1 lð Þ; we have, for every n ¼ 1; 2; . . .;

fnk k12 0;1½ Þ; and

l Anð Þ ¼ l x : fnk k1\ fn xð Þj j� �� ¼ l fnj j�1 fnk k1;1� �� ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

where An � x : fnk k1\ fn xð Þj j� � 2 ℳð Þ: Since for every m; n ¼ 1; 2; . . .; fm; fn 2L1 lð Þ; and L1 lð Þ is a complex linear space, we have, for every m; n ¼ 1; 2; . . .;fm � fnð Þ 2 L1 lð Þ; and hence for every m; n ¼ 1; 2; . . .; fm � fnk k12 0;1½ Þ: Also,for every m; n ¼ 1; 2; . . .;

276 2 Lp-Spaces

l Bm;n� � ¼ l x : fm � fnk k1\ fm xð Þ � fn xð Þj j� ��

¼ l fm � fnj j�1 fm � fnk k1;1� �� ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

where

Bm;n � x : fm � fnk k1\ fm xð Þ � fn xð Þj j� � 2 ℳð Þ:

Put

E � [1n¼1An

� �[ [ m;nð Þ2NNBm;n� � 2 ℳð Þ:

Now, since each l Anð Þ ¼ 0; and each l Bm;n� � ¼ 0; we have l Eð Þ ¼ 0:

Problem 2.34 fnf g converges uniformly over Ec:

(Solution It suffices to show that for every e[ 0; there exists a positive integer Nsuch that, for every m; n�N; and for every x 2 Ec; fm xð Þ � fn xð Þj j\e: For thispurpose, let us take any e[ 0: Since fnf g is a Cauchy sequence in the metric spaceL1 lð Þ; dð Þ; there exists a positive integer N such that, for every m; n�N; we havefm � fnk k1\e: Let us fix any

y 2 Ec|fflfflffl{zfflfflffl} ¼ [1n¼1An

� �[ [ m;nð Þ2NNBm;n� �� c

¼ \1n¼1 Anð Þc� �\ \ m;nð Þ2NN Bm;n

� �c� �:

It suffices to show that, for every m; n�N; fm yð Þ � fn yð Þj j\e: Since for everym; n ¼ 1; 2; . . .;

y 2 Bm;n� �c ¼ x : fm � fnk k1\ fm xð Þ � fn xð Þj j� �c

¼ x : fm xð Þ � fn xð Þj j � fm � fnk k1� �

;

we have

fm yð Þ � fn yð Þj j � fm � fnk k1:

Since for every m; n�N;

fm yð Þ � fn yð Þj j � fm � fnk k1\e|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl};we have, for every m; n�N; fm yð Þ � fn yð Þj j\e: ∎)


Since fnf g converges uniformly over Ec; there exists a function f : Ec ! C suchthat fnf g converges uniformly to f on Ec 2 ℳð Þ: Thus, for every x 2 Ec; we havelimn!1 fn xð Þ ¼ f xð Þ: Let us define a function f : X ! C as follows: For everyx 2 X;

f xð Þ � f xð Þ if x 2 Ec

0 if x 2 E:

�Now, since each fn is a measurable function on X; Ec 2 ℳ; and for every

x 2 Ec; limn!1 fn xð Þ ¼ f xð Þ; by Lemma 1.48, f : X ! C is a measurable function.Since l Eð Þ ¼ 0; and for every x 2 Ec; limn!1 fn xð Þ ¼ f xð Þ ¼ f xð Þð Þ uniformly onEc; limn!1 fn ¼ f uniformly a.e. on X:

Problem 2.35 f 2 L1 lð Þ:(Solution Since limn!1 fn ¼ f uniformly a.e. on X; there exists a positive integerN such that fN � fj j � 1 holds a.e. on X: Now, by Conclusion 2.20, fN � fð Þ 2L1 lð Þ: Since fN � fð Þ; fN 2 L1 lð Þ; and L1 lð Þ is a complex linear space, we havef ¼ð Þ fN � fN � fð Þð Þ 2 L1 lð Þ; and hence f 2 L1 lð Þ: ∎)We show try to show that

limn!1 fn � fk k1¼ 0:

For this purpose, let us take any e[ 0: Since limn!1 fn ¼ f uniformly a.e. on X;there exists a positive integer N1 such that for every n�N1; fn � fj j\ e

3 holds a.e.on X: Since fnf g is a Cauchy sequence in the metric space L1 lð Þ; dð Þ; there exists apositive integer N[N1 such that m; n�N implies fm � fnk k1\ e

3 ; and hence forevery m; n�N; fm � fnj j � e

3 holds a.e. on X: It follows that, for every n�N;fn � fNj j � e

3 holds a.e. on X: Clearly, fN � fj j\ e3 holds a.e. on X:

Since fN � fj j\ e3 holds a.e. on X; and, for every n�N; fn � fNj j � e

3 holds a.e.on X; we have, for every n�N; fn � fj j\ 2e

3 holds a.e. on X; and hence for everyn�N; fn � fk k1 � 2e

3 \eð Þ: Thus,

limn!1 fn � fk k1¼ 0:

Conclusion 2.36

1. The metric space L1 lð Þ; dð Þ is complete.2. If fnf g is a Cauchy sequence in the metric space L1 lð Þ; dð Þ; which converges to

f 2 L1 lð Þ; then limn!1 fn ¼ f a.e. on X:

Note 2.37 Let p 2 1;1½ Þ: Let X be any nonempty set. Let ℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let s : X ! C be a measurablesimple function. Let l x : s xð Þ 6¼ 0f gð Þ\1:

278 2 Lp-Spaces

Problem 2.38 s 2 Lp lð Þ:(Solution Case I: when s ¼ 0: Since Lp lð Þ is a complex linear space, 0 2 Lp lð Þ;and hence s 2 Lp lð Þ:

Case II: when s 6¼ 0: Since s : X ! C is a measurable simple function, we canwrite

s ¼ a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ

for some a1; . . .; an 2 C� 0f g such that a1; . . .; an are distinct, ands�1 a1ð Þ; . . .; s�1 anð Þ 2 ℳ: Since each ai 2 C� 0f g; each s�1 aið Þ �x : s xð Þ 6¼ 0f g; and hence each l s�1 aið Þð Þ� l x : s xð Þ 6¼ 0f gð Þ \1ð Þ: It follows

that each l s�1 aið Þð Þ\1: Now, since

sj jp¼ a1j jpv s�1 a1ð Þð Þ þ � � � þ anj jpv s�1 anð Þð Þ;

we have ZX

sj jpdl ¼ a1j jp l s�1 a1ð Þ� �� þ � � � þ anj jp l s�1 anð Þ� �� \1:

Thus, s 2 Lp lð Þ: ∎)

Conclusion 2.39 Let p 2 1;1½ Þ: Let X be any nonempty set. Letℳ be a r-algebrain X: Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let s : X ! C be a mea-surable simple function. Let l x : s xð Þ 6¼ 0f gð Þ\1: Then s 2 Lp lð Þ:Note 2.40 Let p 2 1;1½ Þ: Let X be any nonempty set. Let ℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let S be the collection of allmeasurable simple functions s : X ! C such that l x : s xð Þ 6¼ 0f gð Þ\1: ByConclusion 2.39, S � Lp lð Þ: Let f : X ! 0;1½ Þ be a measurable function such thatf 2 Lp lð Þ:Problem 2.41 f is an adherent point of S; that is, every open neighborhood of f inLp lð Þ contains some point of S:

(Solution By Lemma 1.98, there exists a sequence snf g of simple measurablefunctions sn : X ! 0;1½ Þ such that for every x in X;0� s1 xð Þ� s2 xð Þ� � � � � f xð Þ; and limn!1 sn xð Þ ¼ f xð Þ: Since, for each n;0� sn � f ; we have, for each n;Z

X

snj jpdl�ZX

fj jpdl;


and hence, for each n;

ZX

snj jpdl0@ 1A1

p

�ZX

fj jpdl0@ 1A1

p

\1ð Þ:

Thus, each sn 2 Lp lð Þ:Problem 2.42 Let n0 be a positive integer. Then sn0 2 S:

(Solution Case I: when sn0 ¼ 0: In this case, x : sn0 xð Þ 6¼ 0f g ¼ ;; and hence,l x : sn0 xð Þ 6¼ 0f gð Þ ¼ 0 \1ð Þ: It follows that sn0 2 S:

Case II: when sn0 6¼ 0: Since sn0 : X ! C is a measurable simple function, wecan write

sn0 ¼ a1v s�1n0

a1ð Þð Þ þ � � � þ anv s�1n0

anð Þð Þ

for some a1; . . .; an 2 C� 0f g such that a1; . . .; an are distinct, and

s�1n0 a1ð Þ; . . .; s�1

n0 anð Þ 2 ℳ:


l s�1n0 a1ð Þ

þ � � � þ l s�1

n0 anð Þ

¼ l x : sn0 xð Þ 6¼ 0f gð Þ\1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :It suffices to show that each l s�1

n0 aið Þ

\1: Since sn0 2 Lp lð Þ;

a1j jp l s�1 a1ð Þ� �� þ � � � þ anj jp l s�1 anð Þ� �� 1p¼

ZX

sn0j jpdl0@ 1A1

p

\1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

;

and hence each l s�1n0 aið Þ

\1: ∎)

By Problem 2.42, each sn 2 S: Since for each n; 0� sn � f ; and p 2 1;1½ Þ; wehave f � snð Þpj j � f pð Þ: Since f 2 Lp lð Þ; we have f pð Þ 2 L1 lð Þ: Clearly, eachf � snð Þp is a measurable function. Since limn!1 sn ¼ f ; we havelimn!1 f � snð Þp¼ 0: Now, by Theorem 1.136,

limn!1 f � snk kp p

¼ limn!1 f � snk kp

p¼ lim

n!1

ZX

f � snð Þpdl0@ 1A ¼

ZX

0dl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 0;

280 2 Lp-Spaces

and hence,

limn!1 f � snk kp¼ 0:

It follows that every open neighborhood of f in Lp lð Þ contains some sn 2 Sð Þ:Hence, f is an adherent point of S: ∎)

Conclusion 2.43 Let p 2 1;1½ Þ: Let X be any nonempty set. Letℳ be a r-algebrain X: Let l : ℳ ! 0;1½ � be a positive measure onℳ: Let S be the collection of allmeasurable simple functions s : X ! C such that l x : s xð Þ 6¼ 0f gð Þ\1: ByConclusion 2.39, S � Lp lð Þ: Let f : X ! 0;1½ Þ be a measurable function such thatf 2 Lp lð Þ: Then f is an adherent point of S:

Note 2.44 Let p 2 1;1½ Þ: Let X be any nonempty set. Let ℳ be a r-algebra in X:Let l : ℳ ! 0;1½ � be a positive measure on ℳ: Let S be the collection of allmeasurable simple functions s : X ! C such that l x : s xð Þ 6¼ 0f gð Þ\1: ByProblem 2.38, S � Lp lð Þ: Let f : X ! R be a measurable function such thatf 2 Lp lð Þ:Problem 2.45 f is an adherent point of S:

(Solution Let us take any e[ 0: Since f 2 Lp lð Þ; we have fj j 2 Lp lð Þ: Sincef ; fj j 2 Lp lð Þ; and Lp lð Þ is a complex linear space, we have

f þ ¼ 12

f þ fj jð Þ 2 Lp lð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since f þ 2 Lp lð Þ; and f þ : X ! 0;1½ Þ; by Note 2.11, there exists s 2 S such

that f þ � sk kp\ e2 : Similarly, there exists t 2 S such that f� � tk kp\ e

2 : It followsthat

f � s� tð Þk kp ¼ f þ � f�ð Þ � s� tð Þk kp¼ f þ � sð Þþ t � f�ð Þk kp� f þ � sk kp þ t � f�k kp¼ f þ � sk kp þ f� � tk kp\

e2þ e

2¼ e:

Thus, f � s� tð Þk kp\e: It suffices to show that s� tð Þ 2 S: Since s; t 2S � Lp lð Þð Þ; and Lp lð Þ is a complex linear space, s� tð Þ 2 Lp lð Þ: Since s 2 S;l x : s xð Þ 6¼ 0f gð Þ\1: Similarly, l x : t xð Þ 6¼ 0f gð Þ\1: Since

x : s� tð Þ xð Þ 6¼ 0f g ¼ x : s xð Þ 6¼ t xð Þf g � x : s xð Þ 6¼ 0f g[ x : t xð Þ 6¼ 0f g;


we have

l x : s� tð Þ xð Þ 6¼ 0f gð Þ� l x : s xð Þ 6¼ 0f g[ x : t xð Þ 6¼ 0f gð Þ� l x : s xð Þ 6¼ 0f gð Þþ l x : t xð Þ 6¼ 0f gð Þ\1;

and hencel x : s� tð Þ xð Þ 6¼ 0f gð Þ\1:

It follows that s� tð Þ 2 S: ∎)

Conclusion 2.46 Let p 2 1;1½ Þ: Let X be any nonempty set. Letℳ be a r-algebrain X: Let l : ℳ ! 0;1½ � be a positive measure onℳ: Let S be the collection of allmeasurable simple functions s : X ! C such that l x : s xð Þ 6¼ 0f gð Þ\1: ByConclusion 2.39, S � Lp lð Þ: Let f : X ! R be a measurable function such thatf 2 Lp lð Þ: Then f is an adherent point of S:

Theorem 2.47 Let p 2 1;1½ Þ: Let X be any nonempty set. Let ℳ be a r-algebrain X: Let l : ℳ ! 0;1½ � be a positive measure onℳ: Let S be the collection of allmeasurable simple functions s : X ! C such that l x : s xð Þ 6¼ 0f gð Þ\1: ByConclusion 2.39, S � Lp lð Þ: Then, S is dense.

Proof Let f : X ! C be a measurable function such that f 2 Lp lð Þ: We have toshow that f is an adherent point of S: For this purpose, let us take any e[ 0: Sincef 2 Lp lð Þ; we haveZ

X

Re fð Þj jpdl�ZX

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiRe fð Þð Þ2 þ Im fð Þð Þ2

q� �p

dl ¼ZX

fj jpdl\1|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

;

and hence Re fð Þ 2 Lp lð Þ: Since Re fð Þ 2 Lp lð Þ; and Re fð Þ : X ! R; by Note2.44, there exists s 2 S such that Re fð Þ � sk kp\ e

2 : Similarly, there exists t 2 Ssuch that Im fð Þ � tk kp\ e

2 : It follows that

f � sþ itð Þk kp ¼ Re fð Þþ iIm fð Þð Þ � sþ itð Þk kp¼ Re fð Þ � sð Þþ i Im fð Þ � tð Þk kp� Re fð Þ � sk kp þ ij j Im fð Þ � tk kp� Re fð Þ � sk kp þ Im fð Þ � tk kp\

e2þ e

2¼ e:

Thus, f � sþ itð Þk kp\e: It suffices to show that sþ itð Þ 2 S: Since s; t 2S � Lp lð Þð Þ; and Lp lð Þ is a complex linear space, sþ itð Þ 2 Lp lð Þ: Since s 2 S;we have l x : s xð Þ 6¼ 0f gð Þ\1: Similarly, l x : t xð Þ 6¼ 0f gð Þ\1: Since

x : sþ itð Þ xð Þ 6¼ 0f g � x : s xð Þ 6¼ 0f g[ x : t xð Þ 6¼ 0f g;

282 2 Lp-Spaces

we have

l x : sþ itð Þ xð Þ 6¼ 0f gð Þ� l x : s xð Þ 6¼ 0f g[ x : t xð Þ 6¼ 0f gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}� l x : s xð Þ 6¼ 0f gð Þþ l x : t xð Þ 6¼ 0f gð Þ\1;

and hence

l x : sþ itð Þ xð Þ 6¼ 0f gð Þ\1:

It follows that sþ itð Þ 2 S: ∎

Note 2.48 Let p 2 1;1½ Þ: Let X be a locally compact Hausdorff space. Let K be apositive linear functional on Cc Xð Þ: By Theorem 1.225, there exists a r-algebra ℳin X that contains all Borel sets in X; and there exists a unique positive measure lon ℳ such that conditions (1) to (5) are satisfied, and, for every f 2 Cc Xð Þ;

K fð Þ ¼ZX

f dl:

Problem 2.49 Cc Xð Þ � Lp lð Þ:(Solution Let us take any f 2 Cc Xð Þ: It follows that f : X ! C is continuous, andhence fj jp: X ! 0;1½ Þ is continuous. Since

f�1 0f gð Þ ¼ x : f xð Þ ¼ 0f g ¼ x : f xð Þj jp¼ 0f g ¼ fj jpð Þ�1 0f gð Þ;

we have

f�1 C� 0f gð Þ ¼ f�1 0f gð Þ� �c¼ fj jpð Þ�1 0f gð Þ c

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fj jpð Þ�1 C� 0f gð Þ;

and hence,

supp fð Þ ¼ f�1 C� 0f gð Þ� ��¼ fj jpð Þ�1 C� 0f gð Þ �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ supp fj jpð Þ:

Since f 2 Cc Xð Þ; supp fj jpð Þ ¼ð Þsupp fð Þ is compact, and hence supp fj jpð Þ iscompact. It follows that fj jp2 Cc Xð Þ; and hence

C 3ð ÞK fj jpð Þ ¼ZX

fj jpdl � 0ð Þ:

It follows thatRX fj jpdl 2 0;1½ Þ; and hence f 2 Lp lð Þ: Thus,

Cc Xð Þ � Lp lð Þ: ∎)


We shall try to show that Cc Xð Þ is dense in Lp lð Þ:For this purpose, let us take any f 2 Lp lð Þ: We have to show that f is an

adherent point of Cc Xð Þ:For the purpose, let us take any e[ 0: By Theorem 2.47, there exists t 2

S � Lp lð Þð Þ such that

t � fk kp\e;

where S denotes the collection of all measurable simple functions s : X ! C such that

l x : s xð Þ 6¼ 0f gð Þ\1:

Since t : X ! C is a simple function, its image set t Xð Þ is a finite set, and hencet1k k ¼ sup t xð Þj j : x 2 Xf g: Here t 2 S; so l x : t xð Þ 6¼ 0f gð Þ\1: Now, by

Theorem 1.273, there exists g 2 Cc Xð Þ � Lp lð Þð Þ such that

1. l x : t xð Þ 6¼ g xð Þf gð Þ\e;2. sup g xð Þj j : x 2 Xf g� sup t xð Þj j : x 2 Xf g ¼ tk k1

� �:

Since g; t 2 Lp lð Þ; and Lp lð Þ is a complex linear space, we have g� tð Þ 2Lp lð Þ: Next,

g� tk kp ¼ZX

g� tj jpdl0@ 1A1

p

¼Z

x:t xð Þ6¼g xð Þf g

g� tj jpdl

0B@1CA

1p

�Z


gj j þ tj jð Þpdl

0B@1CA

1p

�Z


sup g xð Þj j : x 2 Xf gþ tj jð Þpdl

0B@1CA

1p

�Z


tk k1 þ tj j� �pdl

0B@1CA

1p

�Z


tk k1 þ sup t xð Þj j : x 2 Xf g� �pdl

0B@1CA

1p

¼Z


tk k1 þ tk k1� �pdl

0B@1CA

1p

¼ tk k1 þ tk k1� �p

l x : t xð Þ 6¼ g xð Þf gð Þ� �1p

\ tk k1 þ tk k1� �p

e� �1

p¼ 2 tk k1e1p:

284 2 Lp-Spaces

Since t � fk kp\e; and

g� tk kp\2 tk ke1p;

we have

g� fk kp\ eþ 2 tk ke1p

! 0 as e ! 0þð Þ:

Conclusion 2.50 Let p 2 1;1½ Þ: Let X be a locally compact Hausdorff space. LetKbe a positive linear functional on Cc Xð Þ: By Theorem 1.225, there exists a r-algebraℳ in X that contains all Borel sets in X; and there exists a unique positive measure lon ℳ such that conditions (1) to (5) are satisfied, and, for every f 2 Cc Xð Þ;

K fð Þ ¼ZX

f dl:

Further, Cc Xð Þ is dense in Lp lð Þ:

2.3 Inner Products

Although, Banach space is more general than Hilbert space, we shall study Hilbertspace first, owing to its simpler nature.

Note 2.51

Definition Let H be a complex linear space. Suppose that, to every ordered pair ofvectors x and y in H; there is associated a complex number x; yð Þ: If1. for every x; y 2 H; y; xð Þ ¼ x; yð Þ�;2. for every x; y; z 2 H; xþ y; zð Þ ¼ x; zð Þþ y; zð Þ;3. for every x; y 2 H; and for every a 2 C; ax; yð Þ ¼ a x; yð Þ;4. for every x 2 H; x; xð Þ is a nonnegative real number,5. x; xð Þ ¼ 0 implies x ¼ 0;

then we say that H is an inner product space. Here, x; yð Þ is called the innerproduct of x and y:

Note 2.52 Let H be an inner product space.

Problem 2.53 For every y 2 H; 0; yð Þ ¼ 0:

(Solution Let us fix any y 2 H: We have to show that 0; yð Þ ¼ 0:

LHS ¼ 0; yð Þ ¼ 0 0ð Þ; yð Þ ¼ 0 � 0; yð Þ ¼ 0 ¼ RHS: ∎)


Problem 2.54 For every x; y 2 H; and, for every a 2 C; x; ayð Þ ¼ �a x; yð Þ:(Solution Let us take any x; y 2 H; and a 2 C: We have to show that x; ayð Þ ¼�a x; yð Þ:

LHS ¼ x; ayð Þ ¼ ay; xð Þ�¼ a y; xð Þð Þ�¼ �a y; xð Þ�ð Þ ¼ �a x; yð Þ ¼ RHS: ∎)

Problem 2.55 For every x; y; z 2 H; x; yþ zð Þ ¼ x; yð Þþ x; zð Þ:(Solution Let us take any x; y; z 2 H: We have to show thatx; yþ zð Þ ¼ x; yð Þþ x; zð Þ:

LHS ¼ x; yþ zð Þ ¼ yþ z; xð Þ�¼ y; xð Þþ z; xð Þð Þ�¼ y; xð Þ� þ z; xð Þ�¼ x; yð Þþ x; zð Þ ¼ RHS: ∎)

Conclusion 2.56 Let H be an inner product space. Then 1. for every y 2 H;0; yð Þ ¼ 0; 2. for every x; y 2 H; and, for every a 2 C; x; ayð Þ ¼ �a x; yð Þ; and 3. forevery x; y; z 2 H; x; yþ zð Þ ¼ x; yð Þþ x; zð Þ:Definition Let H be an inner product space. Let x 2 H: Since x; xð Þ is a nonneg-ative real number,

ffiffiffiffiffiffiffiffiffiffix; xð Þp

is a nonnegative real number. The nonnegative real

numberffiffiffiffiffiffiffiffiffiffix; xð Þp

is denoted by xk k; and xk k is called the norm of x: Thus, for every

x 2 H; xk k2¼ x; xð Þ:Let H be an inner product space.

Problem 2.57 0k k ¼ 0:

(Solution LHS ¼ 0k k ¼ ffiffiffiffiffiffiffiffiffiffiffi0; 0ð Þp ¼ ffiffiffi

0p ¼ 0 ¼ RHS: ∎)

Problem 2.58 xk k ¼ 0 implies x ¼ 0:

(Solution Letffiffiffiffiffiffiffiffiffiffix; xð Þp ¼� �

xk k ¼ 0:Wehave to show that x ¼ 0: Sinceffiffiffiffiffiffiffiffiffiffix; xð Þp ¼ 0;

we have x; xð Þ ¼ 0; and hence x ¼ 0: ∎)

Problem 2.59 For every x 2 H; and for every a 2 C; axk k ¼ aj j xk k:(Solution Here,

axk k2 ¼ ax; axð Þ ¼ a x; axð Þ ¼ a �a x; xð Þð Þ ¼ a�að Þ x; xð Þ¼ a�að Þ xk k2

¼ aj j2 xk k2¼ aj j xk kð Þ2;

so axk k2¼ aj j xk kð Þ2; and hence, axk k ¼ aj j xk k: ∎)

286 2 Lp-Spaces

Problem 2.60 For every x; y 2 H; x; yð Þj j � xk k yk k:(Solution If x ¼ 0; then the inequality is trivially true. Similarly, if y ¼ 0; then theinequality is trivially true. So, we consider the case when x and y both are nonzero.Thus, xk k and yk k are positive real numbers.

Put ~x � 1xk k x 2 Hð Þ; and ~y � 1

yk k y 2 Hð Þ: It suffices to show that ~x;~yð Þj j � 1:

Clearly, ~xk k ¼ 1 ¼ ~yk k: Thus, ~x; ~y are nonzero vectors in H: If ~x;~yð Þ ¼ 0; thenthe inequality ~x;~yð Þj j � 1 is trivially true. So, we consider the case when ~x;~yð Þ is anonzero complex number. It follows that ~x;~yð Þj j is nonzero. Since ~x;~yð Þ; ~x;~yð Þj j 2C� 0f gð Þ; and C� 0f gð Þ is a multiplicative group, there exists a 2 C� 0f gð Þsuch that ~x;~yð Þj j ¼ a ~x;~yð Þ: It follows that aj j ¼ 1: Since

0� a~x� ~y; a~x� ~yð Þ ¼ a�a ~x;~xð Þ � a ~x;~yð Þ � �a ~y;~xð Þþ ~y;~yð Þ¼ a�a ~xk k2�a ~x;~yð Þ � �a ~y;~xð Þþ ~yk k2

¼ aj j2 ~xk k2�a ~x;~yð Þ � �a ~y;~xð Þþ ~yk k2

¼ 12 ~xk k2�a ~x;~yð Þ � �a ~x;~yð Þ�ð Þþ ~yk k2

¼ ~xk k2�a ~x;~yð Þ � a ~x;~yð Þð Þ� þ ~yk k2

¼ ~xk k2� ~x;~yð Þj j � ~x;~yð Þj jð Þ� þ ~yk k2

¼ ~xk k2� ~x;~yð Þj j � ~x;~yð Þj j þ ~yk k2¼ 12 � ~x;~yð Þj j � ~x;~yð Þj j þ 12 ¼ 2 1� ~x;~yð Þj jð Þ;

we have 0� 2 1� ~x;~yð Þj jð Þ: It follows that ~x;~yð Þj j � 1: ∎)This conclusion, known as the Schwarz inequality, is due to H. A. Schwarz

(25.01.1843–30.11.1921, German). He developed a special case of Cauchy-Schwarz inequality. He proved that a ball has less surface area than any other bodyof equal volume.

Problem 2.61 For every x; y 2 H; xþ yk k� xk kþ yk k:(Solution Let us take any x; y 2 H: We have to show that

xþ yk k� xk kþ yk k;

that is

xþ yk k2 � xk kþ yk kð Þ2:

2.3 Inner Products 287

Here

xþ yk k2 ¼ xþ y; xþ yð Þ¼ x; xð Þþ x; yð Þþ y; xð Þþ y; yð Þ¼ xk k2 þ x; yð Þþ y; xð Þþ yk k2

¼ xk k2 þ x; yð Þþ x; yð Þ� þ yk k2

¼ xk k2 þ 2 Re x; yð Þð Þþ yk k2

� xk k2 þ 2 x; yð Þj j þ yk k2

� xk k2 þ 2 xk k yk kþ yk k2

¼ xk kþ yk kð Þ2;

so

xþ yk k2 � xk kþ yk kð Þ2: ∎)

This conclusion is known as the triangle inequality.

Problem 2.62 For every x; y 2 H; xk k � yk kj j � x� yk k:(Solution Let us take any x; y 2 H: It suffices to show that

xk k � yk k� x� yk k; and � xk k � yk kð Þ� x� yk k:

Since

xk k ¼ x� yð Þþ yk k� x� yk kþ yk k;

we have

xk k � yk k� x� yk k:

Similarly,

� xk k � yk kð Þ ¼ yk k � xk k� y� xk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ �1ð Þ x� yð Þk k ¼ �1j j x� yk k

¼ x� yk k;

and hence

� xk k � yk kð Þ� x� yk k: ∎)

288 2 Lp-Spaces

Problem 2.63 For every x; y 2 H;

1. xþ yk k2 þ x� yk k2¼ 2 xk k2 þ 2 yk k2;2. x; yð Þ ¼ 1

4 xþ yk k2� x� yk k2 þ i xþ iyk k2�i x� iyk k2

(Solution

1. Let us take any x; y 2 H: We have to show that

xþ yk k2 þ x� yk k2¼ 2 xk k2 þ 2 yk k2:

LHS ¼ xþ yk k2 þ x� yk k2¼ xþ y; xþ yð Þþ x� y; x� yð Þ¼ x; xð Þþ x; yð Þþ y; xð Þþ y; yð Þð Þþ x; xð Þ � x; yð Þ � y; xð Þþ y; yð Þð Þ¼ 2 x; xð Þþ 2 y; yð Þ ¼ 2 xk k2 þ 2 yk k2¼ RHS:

This result is known as the parallelogram law.

2. Let us take any x; y 2 H: We have to show that

x; yð Þ ¼ 14

xþ yk k2� x� yk k2 þ i xþ iyk k2�i x� iyk k2

:

RHS ¼ 14

xþ yk k2� x� yk k2 þ i xþ iyk k2�i x� iyk k2

¼ 14

xþ y; xþ yð Þ � x� y; x� yð Þþ i xþ iy; xþ iyð Þ � i x� iy; x� iyð Þð Þ

¼ 14

xk k2 þ yk k2 þ x; yð Þþ x; yð Þ�

� xk k2 þ yk k2� x; yð Þ � x; yð Þ�

þ i xk k2 þ yk k2�i x; yð Þþ i x; yð Þ�

� i xk k2 þ yk k2 þ i x; yð Þ � i x; yð Þ�

¼ 14

2 x; yð Þþ x; yð Þ�ð Þþ i2 �i x; yð Þþ i x; yð Þ�ð Þð Þ ¼ x; yð Þ ¼ LHS:

This result is known as the polarization identity. ∎)

Conclusion 2.64 Let H be an inner product space. Then1. 0k k ¼ 0; 2. xk k ¼ 0 implies x ¼ 0; 3. for every x 2 H; and, for every a 2 C;

axk k ¼ aj j xk k; 4. for every x; y 2 H; x; yð Þj j � xk k yk k; 5. for every x; y 2 H;

xþ yk k� xk kþ yk k; 6. for every x; y 2 H; xk k � yk kj j � x� yk k; 7. for every x; y 2H; xþ yk k2 þ x� yk k2¼ 2 xk k2 þ 2 yk k2; and 8. x; yð Þ ¼ 1

4 xþ yk k2� x� yk k2 þ

i xþ iyk k2�i x� iyk k2Þ:


Definition Let H be an inner product space. If we define the ‘distance’ between xand y to be x� yk k; all the axioms for a metric space are satisfied. Thus, H becomesa metric space. If this metric space is complete, then H is called a Hilbert space.

D. Hilbert (23.01.1862–14.02.1943, German) was a great mathematician. He isrecognized as one of the most influential and universal mathematicians of nine-teenth and early twentieth centuries. He is also famous for his collaboration withEinstein during the birth of general theory of relativity.

Example 2.65 Let n be a positive integer. For every n1; . . .; nnð Þ; g1; . . .; gnð Þ 2 Cn;put

n1; . . .; nnð Þ; g1; . . .; gnð Þð Þ � n1g1 þ � � � þ nngn:

Then Cn becomes a Hilbert space

Example 2.66 Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Let us take any f ; g 2 L2 lð Þ:

Since g 2 L2 lð Þ; g : X ! C is a measurable function, and hence �g : X ! C is ameasurable function. Since g 2 L2 lð Þ; we have

ZX

�gj j2dl0@ 1A1

2

¼ZX

gj j2dl0@ 1A1

2

6¼ 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

;

and hence

ZX

�gj j2dl0@ 1A1

2

6¼ 1:

Since �g : X ! C is a measurable function, and

ZX

�gj j2dl0@ 1A1

2

6¼ 1;

we have �g 2 L2 lð Þ: Since f ; �g 2 L2 lð Þ; by Lemma 2.21 f � �gð Þ 2 L1 lð Þ; and

f � �gk k1 � fk k2 �gk k2:

Since f � �gð Þ 2 L1 lð Þ; by Note 1.133,RX f � �gð Þdl 2 C:

290 2 Lp-Spaces

Notation By f ; gð Þ; we shall meanRX f � �gð Þdl:

Clearly, L2 lð Þ is an inner product space. For every f 2 L2 lð Þ; we have

fk k ¼ffiffiffiffiffiffiffiffiffiffif ; fð Þ

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZX

f � �fð Þdlvuut ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZX

fj j2dlvuut ¼

ZX

fj j2dl0@ 1A1

2

¼ kf k2:

Thus, for every f 2 L2 lð Þ; fk k ¼ fk k2: By Conclusion 2.33, L2 lð Þ is complete,and hence L2 lð Þ is a Hilbert space.

Note 2.67 Let H be a Hilbert space.

Problem 2.68 For fixed y 2 H; the mapping x 7! x; yð Þ is uniformly continuousfrom H to C:

(Solution Let us fix any y 2 H: We have to show that the mapping x 7! x; yð Þ isuniformly continuous from H to C:

Case I: when y 6¼ 0: It follows that yk k[ 0: Let us take any e[ 0: Let x1; x2 2H; and x1 � x2k k\ e

yk k : It suffices to show that

x1; yð Þ � x2; yð Þj j\e:

Since

x1; yð Þ � x2; yð Þj j ¼ x1 � x2; yð Þj j � x1 � x2k k yk k\ eyk k yk k ¼ e;

we have x1; yð Þ � x2; yð Þj j\e:Case II: when y ¼ 0: In this case, the mapping x 7! x; yð Þ becomes the constant

mapping x 7! 0; which is uniformly continuous from H to C. ∎)

Problem 2.69 For fixed y 2 H; the mapping x 7! y; xð Þ is uniformly continuousfrom H to C:

(Solution Similar as (i). ∎)

Problem 2.70 The mapping x 7! xk k is uniformly continuous from H to 0;1½ Þ:(Solution: Let us take any e[ 0: Let x1; x2 2 H and x1 � x2k k\e: It suffices toshow that

x1k k � x2k kj j\e:

Since

x1k k � x2k kj j � x1 � x2k k\e;


we have

x1k k � x2k kj j\e: ∎)

Conclusion 2.71 Let H be a Hilbert space. Then, 1. for fixed y 2 H; the mappingx 7! x; yð Þ is uniformly continuous from H to C; 2. for fixed y 2 H; the mappingx 7! y; xð Þ is uniformly continuous from H to C; and 3. the mapping x 7! xk k isuniformly continuous from H to 0;1½ Þ;Lemma 2.72 Let H be a Hilbert space. Let E be a nonempty, closed, convex subsetof H: There exists a unique y0 2 E such that for every x 2 E; y0k k� xk k:Proof Existence: Since E is nonempty, zk k : z 2 Ef g is a nonempty set of non-negative real numbers, and hence inf zk k : z 2 Ef g exists. Clearly,0� inf zk k : z 2 Ef g: There exists a sequence xnf g in E such that

limn!1 xnk k ¼ inf zk k : z 2 Ef g:

Since E is convex, and xnf g is a sequence in E; for every positive integer m; n;12 xm þ xnð Þ 2 E; and hence

inf zk k : z 2 Ef g� 12

xm þ xnð Þ�� :

Since, for every positive integer m; n;

inf zk k : z 2 Ef gð Þ2 þ 14

xm � xnk k2 � 12

xm þ xnð Þ�� 2 þ 1

4xm � xnk k2

¼ 12xm þ 1

2xn

�� 2 þ 12xm � 1

2xn

�� 2¼ 2

12xm

�� 2 þ 212xn

�� 2¼ 1

2xmk k2 þ 1

2xnk k2;

we have, for every positive integer m; n;

xm � xnk k2 � 2 xmk k2� inf zk k : z 2 Ef gð Þ2

þ 2 xnk k2� inf zk k : z 2 Ef gð Þ2

:

Now, since

limn!1 xnk k ¼ inf zk k : z 2 Ef g;

292 2 Lp-Spaces

xnf g is a Cauchy sequence in H: Since xnf g is a Cauchy sequence in H; and H isa Hilbert space, there exists y0 in H such that limn!1 xn ¼ y0: Since limn!1 xn ¼y0; we have

inf zk k : z 2 Ef g ¼ limn!1 xnk k ¼ y0k k|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl};

and hence inf zk k : z 2 Ef g ¼ y0k k: Since limn!1 xn ¼ y0; xnf g is a sequence in E;

and E is closed, we have y0 2 E: Since y0 2 E; and inf zk k : z 2 Ef g ¼ y0; we havemin zk k : z 2 Ef g ¼ y0k k: Thus, for every x 2 E; y0k k� xk k:

Uniqueness: If not, otherwise, suppose that there exist y1; y2 2 E such that, forevery x 2 E; y1k k� xk k; y2k k� xk k; and y1 6¼ y2: We have to arrive at acontradiction.

Clearly, y1k k ¼ y2k k: Since E is convex, and y1; y2 2 E; 12 y1 þ y2ð Þ 2 E; and

hence

y1k k� 12

y1 þ y2ð Þ�� :

Also,

y2k k� 12

y1 þ y2ð Þ�� :

Since

y1k kð Þ2 þ 14

y1 � y2k k2 � 12

y1 þ y2ð Þ�� 2 þ 1

4y1 � y2k k2

¼ 12y1 þ 1

2y2

�� 2 þ 12y1 � 1

2y2

�� 2¼ 2

12y1

�� 2 þ 212y2

�� 2¼ 1

2y1k k2 þ 1

2y2k k2

¼ 12

y1k k2 þ 12

y1k k2¼ y1k k2;

we have,

y1k kð Þ2 þ 14

y1 � y2k k2 � y1k k2;

and hence y1 ¼ y2: This is a contradiction. ∎


Definition Let H be a Hilbert space. Let x; y 2 H: If x; yð Þ ¼ 0; then we say that xis orthogonal to y; and we write x? y:

Notation Let H be a Hilbert space. Let a 2 H: The set x : x 2 H; and x?af g isdenoted by a?: It is clear that, for every a 2 H; a? is a closed linear subspace of H:

Notation Let H be a Hilbert space. Let M be a linear subspace of H: The setx : x 2 H; and for everym 2 M; x?mf g is denoted by M?: Clearly, M? is a closed

linear subspace of H:

Note 2.73 Let H be a Hilbert space. Let M be a closed linear subspace of H: Leta 2 H:

Since M is a linear subspace of H; aþM ¼ aþm : m 2 Mf gð Þ is a nonemptyconvex subset of H: Since x 7! aþ xð Þ is a homeomorphism from H onto H; and Mis a closed subset of H; aþM is a closed subset of H: Now, by Lemma 2.72, thereexists Q að Þ 2 M such that for every m 2 M;

aþQ að Þk k� aþmk k:Problem 2.74 aþQ að Þð Þ 2 M?:

(Solution Let us fix any m 2 M satisfying mk k ¼ 1: It suffices to show thataþQ að Þ;mð Þ ¼ 0: Here, for every a 2 C;

aþQ að Þk k2 � aþ Q að Þ � amð Þk k2

¼ aþQ að Þð Þ � amk k2¼ aþQ að Þð Þ � am; aþQ að Þð Þ � amð Þ¼ aþQ að Þ; aþQ að Þð Þ � �a aþQ að Þ;mð Þ � a m; aþQ að Þð Þþ a�a m;mð Þ¼ aþQ að Þk k2��a aþQ að Þ;mð Þ � a m; aþQ að Þð Þþ aj j2 mk k2

¼ aþQ að Þk k2��a aþQ að Þ;mð Þ � a m; aþQ að Þð Þþ aj j212¼ aþQ að Þk k2��a aþQ að Þ;mð Þ � a aþQ að Þ;mð Þ� þ aj j2;

so, for every a 2 C;

0� � �a aþQ að Þ;mð Þ � a aþQ að Þ;mð Þ� þ aj j2:

Now, let us take aþQ að Þ;mð Þ for a: We get

0� � �aa� a�aþ aj j2;

that is 0�ð Þ aj j2 � 0; that is, aþQ að Þ;mð Þ ¼ 0: ∎)

294 2 Lp-Spaces

Since Q að Þ 2 M; and M is a linear subspace of H; � Q að Þð Þ 2 M: Here a ¼� Q að Þð Þð Þþ aþQ að Þð Þ; where � Q að Þð Þ 2 M; and aþQ að Þð Þ 2 M?:

Conclusion 2.75 Let H be a Hilbert space. Let M be a closed linear subspace of H:

Let a 2 H: Then there exist unique x; y such that x 2 M; y 2 M?; and a ¼ xþ y:

Proof of uniqueness Let x; x0 2 M; y; y0 2 M?; and a ¼ xþ y ¼ x0 þ y0: We have toshow that x ¼ x0 and y ¼ y0: Here,

M3ð Þ x� x0ð Þ ¼ y0 � yð Þ 2 M?� �;

so

x� x0; x� x0ð Þ ¼ x� x0; y0 � yð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence x� x0; x� x0ð Þ ¼ 0: It follows that x� x0 ¼ 0; and hence x ¼ x0:Similarly, y ¼ y0: ∎

Note 2.76 Let H be a Hilbert space. Let M be a closed linear subspace of H: ByConclusion 2.75, for every x 2 H; there exist unique P xð Þ;Q xð Þ such that P xð Þ 2M; Q xð Þ 2 M?; and x ¼ P xð ÞþQ xð Þ: Thus, P : H ! M; and Q : H ! M?:

Problem 2.77 P : H ! M is linear, and Q : H ! M? is linear.

(Solution Let x; y 2 H: We have to show that

P xþ yð Þ ¼ P xð ÞþP yð Þ; and Q xþ yð Þ ¼ Q xð ÞþQ yð Þ:

Here,

x ¼ P xð ÞþQ xð Þ; y ¼ P yð ÞþQ yð Þ;

and

P xð ÞþP yð Þð Þþ Q xð ÞþQ yð Þð Þ ¼ P xð ÞþQ xð Þð Þþ P yð ÞþQ yð Þð Þ¼ xþ y ¼ P xþ yð ÞþQ xþ yð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Since

P xð ÞþP yð Þð Þþ Q xð ÞþQ yð Þð Þ ¼ P xþ yð ÞþQ xþ yð Þ;

P xð ÞþP yð Þð Þ;P xþ yð Þ 2 M;


and

Q xð ÞþQ yð Þð Þ;Q xþ yð Þ 2 M?;

we have

P xþ yð Þ ¼ P xð ÞþP yð Þ; and Q xþ yð Þ ¼ Q xð ÞþQ yð Þ:

Let a 2 C: We have to show that

P axð Þ ¼ a P xð Þð Þ; and Q axð Þ ¼ a Q xð Þð Þ:

Here, x ¼ P xð ÞþQ xð Þ; and

a P xð Þð Þþ a Q xð Þð Þ ¼ a P xð ÞþQ xð Þð Þ ¼ ax ¼ P axð ÞþQ axð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since

a P xð Þð Þþ a Q xð Þð Þ ¼ P axð ÞþQ axð Þ;

a P xð Þð Þ;P axð Þ 2 M and a Q xð Þð Þ;Q axð Þ 2 M?; we have

P axð Þ ¼ a P xð Þð Þ; and Q axð Þ ¼ a Q xð Þð Þ: ∎)

Problem 2.78 P : H ! M is onto, and Q : H ! M? is onto.

(Solution Let us take any a 2 M: Since

P að ÞþQ að Þ ¼ a ¼ aþ 0|fflfflfflfflfflffl{zfflfflfflfflfflffl};a 2 M and 0 2 M?; we have P að Þ ¼ a: It follows that P : H ! M is onto.Similarly, Q : H ! M? is onto. ∎)

Definition Here, P;Q are called the orthogonal projections of H onto M and M?:


Then, there exist orthogonal projections of H onto M and M?:

Note 2.80 Let H be a Hilbert space. Let M be a closed linear subspace of H: LetP;Q be the orthogonal projections of H onto M and M?: Let a 2 H:

296 2 Lp-Spaces

Problem 2.81 For every x 2 M; a� P að Þk k� a� xk k:(Solution Let us fix any x 2 M: It suffices to show that

Q að Þk k2 ¼ a� P að Þk k2 � a� xk k2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ P að ÞþQ að Þð Þ � xk k2¼ Q að Þþ P að Þ � xð Þk k2

¼ Q að Þþ P að Þ � xð Þ;Q að Þþ P að Þ � xð Þð Þ¼ Q að Þ;Q að Þð Þþ Q að Þ;P að Þ � xð Þþ P að Þ � x;Q að Þð Þþ P að Þ � x;P að Þ � xð Þ¼ Q að Þk k2 þ Q að Þ;P að Þ � xð Þþ P að Þ � x;Q að Þð Þþ P að Þ � xk k2

¼ Q að Þk k2 þ 0þ 0þ P að Þ � xk k2¼ Q að Þk k2 þ P að Þ � xk k2;

that is 0�P að Þ � x2: This is clearly true. ∎)


Let P;Q be the orthogonal projections of H onto M and M?: Let a 2 H: Then, forevery x 2 M; a� P að Þk k� a� xk k:Note 2.83 Let H be a Hilbert space. Let M be a closed linear subspace of H: LetP;Q be the orthogonal projections of H onto M and M?: Let a 2 H:

Problem 2.84 For every x 2 M?; a� Q að Þk k� a� xk k:(Solution Let us fix any x 2 M?: It suffices to show that

P að Þk k2 ¼ a� Q að Þk k2 � a� xk k2¼ P að ÞþQ að Þð Þ � xk k2¼ P að Þþ Q að Þ � xð Þk k2¼ P að Þþ Q að Þ � xð Þ;P að Þþ Q að Þ � xð Þð Þ¼ P að Þ;P að Þð Þþ P að Þ;Q að Þ � xð Þþ Q að Þ � x;P að Þð Þþ Q að Þ � x;Q að Þ � xð Þ¼ P að Þk k2 þ P að Þ;Q að Þ � xð Þþ Q að Þ � x;P að Þð Þþ Q að Þ � xk k2

¼ P að Þk k2 þ 0þ 0þ Q að Þ � xk k2¼ P að Þk k2 þ Q að Þ � xk k2;

that is, 0� Q að Þ � xk k2: This is clearly true. ∎)


Let P;Q be the orthogonal projections of H onto M and M?: Let a 2 H: Then, forevery x 2 M?; a� Q að Þk k� a� xk k:Note 2.86 Let H be a Hilbert space. Let M be a closed linear subspace of H: LetP;Q be the orthogonal projections of H onto M and M?: Let a 2 H:

Problem 2.87 ak k2¼ P að Þk k2 þ Q að Þk k2:(Solution

LHS ¼ ak k2¼ P að ÞþQ að Þk k2¼ P að ÞþQ að Þ;P að ÞþQ að Þð Þ¼ P að Þ;P að Þð Þþ P að Þ;Q að Þð Þþ Q að Þ;P að Þð Þþ Q að Þ;Q að Þð Þ¼ P að Þk k2 þ P að Þ;Q að Þð Þþ Q að Þ;P að Þð Þþ Q að Þk k2

¼ P að Þk k2 þ 0þ 0þ Q að Þk k2¼ P að Þk k2 þ Q að Þk k2¼ RHS: ∎)



Let P;Q be the orthogonal projections of H onto M and M?: Let a 2 H: Thenak k2¼ P að Þk k2 þ Q að Þk k2:

Note 2.89 Let H be a Hilbert space. Let M be a closed linear subspace of H: LetM 6¼ H: Let P;Q be the orthogonal projections of H onto M and M?: SinceM 6¼ H; and M is a linear subspace of H; there exists a nonzero y 2 H such thatP yð ÞþQ yð Þ ¼ð Þy 62 M: It follows that Q yð Þ is a nonzero member of M?: Thus,Q yð Þ 6¼ 0; and Q yð Þ ? M:

Conclusion 2.90 Let H be a Hilbert space. Let M be a closed linear subspace of H:Let M 6¼ H: Then there exists a nonzero z in H such that z ? M:

Note 2.91 Let H be a Hilbert space. Let L : H ! C be a linear functional. Let L becontinuous.

Since L : H ! C is linear, N Lð Þ ¼ x : L xð Þ ¼ 0f gð Þ is a linear subspace of H:

Problem 2.92 N Lð Þ is a closed subset of H:

(Solution Let xnf g be any convergent sequence in N Lð Þ: Let limn!1 xn ¼ x forsome x 2 H:We have to show that x 2 N Lð Þ; that is, L xð Þ ¼ 0: Since L : H ! C iscontinuous, and limn!1 xn ¼ x; we have

0 ¼ limn!1 0 ¼

limn!1 L xnð Þ ¼ L xð Þ;

and hence L xð Þ ¼ 0: ∎)Case I: when N Lð Þ 6¼ H: By Conclusion 2.89, there exists a nonzero z in H such

that z?N Lð Þ: Let us take any x 2 H: Observe that

L xð Þð Þ zzk k � L

zzk k

� �� x

� �2 N Lð Þ:

Now, since z?N Lð Þ; we have

L xð Þð Þ�� Lzzk k

� �� zzk k ; x

� �¼ L xð Þð Þ�1� L

zzk k

� �� zzk k ; x

� �¼ L xð Þð Þ� z

zk k ;zzk k

� �� L

zzk k

� �� zzk k ; x

� �¼ z

zk k ; L xð Þð Þ zzk k � L

zzk k

� �� x

� �¼ 0;

and hence

L xð Þð Þ�¼ Lzzk k

� �� zzk k ; x

� �:

298 2 Lp-Spaces

It follows that

L xð Þ ¼ Lzzk k

� �� x;

zzk k

� �¼ x; L

zzk k

� �� zzk k

� �� :

Thus, for every x 2 H;

L xð Þ ¼ x; Lzzk k

� �� zzk k

� �:

Case II: when N Lð Þ ¼ H: Here, for every x 2 H; we have L xð Þ ¼ x; 0ð Þ:Conclusion 2.93 Let H be a Hilbert space. Let L : H ! C be a linear functional.Let L be continuous. Then there exists a unique y0 2 H such that for every x 2 H;L xð Þ ¼ x; y0ð Þ:Proof of uniqueness Let y0; y00 2 H such that for every x 2 H; L xð Þ ¼ x; y0ð Þ; andL xð Þ ¼ x; y00ð Þ: We have to show that y0 ¼ y00: Since

L y0 � y00ð Þ ¼ y0 � y00; y0ð Þ; and L y0 � y00ð Þ ¼ y0 � y00; y00ð Þ;

we have

y0 � y00; y0 � y00ð Þ ¼ y0 � y00; y0ð Þ � y0 � y00; y00ð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence y0 � y00 ¼ 0: Thus, y0 ¼ y00: ∎)

2.4 Orthogonal Sets

Using orthogonal sets, we shall introduce some geometrical ideas in Hilbert spaces.

Note 2.94

Definition Let H be a Hilbert space. Let ui : i 2 If g be any collection of vectorsin H: If

ui; uj� � � 0 if i 6¼ j

1 if i ¼ j;

�then we say that ui : i 2 If g is an orthonormal set in H:

Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:


Problem 2.95 Each ui is nonzero.

(Solution Let us fix any i 2 I: We have to show that ui 6¼ 0: Since uk : k 2 If g is

an orthonormal set in H; uik k2¼

ui; uið Þ ¼ 1; and hence uik k ¼ 1: This shows

that ui 6¼ 0: ∎)

Problem 2.96 If i 6¼ j; then ui 6¼ uj:

(Solution Let i 6¼ j:We have to show that ui 6¼ uj: If not, otherwise, let ui ¼ uj:Wehave to arrive at a contradiction. Since i 6¼ j; and uk : k 2 If g is an orthonormal setin H; we have

uik k2¼ ui; uið Þ ¼ ui; uj� � ¼ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl};

and hence, uk ki¼ 0: This shows that ui ¼ 0: Since uk : k 2 If g is an orthonormalset in H; we have ui 6¼ 0: This is a contradiction. ∎)

Definition Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let x 2 H: By x ið Þ we mean the complex number x; uið Þ; x ið Þ are called the Fouriercoefficients of x; relative to uk : k 2 If g: Thus, for every x 2 H; x : I ! C:

Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H: Let F be anonempty finite subset of I: Let MF be the linear span of uk : k 2 Ff g: Let u :I ! C be a function such that for every i 2 I � Fð Þ; u ið Þ ¼ 0:

Problem 2.97 For every i 2 I; u ið Þ ¼ Pj2F u jð Þð Þuj; ui

:

(Solution Let us fix any i0 2 I: Let F ¼ i1; . . .; inf g; where i1; . . .; in are distinct.We have to show that

u i0ð Þ ¼ u i1ð Þð Þui1 þ � � � þ u inð Þð Þuin ; ui0ð Þ:

Case I: when i0 2 F: It follows that i0 2 i1; . . .; inf g: For simplicity, let i0 ¼ i1:

RHS ¼ u i1ð Þð Þui1 þ � � � þ u inð Þð Þuin ; ui0ð Þ ¼ u i1ð Þð Þui1 þ � � � þ u inð Þð Þuin ; ui1ð Þ¼ u i1ð Þð Þ ui1 ; ui1ð Þþ u i2ð Þð Þ ui2 ; ui1ð Þþ � � � þ u inð Þð Þ uin ; ui1ð Þ¼ u i1ð Þð Þ 1ð Þþ u i2ð Þð Þ 0ð Þþ � � � þ u inð Þð Þ 0ð Þ ¼ u i1ð Þ ¼ u i0ð Þ ¼ LHS:

Case II: when i0 62 F: It follows that u i0ð Þ ¼ 0; and i0 62 i1; . . .; inf g:

RHS ¼ uði1Þð Þui1 þ � � � þ uði1Þð Þuin ; uioð Þ¼ ðuði1ÞÞðui1 ; uioÞþ ðuði2ÞÞðui2 ; uioÞþ � � � þ ðuðinÞÞðuin ; uioÞ¼ ðuði1ÞÞð0Þþ ðuði2ÞÞð0Þþ � � � þ ðuðinÞÞð0Þ ¼ 0 ¼ uði0Þ ¼ LHS:

300 2 Lp-Spaces

Thus, in all cases,

u i0ð Þ ¼ u i1ð Þð Þui1 þ � � � þ u inð Þð Þuin ; ui0ð Þ: ∎)

Clearly,P

j2F u jð Þð Þuj 2 MF : Also,

Xj2F

u jð Þð Þuj��

��2

¼Xj2F

u jð Þð Þuj;Xj2F

u jð Þð Þuj !

¼Xj2F

u jð Þð Þ u jð Þð Þ� uj; uj� �

¼Xj2F

u jð Þj j2:

Conclusion 2.98 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let F be a nonempty finite subset of I: Let MF be the linear span ofuk : k 2 Ff g: Let u : I ! C be a function such that, for every i 2 I � Fð Þ; u ið Þ ¼

0: Then there exists y 2 MF such that for every i 2 I; u ið Þ ¼ y ið Þ: Also,yk k2¼Pj2F y jð Þj j2:

Note 2.99 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let F be a nonempty finite subset of I: Let MF be the linear span of uk : k 2 Ff g:Let a 2 H: Let s 2 MF ; and

s 6¼Xj2F

a jð Þð Þuj:

Problem 2.100 ka�Pj2F a jð Þð Þujk\ a� sk k:(Solution Observe that, if k 2 F; then

a�Xj2F

a jð Þð Þuj; uk !

¼ a; ukð Þ � a kð Þð Þ uk; ukð Þ

¼ a; ukð Þ � a kð Þð Þ1 ¼ a kð Þ � a kð Þ ¼ 0:

Thus, for every k 2 F;

a�Xj2F


¼ 0:

2.4 Orthogonal Sets 301

Since s 2 MF ; for every k 2 F; there exists ak 2 C such that s ¼Pk2F akuk:Observe that,

a�Xj2F

a jð Þð Þuj;Xj2F

a jð Þð Þuj � s

!¼Xk2F

a kð Þð Þ�ð Þ a�Xj2F


� a�Xj2F

a jð Þð Þuj; s !

¼Xk2F

a kð Þð Þ�ð Þ0� a�Xj2F


¼ � a�Xj2F


¼ � a�Xj2F

a jð Þð Þuj;Xk2F

akuk

!

¼ �Xk2F

akð Þ�ð Þ a�Xj2F


¼ �Xk2F

akð Þ�ð Þ0 ¼ 0;

so

a�Xj2F



!¼ 0:

Now,

a� sk k2 ¼ a�Xj2F

a jð Þð Þuj !

þXj2F


!��2

¼ a�Xj2F

a jð Þð Þuj��

��2

þXj2F

a jð Þð Þuj � s2��

��2

þ a�Xj2F



!þ a�

Xj2F



!�

¼ a�Xj2F


��2

þXj2F


��2

þ 0ð Þþ 0ð Þ�

¼ a�Xj2F


��2

þXj2F


��2

[ a�Xj2F


��2

;

so

a�Xj2F


��2

\ a� sk k2;

302 2 Lp-Spaces

and hence

a�Xj2F


��\ a� sk k:∎)

Conclusion 2.101 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let F be a nonempty finite subset of I: Let MF be the linear span ofuk : k 2 Ff g: Let a 2 H: Let s 2 MF ; and s 6¼Pj2F a jð Þð Þuj: Then

a� Pj2F a jð Þð Þuj

�� \ a� sk k:

Note 2.102 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let F be a nonempty finite subset of I: Let a 2 H:

Problem 2.103P

j2F a jð Þj j2 � ak k2:(Solution Observe that, if k 2 F; then

a�Xj2F


¼ a; ukð Þ � a kð Þð Þ uk; ukð Þ

¼ a; ukð Þ � a kð Þð Þ1¼ a kð Þ � a kð Þ ¼ 0:

Thus, for every k 2 F;

a�Xj2F


¼ 0:

Observe that,

a�Xj2F


a jð Þð Þuj !

¼Xk2F

a kð Þð Þ�ð Þ a�Xj2F


¼Xk2F

a kð Þð Þ�ð Þ0 ¼ 0;

so

a�Xj2F


a jð Þð Þuj !

¼ 0:


Now,

ak k2 ¼ a�Xj2F

a jð Þð Þuj !

þXj2F


��2

¼ a�Xj2F


��2

þXj2F


��2

þ a�Xj2F


a jð Þð Þuj !

þ a�Xj2F


a jð Þð Þuj !�

¼ a�Xj2F


��2

þXj2F


��2

þ 0ð Þþ 0ð Þ�

¼ a�Xj2F


��2

þXj2F


��2

�Xj2F


��2

¼Xj2F


a jð Þð Þuj !

¼Xj2F

a jð Þð Þ a jð Þð Þð Þ� uj; uj� �

¼Xj2F

a jð Þj j2 1ð Þ ¼Xj2F

a jð Þj j2;

so Xj2F

a jð Þj j2 � ak k2:∎)

Conclusion 2.104 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let F be a nonempty finite subset of I: Let a 2 H: Then,

Pj2F a jð Þj j2 � ak k2:

Lemma 2.105 Let X; dð Þ; Y ; qð Þ be metric spaces. Let X be complete. Let f : X !Y be continuous. Let X0 be a dense subset of X: Let f be an ‘isometry’ on X0; in thesense that, for every a; b 2 X0; d a; bð Þ ¼ q f að Þ; f bð Þð Þ: Let f X0ð Þ be dense in Y :Then, f is an isometry from X onto Y :

Proof

1. f : X ! Y is an isometry: Let a; b 2 X: We have to show that d a; bð Þ ¼q f að Þ; f bð Þð Þ: Since X0 is a dense subset of X; and a 2 X; there exists a sequencexnf g in X0 such that limn!1 xn ¼ a: Similarly, there exists a sequence ynf g in

X0 such that limn!1 yn ¼ b: Since xnf g is a sequence in X0 such that lim

n!1 xn ¼ a;

and f : X ! Y is continuous, limn!1 f xnð Þ ¼ f að Þ: Similarly, lim

n!1 f ynð Þ ¼ f bð Þ:

304 2 Lp-Spaces

LHS ¼ d a; bð Þ ¼ d limn!1 xn; lim

n!1 yn

¼ limn!1 d xn; ynð Þ

¼ limn!1 q f xnð Þ; f ynð Þð Þ ¼ q lim

n!1 f xnð Þ; limn!1 f ynð Þ

¼ q f að Þ; f bð Þð Þ ¼ RHS:

2. f : X ! Y is onto: Let y 2 Y : Since y 2 Y ¼ f X0ð Þð Þ�ð Þ; there exists a sequencexnf g in X0 such that lim

n!1 f xnð Þ ¼ y: It follows that f xnð Þf g is a Cauchy sequencein f X0ð Þ: Since f xnð Þf g is a Cauchy sequence in f X0ð Þ; xnf g is a sequence in X0;and f is an ‘isometry’ on X0; xnf g is a Cauchy sequence in X0: Since xnf g is aCauchy sequence in X0 � Xð Þ; and X is complete, there exists x 2 X such thatlimn!1 xn ¼ x: It suffices to show that

f limn!1 xn

¼ f xð Þ ¼ y|fflfflfflffl{zfflfflfflffl} ¼ limn!1 f xnð Þ;

that is

f limn!1 xn

¼ limn!1 f xnð Þ:

Since f : X ! Y is continuous, and limn!1 xn ¼ x; we have

f limn!1 xn

¼ limn!1 f xnð Þ: ∎

Note 2.106 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let l : P Ið Þ ! 0;1½ � be the counting measure on I: Let u : I ! C be any mea-surable function. Let p 2 1;1½ Þ: Let u 2 ‘p Ið Þ: It follows that

sup u k1ð Þj jp þ � � � þ u knð Þj jp: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of If g

¼ZIuj jpdl

� �2 0;1½ �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Notation Here,

sup u k1ð Þj jp þ � � � þ u knð Þj jp: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of If g

is denoted byP

k2I u kð Þj jp:


Thus,

u 2 ‘2 Ið Þ if and only ifXk2I

u kð Þj j2\1:

By Example 2.66, ‘2 Ið Þ is a Hilbert space, where for every u;w 2 ‘2 Ið Þ;

u;wð Þ ¼ZI

u � �w� �dl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ sup u k1ð Þð Þ w k1ð Þð Þ� þ � � � þ u knð Þð Þ w knð Þð Þ�: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of If g

Notation Here,

sup u k1ð Þð Þ w k1ð Þð Þ� þ � � � þ u knð Þð Þ w knð Þð Þ�: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of If g

is denoted by Xk2I

u kð Þð Þ w kð Þð Þ�:

Thus, for every u;w 2 ‘2 Ið Þ;

u;wð Þ ¼Xk2I

u kð Þð Þ w kð Þð Þ�:

By Lemma 2.24, if u;w 2 ‘2 Ið Þ; then u � �w� � 2 ‘1 Ið Þ:ByTheorem 2.47, the collection of all measurable simple functions s : I ! C such

that l k : s kð Þ 6¼ 0f gð Þ\1; is dense in ‘2 Ið Þ, that is, the collection of all functionss : I ! C such that s is zero except on some finite subset of I; is dense in ‘2 Ið Þ:

Let u 2 ‘2 Ið Þ:Problem 2.107 k : u kð Þ 6¼ 0f g is countable.

(Solution Since u 2 ‘2 Ið Þ; we havePk2I

u kð Þj j2\1: Since

k : u kð Þ 6¼ 0f g ¼ [ n2N k :1n\ u kð Þj j

� �¼ [ n2N k :

1n2

\ u kð Þj j2� �

;

it suffices to show that each

k :1n2

\ u kð Þj j2� �

¼ k : 1\ n2u2� �kð Þ��

306 2 Lp-Spaces

is finite. Let us fix any positive integer n: It suffices to show that the number ofelements in

k : 1\ n2u2� �kð Þ��

is finite. Since

the number of elements in k : 1\ n2u2� �kð Þ��

\X

1n\ u kð Þj j

n2u2� �kð Þ��

�X

u kð Þ6¼0

n2u2� �kð Þ�� ¼ n2

Xk2I

u kð Þj j2\1;

the number of elements in k : 1\ n2u2ð Þ kð Þ�� is finite. ∎)

Conclusion 2.108 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let u : I ! C be any measurable function. Let u 2 ‘2 Ið Þ: Thenk : u kð Þ 6¼ 0f g is countable.

Note 2.109 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let a 2 H: By Conclusion 2.104, for every nonempty finite subset F of I;X

j2Fa jð Þj j2 � ak k2;

and hence

Xk2I

a kð Þj j2 ¼ !

sup a k1ð Þj j2þ � � � þ a knð Þj j2: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of In o

� ak k2:

Thus Xk2I

a kð Þj j2 � ak k2:

Conclusion 2.110 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let a 2 H: Then X

k2Ia kð Þj j2 � ak k2:

This inequality, known as the Bessel inequality, is due to F. W. Bessel(22.07.1784–17.03.1846, German). He was an astronomer, mathematician andphysicist. He was the first astronomer to calculate a reliable value for the distancefrom the sun to another star by the method of parallax.


Note 2.111 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let x 2 H: By Conclusion 2.110,X

k2Ix kð Þj j2 � xk k2 \1ð Þ;

and hence Xk2I

x kð Þj j2\1:

It follows that x 2 ‘2 Ið Þ: Thus, x 7! x is a mapping from Hilbert space H toHilbert space ‘2 Ið Þ:Problem 2.112 x 7! x is a linear map from H to ‘2 Ið Þ: Also, x 7! x is a continuousmap from H to ‘2 Ið Þ:(Solution Let x; y 2 H; and a; b 2 C: We have to show that

axþ byð Þ^¼ aðxÞþ b yð Þ:

For this purpose, let us take any k 2 I: We have to show that

axþ byð Þ^ kð Þ ¼ aðxÞþ b yð Þð Þ kð Þ:

LHS ¼ axþ byð Þ^ kð Þ ¼ axþ by; ukð Þ ¼ a x; ukð Þþ b y; ukð Þ¼ a x kð Þð Þþ b y kð Þð Þ ¼ aðxÞþ b yð Þð Þ kð Þ ¼ RHS:

Let us fix any a 2 H:We have to show that the map x 7! x is continuous at a. Forthis purpose, let us take any e[ 0: Let x� ak k\e; where x 2 H: It suffices to showthat x� a; x� að Þ\e2: On using Conclusion 2.110, we get

x� a; x� að Þ ¼ x� að Þ^; x� að Þ^� �¼Xk2I

x� að Þ^ kð Þ� �x� að Þ^ kð Þ� ��

¼Xk2I

x� að Þ^ kð Þ�� 2 � x� a2\e2;

and hence x� a; x� að Þ\e2: ∎)

Conclusion 2.113 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: By Problem 2.112, the mapping x 7! x from Hilbert space H to Hilbert space‘2 Ið Þ is linear and continuous.

308 2 Lp-Spaces

Note 2.114 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:

Problem 2.115 For every k1; k2 2 I;

uk1 ; uk2ð Þ ¼ uk1ð Þ^; uk2ð Þ^� �:

(Solution Case I: when k1 6¼ k2:

RHS ¼ uk1ð Þ^; uk2ð Þ^� � ¼Xk2I

uk1ð Þ^ kð Þ� �uk2ð Þ^ kð Þ� ��

¼Xk2I

uk1 ; ukð Þ uk2 ; ukð Þ�¼Xk2I

0 ¼ 0 ¼ uk1 ; uk2ð Þ ¼ LHS:

Case II: when k1 ¼ k2:

RHS ¼ uk1ð Þ^; uk2ð Þ^� � ¼Xk2I

uk1ð Þ^ kð Þ� �uk2ð Þ^ kð Þ� ��

¼Xk2I

uk1 ; ukð Þ uk2 ; ukð Þ�¼Xk2I

uk1 ; ukð Þ uk1 ; ukð Þ�

¼Xk2I

uk1 ; ukð Þj j2 ¼ uk1 ; uk1ð Þj j2

¼ 1j j2¼ 1 ¼ uk1 ; uk1ð Þ ¼ uk1 ; uk2ð Þ ¼ LHS: ∎)

Conclusion 2.116 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Then, for every k1; k2 2 I;

uk1 ; uk2ð Þ ¼ uk1ð Þ^; uk2ð Þ^� �:

Note 2.117 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let P be the linear span of uk : k 2 If g: Let a1uk1 þ � � � þ anukn ; b1ul1 þ � � �þ bmulm 2 P; where a1; . . .; an; b1; . . .; bm 2 C; and k1; . . .; kn; l1; . . .; lm 2 I:

Problem 2.118 a1uk1 þ � � � þ anukn ; b1ul1 þ � � � þ bmulmð Þ ¼ a1uk1 þ � � � þ anuknð Þ^;�b1ul1 þ � � � þ bmulmð Þ^Þ:(Solution

RHS ¼ a1uk1 þ � � � þ anuknð Þ^; b1ul1 þ � � � þ bmulmð Þ^� �¼ a1 uk1ð Þ^ þ � � � þ an uknð Þ^; b1 ul1ð Þ^ þ � � � þ bm ulmð Þ^� �¼ a1b1 uk1ð Þ^; ul1ð Þ^� �þ � � � þ anbm uknð Þ^; ulmð Þ^� �¼ a1b1 uk1 ; ul1ð Þþ � � � þ anbm ukn ; ulmð Þ¼ a1uk1 þ � � � þ anukn ; b1ul1 þ � � � þ bmulmð Þ ¼ LHS: ∎)


Conclusion 2.119 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let P be the linear span of uk : k 2 If g: Leta1uk1 þ � � � þ anukn ; b1ul1 þ � � � þ bmulm 2 P; where a1; . . .; an; b1; . . .; bm 2 C;and k1; . . .; kn; l1; . . .; lm 2 I: Then,

a1uk1 þ � � � þ anukn ; b1ul1 þ � � � þ bmulmð Þ¼ a1uk1 þ � � � þ anuknð Þ^; b1ul1 þ � � � þ bmulmð Þ^� �

:

Note 2.120 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let P be the linear span of uk : k 2 If g: By Conclusion 2.119, it is clear that x 7! xis an isometry from P to ‘2 Ið Þ:

The image set of the mapping x 7! x from P to ‘2 Ið Þ is

a1uk1 þ � � � þ anuknð Þ^: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of I; and a1; . . .; an 2 C� �¼ a1 uk1ð Þ^ þ � � � þ an uknð Þ^: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of I; and a1; . . .; an 2 C� �

¼ the collection of all functions s : I ! C such that s is zero except on some finite subset of Ið Þ;

so, the image set of the mapping x 7! x from P to ‘2 Ið Þ is equal to the collection of allfunctions s : I ! C such that s is zero except on some finite subset of I; and hence, byConclusion 2.108, the image set of themapping x 7! x fromP to ‘2 Ið Þ is dense in ‘2 Ið Þ:

Since P is a linear subspace of H; the closure �P is a linear subspace of H: Since �Pis a closed subset of the complete space H; �P is complete. Now, we can applyLemma 2.105 with X ¼ �P; X0 ¼ P; Y ¼ ‘2 Ið Þ; and f is the continuous map x 7! xfrom �P to ‘2 Ið Þ: Since the map x 7! x is an ‘isometry’ from P to ‘2 Ið Þ; by Lemma2.105, x 7! x is an isometry from �P � Hð Þ onto ‘2 Ið Þ; and hence x 7! x is a linearcontinuous mapping from H onto ‘2 Ið Þ:Conclusion 2.121 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let P be the linear span of uk : k 2 If g: Then1. x 7! x is a linear continuous mapping from H onto ‘2 Ið Þ;2. the restriction of x 7! x to �P is an isometry from �P onto ‘2 Ið Þ:

Here, the Conclusion (1), known as the Riesz-Fischer theorem, is due to F.Riesz, and E. S. Fischer (12.07.1875–14.11.1954, Austrian). Fischer’s main area ofresearch was orthogonal sequence of functions. This laid the foundation for theconcept of Hilbert space.


Problem 2.123 uk : k 2 If g is a maximal orthonormal set in H if and only if theredoes not exist any nonzero v in H such that v is orthogonal to each uk:

(Solution Let uk : k 2 If g be a maximal orthonormal set in H: We claim that theredoes not exist any nonzero v in H such that v is orthogonal to each uk: If not,otherwise, let there exist a nonzero v in H such that v is orthogonal to each uk: Wehave to arrive at a contradiction.

310 2 Lp-Spaces

It follows that

uk : k 2 If g[ 1kvk v� �

is an orthonormal set in H: Now, since uk : k 2 If g is a maximal orthonormalset in H;

uk : k 2 If g ¼ uk : k 2 If g[ 1kvk v� �

;

and hence

1kvk v 2 uk : k 2 If g:

It follows that there exists k0 2 I such that 1kvk v ¼ uk0 : Since v is orthogonal to

each uk;

1 ¼ uk0 ; uk0ð Þ ¼ 1kvk v; uk0� �

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, we get a contradiction.Conversely, suppose that there does not exist any nonzero v in H such that v is

orthogonal to each uk: We have to show that uk : k 2 If g is a maximal orthonormalset in H: If not, otherwise, suppose that there exists v 62 uk : k 2 If g such thatuk : k 2 If g[ vf g is an orthonormal set in H: We have to arrive at a contradiction.It follows that, for every k 2 I; uk; vð Þ ¼ 0; and hence v is orthogonal to each uk:

Since uk : k 2 If g[ vf g is an orthonormal set in H; v; vð Þ ¼ 1 6¼ 0ð Þ; and hence v isnonzero. Thus, v is a nonzero vector in H such that v is orthogonal to each uk: Thisis a contradiction. ∎)

Conclusion 2.124 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: uk : k 2 If g is a maximal orthonormal set in H if and only if there does notexist any nonzero v in H such that v is orthogonal to each uk:

Note 2.125 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let P be the linear span of uk : k 2 If g: Let uk : k 2 If g be a maximal orthonormalset in H:

Problem 2.126 P is dense in H:

(Solution If not, otherwise, let P not be dense in H: We have to arrive at acontradiction.

Since P is not dense in H; �P 6¼ H: Since P is a linear subspace of H; �P is a closedlinear subspace of H: Now, by Conclusion 2.90, there exists a nonzero z in H suchthat z ? �P, and hence, for every k 2 I; uk; zð Þ ¼ 0: Since uk : k 2 If g is a maximal


orthonormal set in H; and z is a nonzero vector in H; by Conclusion 2.124, z is notorthogonal to some uk: Thus, there exists k0 2 I such that uk0 ; zð Þ 6¼ 0: This is acontradiction. ∎)

Conclusion 2.127 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let P be the linear span of uk : k 2 If g: Let uk : k 2 If g be a maximalorthonormal set in H: Then P is dense in H:

Note 2.128 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let P be the linear span of uk : k 2 If g: Let P be dense in H: Let x 2 H:

Problem 2.129P

k2I xðkÞ2�� ¼ xk k2:(Solution By Conclusion 2.121, the restriction of x 7! x to �P ¼ Hð Þ is an isometryfrom �P ¼ Hð Þ onto ‘2 Ið Þ: Thus, x 7! x is an isometry from H onto ‘2 Ið Þ: Now, sincex 2 H;

x ¼ bx2 ¼ ffiffiffiffiffiffiffiffiffiffix; xð Þ

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisup x k1ð Þð Þ x k1ð Þð Þ� þ � � � þ x knð Þð Þ x knð Þð Þ�: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of If g

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisup x k1ð Þj j2 þ � � � þ x knð Þj j2: n ¼ 1; 2; . . .; k1; . . .; kn are distinct members of In or

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiXk2I

x kð Þj j2s

;

and hence xk k2¼Pk2I x kð Þj j2: ∎)

Conclusion 2.130 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let P be the linear span of uk : k 2 If g: Let P be dense in H: Let x 2 H: Then,P

k2I x kð Þj j2 ¼ xk k2:Note 2.131 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:

Suppose that, for every x 2 H; ððkxk2Þ2 ¼ÞPk2I jxðkÞj2 ¼ kxk2: Let a; b 2 H:

Problem 2.132Pk2I

a kð Þð Þ b kð Þ� ��¼ a; bð Þ:

(Solution

RHS ¼ a; bð Þ ¼ 14

aþ bk k2� a� bk k2 þ i aþ ibk k2�i a� ibk k2

¼ 14

aþ bð Þ^�� 2

2� a� bð Þ^��

2

2þ i aþ ibð Þ^��

2

2�i a� ibð Þ^��

2

2� �¼ 1

4aþ b��

2

2� a� b��

2

2þ i aþ ib

�� 2

2�i a� ib

�� 2

2� �¼ a; b� � ¼X

k2Ia kð Þð Þ b kð Þ� ��¼ LHS:

∎)

312 2 Lp-Spaces

Conclusion 2.133 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Suppose that, for every x 2 H; ððkxk2Þ2 ¼ÞPk2I jxðkÞj2 ¼ kxk2: Let a; b 2H: Then,

Pk2I a kð Þð Þ b kð Þ� ��¼ a; bð Þ:


Suppose that, for every a; b 2 H;P

k2I a kð Þð Þ b kð Þ� ��¼ a; bð Þ:Problem 2.135 uk : k 2 If g is a maximal orthonormal set in H:

(Solution If not, otherwise, let uk : k 2 If g be not a maximal orthonormal set in H:We have to arrive at a contradiction. Since uk : k 2 If g is not a maximalorthonormal set in H; by Conclusion 2.124, there exists any any nonzero v in Hsuch that v is orthogonal to each uk: It follows that

vk k2 ¼ v; vð Þ ¼Xj2I

v jð Þð Þ v jð Þð Þ�|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Xj2I

v jð Þj j2 ¼Xj2I

v; ukð Þj j2 ¼Xj2I

0j j2 ¼ 0;

and hence vk k2¼ 0 Thus, v ¼ 0: This is a contradiction. ∎)If we recollect the conclusions in Note 2.122 to Note 2.131, we get the following

Conclusion 2.136 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Let P be the linear span of uk : k 2 If g: Then the following four statementsare equivalent:

1. uk : k 2 If g is a maximal orthonormal set in H;2. P is dense in H;

3. for every x 2 H;P

k2I x kð Þj j2 ¼ xk k2;4. for every x; y 2 H;

Pk2I x kð Þð Þ y kð Þð Þ�¼ x; yð Þ:

Here, the formula (4), known as the Parseval’s identity, is due to M.A. Parseval (27.04.1755–16.08.1836, French). He proved that the integral of thesquare of a function is equal to the integral of the square of its transform.

Definition Maximal orthonormal sets are also known as complete orthonormalsets or orthonormal bases.

Note 2.137 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H:Let P be the collection of all orthonormal sets B in H such that uk : k 2 If g � B:

Clearly, uk : k 2 If g 2 P; and hence P is nonempty. Obviously, P;�ð Þ is a par-tially ordered set. Now, by the Hausdorff maximality axiom of set theory, thereexists a maximal linearly ordered set B such that B � P: It follows thatuk : k 2 If g � [Bð Þ:


Problem 2.138 [B is an orthonormal set in H:

(Solution For this purpose, let us take any x; y 2 [Bð Þ: We have to show that

x; yð Þ ¼ 0 if x 6¼ y1 if x ¼ y:

�Since x 2 [Bð Þ; there exists A1 2 B such that x 2 A1: Similarly, there exists

A2 2 B such that y 2 A2: Since A1;A2 2 B; and B is a linearly ordered set, we haveA1 � A2 or A2 � A1: For definiteness, let A1 � A2: It follows that x; y 2 A2: SinceA2 2 B � Pð Þ; we have A2 2 P; and hence A2 is an orthonormal set in H: Since A2

is an orthonormal set in H; and x; y 2 A2; we have

x; yð Þ ¼ 0 if x 6¼ y1 if x ¼ y:

�∎)

Problem 2.139 [B is a maximal orthonormal set in H:

(Solution If not, otherwise, suppose that there exists an orthonormal set C in Hsuch that [Bð Þ � C; and [Bð Þ 6¼ C: We have to arrive at a contradiction. Itfollows that B[ Cf g is a linearly ordered set such that B � B[ Cf gð Þ: Now, sinceB is a maximal linearly ordered set, B ¼ B[ Cf gð Þ; and hence C 2 B: SinceC 2 B; we have C � [Bð Þ � Cð Þ; and hence [Bð Þ ¼ C: This is a contradiction.∎)

Thus, [B is a maximal orthonormal set in H containing uk : k 2 If g:Conclusion 2.140 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal setin H: Then there exists a maximal orthonormal set in H that contains uk : k 2 If g:Note 2.141 LetH be a nontrivial Hilbert space. There exists a nonzero vector a 2 H:

It follows that 1ak k a

n ois an orthonormal set in H: Now, by Conclusion 2.140,

there exists a maximal orthonormal set in H which contains 1ak k a

n o:

Conclusion 2.142 Let H be a nontrivial Hilbert space. Then, there exists a max-imal orthonormal set in H:

Let H be a Hilbert space. Let uk : k 2 If g be a maximal orthonormal set in H:

Problem 2.143 The mapping x 7! x from Hilbert space H onto Hilbert space ‘2 Ið Þis 1-1, linear and, for every x; y 2 H; x; yð Þ ¼ x; yð Þ:(Solution Since uk : k 2 If g is a maximal orthonormal set in H; by Conclusion2.136 P is dense in H; where P is the linear span of uk : k 2 If g: Also, byConclusion 2.121, x 7! x is a linear continuous mapping from H onto ‘2 Ið Þ; and therestriction of x 7! x to �P ¼ Hð Þ is an isometry from �P ¼ Hð Þ onto ‘2 Ið Þ: Thus, x 7! xis a linear mapping from Hilbert space H onto Hilbert space ‘2 Ið Þ; and is anisometry. Since x 7! x is a linear mapping, and is an isometry, x 7! x is 1-1. ByConclusion 2.136, for every x; y 2 H;

314 2 Lp-Spaces

x; yð Þ ¼Xk2I

x kð Þð Þ y kð Þð Þ�¼ x; yð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence for every x; y 2 H; x; yð Þ ¼ x; yð Þ: ∎)

Conclusion 2.144 The mapping x 7! x from Hilbert space H to Hilbert space ‘2 Ið Þis a Hilbert space isomorphism from H onto ‘2 Ið Þ:

2.5 Riesz-Fischer Theorem

Note 2.145 Let F : z : z 2 C; and zj j ¼ 1f g ! C be a function. Let E : t 7! eit be amapping from R to z : z 2 C; and zj j ¼ 1f g: Thus, F � Eð Þ : R ! C:

Problem 2.146 F � Eð Þ is 2p-periodic, in the sense that, for every t 2 R;F � Eð Þ tþ 2pð Þ ¼ F � Eð Þ tð Þ:(Solution Let us take any t 2 R: We have to show that

F � Eð Þ tþ 2pð Þ ¼ F � Eð Þ tð Þ:

LHS ¼ F � Eð Þ tþ 2pð Þ ¼ F E tþ 2pð Þð Þ ¼ F ei tþ 2pð Þ

¼ F eit � ei 2pð Þ

¼ F eit � 1� � ¼ F eit� � ¼ F E tð Þð Þ ¼ F � Eð Þ tð Þ ¼ RHS: ∎)

Conclusion 2.147 Let F : z : z 2 C; and zj j ¼ 1f g ! C be a function. Let E :

t 7! eit be a mapping from R to z : z 2 C; and zj j ¼ 1f g: Then F � Eð Þ : R ! C is2p-periodic.

Note 2.148 Let f : R ! C be a 2p-periodic function. Let E : t 7! eit be a mappingfrom R to z : z 2 C; and zj j ¼ 1f g: It follows that Ej 0;2p½ Þ is a 1-1 mapping from0; 2p½ Þ onto z : z 2 C; and zj j ¼ 1f g; and hence

ðEj 0;2p½ ÞÞ�1 : z : z 2 C; and zj j ¼ 1f g ! 0; 2p½ Þ

is a 1-1 and onto mapping. Thus,

f � ððEj 0;2p½ ÞÞ�1Þ : z : z 2 C; and zj j ¼ 1f g ! C:


Let us put

F � f � Ej 0;2p½ Þ �1� �

:

It follows that

F : z : z 2 C; and zj j ¼ 1f g ! C:

Since F ¼ f � ððEj½0;2pÞÞ�1Þ; we have

f j 0;2p½ Þ¼ F � Ej 0;2p½ Þ

:

Problem 2.149 f ¼ F � E:(Solution Let us take any t 2 R: We have to show that f tð Þ ¼ F � Eð Þ tð Þ: Thereexists an integer n; and t1 2 0; 2p½ Þ such that t ¼ t1 þ 2np:

LHS ¼ f tð Þ ¼ f t1 þ 2npð Þ ¼ f t1ð Þ ¼ f j 0;2p½ Þ

t1ð Þ

¼ F � Ej 0;2p½ Þ

t1ð Þ ¼ F Ej 0;2p½ Þ

t1ð Þ

¼ F E t1ð Þð Þ ¼ F eit1� � ¼ F eit1 � 1� �

¼ F eit1 � ei2np� � ¼ F ei t1 þ 2npð Þ

¼ F eit� �

¼ F E tð Þð Þ ¼ F � Eð Þ tð Þ ¼ RHS: ∎)

Conclusion 2.150 Let f : R ! C be a 2p-periodic function. Let E : t 7! eit be amapping from R to z : z 2 C; and zj j ¼ 1f g: Then there exists z : z 2 C;f and zj j ¼1g ! C such that f ¼ F � E: In short, we can identify any 2p-periodic functionf : R ! C with a function z : z 2 C; and zj j ¼ 1f g: such that for every t 2 R;f tð Þ ¼ F eitð Þ:Note 2.151 Let d 2 0; pð �: Let c1; c2; c3; . . . be any positive real numbers. Let

Q1 : t 7! c1 12 1þ cos tð Þ� �1

;

Q2 : t 7! c2 12 1þ cos tð Þ� �2

;

Q3 : t 7! c3 12 1þ cos tð Þ� �3

;

..

.

be 2p-periodic functions from R to 0;1½ Þ: Clearly, each Qn is a decreasingfunction over 0; p½ �: Observe that, for every positive integer n;

316 2 Lp-Spaces

2pR p�p

12 1þ cos tð Þ� �ndt ¼ 2pR p

�p cos 12 t� �� 2n

dt

¼ 2p

2R p0 cos 1

2 t� �� 2ndt ¼ pR p

20 cos hð Þ2n 2dhð Þ

¼p2R p

20 cos hð Þ2ndh

¼p2

2n�12n � 2n�3

2n�2 � � � 12 � p2¼ 2n

2n� 1� 2n� 22n� 3

� � � 21

so, for every positive integer n;

1 ¼ 12p

Zp�p

2n2n� 1

� 2n� 22n� 3

� � � 21

� �12

1þ cos tð Þ� �n

dt:

Put c1 ¼ 21 ; c2 ¼ 2

1 � 43 ; c3 ¼ 21 � 43 � 65 ; etc. It follows that, for every positive

integer n;

1 ¼ 12p

Zp�p

Qn tð Þdt:

Since, each Qn : R ! 0;1½ Þ is continuous, and

t : t 2 R; and d� tj j � pf g ¼ �p;�d½ � [ d; p½ �ð Þ

is a compact subset of R; the image set

Qn t : t 2 R; and d� tj j � pf gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ Qn tð Þ : t 2 R; and d� tj j � pf g ¼ Qn �p;�d½ � [ d; p½ �ð Þ

¼ Qn �p;�d½ �ð Þ [Qn d; p½ �ð Þ ¼ Qn d;p½ �ð Þ [Qn d; p½ �ð Þ¼ Qn d; p½ �ð Þ ¼ Qn tð Þ : d� t� pf g

is compact, and hence Qn tð Þ : d� t� pf g is closed and bounded. It follows that

cn12

1þ cos dð Þ� �n

¼ Qn dð Þ ¼ sup Qn tð Þ : d� t� pf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}exists.

2.5 Riesz-Fischer Theorem 317

Problem 2.152 limn!1 sup Qn tð Þ : t 2 R; and d� tj j � pf gð Þ ¼ 0:

(Solution Observe that, for every positive integer n;

1 ¼ 12p

Zp�p

Qn tð Þdt¼ 12p

Zp�p

cn12


dt

¼ 12p

2Zp0

cn12


dt

0@ 1A;

so, for every positive integer n;

2nþ 1

¼ � 12n

1nþ 1

�2nþ 1� �¼ � 1

2n1

nþ 1ð 1þ cos pð Þnþ 1� 1þ cos 0ð Þnþ 1Þ

¼ � 12n

1nþ 1

1þ cos tð Þnþ 1��t¼p

t¼0

¼ �Zp0

12


d 1þ cos tð Þ

¼Zp0

12


sin tdt

�Zp0

12


dt ¼ pcn

;

and hence for every positive integer n; cn � p nþ 1ð Þ2 : Here

0� sup Qn tð Þ : t 2 R; and d� tj j � pf g ¼ cn12


� p nþ 1ð Þ2

12


¼ p2

1cos2 1

2 d� � !

nþ 1ð Þ cos212d

� �� nþ 1 !

;

318 2 Lp-Spaces

so

0� sup Qn tð Þ : t 2 R; and d� tj j � pf g

� p2

1cos2 1

2 d� � !

nþ 1ð Þ cos212d

� �� nþ 1 !

:

Since d 2 0;pð �; we have

cos212d

� �2 0; 1½ Þ;

and hence

limn!1 nþ 1ð Þ1 cos2

12d

� �� nþ 1

¼ 0:

Since

limn!1 nþ 1ð Þ1 cos2

12d

� �� nþ 1

¼ 0;

and

0� sup Qn tð Þ : t 2 R; and d� tj j � pf g

� p2

1cos2 1

2 d� � !

nþ 1ð Þ cos212d

� �� nþ 1 !

;

we havelimn!1 sup Qn tð Þ : t 2 R; and d� tj j � pf gð Þ ¼ 0: ∎)

Conclusion 2.153 Let d 2 0; pð �: Then there exist positive real numbersc1; c2; c3; . . ., and trigonometric polynomials Q1 : R ! 0;1½ Þ; Q2 : R ! 0;1½ Þ;Q3 : R ! 0;1½ Þ; � � � such that

1. 1 ¼ 12p

R p�p Qn tð Þdt for every positive integer n,

2. limn!1 sup Qn tð Þ : t 2 R; and d� tj j � pf gð Þ ¼ 0;3. limn!1 Qn ¼ 0 uniformly on �p;�d½ � [ d; p½ �:

Proof of 3 Let us take any e[ 0: From 2, there exists a positive integer N such thatn�N implies

sup Qn tð Þ � 0j j : t 2 �p;�d½ � [ d; p½ �f g ¼ sup Qn tð Þ : t 2 �p;�d½ � [ d; p½ �f g¼ sup Qn tð Þ : t 2 R; and d� tj j � pf g¼ sup Qn tð Þ : t 2 R; and d� tj j � pf gj j¼ sup Qn tð Þ : t 2 R; and d� tj j � pf gð Þ � 0j j\e;


and hence

Qn tð Þ � 0j j\e for every t 2 �p;�d½ � [ d; p½ �:

Hence,

limn!1Qn ¼ 0 uniformly on �p;�d½ � [ d; p½ �: ∎)

Definition Let N be a positive integer, and a0; a1; . . .; aN ; b1; b2; . . .; bN be anycomplex numbers. The function

P : t 7! a0 þ a1 cos 1tþ b1 sin 1tð Þþ � � � þ aN cosNtþ bN sinNtð Þ

from R to C is called a trigonometric polynomial. Here, for every real t;

a0 þ a1 cos 1tþ b1 sin 1tð Þþ � � � þ aN cosNtþ bN sinNtð Þ

¼ a0 þXNn¼1

an12

eint þ ei �nð Þt

þ bn12i

eint � ei �nð Þt � �

¼ a0ei0t þ

XNn¼1

12

an � ibnð Þeint þ 12

an þ ibnð Þei �nð Þt� �

¼XNn¼�N

cneint;

where

c0 � a0; c1 � 12

a1 � ib1ð Þ; c2 � 12

a2 � ib2ð Þ; � � � ; and

c�1 � 12

a1 þ ib1ð Þ; c�2 � 12

a2 þ ib2ð Þ; � � � :

Thus, every trigonometric polynomial can be expressed as

t 7!XNn¼�N

cneint:

Observe that, for every real t;XNn¼�N

cneint ¼ c0 þ c1e

i1t þ � � � þ cNeiNt

� �þ c�1ei �1ð Þt þ � � � þ c�Ne

i �Nð Þt

¼ c0 þ c1 cos 1tþ i sin 1tð Þþ � � � þ cN cosNtþ i sinNtð Þð Þþ c�1 cos 1t � i sin 1tð Þþ � � � þ c�N cosNt � i sinNtð Þð Þ

¼ c0 þ c1 þ c�1ð Þ cos 1tþ i c1 � c�1ð Þ sin 1tð Þþ � � � þ cN þ c�Nð Þ cosNtþ i cN � c�Nð Þ sinNtð Þ

¼ a0 þ a1 cos 1tþ b1 sin 1tð Þþ � � � þ aN cosNtþ bN sinNtð Þ;

320 2 Lp-Spaces

where

a0 � c0; a1 � c1 þ c�1; a2 � c2 þ c�2; . . .; and

b1 � i c1 � c�1ð Þ; b2 � i c2 � c�2ð Þ; . . .:

Thus, every function of the form t 7! PNn¼�N cneint from R to C is a trigono-

metric polynomial.

Conclusion 2.154 Every function of the form t 7! PNn¼�N cneint from R to C is a

trigonometric polynomial.

Note 2.155 Let f : R ! C be a continuous, 2p-periodic function. Let e[ 0:Since f : R ! C is continuous and 2p-periodic function, fj j : R ! 0;1½ Þ is

continuous and 2p-periodic function, and hence

sup f tð Þj j : t 2 Rf g ¼ sup f tð Þj j : t 2 0; 2p½ �f g 2 0;1½ Þ:

Thus,0� sup f tð Þj j : t 2 Rf g\1:

Since f : R ! C is a 2p-periodic function, by Conclusion 2.150, there existsF : z : z 2 C; and zj j ¼ 1f g ! C such that f ¼ F � E; where E : t 7! eit is a map-ping from R onto z : z 2 C; and zj j ¼ 1f g: Now, since f : R ! C is continuous,F : z : z 2 C; and zj j ¼ 1f g ! C is continuous. Since F : z : z 2 C; andf zj j ¼1g ! C is continuous, and z : z 2 C; and zj j ¼ 1f g is compact, F :z : z 2 C; and zj j ¼ 1f g ! C is uniformly continuous, and hence there exists d 20; pð Þ such that for every real s; t 2 R satisfying s� tj j\d; we have

f sð Þ � f tð Þj j ¼ F � Eð Þ sð Þ � F � Eð Þ tð Þj j ¼ F E sð Þð Þ � F E tð Þð Þj j¼ F eis

� �� F eit� �� \ e

2:

Thus, for every real s; t 2 R satisfying s� tj j\d; we have f sð Þ � f tð Þj j\ e2 :

By Conclusion 2.153, there exist positive real numbers c1; c2; c3; . . ., and trigono-metric polynomials Q1 : R ! 0;1½ Þ; Q2 : R ! 0;1½ Þ; Q3 : R ! 0;1½ Þ; . . . suchthat

1. 1 ¼ 12p

R p�p Qn tð Þdt for every positive integer n,

2. limn!1 sup Qn tð Þ : t 2 �p;�d½ � [ d; p½ �f gð Þ ¼ 0;

3. limn!1Qn ¼ 0 uniformly on �p;�d½ � [ d; p½ �:

Let

P1 : t 7! 12p

Rp�p

f t � sð ÞQ1 sð Þds;

P2 : t 7! 12p

Rp�p


P3 : t 7! 12p

Rp�p


..

.


be 2p-periodic functions from R to R: Since each Qn is 2p-periodic, and f is 2p-periodic, for every positive integer n; and, for every t 2 R; we have

Zp�p

f t � sð ÞQn sð Þds¼Zt�p

tþ p

f s1ð ÞQn t � s1ð Þ �ds1ð Þ ¼Ztþ p

t�p

f s1ð ÞQn t � s1ð Þds1

¼Zp�p

f s1ð ÞQn t � s1ð Þds1 ¼Zp�p

f sð ÞQn t � sð Þds:

Hence,

P1 : t ! 12p

Rp�p

f sð ÞQ1 t � sð Þds;

P2 : t ! 12p

Rp�p


P3 : t ! 12p

Rp�p


..

.:

Since Q1 : R ! 0;1½ Þ is a trigonometric polynomial, there exists a positiveinteger N; and complex numbers a0; a1; . . .; aN ; b1; b2; . . .; bN such that

Q1 : t 7! a0 þ a1 cos 1tþ b1 sin 1tð Þþ � � � þ aN cosNtþ bN sinNtð Þ:

So, for every t 2 R;

P1 tð Þ ¼ 12p

Zp�p

f sð ÞQ1 t � sð Þds

¼ 12p

Zp�p

f sð Þ a0 þ a1 cos 1 t � sð Þþ b1 sin 1 t � sð Þð Þþ � � � þ aN cosN t � sð Þþ bN sinN t � sð Þð Þð Þds

¼ a02p

Zp�p

f sð Þdsþ 12p

XNn¼1

Zp�p

f sð Þ an cos n t � sð Þþ bn sin n t � sð Þð Þds

¼ a02p

Zp�p

f sð Þdsþ 12p

XNn¼1

Zp�p

f sð Þ an cos ntþ bn sin ntð Þ cos nsþ an sin nt � bn cos ntð Þ sin nsð Þds

¼ a02p

Zp�p

f sð Þdsþ 12p

XNn¼1

Zp�p

f sð Þ cos ns ds0@ 1A an cos ntþ bn sin ntð Þ

þ 12p

XNn¼1

Zp�p

f sð Þ sin ns ds0@ 1A an sin nt � bn cos ntð Þ

¼ a02p

Zp�p

f sð Þdsþ 12p

XNn¼1

an

Zp�p

f sð Þ cos ns ds� bn

Zp�p

f sð Þ sin ns ds0@ 1A cos nt

0@þ bn

Zp�p

f sð Þ cos ns dsþ an

Zp�p

f sð Þ sin ns ds0@ 1A sin nt:

322 2 Lp-Spaces

Thus, P1 : t 7! 12p

R p�p f t � sð ÞQ1 sð Þds is a trigonometric polynomial. Similarly,

P2 : t 7! 12p

R p�p f t � sð ÞQ2 sð Þds is a trigonometric polynomial, P3 : t 7! 1

2p

R p�p f

t � sð ÞQ3 sð Þds is a trigonometric polynomial, etc. Thus, each Pn : R ! C is con-tinuous, and 2p-periodic. Since, for every positive integer n; and for every t 2 R;

Pn tð Þ � f tð Þj j ¼ 12p

Zp�p

f t � sð ÞQn sð Þds� f tð Þ � 1��

��¼ 1

2p

Zp�p

f t � sð ÞQn sð Þds� f tð Þ � 12p

Zp�p

Qn sð Þds��

��¼ 1

2p

Zp�p

f t � sð ÞQn sð Þds� 12p

Zp�p

f tð ÞQn sð Þds��

��¼ 1

2p

Zp�p

f t � sð Þ � f tð Þð ÞQn sð Þds��

��¼ 1

2p

Zp�p

f t � sð Þ � f tð Þð ÞQn sð Þds��

�� 12p

Zp�p

f t � sð Þ � f tð Þð ÞQn sð Þj jds

¼ 12p

Zp�p

f t � sð Þ � f tð Þj j Qn sð Þj jds ¼ 12p

Zp�p

f t � sð Þ � f tð Þj jQn sð Þds

¼ 12p

Zd�d

f t � sð Þ � f tð Þð Þj jQn sð Þdsþ 12p

Z�p;�d½ � [ d;p½ �


� 12p

Zd�d

e2� Qn sð Þdsþ 1

2p

Z�p;�d½ � [ d;p½ �


¼ e2

12p

Zd�d

Qn sð Þds0@ 1Aþ 1

2p

Z�p;�d½ � [ d;p½ �


� e2

12p

Zp�p

Qn sð Þds0@ 1Aþ 1

2p

Z�p;�d½ � [ d;p½ �


¼ e2� 1þ 1

2p

Z�p;�d½ � [ d;p½ �


� e2þ 1

2p

Z�p;�d½ � [ d;p½ �

f t � sð Þ � f tð Þj j

sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þds

� e2þ 1

2p

Z�p;�d½ � [ d;p½ �

f t � sð Þj j þ f tð Þj jð Þ


sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þds

� e2þ 1

2p

Z�p;�d½ � [ d;p½ �

sup f uð Þj j : u 2 Rf gþ sup f uð Þj j : u 2 Rf gð Þ

sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þds¼ e

2þ 1

2p2 sup f uð Þj j : u 2 Rf gð Þ sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þ 2 p� dð Þð Þ

� e2þ 2 sup f uð Þj j : u 2 Rf gð Þ sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þ;

we have, for every positive integer n; and, for every t 2 R;

Pn tð Þ � f tð Þj j � e2þ 2 sup f uð Þj j : u 2 Rf gð Þ sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þ:

By (2), there exists a positive integer N1 such that n�N1 implies

2 sup f uð Þj j : u 2 Rf gð Þ sup Qn uð Þ : u 2 �p;�d½ � [ d; p½ �f gð Þ\ e2:

It follows that, for every positive integer n�N1; and, for every t 2 R;

Pn tð Þ � f tð Þj j\ e2þ e

2¼ eð Þ:

Thus, limn!1 Pn ¼ f uniformly on R: Now, since each Pn : R ! C is contin-uous 2p-periodic, and f : R ! C is a continuous 2p-periodic function,Pn � fk k1! 0 as n ! 1:

Conclusion 2.156 Let f : R ! C be a continuous, 2p-periodic function. Let e[ 0:Then there exists a trigonometric polynomial P : R ! C such that, for every t 2 R;P tð Þ � f tð Þj j\e:

Note 2.157 Let p 2 1;1½ Þ: Now we construct an Lp lð Þ-space. Let us take

z : z 2 C; and zj j ¼ 1f g

for the measure space X: Let us identify

z : z 2 C; and zj j ¼ 1f g

with �p; pð �: Let us take

12p

Lebesgue measure on �p; pð �ð Þ

324 2 Lp-Spaces

for l: Then Lp lð Þ is the collection of all Lebesgue measurable functions

F : z : z 2 C; and zj j ¼ 1f g ! C

for which

kFkp ¼ Z

X

Fj jpdl0@ 1A1

p

2 0;1½ Þ:

By Conclusion 2.150, we can identify function

F : z : z 2 C; and zj j ¼ 1f g ! C

with a 2p-periodic function f : R ! C such that, for every t 2 R; f tð Þ ¼ F eitð Þ:Thus, Lp lð Þ can be thought of as the collection of all Lebesgue measurable2p-periodic functions f : R ! C such that

12p

Zp�p

f tð Þj jpdt0@ 1A1

p

¼Zp�p

F eit� �� p 1

2pdt

� �0@ 1A1p

¼ZX

Fj jpdl0@ 1A1

p

2 0;1½ Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

;

that is Lp lð Þ can be thought of as the collection of all Lebesgue measurable2p-periodic functions f : R ! C such that

12p

Zp�p


p

2 0;1½ Þ:

Notation Here, Lp lð Þ is denoted by

Lp z : z 2 C; and zj j ¼ 1f gð Þ:Problem 2.158 l z : z 2 C; and zj j ¼ 1f gð Þ ¼ 1:

(Solution

LHS ¼ l z : z 2 C; and zj j ¼ 1f gð Þ¼ 1

2p Lebesgue measure of z : z 2 C; and zj j ¼ 1f gð Þ

¼ 12p

arc length of the unit circleð Þ

¼ 12p

2p � 1ð Þ ¼ 1 ¼ RHS: ∎)


Problem 2.159 L2 z : z 2 C; and zj j ¼ 1f gð Þ � L1 z : z 2 C; and zj j ¼ 1f gð Þ:(Solution Let f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ: We have to show that

f 2 L1 z : z 2 C; and zj j ¼ 1f gð Þ;

that is f : R ! C is a Lebesgue measurable 2p-periodic function satisfying

12p

Zp�p

f tð Þj j1dt0@ 1A1

1

2 0;1½ Þ;


Zp�p

f tð Þj jdt 2 0;1½ Þ:

Since


f : R ! C is a Lebesgue measurable 2p-periodic function satisfying

12p

Zp�p


2

2 0;1½ Þ;


Zp�p

f tð Þj j2dt 2 0;1½ Þ:


Zp�p

f tð Þj jdt 2 0;1½ Þ:

If not, otherwise, let Zp�p

f tð Þj jdt ¼ 1:

326 2 Lp-Spaces

We have to arrive at a contradiction. Here,

1 ¼Zp�p

f tð Þj jdt ¼Z

t:t2 �p;p½ �; and f tð Þj j � 1f g[ t:t2 �p;p½ �; and 1\ f tð Þj jf g

fj jdm

¼Z

t:t2 �p;p½ �; and 1\ f tð Þj jf g

fj jdmþZ

t:t2 �p;p½ �; and f tð Þj j � 1f g

fj jdm

�Z


fj jdmþZ

t:t2 �p;p½ �; and f tð Þj j � 1f g

1dm

�Z


fj jdmþ 2p�Z


f tð Þj j2dtþ 2p

�Z

t:t2 �p;p½ �f g

f tð Þj j2dtþ 2p ¼Zp�p

f tð Þj j2dtþ 2p;

so,

1�Zp�p

f tð Þj j2dtþ 2p;

and henceR p�p f tð Þj j2dt ¼ 1: This contradicts

R p�p f tð Þj j2dt 2 0;1½ Þ: ∎)

For every integer n; Un : z 7! zn is a continuous function fromz : z 2 C; and zj j ¼ 1f g to C; so, for every integer n; Un : z 7! zn is a Lebesgue

measurable function from z : z 2 C; and zj j ¼ 1f g to C: Also, for every integer n;

ZX

Unj jpdl0@ 1A1

p

¼Zp�p

Un eit� �� p 1

2pdt

� �0@ 1A1p

¼Zp�p

eit� �n�� p 1

2pdt

� �0@ 1A1p

¼Zp�p

112p

dt� �0@ 1A1

p

¼ 1 2 0;1½ Þ;


so, for every integer n; Un 2 Lp z : z 2 C; and zj j ¼ 1f gð Þ: Since Lp lð Þ is complete,Lp z : z 2 C; and zj j ¼ 1f gð Þ is complete with the norm given by

fk kp�12p

Zp�p


p

:

It follows that L2 z : z 2 C; and zj j ¼ 1f gð Þ is a Hilbert space with the ‘innerproduct’ given by

f ; gð Þ ¼ F;Gð Þ ¼ZX

F � �Gð Þdl

¼Zp�p

F eit� ��

G eitð Þð Þ 12p

dt� �

¼ 12p

Zp�p

f tð Þð Þ g tð Þð Þdt:

For any distinct integers m; n;

um; unð Þ ¼ Um;Unð Þ ¼Zp�p

Um eit� ��

Un eitð Þð Þ 12p

dt� �

¼Zp�p

eit� �m

eitð Þm 12p

dt� �

¼ 12p

Zp�p

ei m�nð Þtdt ¼ 12p

1i m� nð Þ e

i m�nð Þt��p�p

¼ 12p

1i m� nð Þ ei m�nð Þp � ei m�nð Þ �pð Þ

¼ 1

2p1

i m� nð Þ 2i sin m� nð Þp

¼ 1p

1m� nð Þ sin m� nð Þp

¼ 1p

1m� nð Þ 0 ¼ 0:

Thus, Un : n 2 Zf g is an orthonormal set in the Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ:

328 2 Lp-Spaces

Conclusion 2.160 Un : n 2 Zf g is an orthonormal set in the Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ; where, for every integer n; Un : z 7! zn is a functionfrom z : z 2 C; and zj j ¼ 1f g to C:

Note 2.161

Problem 2.162 un : n 2 Zf g is a maximal orthonormal set in the Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ:(Solution In view of Conclusion 2.160, it remains to show that un : n 2 Zf g ismaximal. By Conclusion 2.136, it suffices to show that P is dense in

L2 z : z 2 C; and zj j ¼ 1f gð Þ;

where P is the linear span of un : n 2 Zf g: Clearly, P is equal to the collection C ofall trigonometric polynomials. So, it suffices to show that C is dense in

L2 z : z 2 C; and zj j ¼ 1f gð Þ:

For this purpose, let us take any f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ:Thus, f : R ! C is a Lebesgue measurable 2p-periodic function such that

12p

Zp�p


2

2 0;1½ Þ:

Let us take any e[ 0: We have to find a polynomial P : R ! C such that

12p

Zp�p

P tð Þ � f tð Þj j2dt0@ 1A1

2

¼ P� fk k2\e|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};that is, we have to find a polynomial P : R ! C such that

Zp�p

P tð Þ � f tð Þj j2dt\e22p:

By Conclusion 2.156, there exists a trigonometric polynomial P : R ! C suchthat for every t 2 R; P tð Þ � f tð Þj j\ e

2 : It follows that

Zp�p

P tð Þ � f tð Þj j2dt�Zp�p

e2

2dt ¼ e

2

22p\e22p:

∎)


Definition Let f 2 L1 z : z 2 C; and zj j ¼ 1f gð Þ: Thus, f : R ! C is a Lebesguemeasurable 2p-periodic function satisfying

12p

Zp�p

f sð Þj jds 2 0;1½ Þ:

Since, for every integer n; the mapping s 7! e�ins from R to C is continuous, themapping s 7! e�ins from R to C is Lebesgue measurable. It follows that, for everyinteger n; the mapping s 7! f sð Þe�ins from R to C is Lebesgue measurable. Also, themapping s 7! f sð Þe�ins from R to C is 2p-periodic function. Next,

12p

Zp�p

f sð Þe�ins�� ds¼ 1

2p

Zp�p

f sð Þj j e�ins�� ds

¼ 12p

Zp�p

f sð Þj j � 1ds

¼ 12p

Zp�p

f sð Þj jds 2 0;1½ Þ;

so,

12p

Zp�p

f sð Þe�ins�� ds 2 0;1½ Þ;

and hence the mapping s 7! f sð Þe�ins from R to C is a member of

L1 z : z 2 C; and zj j ¼ 1f gð Þ:Definition For every real t; the complex number

limN!1

XNn¼�N

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1Ais denoted by

X1n¼�1

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1A;

330 2 Lp-Spaces

and is called the Fourier series of f : Here, for every positive integer N; the mapping

t 7!XNn¼�N

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1Afrom R to C is denoted by sN ; and is called the Nth partial sum. For every integer n;

12p

Zp�p

f sð Þe�insds

is denoted by f nð Þ; and is called the nth Fourier coefficient of f : Thus, for everyreal t;

X1n¼�1

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1A ¼X1

n¼�1f nð Þ� �

eint� � ¼ lim

N!1sN tð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Clearly, each sN is a trigonometric polynomial.

Conclusion 2.163 a. un : n 2 Zf g is a maximal orthonormal set in the Hilbertspace L2 z : z 2 C; and zj j ¼ 1f gð Þ: b. each sN is a trigonometric polynomial.

Note 2.164 Let cn : n 2 Zf g be any collection of complex numbers. LetPn2Z cnj j22 0;1½ Þ:

Problem 2.165 There exists f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ such that for everyinteger n;

cn ¼ 12p

Zp�p

f sð Þe�insds:

(Solution From Problem 2.162, un : n 2 Zf g is a maximal orthonormal set in theHilbert space

L2 z : z 2 C; and zj j ¼ 1f gð Þ;

where, for every integer n; un : t 7! eint: Now, by Conclusion 2.144, the mappingf 7! f from Hilbert space

L2 z : z 2 C; and zj j ¼ 1f gð Þ


to ‘2 Zð Þ is 1-1, onto, linear, preserves inner product, and preserves norms. SincePn2Z cnj j22 0;1½ Þ; the mapping u : n 7! cn from Z to C is a member of ‘2 Zð Þ:

Now, since f 7! f is a mapping from Hilbert space L2 z : z 2 C; and zj j ¼ 1f gð Þ onto‘2 Zð Þ; and u is a member of ‘2 Zð Þ; there exists

f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ

such that f ¼ u: Since

f : n 7! f ; unð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ 12p

Zp�p

f sð Þ � un sð Þ

ds

¼ 12p

Zp�p

f sð Þ � eins

ds

¼ 12p

Zp�p

f sð Þ � e�ins� �

ds

is a mapping from Z to C;

u : n 7! cn

is a mapping from Z to C; and f ¼ u: Hence, for every integer n;

12p

Zp�p

f sð Þ � e�ins� �

ds ¼ f nð Þ ¼ u nð Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} ¼ cn:

Thus, for every integer n;

cn ¼ 12p

Zp�p

f sð Þe�insds:∎)

This result, known as the Riesz–Fischer theorem is due to F. Riesz, andE. S. Fischer (12.07.1875–14.11.1954),

Note 2.166

Problem 2.167 For every f ; g 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ;

12p

Zp�p

f sð Þg sð Þds ¼Xn¼1

n¼�1f nð Þg nð Þ:

332 2 Lp-Spaces

(Solution From Problem 2.162, un : n 2 Zf g is a maximal orthonormal set in theHilbert space L2 z : z 2 C; and zj j ¼ 1f gð Þ; where for every integer n; un : t 7! eint:Now, by Conclusion 2.136, for every f ; g 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ;Xn¼1

n¼�1f nð Þg nð Þ ¼

!Xn2Z

f nð Þg nð Þ ¼ f ; gð Þ ¼ 12p

Zp�p

f sð Þg sð Þds0@ 1A:

∎)

Conclusion 2.168 For every f ; g 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ;12p

R p�p f sð Þg sð Þds ¼Pn¼1

n¼�1 f nð Þg nð Þ:This result is known as the Parseval theorem.Let f ; g 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ:

Problem 2.169Pn¼1

n¼�1 f nð Þg nð Þ�� \1:

(Solution By Conclusion 2.144, f ; g are the members of the Hilbert space ‘2 Zð Þ;and hence, by Lemma 2.24,

Xn¼1

n¼�1f nð Þg nð Þ�� ¼ Xn¼1

n¼�1f nð Þg nð Þ�� 1 !1

1

�Xn¼1

n¼�1f nð Þ�� 2 !1

2 Xn¼1

n¼�1g nð Þj j2

!12

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus,

Xn¼1

n¼�1f nð Þg nð Þ�� Xn¼1

n¼�1f nð Þ�� 2 !1

2 Xn¼1

n¼�1g nð Þj j2

!12

:

Since f is a member of the Hilbert space ‘2 Zð Þ;

Xn¼1

n¼�1f nð Þ�� 2 !1

2

2 0;1½ Þ:

Similarly, Xn¼1

n¼�1g nð Þj j2

!12

2 0;1½ Þ:

It follows that

0�Xn¼1

n¼�1f nð Þg nð Þ�� Xn¼1

n¼�1f nð Þ�� 2 !1

2 Xn¼1

n¼�1g nð Þj j2

!12

2 0;1½ Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};


and hence Xn¼1

n¼�1f nð Þg nð Þ�� \1: ∎)

Conclusion 2.170 Let f ; g 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ: Then Pn¼1n¼�1 f nð Þg nð Þ

converges absolutely.

Note 2.171

Problem 2.172 For every f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ; limN!1 f � sNk k2¼ 0:

(Solution Let us take any f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ: Let us take any e[ 0:Since un : n 2 Zf g is a maximal orthonormal set in the Hilbert space

L2 z : z 2 C; and zj j ¼ 1f gð Þ; by Conclusion 2.136, the collection P of all linearcombinations of members in un : n 2 Zf g is dense in L2 z : z 2 C; and zj j ¼ 1f gð Þ:Clearly, P is equal to the collection C of all trigonometric polynomials. Hence, thecollection C of all trigonometric polynomials is dense in L2 z : z 2 C;fð and zj j ¼1gÞ: Now, since each sN is a trigonometric polynomial, each sN 2 L2 z : z 2 C;fðand zj j ¼ 1gÞ:

Since each sN 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ; f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ;and L2 z : z 2 C; and zj j ¼ 1f gð Þ is a Hilbert space, each f � sNk k2 is meaningful.Here, for every positive integer N;

f � sNk k2� �2¼ f � sN ; f � sNð Þ ¼ f ; fð Þ � f ; sNð Þ � f ; sNð Þþ sN ; sNð Þ;

so, for every positive integer N;

f � sNk k2� �2¼ f ; fð Þ � f ; sNð Þ � f ; sNð Þþ sN ; sNð Þ:

For every positive integer N; and for every real t;

sN tð Þ ¼XNn¼�N

f nð Þ� �eint

� � ¼ Xnj j �N

f nð Þ� �eint ¼

Xnj j �N

f nð Þ� �un tð Þ:

It follows that, for every positive integer N; sN ¼P nj j �N f nð Þ� �un:

Now, since un : n 2 Zf g is an orthonormal set in the Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ; we have

sN ; sNð Þ ¼�X

nj j �N

f nð Þ� �un;

Xnj j �N

f nð Þ� �un

�¼Xnj j �N

f nð Þf nð Þ

1 ¼Xnj j �N

f nð Þ� �2;

334 2 Lp-Spaces

and hence

sN ; sNð Þ ¼Xnj j �N

f nð Þ�� 2:Next,

f ; sNð Þ ¼�f ;Xnj j �N

f nð Þ� �un

�¼Xnj j �N

f nð Þ f ; unð Þ

¼Xnj j �N

f nð Þf nð Þ

¼Xnj j �N

f nð Þ�� 2;so

f ; sNð Þ ¼Xnj j �N

f nð Þ�� 2:It follows that


f nð Þ�� 20@ 1A�

¼Xnj j �N

f nð Þ�� 2;and hence


f nð Þ�� 2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f ; sNð Þ:

This shows that f ; sNð Þ is a real number. Next

f ; fð Þ ¼Xn¼1

n¼�1f nð Þf nð Þ ¼

Xn¼1

n¼�1f nð Þ�� 2;

so

f ; fð Þ ¼Xn¼1

n¼�1f nð Þ�� 2:


It follows that

f � sNk k2� �2 ¼ f ; fð Þ � f ; sNð Þ � f ; sNð Þþ sN ; sNð Þ

¼ f ; fð Þ � f ; sNð Þ � f ; sNð Þþ sN ; sNð Þ¼ f ; fð Þ � 2 f ; sNð Þþ sN ; sNð Þ

¼Xn¼1

n¼�1f nð Þ�� 2 � 2

Xnj j �N

f nð Þ�� 2 þ Xnj j �N

f nð Þ2

¼Xn¼1

n¼�1f nð Þ�� 2 � X

nj j �N

f nð Þ�� 2 ¼ XN\ nj j

f nð Þ�� 2;so

f � sNk k2� �2¼ X

N\ nj jf nð Þ�� 2:

Since,


by Problem 2.167,

0;1½ Þ 3 12p

Zp�p

f sð Þj j2ds

¼ 12p

Zp�p

f sð Þf sð Þds ¼Xn¼1

n¼�1f nð Þf nð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Xn¼1

n¼�1f nð Þ�� 2;

and hence

Xn¼1

n¼�1f nð Þ�� 2\1:

Since

Xn¼1

n¼�1f nð Þ�� 2\1;

336 2 Lp-Spaces

there exists a positive integer N1 such that N�N1 implies

f � sNk k2� �2¼ X

N\ nj jf nð Þ�� 2\e2|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl};

and hence N �N1 implies f � sNk k2\e: This proves that

limN!1

f � sNk k2¼ 0: ∎)

In other words, for every f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ; limN!1 sN ¼ f in thetopology of Hilbert space L2 z : z 2 C; and zj j ¼ 1f gð Þ:

By Conclusion 2.144, f 7! f is a Hilbert isomorphism from Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ onto Hilbert space ‘2 Zð Þ:Conclusion 2.173 f 7! f is a Hilbert isomorphism from Hilbert spaceL2 z : z 2 C; and zj j ¼ 1f gð Þ onto Hilbert space ‘2 Zð Þ:

2.6 Baire’s Category Theorem

Note 2.174 Let X; dð Þ be a complete metric space. Let Vnf g be any sequence ofopen dense subsets of X: Let a 2 X: Let e[ 0:

Since V1 is dense, and x : d x; að Þ\ef g is an open neighborhood of a; we haveV1 \ x : d x; að Þ\ef g 6¼ ;: Now, since V1 is open, V1 \ x : d x; að Þ\ef g is a non-empty open set, and hence there exists x1 such that x1 is an interior point ofV1 \ x : d x; að Þ\ef g: It follows that there exists a real number r1 2 0; 1ð Þ such that

x : d x; x1ð Þ� r1f g � V1 \ x : d x; að Þ\ef g:

Since V2 is dense, and x : d x; x1ð Þ\r1f g is an open neighborhood of x1; we haveV2 \ x : d x; x1ð Þ\r1f g 6¼ ;: Now, since V2 is open, V2 \ x : d x; x1ð Þ\r1f g is anonempty open set, and hence there exists x2 such that x2 is an interior point ofV2 \ x : d x; x1ð Þ\r1f g: Thus, d x2; x1ð Þ\r1: It follows that

min12; r1 � d x2; x1ð Þ

� �[ 0:

Also, there exists a positive real number r2 such that

r2\min12; r1 � d x2; x1ð Þ

� �;


and x : d x; x2ð Þ� r2f g � V2 \ x : d x; x1ð Þ\r1f g: Clearly, 0\r2\r1:Since V3 is dense, and x : d x; x2ð Þ\r2f g is an open neighborhood of x2; we have

V3 \ x : d x; x2ð Þ\r2f g 6¼ ;: Now, since V3 is open,

V3 \ x : d x; x2ð Þ\r2f g

is a nonempty open set, and hence there exists x3 such that x3 is an interior point of

V3 \ x : d x; x2ð Þ\r2f g:

It follows that

d x3; x2ð Þ\r2 \r1 � d x2; x1ð Þð Þ;

and hence

d x3; x1ð Þ� d x3; x2ð Þþ d x2; x1ð Þ\r1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, d x3; x1ð Þ\r1; and d x3; x2ð Þ\r2: Hence,

min13; r1 � d x3; x1ð Þ; r2 � d x3; x2ð Þ

� �[ 0:

Also, there exists a positive real number r3 such that

r3\min13; r1 � d x3; x1ð Þ; r2 � d x3; x2ð Þ

� �;

and

x : d x; x3ð Þ� r3f g � V3 \ x : d x; x2ð Þ\r2f g:

Clearly,

0\r3\r2\r1:

Since V4 is dense, and x : d x; x3ð Þ\r3f g is an open neighborhood of x3; we haveV4 \ x : d x; x3ð Þ\r3f g 6¼ ;: Now, since V4 is open, V4 \ x : d x; x3ð Þ\r3f g is anonempty open set, and hence there exists x4 such that x4 is an interior point ofV4 \ x : d x; x3ð Þ\r3f g: It follows that d x4; x3ð Þ\r3 \r1 � d x3; x1ð Þð Þ; and hence

d x4; x1ð Þ� d x4; x3ð Þþ d x3; x1ð Þ\r1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

338 2 Lp-Spaces

Since,

d x4; x3ð Þ\r3|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}\r2 � d x3; x2ð Þ;

we have

d x4; x2ð Þ� d x4; x3ð Þþ d x3; x2ð Þ\r2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus,

d x4; x1ð Þ\r1; d x4; x2ð Þ\r2 and d x4; x3ð Þ\r3:

Hence,

min14; r1 � d x4; x1ð Þ; r2 � d x4; x2ð Þ; r3 � d x4; x3ð Þ

� �[ 0:

Similarly, there exists a positive real number r4 such that

r4\min14; r1 � d x4; x1ð Þ; r2 � d x4; x2ð Þ; r3 � d x4; x3ð Þ

� �;

and

x : d x; x4ð Þ� r4f g � V4 \ x : d x; x3ð Þ\r3f g; etc:

Clearly,

0\ � � �\r4\r3\r2\r1:

Here, we get a sequence xnf g in the metric space X such that

1. for every positive integer n; xn 2 Vn;2. for every positive integer m; n satisfying m\n;

d xn; xmð Þ\rm and hence; xn is in the closed set x : d x; xmð Þ� rmf gð Þ;

3. for every positive integer n; x : d x; xnþ 1ð Þ� rnþ 1f g � Vnþ 1 \x : d x; xnð Þ\rnf g:

2.6 Baire’s Category Theorem 339

Problem 2.175 xnf g is a Cauchy sequence.

(Solution Let us take any e1 [ 0: Since each rn 2 0; 1n� �

; limn!1 rn ¼ 0; andhence, by (2), there exists a positive integer N such that, for every integer nsatisfying n[N;

d xnxNð Þ\rN ¼ rNj j ¼ rN � 0j j\ e12|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} :

Thus, n[N implies d xnxNð Þ\ e12 : It follows that m; n[N implies

d xn; xmð Þ� d xn; xNð Þþ d xN ; xmð Þ\ e12þ d xN ; xmð Þ

¼ e12þ d xm; xNð Þ\ e1

2þ e1

2¼ e1:

Thus, m; n[N implies d xn; xmð Þ\e1: This shows that xnf g is a Cauchysequence. ∎)Now, since X; dð Þ is a complete metric space, xnf g is convergent in X; and hencethere exists y 2 X such that limn!1 xn ¼ y:

Problem 2.176 y 2 x : d x; að Þ\ef g\V1 \V2 \V3 \ � � � :(Solution By (2), x2; x3; x4; . . . are in the closed set x : d x; x1ð Þ� r1f g; so

y ¼ limn!1 xn ¼ lim

n!1 xnþ 1

2 x : d x; x1ð Þ� r1f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � V1 \ x : d x; að Þ\ef g;

and hence

y 2 x : d x; að Þ\ef g\V1:

Again by (2), x3; x4; x5; . . . are in the closed set x : d x; x2ð Þ� r2f g; so

y ¼ limn!1 xn ¼ lim

n!1 xnþ 2

2 x : d x; x2ð Þ� r2f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � V2 \ x : d x; x1ð Þ\r1f g � V2;

and hence, y 2 V2: Similarly, y 2 V3; y 2 V4; . . .: Thus,

y 2 x : d x; að Þ\ef g\V1 \V2 \V3 \ � � � : ∎)

It follows that

x : d x; að Þ\ef g\ V1 \V2 \V3 \ � � �ð Þ 6¼ ;:

340 2 Lp-Spaces

Conclusion 2.177 Let X; dð Þ be a complete metric space. Let Vnf g be any sequenceof open dense subsets of X: Then V1 \V2 \V3 \ � � � is a dense subset of X:

This conclusion, known as the Baire’s category theorem, is due to R. L. Baire(21.01.1874–05.07.1932, French). He was famous for his Baire’s category theorem,which helped to generalize future theorems.

Note 2.178

Definition Let X be a complex linear space. Suppose that, to every x 2 X; there isassociated a nonnegative real number kxk: We say that X is a normed linear space,if the following conditions are satisfied:

1. if xk k ¼ 0 then x ¼ 0;2. for every x 2 X; and for every a 2 C; axk k ¼ aj j xk k;3. for every x; y 2 X; xþ yk k� xk kþ yk k:

Here, xk k is called the norm of x: Clearly, if for every x; y 2 X; d x; yð Þ �x� yk k; then X; dð Þ is a metric space. If X; dð Þ is a complete metric space, then the

normed linear space X is called a Banach space.

Examples

1. Every Hilbert space is a Banach space.2. If p 2 1;1½ Þ; then Lp lð Þ is a Banach space.

Definition Let X and Y be normed linear spaces. Let K : X ! Y be a lineartransformation. Suppose that K xð Þk k : xk k� 1f g is a bounded above set of non-negative real numbers. Then

sup K xð Þk k : xk k� 1f g 2 0;1½ Þð Þ

exists. Here,

sup K xð Þk k : xk k� 1f g

is denoted by Kk k; and K is called a bounded linear transformation.Let X and Y be normed linear spaces. Let K : X ! Y be a bounded linear

transformation.

Problem 2.179 Kk k ¼ sup K xð Þk k : xk k ¼ 1f g:(Solution: Since

K xð Þk k : xk k ¼ 1f g � K xð Þk k : xk k� 1f g;

we have

sup K xð Þk k : xk k ¼ 1f g� sup K xð Þk k : xk k� 1f g ¼ kKkð Þ;


and hence

sup K xð Þk k : xk k ¼ 1f g� Kk k:


sup K xð Þk k : xk k ¼ 1f g ¼ Kk k:


sup K xð Þk k : xk k ¼ 1f g\ Kk k:

We have to arrive at a contradiction. Since

0� sup K xð Þk k : xk k ¼ 1f g\ Kk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};there exists a positive real number t such that

sup K xð Þk k : xk k ¼ 1f g\t\ Kk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ sup K xð Þk k : xk k� 1f g:

Since

t\sup K xð Þk k : xk k� 1f g;

there exists y 2 X such that yk k� 1; and 0\ t\ K yð Þk k|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} : Since 0\ K yð Þk k; we

have y 6¼ 0; and hence 1yk k y

�� ¼ 1: Since 1yk k y

�� ¼ 1; we have

1yk k K yð Þk k ¼ K

1yk k y

� �� sup K xð Þk k : xk k ¼ 1f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\t\ K yð Þk k;

and hence

1yk k K yð Þk k\ K yð Þk k:

Now, since 0\ K yð Þk k; we have 1yk k\1: It follows that 1\ yk k; that is yk k�1:

This is a contradiction. ∎)Let X and Y be normed linear spaces. Let K : X ! Y be a bounded linear

transformation.

342 2 Lp-Spaces

Problem 2.180

1. For every x 2 X; K xð Þk k� Kk k xk k;2. If k is a nonnegative real number such that, for every x 2 X; K xð Þk k� k xk k;

then Kk k� k:

(Solution

1. If x ¼ 0; then K xð Þk k� Kk k xk k is trivially true. So, we consider the case when

x 6¼ 0: In this case, 1xk k x

�� ¼ 1; and hence

1xk k K xð Þk k ¼ K

1xk k x

� �� Kk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :This shows that 1

xk k K xð Þk k� Kk k; and hence K xð Þk k� Kk k xk k:2. Let k be a nonnegative real number such that, for every x 2 X; K xð Þk k� k xk k:

We have to show that Kk k� k: If not, otherwise, let k\ Kk k: We have to arriveat a contradiction. Since

k\ Kk k|fflfflfflffl{zfflfflfflffl} ¼ sup K xð Þk k : xk k ¼ 1f g;

there exists y 2 X such that yk k ¼ 1; and k\ K yð Þk k: It is given thatK yð Þk k� k yk k|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} ¼ k � 1 ¼ k; so

K yð Þk k� k; and hence k¥ K yð Þk k: This is a contradiction. ∎)Let X and Y be normed linear spaces. Let K : X ! Y be a bounded linear

transformation.

Problem 2.181 K x : xk k� 1f gð Þ � y : yk k� Kk kf g:(Solution Let us take any x 2 X satisfying xk k� 1: We have to show thatK xð Þk k� Kk k: Since

K xð Þk k� Kk k xk k|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � Kk k � 1 ¼ Kk k;

we have K xð Þk k� Kk k: ∎)Let X and Y be normed linear spaces. Let K : X ! Y be a bounded linear

transformation.


Problem 2.182 K : X ! Y is continuous.

(Solution Let us take any a 2 X:We have to show that K : X ! Y is continuous ata: For this purpose, let us take any e[ 0:

Case I: when Kk k 6¼ 0: In this case, let us take any x 2 X satisfyingx� ak k\ e

Kk k : It follows that

K xð Þ � K að Þk k ¼ K x� að Þk k� Kk k x� ak k\ Kk k eKk k

� �¼ e;

and hence K xð Þ � K að Þk k\e:Case II: when Kk k ¼ 0: In this case, let us take any x 2 X satisfying x� ak k\e:

It follows that

K xð Þ � K að Þk k ¼ K x� að Þk k� Kk k x� ak k ¼ 0 � x� ak k ¼ 0\e:

Thus, in all cases, there exists d[ 0 such that, for every x 2 X satisfyingx� ak k\d; we have K xð Þ � K að Þk k\e: Thus, K : X ! Y is continuous at a: ∎)Let X and Y be normed linear spaces. Let K : X ! Y be a linear transformation.

Let a 2 X: Let K : X ! Y be continuous at a:

Problem 2.183 K : X ! Y is bounded.

(Solution Here, we have to show that set K xð Þk k : xk k� 1f g of nonnegative realnumbers is bounded above. Since K : X ! Y is continuous at a; there exists d[ 0such that, for every x 2 X satisfying x� ak k\d; we have K xð Þ � K að Þk k\1:

Let us take any b 2 X satisfying bk k� 1: It follows that

d2bþ a

� �� a

�� ¼ d2b

�� d2|fflfflfflfflfflffl{zfflfflfflfflfflffl}\d;

and hence

d2

K bð Þk k ¼ Kd2bþ a

� �� K að Þ

�� \1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus K bð Þk k� 2

d : Hence,2d is an upper bound of K xð Þj j : xk k� 1f g: This shows

that K xð Þk k : xk k� 1f g is bounded above. ∎)If we recollect the above results, we get the following

Conclusion 2.184 Let X and Y be normed linear spaces. Let K : X ! Y be a lineartransformation. Then the following three statements are equivalent:

344 2 Lp-Spaces

1. K is bounded,2. K is continuous,3. K is continuous at one point of X.

Note 2.185 Let X be a Banach space, and Y be a normed linear space. Let I be anonempty set. Suppose that, for every i 2 I; Ki : X ! Y is a bounded lineartransformation.

For every x 2 X; Ki xð Þk k : i 2 If g is a set of nonnegative real numbers. Clearly,for every x 2 X;

sup Ki xð Þk k : i 2 If g 2 0;1½ �:

For every x 2 X; let us denote

sup Ki xð Þk k : i 2 If g

by u xð Þ: Thus,

u : x 7! sup Ki xð Þk k : i 2 If g

is a function from X to 0;1½ �: Since each Ki : X ! Y is a bounded linear trans-formation, by Conclusion 2.184, each Ki : X ! Y is continuous. Also, y 7! yk kfrom Y to 0;1½ Þ is continuous. Since each Ki : X ! Y is continuous, and y 7! yk kfrom Y to 0;1½ Þ is continuous, each composite x 7! Ki xð Þk k from X to 0;1½ Þ iscontinuous.

Problem 2.186 u : X ! 0;1½ � is lower semicontinuous.

(Solution Let us take any a 2 R: We have to show that

[ i2I x : a\ Ki xð Þk kf g ¼ x : there exists i 2 I such that a\ Ki xð Þk kf g¼ x : a\sup Ki xð Þk k : i 2 If gf g ¼ x : a\u xð Þf g ¼ u�1 a;1ð �ð Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

is open in X; that is, [ i2I x : a\Ki xð Þf g is open in X: Since each x 7! Ki xð Þk k fromX to 0;1½ Þ is continuous, each x : a\ Ki xð Þk kf g is open in X; and hence[ i2I x : a\ Ki xð Þk kf g is open in X: ∎)

Case I: For every positive integer n; u�1 n;1ð �ð Þ is dense in X: Since u : X !0;1½ � is lower semicontinuous, for every positive integer n; u�1 n;1ð �ð Þ is open inX: By Baire’s category theorem,

u�1 1ð Þ ¼ u�1 1f gð Þ ¼ u�1 1;1ð �\ 2;1ð �\ 3;1ð �\ � � �ð Þ¼ u�1 1;1ð �ð Þ \u�1 2;1ð �ð Þ \u�1 3;1ð �ð Þ \ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}


is a dense subset of X; and hence, u�1 1ð Þ is a dense subset of X: Since eachu�1 n;1ð �ð Þ is open in X;

u�1 1ð Þ ¼ u�1 1;1ð �ð Þ \u�1 2;1ð �ð Þ \u�1 3;1ð �ð Þ \ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}is a Gd; and hence u�1 1ð Þ is a Gd: Thus u�1 1ð Þ is a dense Gd: Also, if x 2u�1 1ð Þ; then

sup Ki xð Þk k : i 2 If g ¼ u xð Þ ¼ 1|fflfflfflfflfflffl{zfflfflfflfflfflffl} :Thus, u�1 1ð Þ is a dense Gd such that for every x 2 u�1 1ð Þ;

sup Ki xð Þk k : i 2 If g ¼ 1:

Case II: There exists a positive integer n0 such that u�1 n0;1ð �ð Þ is not dense inX: In this case, there exists x0 2 X such that x0 is not an adherent point ofu�1 n0;1ð �ð Þ: It follows that there exists a positive real r0 such that

x0 þ x : xk k\r0f gð Þ \u�1 n0;1ð �ð Þ ¼ ;:

It follows that, if xk k\r0; then

sup Ki x0 þ xð Þk k : i 2 If g ¼ u x0 þ xð Þ� n0|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} :Thus, for every i 2 I; and for every x 2 X satisfying kxk\r0; we have

Ki x0 þ xð Þk k� n0: It follows that Ki x0ð Þk k ¼ Ki x0 þ 0ð Þk k� n0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}; and hence

Ki x0ð Þk k� n0: Now, for every i 2 I; and, for every x 2 X satisfying xj j\r0;

Ki xð Þk k ¼ Ki x0 þ xð Þ � x0ð Þk k ¼ Ki x0 þ xð Þ � Ki x0ð Þk k� Ki x0 þ xð Þk kþ Ki x0ð Þk k� n0 þ n0 ¼ 2n0:

Thus, for every i 2 I; and, for every x 2 X satisfying xk k\r0; we haveKi xð Þk k� 2n0:Let us fix any i 2 I: Let us take any x 2 X satisfying xk k� 1: It follows that

r02 x�� r0

2 \r0ð Þ: Hence,r02� Ki xð Þk k ¼ Ki

r02x

�� 2n0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} :This shows that, for every x 2 X satisfying xk k� 1; Ki xð Þk k� 4n0

r0; and hence

Kik k� 4n0r0:

346 2 Lp-Spaces

Conclusion 2.187 Let X be a Banach space, and Y be a normed linear space. Let Ibe a nonempty set. Suppose that, for every i 2 I; Ki : X ! Y is a bounded lineartransformation. Then, either

there exists a dense Gd-set V such that for every x 2 V ;

sup Ki xð Þk k : i 2 If g ¼ 1;

or

there exists a positive real number M such that for every i 2 I; Kik k�M:

This conclusion, known as the Banach-Steinhaus theorem, is due to S. Banach(30.03.1892–31.08.1945, Polish), and H. D. Steinhaus (14.01.1887–15.02.1972,Polish).

Banach is considered one of the world’s most influential 20th century mathe-maticians. He is one of the founder of modern analysis. There are more than eleventhousand publications with the word ‘Banach’.

Steinhaus was a student of Hilbert, and a pioneer of the foundations of proba-bility and game theory. He also contributed to functional analysis.

Note 2.188 Let X; Y be Banach spaces. Let K be a bounded linear transformationfrom X onto Y :

Problem 2.189 Y ¼ K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � � :(Solution It suffices to show that

Y � K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � � :

For this purpose, let us take any y 2 Y : We have to show that

y 2 K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � � :

Since y 2 Y ; and K : X !onto

Y ; there exists x0 2 X such that K x0ð Þ ¼ y: There

exists a positive integer k such that x0k k\k: It follows that

y ¼ K x0ð Þ 2 K xð Þ : xk k\kf g � K xð Þ : xk k\kf g�� K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � � ;

and hence

y 2 K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � � : ∎)


It follows that

; ¼ Yc ¼ K xð Þ : xk k\1f g� [ K xð Þ : xk k\2f g� [ K xð Þ : xk k\3f g� [ � � �ð Þc¼ K xð Þ : xk k\1f g�ð Þc \ K xð Þ : xk k\2f g�ð Þc \ K xð Þ : xk k\3f g�ð Þc \ � � � :

Thus,

K xð Þ : xk k\1f g�ð Þc \ K xð Þ : xk k\2f g�ð Þc \ K xð Þ : xk k\3f g�ð Þc \ � � � ¼ ;:

Now, since ; is not a dense set,

K xð Þ : xk k\1f g�ð Þc \ K xð Þ : xk k\2f g�ð Þc \ K xð Þ : xk k\3f g�ð Þc \ � � �

is not a dense set. Since Y is a Banach space, Y is a complete metric space. Also, forevery positive integer n; K xð Þ : xk k\nf g�ð Þc is an open subset of Y : Now, byBaire’s category theorem, there exists a positive integer n0 such thatK xð Þ : xk k\n0f g�ð Þc is not dense, that is

K xð Þ : xk k\n0f g�ð Þcð Þ�6¼ Y :

It follows that there exist y0 2 Y ; and a positive real number r0 such that

K xð Þ : xk k\n0f g�ð Þc \ y : y� y0k k\r0f g ¼ ;

that is

y0 þ y : yk k\r0f g ¼ y : y� y0k k\r0f g � K xð Þ : xk k\n0f g�|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is

y0 ¼ y0 þ 0ð Þ 2 y0 þ y : yk k\r0f gð Þ � K xð Þ : xk k\n0f g�|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since y0 2 K xð Þ : xk k\n0f g�; there exists a convergent sequence K x0n

� �� in

K xð Þ : xk k\n0f g such that each x0n�� \n0; and limn!1 K x0n

� � ¼ y0: Next, let y1 2Y such that y1k k\r0: It follows that y0 þ y1ð Þ 2 K xð Þ : xk k\n0f g�; and hencethere exists a convergent sequence K x00n

� �� in K xð Þ : xk k\n0f g such that each

x00n�� \n0; and

limn!1K x00n

� � ¼ y0 þ y1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ limn!1K x0n

� � þ y1:

348 2 Lp-Spaces

It follows that

limn!1K x00n � x0n

� � ¼ limn!1 K x00n

� �� K x0n� �� ¼ lim

n!1K x00n� �� lim

n!1K x0n� � ¼ y1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

For every positive integer n; put xn � x00n � x0n: We have limn!1K xnð Þ ¼ y1: Also,

for every positive integer n;

xnk k ¼ x00n � x0n�� x00n

�� þ x0n�� \n0 þ x0n

�� \n0 þ n0 ¼ 2n0:

Thus, for every y1 2 Y satisfying y1k k\r0; there exists a sequence xnf g in Xsuch that for every positive integer n; xnk k\2n0; and limn!1 K xnð Þ ¼ y1: Let usfix any y0 2 Y : Let e[ 0:

Let us consider the case when y0 6¼ 0: Clearly, r02 y0k k y0�� \r0; and hence there

exists a sequence xnf g in X such that for every positive integer n; xnk k\2n0; and

limn!1 K xnð Þ ¼ r02 y0k k y0: It follows that limn!1 K 2 y0k k

r0xn

¼ y0: Also, for every

positive integer n;

2 y0k kr0

xn

�� \ 2 y0k kr0

� 2n0:

For every positive integer n; put zn � 2 y0k kr0

xn: Thus, limn!1 K znð Þ ¼ y0; and, for

every positive integer n; znk k\ 4n0r0ky0k: Since limn!1 K znð Þ ¼ y0; and e[ 0;

there exists a positive integer N such that K zNð Þ � y0k k\e: Also, zNk k\ 1d y0k k;

where d � r04n0

2 0;1ð Þð ÞConclusion 2.190 Let X; Y be Banach spaces. Let K be a bounded linear trans-formation from X onto Y : Then there exists d[ 0 such that for every y 2 Y ; and,for every e[ 0; there exists x 2 X satisfying

K xð Þ � yk k\e; and xk k� 1d

yk k:

Theorem 2.191 Let X; Y be Banach spaces. Let K be a bounded linear transfor-mation from X onto Y : Then, K is open, in the sense that, for every open set V in X;K Vð Þ is open in Y :

Proof If V ¼ ;; thenK Vð Þ ¼ ;ð Þ is open in Y : So, we consider the case when V 6¼ ;:We have to show that K Vð Þ is open in Y : For this purpose, let us take any K x0ð Þ 2K Vð Þ; where x0 2 V : We have to show that K x0ð Þ is an interior point of K Vð Þ:

By Conclusion 2.190, there exists d[ 0 such that, for every y 2 Y ; and, forevery e[ 0; there exists x 2 X satisfying y� K xð Þk k\e; and xk k� 1

d yk k:


Let us fix any y0 2 Y satisfying y0k k\d: Let us fix any e0 [ 0: It follows thatthere exists x1 2 X satisfying

y0 � K x1ð Þk k\ 12de0; and x1k k� 1

dy0k k \

1d� d ¼ 1

� �:

Again, it follows that there exists x2 2 X satisfying

y0 � K x1ð Þð Þ � K x2ð Þk k\ 122

de0;

and

x2k k� 1d

y0 � K x1ð Þk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\ 1d� 12de0 ¼ 1

2e0:

Again, it follows that there exists x3 2 X satisfying

y0 � K x1ð Þ � K x2ð Þð Þ � K x3ð Þk k\ 123

de0;

and

x3k k� 1d

y0 � K x1ð Þ � K x2ð Þk k\ 1d� 122

de0 ¼ 122

e0; etc:

Thus, we get a sequence xnf g in X such that for every positive integer n;

y0 � K x1 þ x2 þ � � � þ xnð Þk k ¼ y0 � K x1ð Þ � K x2ð Þ � � � � � K xnð Þk k\ 12n

de0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and xnþ 1k k\ 1

2n e0: Also, x1k k\1:

Problem 2:192 x1 þ x2 þ � � � þ xnf g is a Cauchy sequence in X:

(Solution Let us take any e[ 0: Since the geometric seriesP1

n¼112n e0 is conver-

gent, there exists a positive integer N such that n[m�N implies

Xnþ 1

k¼1

xk �Xmk¼1

xk

�� ¼ xmþ 1 þ xmþ 2 þ � � � þ xnþ 1k k

� xmþ 1k kþ xmþ 2k kþ � � � þ xnþ 1k k\

12m

e0 þ 12mþ 1 e0 þ � � � þ 1

2ne0\e;

350 2 Lp-Spaces

and hence n[m�N impliesPnþ 1

k¼1 xk �Pm

k¼1 xk�� \e: Therefore,

x1 þ x2 þ � � � þ xnf g is a Cauchy sequence in X: ∎)Since x1 þ x2 þ � � � þ xnf g is a Cauchy sequence in X; and X is a complete

metric space, there exists x 2 X such that

limn!1 x1 þ x2 þ � � � þ xnð Þ ¼ x:

It follows that

xk k ¼ limn!1 x1 þ x2 þ � � � þ xnk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}� lim

n!1 x1k kþ x2k kþ � � � þ xnk kð Þ

\ limn!1 1þ 1

21e0 þ � � � þ 1

2n�1 e0

� �¼ 1þ e0;

and hence xk k\1þ e0: It follows that x 2 z : z 2 X and zk k\1þ e0f g:Since K is a bounded linear transformation from X onto Y , K is a continuous

linear transformation from X onto Y : Since K is continuous, andlimn!1 x1 þ x2 þ � � � þ xnð Þ ¼ x; we have

limn!1K x1 þ x2 þ � � � þ xnð Þ ¼ K xð Þ:

Since for every positive integer n;

y0 � K x1 þ x2 þ � � � þ xnð Þk k\ 12n

de0;

and limn!1 12n de0 ¼ 0; we have

K xð Þ ¼ limn!1K x1 þ x2 þ � � � þ xnð Þ ¼ y0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

K z : z 2 X and zk k\1þ e0f gð Þ3ð Þ K xð Þ ¼ y0:

Thus,

y0 2 K z : z 2 X and zk k\1þ e0f gð Þ:

Now, since y0 is any element of Y satisfying y0k k\d; we have

d w : w 2 Y and wk k\1f g ¼ w : w 2 Y and wk k\df g� K z : z 2 X and zk k\1þ e0f gð Þ ¼ 1þ e0ð Þ K z : z 2 X and zk k\1f gð Þð Þð Þ:


Hence

l � w : w 2 Y and wk k\1f g � K z : z 2 X and zk k\1f gð Þ;

where l � d1þ e0

[ 0ð Þ: This shows that

l � w : w 2 Y and wk k\1f g � K z : z 2 X and zk k\1f gð Þ;

where l is a positive real number. Since x0 2 V ; and V is open in X; there existsg[ 0 such that

x0 þ g � z : z 2 X and zk k\1f g � V :

Now

K Vð Þ � K x0 þ g � z : z 2 X and zk k\1f gð Þ ¼ K x0ð Þþ g � K z : z 2 X and zk k\1f gð Þ� K x0ð Þþ g � l � w : w 2 Y and wk k\1f g ¼ K x0ð Þþ w : w 2 Y and wk k\glf g:

Since

K x0ð Þþ w : w 2 Y and wk k\glf g � K Vð Þ;

and

K x0ð Þþ w : w 2 Y and wk k\glf g

is an open neighborhood of K x0ð Þ; K x0ð Þ is an interior point of K Vð Þ: ∎Theorem 2.191 is known as the open mapping theorem.

Note 2.193 Let X; Y be Banach spaces. Let K be a bounded linear transformationfrom X onto Y : Let K be 1-1.

Then, by Theorem 2.191, 0 ¼ K 0ð Þ 2 K z : z 2 X and zk k\1f gð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} is open in Y ;

and hence there exists d[ 0 such that

w : w 2 Y and wk k\df g � K z : z 2 X and zk k\1f gð Þ:Problem 2.194 For every x 2 X; d xk k� K xð Þk k:(Solution Let us fix any x 2 X: If x ¼ 0; then d xk k� K xð Þk k is trivially true. So,we consider the case, when x 6¼ 0: We claim that d xk k� K xð Þk k: If not, otherwise,let K xð Þk k\d xk k: We have to arrive at a contradiction. Since K xð Þk k\d xk k; andx 6¼ 0; K 1

xk k x �� \d; and hence

K1xk k x

� �2 w : w 2 Y and w\df g � K z : z 2 X and z\1f gð Þð Þ:

352 2 Lp-Spaces

It follows that there exists z0 2 X satisfying z0\1; and K z0ð Þ ¼ K 1xk k x

: Since,

K z0ð Þ ¼ K 1xk k x

; and K is 1-1, we have z0 ¼ 1

xk k x; and hence,

1 [ z0k k ¼ 1xk k x

�� |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} ¼ 1: This is a contradiction. ∎)

Since K is a 1-1 linear transformation from X onto Y ; K�1 : Y ! X is a lineartransformation.

Problem 2.195 K�1 is bounded, that is K�1 yð Þ�� : yk k� 1� �

is bounded above.

(Solution For this purpose, let us fix any y 2 Y such that yk k� 1: It follows thatK�1 yð Þ 2 X: Now, by Problem 2.194,

d K�1 yð Þ�� K K�1 yð Þ� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ yk k� 1;

and hence K�1 yð Þ�� 1d : Thus,

1d is an upper bound of K�1 yð Þ�� : yk k� 1

� �; and

hence K�1 yð Þ�� : yk k� 1� �

is bounded above. ∎)

Conclusion 2.196 Let X; Y be Banach spaces. Let K be a bounded linear trans-formation from X onto Y : Let K be 1-1. Then K�1 is a bounded linear transfor-mation from Y onto X:

2.7 Hahn-Banach Theorem

Note 2.197 Let V be a complex linear space. Let f : V ! C be a linear functional.

Problem 2.198 For every x 2 V ; f xð Þ ¼ Re f xð Þð Þ � iRe f ixð Þð Þ:(Solution Let us take any x 2 V : We have to show that f xð Þ ¼ Re f xð Þð Þ �iRe f ixð Þð Þ: Since f xð Þ ¼ Re f xð Þð Þþ iIm f xð Þð Þ; it suffices to show thatIm f xð Þð Þ ¼ �Re f ixð Þð Þ:

RHS ¼ �Re f ixð Þð Þ ¼ �Re i f xð Þð Þð Þ ¼ �Re i Re f xð Þð Þþ iIm f xð Þð Þð Þð Þ¼ �Re iRe f xð Þð Þ � Im f xð Þð Þð Þ ¼ � �Im f xð Þð Þð Þ ¼ Im f xð Þð Þ ¼ LHS: ∎)

Problem 2.199 Re fð Þ : V ! R is a real-linear functional in the sense that, forevery x; y 2 V ; and for every real a; b; Re fð Þð Þ axþ byð Þ ¼a Re fð Þð Þ xð Þð Þþ b Re fð Þð Þ yð Þð Þ: Also, Im fð Þ : V ! R is a real-linear functional.


(Solution Let x; y 2 V ; and a; b 2 R: We have to show that

Re fð Þð Þ axþ byð Þ ¼ a Re fð Þð Þ xð Þð Þþ b Re fð Þð Þ yð Þð Þ:LHS ¼ Re fð Þð Þ axþ byð Þ ¼ Re f axþ byð Þð Þ

¼ Re a f xð Þð Þþb f yð Þð Þð Þ¼ Re a Re f xð Þð Þþ iIm f xð Þð Þð Þþ b Re f yð Þð Þþ iIm f yð Þð Þð Þð Þ¼ Re a Re f xð Þð Þð Þþ b Re f yð Þð Þð Þþ i a Im f xð Þð Þð Þþb Im f yð Þð Þð Þð Þð Þ¼ a Re f xð Þð Þð Þþ b Re f yð Þð Þð Þ¼ a Re fð Þð Þ xð Þð Þþ b Re fð Þð Þ yð Þð Þ ¼ RHS:

∎)

Let V be a complex linear space. Let u : V ! R be a real-linear functional. Letf : x 7! u xð Þ � iu ixð Þð Þ be a mapping from V to C:

Problem 2.200 f : V ! C is a linear functional.

(Solution

1. Let x; y 2 V : We have to show that

u xþ yð Þ � iu i xþ yð Þð Þ ¼ u xð Þ � iu ixð Þð Þþ u yð Þ � iu iyð Þð Þ:LHS ¼ u xþ yð Þ � iu i xþ yð Þð Þ

¼ u xð Þþ u yð Þð Þ � iu i xþ yð Þð Þ¼ u xð Þþ u yð Þð Þ � iu ixþ iyð Þ¼ u xð Þþ u yð Þð Þ � i u ixð Þþ u iyð Þð Þ¼ u xð Þ � iu ixð Þð Þþ u yð Þ � iu iyð Þð Þ ¼ RHS:

2. Let x 2 V : Let sþ itð Þ 2 C; where s; t are reals. We have to show that

u sþ itð Þxð Þ � iu i sþ itð Þxð Þð Þ ¼ sþ itð Þ u xð Þ � iu ixð Þð Þ:LHS ¼ u sþ itð Þxð Þ � iu i sþ itð Þxð Þð Þ

¼ u sxþ t ixð Þð Þ � iu �txþ s ixð Þð Þ¼ s u xð Þð Þþ t u ixð Þð Þð Þ � i �t u xð Þð Þþ s u ixð Þð Þð Þ¼ s u xð Þð Þþ t u ixð Þð Þþ i t u xð Þð Þð Þ � i s u ixð Þð Þð Þ¼ sþ itð Þ u xð Þ � iu ixð Þð Þ ¼ RHS: ∎)

Let V be a normed linear space. Let u : V ! R be a real-linear functional. Letf : x 7! u xð Þ � iu ixð Þð Þ be a mapping from V to C: We have seen that f : V ! C isa linear functional.

354 2 Lp-Spaces

Problem 2.201 sup u xð Þ � iu ixð Þj j : xk k� 1f g ¼ sup u xð Þj j : xk k� 1f g:(Solution Let us take any x 2 V satisfying xk k� 1: Since u xð Þj j � u xð Þ � iu ixð Þj j;we have

sup u xð Þj j : xk k� 1f g� sup u xð Þ � iu ixð Þj j : xk k� 1f g:

Let us fix any y 2 X satisfying yk k� 1: It suffices to show that

u yð Þ � iu iyð Þj j � sup u xð Þj j : xk k� 1f g:

There exists a complex number a such that

u ayð Þ � iu iayð Þ ¼ f ayð Þ ¼ a � f yð Þð Þ ¼ a � u yð Þ � iu iyð Þð Þ ¼ u yð Þ � iu iyð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and aj j ¼ 1: It follows that u iayð Þ ¼ 0; and u yð Þ � iu iyð Þj j ¼ u ayð Þ: Thus,u yð Þ � iu iyð Þj j ¼ u ayð Þj j: Since

ayk k ¼ aj j yk k ¼ 1 yk k ¼ yk k� 1;

we have ayk k� 1; and hence u ayð Þj j 2 u xð Þj j : xk k� 1f g: It follows that

u yð Þ � iu iyð Þj j ¼ u ayð Þj j � sup u xð Þj j : xk k� 1f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence

u yð Þ � iu iyð Þj j � sup u xð Þj j : xk k� 1f g: ∎)

In short, fk k ¼ uk k:Conclusion 2.202

1. Let V be a complex linear space. Let f : V ! C be a linear functional. Then, forevery x 2 V ;

f xð Þ ¼ Re fð Þð Þ xð Þ � i Re fð Þð Þ ixð Þð Þ:

2. Let V be a complex linear space. Let f : V ! C be a linear functional. ThenReðf Þ : V ! R is a real-linear functional, and Im fð Þ : V ! R is a real-linearfunctional.

3. Let V be a complex linear space. Let u : V ! R be a real-linear functional. Letf : x 7! u xð Þ � iu ixð Þð Þ be a mapping from V to C: Then f : V ! C is a linearfunctional.

2.7 Hahn-Banach Theorem 355

4. Let V be a normed linear space. Let u : V ! R be a real-linear functional. Letf : x 7! u xð Þ � iu ixð Þð Þ be a mapping from V to C: Then fk k ¼ uk k:

5. Let V be a normed linear space. Let f : V ! C be a linear functional. Thenfk k ¼ Re fð Þk k:

Note 2.203 Let X be a real normed linear space. Let M be a subspace of X: Letf : M ! R be a bounded real-linear functional. Let M 6¼ X: Let fk k ¼ 1:

Since M 6¼ X; there exists x0 2 X such that x0 62 M: The linear span

M [ x0f g½ �|fflfflfflfflfflffl{zfflfflfflfflfflffl} ¼ xþ tx0 : t 2 R; and x 2 Mf g

is a linear subspace of X containing M [ x0f g: Since M is a subspace of X; we have0 2 M: Now, since x0 62 M; we have x0 6¼ 0: Since x0 62 M; M is a proper subset of

M [ x0f g|fflfflfflfflffl{zfflfflfflfflffl} � M [ x0f g½ � ¼ xþ tx0 : t 2 R; and x 2 Mf g:

Thus, M is a proper subset of xþ tx0 : t 2 R; and x 2 Mf g:Let us fix any a 2 R: Let

F : xþ tx0ð Þ 7! f xð Þþ t � að Þ

be a function from xþ tx0 : t 2 R; and x 2 Mf g to R:

Problem 2.204 F is an extension of f from M to xþ tx0 : t 2 R; and x 2 Mf g; inthe sense that, for every xþ 0x0 ¼ð Þx 2 M; F xð Þ ¼ f xð Þ: Also, F :xþ tx0 : t 2 R; and x 2 Mf g ! R is real-linear.

(Solution

1. Let s; t 2 R; and x; y 2 M: We have to show that

F xþ sx0ð Þþ yþ tx0ð Þð Þ ¼ F xþ sx0ð ÞþF yþ tx0ð Þ:

LHS ¼ F xþ sx0ð Þþ yþ tx0ð Þð Þ ¼ F xþ yð Þþ sþ tð Þx0ð Þ¼ f xþ yð Þþ sþ tð Þ � a ¼ f xð Þþ f yð Þð Þþ sþ tð Þ � a¼ f xð Þþ f yð Þð Þþ s � aþ t � að Þ ¼ f xð Þþ s � að Þþ f yð Þþ t � að Þ¼ F xþ sx0ð Þþ f yð Þþ t � að Þ ¼ F xþ sx0ð ÞþF yþ tx0ð Þ ¼ RHS:

2. Let u; s 2 R; and x 2 M: We have to show that F u � xþ sx0ð Þð Þ ¼u � F xþ sx0ð Þð Þ:

356 2 Lp-Spaces

LHS ¼ F u � xþ sx0ð Þð Þ ¼ F uxþ usð Þx0ð Þ ¼ f uxð Þþ usð Þ � a¼ u � f xð Þð Þþ usð Þ � a ¼ u � f xð Þþ s � að Þ ¼ u � F xþ sx0ð Þð Þ ¼ RHS: ∎)

Since f : M ! R is a bounded real-linear functional, f : M ! R is continuousat 0. Since f : M ! R is continuous at 0; and F is an extension of f ;

F : xþ tx0ð Þ 7! f xð Þþ t � að Þ

from xþ tx0 : t 2 R; and x 2 Mf g to R is continuous at 0: Since F :xþ tx0 : t 2 R; and x 2 Mf g ! R is continuous at 0; and F is a real-linear

functional,

F : xþ tx0 : t 2 R; and x 2 Mf g ! R

is a bounded real-linear functional. Since

f xð Þj j : x 2 M; and xk k� 1f g � f xð Þþ t � aj j : t 2 R; x 2 M; and xþ tx0k k� 1f g;

we have

1 ¼ kf k ¼sup f xð Þj j : x 2 M; and xk k� 1f g� sup f xð Þþ t � aj j : t 2 R; x 2 M; and xþ tx0k k� 1f g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ Fk k;

and hence 1� Fk k:Problem 2.205 Fk k ¼ 1:

(Solution Since 1� Fk k; it suffices to show that Fk k� 1: Since

F : xþ tx0 : t 2 R; and x 2 Mf g ! R

is a bounded real-linear functional, by Problem 2.180, it suffices to show that, forevery t 2 R; and, for every x 2 M satisfying xþ tx0k k� 1;

f xð Þþ t � aj j ¼ F xþ tx0ð Þj j � 1 � xþ tx0k k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since f : M ! R is a bounded real-linear functional, for every x 2 M satisfying

xþ 0 � x0k k ¼ xk k� 1|fflfflfflffl{zfflfflfflffl}; we have

f xþ 0 � x0ð Þj j ¼ f xð Þj j � fk k xk k|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} ¼ 1 xk k ¼ xk k ¼ xþ 0 � x0k k:


Thus, for every x 2 M satisfying xþ 0 � x0k k� 1; we havef xþ 0 � x0ð Þj j � xþ 0 � x0k k: Hence, it suffices to show that for every nonzero real t;and, for every x 2 M satisfying xþ tx0k k� 1; f xð Þþ t � aj j � xþ tx0k k:

For this purpose, let us fix any nonzero real t: It suffices to show that, for everyx 2 M ¼ �tMð Þ satisfying xþ tx0k k� 1; f xð Þþ t � aj j � xþ tx0k k: Thus, itsuffices to show that for every x 2 M satisfying

tj j x� x0k k ¼ �txð Þþ tx0k k� 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};tj j f xð Þ � aj j ¼ �tð Þf xð Þþ t � aj j ¼ f �txð Þþ t � aj j � �txð Þþ tx0k k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ tj j x� x0k k:

Finally, it suffices to show that for every x 2 M satisfying x� x0k k� 1tj j ;

f xð Þ � aj j � x� x0k k: It suffices to show that for every x 2 M satisfying0\ x� x0k k� 1

tj j ;

a 2 f xð Þ � x� x0k k; f xð Þþ x� x0k k½ �:

Thus, it is to enough to show that

\ x2M and 0\ x�x0k k� 1tj jf xð Þ � x� x0k k; f xð Þþ x� x0k k½ �

is nonempty.

Problem 2.206 It suffices to show that for every x 2 M satisfying0\ x� x0k k� 1

tj j ; and, for every y 2 M satisfying 0\ y� x0k k� 1tj j ;

f xð Þ � x� x0k k� f yð Þþ y� x0k k:(Solution Suppose that for every x 2 M satisfying 0\ x� x0k k� 1

tj j ; and, for

every y 2 M satisfying 0\ y� x0k k� 1tj j ;

f xð Þ � x� x0k k� f yð Þþ y� x0k k:



is nonempty.It follows that f 0ð Þþ 0� x0k k is an upper bound of

f x1ð Þ � x1 � x0k k : x1 2 M; and 0\ x1 � x0k k� 1tj j

� �;

358 2 Lp-Spaces

and hence

f x1ð Þ � x1 � x0k k : x1 2 M; and 0\ x1 � x0k k� 1tj j

� �is bounded above. Therefore,

sup f x1ð Þ � x1 � x0k k : x1 2 M; and 0\ x1 � x0k k� 1tj j

� �exists. Also, for every x 2 M satisfying 0\ x� x0k k� 1

tj j ; and, for every y 2 M

satisfying 0\ y� x0k k� 1tj j ;

f xð Þ � x� x0k k� sup f x1ð Þ � x1 � x0k k : x1 2 M; and 0\ x1 � x0k k� 1tj j

� �� f yð Þþ y� x0k k:

Now, it follows that

sup f x1ð Þ � x1 � x0k k : x1 2 M; and 0\ x1 � x0k k� 1tj j

� �� 2 \ x2M and 0\ x�x0k k� 1

tj jf xð Þ � x� x0k k; f xð Þþ x� x0k k½ �;

and hence


is nonempty. ∎)For this purpose, let us fix any x 2 M satisfying 0\ x� x0k k� 1

tj j ; and let us fix

any y 2 M satisfying 0\ y� x0k k� 1tj j : By Problem 2.206, it suffices to show that

f xð Þ � x� x0k k� f yð Þþ y� x0k k;

that is

f xð Þ � f yð Þ� x� x0k kþ y� x0k k;

that is

f x� yð Þ� x� x0k kþ y� x0k k:


Since

f x� yð Þ� f x� yð Þj j � fk k x� yk k ¼ 1 � x� yk k ¼ x� yk k¼ x� x0ð Þ � y� x0ð Þk k� x� x0k kþ y� x0k k;

we have

f x� yð Þ� x� x0k kþ y� x0k k: ∎)

Conclusion 2.207 Let X be a real normed linear space. Let M be a subspace of X:Let f : M ! R be a bounded real-linear functional. Let M 6¼ X: Let f ¼ 1: Thenthere exists a subspace N of X; and a function F : N ! R such that

1. M is a proper subspace of N;2. F : N ! R is an extension of f : M ! R;3. F : N ! R is a bounded real-linear functional,4. F ¼ 1:

Lemma 2.208 Let X be a real normed linear space. Let M be a subspace of X: Letf : M ! R be a bounded real-linear functional. Let M 6¼ X: Then there exists asubspace N of X; and a function F : N ! R such that

1. M is a proper subspace of N;2. F : N ! R is an extension of f : M ! R;3. F : N ! R is a bounded real-linear functional,4. Fk k ¼ fk k:

Proof Case I: when fk k ¼ 0: It follows that

sup f xð Þj j : x 2 M; and xk k� 1f g ¼ 0;

and hence for every x 2 M satisfying xk k� 1; we have 0�ð Þ f xð Þj j � 0: This showsthat for every x 2 M satisfying xk k� 1; f xð Þ ¼ 0: Let us take x 7! 0 from X to R forF: Clearly, M is a proper subspace of X; F : X ! R is an extension of f : M ! R;F : X ! R is a bounded real-linear functional, and Fk k ¼ 0 ¼ fk k:

Case II: when fk k 6¼ 0: Here, since f : M ! R is a bounded real-linear func-tional, 1

fk k f : M ! R is a bounded real-linear functional. Also,

1fk k f

�� ¼ 1fk k � fk k ¼ 1:

Now, by Conclusion 2.207, there exist a subspace N of X; and a functionF1 : N ! R such that

a. M is a proper subspace of N;b. F1 : N ! R is an extension of 1

fk k f : M ! R;

c. F1 : N ! R is a bounded real-linear functional,d. F1k k ¼ 1:

360 2 Lp-Spaces

Since F1 : N ! R is an extension of 1fk k f : M ! R; fk kF1ð Þ : N ! R is an

extension of f : M ! R: Since F1 : N ! R is a bounded real-linear functional,fk kF1ð Þ : N ! R is a bounded real-linear functional. Next,

fk kF1k k ¼ sup fk kF1ð Þ xð Þj j : x 2 N; and xk k� 1f g¼ sup fk k F1 xð Þj j : x 2 N; and xk k� 1f g¼ fk ksup F1 xð Þj j : x 2 N; and xk k� 1f g¼ fk kF1 ¼ fk k � 1 ¼ fk k;

so, fk kF1k k ¼ fk k: ∎)

Lemma 2.209 Let X be a real normed linear space. Let M be a subspace of X: Letf : M ! R be a bounded real-linear functional. Then there exists a function F :X ! R such that

1. F : X ! R is an extension of f : M ! R;2. F : X ! R is a bounded real-linear functional,3. Fk k ¼ fk k:

Proof Case I: when M ¼ X: Here, we can take f for F:Case II: when M 6¼ X: Let P be the collection of all ordered pairs N;Fð Þ such

that N is a subspace of X; and F : N ! R is a function satisfying

a. M is a proper subspace of N;b. F : N ! R is an extension of f : M ! R;c. F : N ! R is a bounded real-linear functional,d. Fk k ¼ fk k:

By Lemma 2.208, P is nonempty. Let N;Fð Þ 2 P; and N1;F1ð Þ 2 P: ByN;Fð Þ4 N1;F1ð Þ; we shall mean N � N1; and for every x 2 N; F xð Þ ¼ F1 xð Þ:Clearly, 4 is a partial ordering over P: Now, by the Hausdorff maximality axiom

of set theory, there exists a maximal linearly ordered set B such that B � P:Problem 2:210 [ N;Fð Þ2BN; [ N;Fð Þ2BF

� � 2 P:(Solution We must prove:

1. [ N;Fð Þ2BN is a subspace of X;2. [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is a function,3. M is a proper subspace of [ N;Fð Þ2BN;4. [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is an extension of f : M ! R;5. [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is a real-linear functional,6. [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is bounded,7. [ N;Fð Þ2BF�� ¼ fk k:


For 1: Let us take any x; y 2 [ N;Fð Þ2BN; and s; t 2 R: We have to show thatsxþ ty 2 [ N;Fð Þ2BN:

Since x 2 [ N;Fð Þ2BN; there exists N1;F1ð Þ 2 B such that x 2 N1: Similarly,there exists N2;F2ð Þ 2 B such that y 2 N2: Since N1;F1ð Þ 2 B; N2;F2ð Þ 2 B; and Bis linearly ordered, N1;F1ð Þ4 N2;F2ð Þ or N2;F2ð Þ4 N1;F1ð Þ: For definiteness, letN1;F1ð Þ4 N2;F2ð Þ: It follows that x 2ð ÞN1 � N2; and hence x 2 N2: Since x 2 N2;

y 2 N2; s; t 2 R; and N2 is a linear subspace of X; we have sxþ ty 2N2 � [ N;Fð Þ2BN� �

; and, hence, sxþ ty 2 [ N;Fð Þ2BN:For 2: Let x; að Þ 2 [ N;Fð Þ2BF; and x; bð Þ 2 [ N;Fð Þ2BF: We have to show that

a ¼ b:Since x; að Þ 2 [ N;Fð Þ2BF; there exists N1;F1ð Þ 2 B such that x; að Þ 2 F1: It

follows that x 2 N1; and F1 xð Þ ¼ a: Similarly, there exists N2;F2ð Þ 2 B such thatx 2 N2; and F2 xð Þ ¼ b: Since N1;F1ð Þ 2 B; N2;F2ð Þ 2 B; and B is linearly ordered,we have N1;F1ð Þ4 N2;F2ð Þ or N2;F2ð Þ4 N1;F1ð Þ: For definiteness, letN1;F1ð Þ4 N2;F2ð Þ: It follows that N1 � N2; and, for every y 2 N1; F1 yð Þ ¼ F2 yð Þ:Now, since x 2 N1; a ¼ð ÞF1 xð Þ ¼ F2 xð Þ ¼ bð Þ; and hence a ¼ b:


dom [ N;Fð Þ2BF� � ¼ [ N;Fð Þ2BN:

Since, for every N;Fð Þ 2 B � Pð Þ; dom Fð Þ ¼ N; we have

dom [ N;Fð Þ2BF� � ¼ [ N;Fð Þ2B dom Fð Þð Þ ¼ [ N;Fð Þ2BN|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus,

dom [ N;Fð Þ2BF� � ¼ [ N;Fð Þ2BN:

3. Let us fix any N1;F1ð Þ 2 B � Pð Þ: It follows that N1;F1ð Þ 2 P; and, hence,M isa proper subspace of N1 � [ N;Fð Þ2BN

� �: It follows that M is a proper subspace

of [ N;Fð Þ2BN:4. Let us take any x 2 M: We have to show that f xð Þ ¼ [ N;Fð Þ2BF

� �xð Þ: Suppose

that [ N;Fð Þ2BF� �

xð Þ ¼ b: We have to show that f xð Þ ¼ b:

Since [ N;Fð Þ2BF� �

xð Þ ¼ b; we have x; bð Þ 2 [ N;Fð Þ2BF; and hence there existsN1;F1ð Þ 2 B � Pð Þ such that x; bð Þ 2 F1: Since N1;F1ð Þ 2 P; F1 : N1 ! R is anextension of f : M ! R: Since x; bð Þ 2 F1; and F1 : N1 ! R; we have F1 xð Þ ¼ b:Since F1 : N1 ! R is an extension of f : M ! R; and x 2 M; we haveb ¼ð ÞF1 xð Þ ¼ f xð Þ; and hence f xð Þ ¼ b:

5. Let us take any x; y 2 [ N;Fð Þ2BN; and s; t 2 R: We have to show that

[ N;Fð Þ2BF� �

sxþ tyð Þ ¼ s [ N;Fð Þ2BF� �

xð Þ� �þ t [ N;Fð Þ2BF� �

yð Þ� �:

362 2 Lp-Spaces

Since x 2 [ N;Fð Þ2BN; there exists N1;F1ð Þ 2 B � Pð Þ such that x 2 N1:

Similarly, there exists N2;F2ð Þ 2 B � Pð Þ such that y 2 N2: Since N1;F1ð Þ 2 B;N2;F2ð Þ 2 B; and B is linearly ordered, we have N1;F1ð Þ4 N2;F2ð Þ orN2;F2ð Þ4 N1;F1ð Þ: For definiteness, let N1;F1ð Þ4 N2;F2ð Þ: It follows thatx 2ð ÞN1 � N2; and, hence, x 2 N2: Since N2;F2ð Þ 2 P; F2 : N2 ! R is a real-linearfunctional. Since F2 : N2 ! R is a real-linear functional, x 2 N2; y 2 N2; s; t 2 R;and N2 is a linear subspace of X; we have sxþ tyð Þ 2 N2; and F2 sxþ tyð Þ ¼s F2 xð Þð Þþ t F2 yð Þð Þ:

Put a � F2 xð Þ; b � F2 yð Þ: It follows that x; að Þ 2 F2 � [ N;Fð Þ2BF� �

; and hence[ N;Fð Þ2BF� �

xð Þ ¼ a|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ F2 xð Þ: Thus, [ N;Fð Þ2BF� �

xð Þ ¼ F2 xð Þ: Similarly,


yð Þ ¼ F2 yð Þ; and[ N;Fð Þ2BF� �

sxþ tyð Þ ¼ F2 sxþ tyð Þ:

Now, since F2 sxþ tyð Þ ¼ s F2 xð Þð Þþ t F2 yð Þð Þ; we have


sxþ tyð Þ ¼ s [ N;Fð Þ2BF� �

xð Þ� �þ t [ N;Fð Þ2BF� �

yð Þ� �:

6. It suffices to show that [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is continuous at 0 2 Mð Þ:Since f : M ! R is a bounded real-linear functional, f : M ! R is continuous at0 2 Mð Þ: Since f : M ! R is continuous at 0; and [ N;Fð Þ2BF : [ N;Fð Þ2BN ! Ris an extension of f : M ! R; [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is continuous at 0:

7. Since [ N;Fð Þ2BF is an extension of f ; we have

f xð Þj j : x 2 M; and xk k� 1f g � [ N;Fð Þ2BF� �

xð Þ�� : x 2 [ N;Fð Þ2BN; and xk k� 1� �

;

and hence

fk k ¼ sup f xð Þj j : x 2 M; and xk k� 1f g� sup [ N;Fð Þ2BF� �

xð Þ�� : x 2 [ N;Fð Þ2BN; and xk k� 1� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ [ N;Fð Þ2BF�� :

Thus, fk k� [ N;Fð Þ2BF�� : It suffices to show that [ N;Fð Þ2BF

�� fk k: ByConclusion 2.184, it suffices to show that, for every x 2 [ N;Fð Þ2BN;[ N;Fð Þ2BF� �

xð Þ�� fk k xk k:For this purpose, let us take any x 2 [ N;Fð Þ2BN: We have to show that[ N;Fð Þ2BF� �

xð Þ�� fk k xk k:Since x 2 [ N;Fð Þ2BN; there exists N1;F1ð Þ 2 B � Pð Þ such that x 2 N1: Since

N1;F1ð Þ 2 P;we have F1k k ¼ fk k; and F1 xð Þj j � F1k k xk k: Since N1;F1ð Þ 2 B; andx 2 N1;we have x;F1 xð Þð Þ 2 F1 � [ N;Fð Þ2BF; and, hence, x;F1 xð Þð Þ 2 [ N;Fð Þ2BF:It follows that [ N;Fð Þ2BF

� �xð Þ ¼ F1 xð Þ: Now, since F1 xð Þj j � F1k k xk k; we have



xð Þ�� F1k k xk k|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fk k xk k:

It follows that


xð Þ�� fk k xk k: ∎)

Problem 2:211 B[ [ N;Fð Þ2BN; [ N;Fð Þ2BF� ��

is a linearly ordered set.

(Solution Since, B is a linearly ordered set, it suffices to show that, for everyN;Fð Þ 2 B; N;Fð Þ4 [ N;Fð Þ2BN; [ N;Fð Þ2BF

� �:

For this purpose, let us fix any N1;F1ð Þ 2 B: We have to show thatN1;F1ð Þ4 [ N;Fð Þ2BN; [ N;Fð Þ2BF

� �; that is,

I. N1 � [ N;Fð Þ2BN;II. for every x 2 N1; F1 xð Þ ¼ [ N;Fð Þ2BF

� �xð Þ:

Here (I) is trivially true. For (II), let us take any x 2 N1: Since N1;F1ð Þ 2 B; andx 2 N1; we have x;F1 xð Þð Þ 2 F1 � [ N;Fð Þ2BF; and hence, x;F1 xð Þð Þ 2 [ N;Fð Þ2BF:It follows that [ N;Fð Þ2BF

� �xð Þ ¼ F1 xð Þ: ∎)

Problem 2:212 [ N;Fð Þ2BN ¼ X:

(Solution If not, otherwise, suppose that [ N;Fð Þ2BN 6¼ X: We have to arrive at acontradiction.

Since [ N;Fð Þ2BN; [ N;Fð Þ2BF� � 2 P; [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is a bounded

real-linear functional. Now, by Lemma 2.208, there exist a subspace N1 of X; and afunction F1 : N1 ! R such that

1′. [ N;Fð Þ2BN is a proper subspace of N1;

2′. F1 : N1 ! R is an extension of [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R;3′. F1 : N1 ! R is a bounded real-linear functional,4′. F1k k ¼ [ N;Fð Þ2BF

�� :Problem 2:213 N1;F1ð Þ 2 P:(Solution We must show

a′. M is a proper subspace of N1;b′. F1 : N1 ! R is an extension of f : M ! R;c′. F1 : N1 ! R is a bounded real-linear functional,d′. F1k k ¼ fk k:

For a0 : Since [ N;Fð Þ2BN; [ N;Fð Þ2BF� � 2 P; M is a proper subspace of

[ N;Fð Þ2BN: By 10; [ N;Fð Þ2BN is a proper subspace of N1: Since M is a proper

364 2 Lp-Spaces

subspace of [ N;Fð Þ2BN; and [ N;Fð Þ2BN is a proper subspace of N1;M is a propersubspace of N1:

For b0 : Since [ N;Fð Þ2BN; [ N;Fð Þ2BF� � 2 P; [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is an

extension of f : M ! R: By 20; F1 : N1 ! R is an extension of [ N;Fð Þ2BF :

[ N;Fð Þ2BN ! R: Since [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R is an extension of f : M !R; and F1 : N1 ! R is an extension of [ N;Fð Þ2BF : [ N;Fð Þ2BN ! R; we haveF1 : N1 ! R is an extension of f : M ! R:

For c0 : By 30; F1 : N1 ! R is a bounded real-linear functional.For d0 : Since [ N;Fð Þ2BN; [ N;Fð Þ2BF

� � 2 P; we have [ N;Fð Þ2BF�� ¼ fk k:

By 40;

F1k k ¼ [ N;Fð Þ2BF�� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fk k;

so F1k k ¼ fk k: Thus, N1;F1ð Þ 2 P: ∎)From 10 and 20; we find that

[ N;Fð Þ2BN; [ N;Fð Þ2BF� �

4 N1;F1ð Þ; and [ N;Fð Þ2BN; [ N;Fð Þ2BF� � 6¼ N1;F1ð Þ:

Since B[ [ N;Fð Þ2BN; [ N;Fð Þ2BF� ��

is a linearly ordered set,

[ N;Fð Þ2BN; [ N;Fð Þ2BF� �

4 N1;F1ð Þ;

and

[ N;Fð Þ2BN; [ N;Fð Þ2BF� � 6¼ N1;F1ð Þ;

we find that B[ N1;F1ð Þf g is a linearly ordered set, and N1;F1ð Þ 62 B: Thus B isnot a maximal linearly ordered set. This is a contradiction. ∎)

Thus, [ N;Fð Þ2BF : X ! R is a function such that [ N;Fð Þ2BF : X ! R is anextension of f : M ! R; [ N;Fð Þ2BF : X ! R is a bounded real-linear functional,and [ N;Fð Þ2BF

�� ¼ fk k:We see, in all cases, that there exists a function F : X ! R such that F : X ! R

is an extension of f : M ! R; F : X ! R is a bounded real-linear functional, andFk k ¼ fk k: ∎)

Theorem 2.214 Let X be a normed linear space. Let M be a subspace of X: Letf : M ! C be a bounded linear functional. Then, there exists a function F : X ! Csuch that

1. F : X ! C is an extension of f : M ! C;2. F : X ! C is a bounded linear functional,3. Fk k ¼ fk k:


Proof By Conclusion 2.202, Re fð Þ : M ! R is a real-linear functional, and fk k ¼Re fð Þk k: Since X is a normed linear space, X is a real normed linear space. Clearly,M

is a subspace of the real linear space X: Since f : M ! C is bounded, Re fð Þ : M ! Ris bounded. Now, by Lemma 2.209, there exists a function u : X ! R such that

a. u : X ! R is an extension of Re fð Þ : M ! R;b. u : X ! R is a bounded real-linear functional,c. uk k ¼ Re fð Þk k:

By Conclusion 2.202, the function F : x 7! u xð Þ � iu ixð Þð Þ from X to C is alinear functional such that Fk k ¼ uk k: Since u : X ! R is bounded, and F :x 7! u xð Þ � iu ixð Þð Þ; F : X ! C is bounded. Thus F : X ! C is a boundedreal-linear functional. For every x 2 M;

F xð Þ ¼ u xð Þ � iu ixð Þ ¼ Re fð Þð Þ xð Þ � i Re fð Þð Þ ixð Þ ¼ f xð Þ;

so F : X ! C is an extension of f : M ! C: Since

Fk k ¼ uk k; uk k ¼ Re fð Þk k; and fk k ¼ Re fð Þk k;

we have Fk k ¼ fk k: ∎The Theorem 2.214, known as the Hahn-Banach theorem, is due to H. Hahn

(27.09.1879–24.07.1935, Austrian) and S. Banach. Hahn made contributions tofunctional analysis, topology, calculus of variations and real analysis.

Note 2.215 Let X be a normed linear space. Let M be a subspace of X:

Problem 2.216 The closure �M is a linear subspace of X:

(Solution Let x; y 2 �M: Let a; b 2 C: We have to show that axþ byð Þ 2 �M:

Since x 2 �M there exists a convergent sequence xnf g in M such thatlimn!1 xn ¼ x: Similarly, there exists a convergent sequence ynf g in M such thatlimn!1 yn ¼ y: Since each xn 2 M; each yn 2 M; a; b 2 C; and M is a subspace ofX; each axn þ byn 2 M: Thus, axn þ bynf g is a sequence in M: Since limn!1 xn ¼x; and limn!1 yn ¼ y; axn þ bynf g is a convergent sequence, and

limn!1 axn þ bynð Þ ¼ axþ byð Þ:

Since axn þ bynf g is a convergent sequence in M; and limn!1 axn þ bynð Þ ¼axþ byð Þ; we have axþ byð Þ 2 �M: ∎)Let x0 2 �M:

Problem 2.217 There does not exist a bounded linear functional f : X ! C suchthat for every x 2 M; f xð Þ ¼ 0; and f x0ð Þ 6¼ 0:

(Solution If not, otherwise, suppose that there exists a bounded linear functionalf : X ! C such that, for every x 2 M; f xð Þ ¼ 0 and f x0ð Þ 6¼ 0: We have to arrive ata contradiction. Since x0 2 �M; there exists a convergent sequence x1; x2; . . .f g in M

366 2 Lp-Spaces

such that limn!1 xn ¼ x0: Since x1; x2; . . . are in M; we have f x1ð Þ ¼ 0; f x2ð Þ ¼ 0;etc., and hence limn!1 f xnð Þ ¼ 0: Since f : X ! C is a bounded linear functional,f : X ! C is continuous at x0: Since f : X ! C is continuous at x0; x1; x2; . . .f g is asequence in M such that limn!1 xn ¼ x0; we have 0 ¼ð Þ lim

n!1 f xnð Þ ¼ f x0ð Þ; andhence, f x0ð Þ ¼ 0: This is a contradiction. ∎)

Conclusion 2.218 Let X be a normed linear space. Then,

1. if M is a subspace of X; then �M is a linear subspace of X;2. if x0 2 �M; then there does not exist a bounded linear functional f : X ! C such

that, for every x 2 M; f xð Þ ¼ 0; and f x0ð Þ 6¼ 0:

Lemma 2.219 Let X be a normed linear space. Let M be a subspace of X: Letx0 2 X: Suppose that there does not exist a bounded linear functional F : X ! Csuch that, for every x 2 M; F xð Þ ¼ 0 and F x0ð Þ 6¼ 0: Then, x0 2 �M:

Proof If not, otherwise, let x0 62 �M: We have to arrive at a contradiction.Since x0 62 �M; there exists a positive real number d such that

x : x 2 X; and x� x0k k\df g\M ¼ ;:

It follows that, for every x 2 M; we have d� x� x0k k: Since x0 62 �M � Mð Þ; wehave x0 62 M: It follows that the linear span M [ x0f g½ � of M [ x0f g is equal toxþ ax0 : x 2 M; and a 2 Cf g: We first show that

f : xþ ax0ð Þ 7! a

is a function from the linear space xþ ax0 : x 2 M; and a 2 Cf g to C: For thispurpose, suppose that xþ ax0 ¼ yþ bx0; where x; y 2 M; and a; b 2 C: We have toshow that a ¼ b: If not, otherwise, let a 6¼ b: We have to arrive at a contradiction.Since xþ ax0 ¼ yþ bx0; we have a� bð Þx0 ¼ y� xð Þ 2 Mð Þ; and hencea� bð Þx0 2 M: Since a� bð Þx0 2 M; and a 6¼ b; we have x0 2 1

a�bð ÞM ¼ Mð Þ; andhence x0 2 M: This is a contradiction. Thus,

f : xþ ax0ð Þ 7! a

is a function from the linear space xþ ax0 : x 2 M; and a 2 Cf g to C: Clearly,

f : xþ ax0 : x 2 M; and a 2 Cf g ! C

is a linear functional such that, for every x 2 M; we have f xð Þ ¼ 0 and f x0ð Þ ¼ 1:We want to show that



is bounded. It suffices to show that


is continuous at 0: Observe that


is an extension of the constant function x 7! 0 from M to C: Now, since the constantfunction x 7! 0 from M to C is continuous at 0;


is continuous at 0: Thus,


is a bounded linear functional. Now, by Theorem 2.214, there exists a functionF : X ! C such that

1. F : X ! C is an extension of f : xþ ax0 : x 2 M; and a 2 Cf g ! C;2. F : X ! C is a bounded real-linear functional,3. Fk k ¼ fk k:

For every x 2 M; by (1),

F xð Þ ¼ F xþ 0x0ð Þ ¼ f xþ 0x0ð Þ ¼ 0:

Thus, for every x 2 M; F xð Þ ¼ 0: Also,

F x0ð Þ ¼ F 0þ 1x0ð Þ ¼ f 0þ 1x0ð Þ ¼ f x0ð Þ ¼ 1 6¼ 0:

Hence F x0ð Þ 6¼ 0: Thus, F : X ! C is a bounded linear functional such that, forevery x 2 M; F xð Þ ¼ 0 and F x0ð Þ 6¼ 0: This is a contradiction. ∎

Note 2.220 Let X be a normed linear space. Let x0 2 X; and x0 6¼ 0:Here, the linear span x0f g½ � of x0f g is a linear subspace of X: Also, x0f g½ � ¼

kx0 : k 2 Cf g: Here, f : kx0 7! k x0k k is a linear functional from kx0 : k 2 Cf g toC: Since, for every k 2 C satisfying kx0k k� 1;

f kx0ð Þj j ¼ k x0k kð Þj j ¼ kj j x0k k ¼ kx0k k� 1;

we have

fk k ¼ sup f kx0ð Þj j : k 2 C; and kx0k k� 1f gð Þ� 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence fk k� 1: Thus, f : kx0 7! k x0k k is a bounded linear functional from

kx0 : k 2 Cf g to C: Since 1x0k k x0

�� 1; we have

368 2 Lp-Spaces

fk k ¼ sup f kx0ð Þj j : k 2 C; and kx0k k� 1f g�

f1x0k k x0

� �� ¼ 1x0k k x0k k

�� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 1j j ¼ 1;

and hence fk k� 1: Since fk k� 1; and fk k� 1; we have fk k ¼ 1: Since f :kx0 7! k x0k k is a bounded linear functional from kx0 : k 2 Cf g to C; byTheorem 2.214, there exists a function F : X ! C such that

1. F : X ! C is an extension of f : kx0 : k 2 Cf g ! C;2. F : X ! C is a bounded real-linear functional,3. Fk k ¼ fk k ¼ 1ð Þ:

By (1),

F x0ð Þ ¼ F 1x0ð Þ ¼ f 1x0ð Þ ¼ 1 x0k k ¼ x0k k:

Thus, F x0ð Þ ¼ x0k k: Also, Fk k ¼ 1:

Conclusion 2.221 Let X be a normed linear space. Let x0 2 X; and x0 6¼ 0: Then,there exists a bounded linear functional F : X ! C such that Fk k ¼ 1; andF x0ð Þ ¼ x0k k:

Let X be a normed linear space. Let X be the collection of all bounded linearfunctionals f : X ! C: Let f ; g 2 X and a 2 C: We define f þ gð Þ :x 7! f xð Þþ g xð Þð Þ; and afð Þ : x 7! a f xð Þð Þ from X to C: Under these ‘vector addi-tion’ and ‘scalar multiplication’, X is a complex linear space.

Problem 2.222 X; kkð Þ becomes a normed linear space.

(Solution We must prove:

1. if f 2 X and fk k ¼ 0, then f ¼ 0;2. for every f 2 X; and for every a 2 C; afk k ¼ aj j fk k;3. for every f ; g 2 X; f þ gk k� fk kþ gk k:

For 1: Let f 2 X; and sup f xð Þj j : xk k ¼ 1f g ¼ fk k ¼ 0|fflfflfflffl{zfflfflfflffl} : We have to show that

f ¼ 0; that is, for every x 2 X; f xð Þ ¼ 0: Since f : X ! C is a linear functional, itsuffices to show that f xð Þ ¼ 0 holds, for every nonzero member x in X: For thispurpose, let us fix any nonzero a 2 X: We have to show that f að Þ ¼ 0: It follows

that ak k[ 0; and 1ak k a

�� ¼ 1: Hence,

0� 1ak k f að Þj j ¼ 1

ak k f að Þð Þ�� ¼ f

1ak k a

� �� fk k|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0:


Thus, 1ak k f að Þj j ¼ 0; that is f að Þj j ¼ 0; that is f að Þ ¼ 0:

2. Let us take any a 2 C; and f 2 X: We have to show that afk k ¼ aj j fk k:If a ¼ 0; then afk k ¼ aj j fk k is trivially true. So, we consider the case when

a 6¼ 0: In this case, aj j[ 0:

LHS ¼ afk k ¼ sup afð Þ xð Þj j : xk k ¼ 1f g ¼ sup aj j f xð Þj j : xk k ¼ 1f g¼ aj j sup f xð Þj j : xk k ¼ 1f gð Þ ¼ aj j fk k ¼ RHS:

3. Let us take any f ; g 2 X: We have to show that f þ gk k� fk kþ gk k: Heref þ gk k ¼ sup f þ gð Þ xð Þj j : xk k ¼ 1f g

¼ sup f xð Þþ g xð Þj j : xk k ¼ 1f g� sup f xð Þj j þ g xð Þj j : xk k ¼ 1f g� sup f xð Þj j : xk k ¼ 1f gþ sup g xð Þj j : xk k ¼ 1f g¼ fk kþ gk k;

so f þ gk k� fk kþ gk k: ∎)

Definition Let X be a normed linear space. Here, the normed linear space X iscalled the dual space of X:

2.8 Banach Algebra

Note 2.223

Definition Let A be a complex linear space. Suppose that, to every x; yð Þ 2 A A;there is associated xy 2 A: We say that A is a complex algebra, if the followingconditions are satisfied:

1. for every x; y; z 2 A; x yzð Þ ¼ xyð Þz;2. for every x; y; z 2 A; x yþ zð Þ ¼ xyþ xz; and yþ zð Þx ¼ yxþ zx;3. for every x; y 2 A; and, for every a 2 C; a xyð Þ ¼ axð Þy ¼ x ayð Þ:

Here, the mapping x; yð Þ 7! xy from A A to A is called the multiplication overA; and xy is called the product of x and y:

Definition Let A be a normed linear space. If A is a complex algebra, and, for everyx; y 2 A; xyk k� xk k yk k; then we say that A is a normed complex algebra.

Definition Let A be a normed complex algebra. If A is a complete metric spacerelative to the norm of the normed complex algebra, then we say that A is a Banachalgebra. Thus, every Banach algebra is a Banach space.

370 2 Lp-Spaces

The simplest example of Banach algebra is the complex field C:Let A be a normed complex algebra.

Problem 2.224 The multiplication over A is continuous.

(Solution Let us fix any a; bð Þ 2 A A: We have to show that the multiplicationx; yð Þ 7! xy from A A to A is continuous at a; bð Þ: For this purpose, let us take anytwo convergent sequences xnf g and ynf g in A such that limn!1 xn ¼ a; andlimn!1 yn ¼ b: It suffices to show that limn!1 xnyn ¼ ab; that is,limn!1 xnyn � abk k ¼ 0:

Since limn!1 xn ¼ a; there exists a positive integer N1 such that n�N1 impliesxn � ak k\ 1

2 : Similarly, there exists a positive integer N2 such that n�N2 implies

ynk k � bk k� yn � bk k\ 12|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} :

Since, for every positive integer n�max N1;N2f g;

xnyn � abk k ¼ xn � að Þyn þ yn � bð Þak k� xn � að Þynk kþ yn � bð Þak k� xn � ak k ynk kþ yn � bð Þak k� xn � ak k ynk kþ yn � bk k ak k

� xn � ak k 12þ bk k

� �þ yn � bk k ak k;

we have n�max N1;N2f g implies

0�ð Þ xnyn � abk k� xn � ak k 12þ bk k

� �þ yn � bk k ak k:

Since limn!1 xn ¼ a; we have limn!1 xn � ak k ¼ 0: Similarly,limn!1 yn � bk k ¼ 0: It follows that

limn!1 xn � ak k 1

2þ bk k

� �þ yn � bk k ak k

� �¼ 0 � 1

2þ bk k

� �þ 0 � ak k ¼ 0

� �:

Since

limn!1 xn � ak k 1

2þ bk k

� �þ yn � bk k ak k

� �¼ 0;

and

n�max N1;N2f g ) xnyn � abk k� xn � ak k 12þ bk k

� �þ yn � bk k ak k;

we have limn!1 xnyn � abk k ¼ 0: ∎)

2.8 Banach Algebra 371

Conclusion 2.225 Let A be a normed complex algebra. Then, the multiplicationover A is continuous.

Note 2.226 Let p 2 1;1½ Þ: Let k be any positive integer. By Conclusion 2.50,Cc Rk� �

is a dense subset of Lp Rk� �

: For every f ; g 2 Cc Rk� � � Lp Rk

� �� ; put

d f ; gð Þ � f � gk kp � 0ð Þ:

Problem 2.227 Cc Rk� �

; d� �

is a metric space.


1. for every f 2 Cc Rk� �

; d f ; fð Þ ¼ 0;2. if d f ; gð Þ ¼ 0 then f ¼ g;3. for every f ; g 2 Cc Rk

� �; d f ; gð Þ ¼ d g; fð Þ;

4. for every f ; g; h 2 Cc Rk� �

; d f ; gð Þ� d f ; hð Þþ d h; gð Þ:For 1: Let us take any f 2 Cc Rk

� �: We have to show that d f ; fð Þ ¼ 0:

LHS ¼ d f ; fð Þ ¼ f � fk kp¼ 0k kp¼ 0 ¼ RHS:

For 2: Let us take any f ; g 2 Cc Rk� �

: Let

ZRk

f � gj jpdm ¼ f � gk kp¼ d f ; gð Þ ¼ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} :We have to show that f ¼ g: If not, otherwise let, there exists a 2 Rk such that

f að Þ 6¼ g að Þ: We have to arrive at a contradiction. SinceRRk f � gj jpdm ¼ 0; by

Lemma 1.151, f � gj jp¼ 0 a.e. on Rk; and hence f � gj j ¼ 0 a.e. on Rk: This showsthat f � g ¼ 0 a.e. on Rk; and hence

m f � gð Þ�1 C� 0f gð Þ

¼ 0:

Since f ; g 2 Cc Rk� �

; by Lemma 1.171, f � g 2 Cc Rk� �

; and hence f � gð Þ :Rk ! C is continuous. Now, since C� 0f g is open in C; f � gð Þ�1 C� 0f gð Þ is anopen set in Rk: Since f að Þ 6¼ g að Þ; we have a 2 f � gð Þ�1 C� 0f gð Þ: Since a 2f � gð Þ�1 C� 0f gð Þ; and f � gð Þ�1 C� 0f gð Þ is an open set in Rk; a is an interior

point of f � gð Þ�1 C� 0f gð Þ; and hence m f � gð Þ�1 C� 0f gð Þ

[ 0: This is a

contradiction.

372 2 Lp-Spaces

For 3: Let us take any f ; g 2 Cc Rk� �

: We have to show that d f ; gð Þ ¼ d g; fð Þ:

LHS ¼ d f ; gð Þ ¼ f � gk kp¼ �1ð Þ g� fð Þk kp¼ �1j j g� fk kp¼ 1 g� fk kp¼ g� fk kp¼ d g; fð Þ ¼ RHS:

For 4: Let us take any f ; g; h 2 Cc Rk� �

: We have to show thatd f ; gð Þ� d f ; hð Þþ d h; gð Þ: Here,

d f ; gð Þ ¼ f � gk kp¼ f � hð Þþ h� gð Þk kp � f � hk kp þ h� gk kp¼ d f ; hð Þþ d h; gð Þ;

so d f ; gð Þ� d f ; hð Þþ d h; gð Þ: ∎)

Conclusion 2.228 Let p 2 1;1½ Þ: Let k be any positive integer. By Conclusion2.14, Cc Rk

� �is a dense subset of Lp Rk

� �: For every f ; g 2 Cc Rk

� � � Lp Rk� ��

; put

d f ; gð Þ � f � gk kp � 0ð Þ:

Then Cc Rk� �

; d� �

is a metric space.

Note 2.229 Let k be any positive integer. Let f 2 Cc Rk� �

: It follows that f : Rk !C is continuous, and hence fj j : Rk ! 0;1½ Þ is continuous. It follows that fj j :Rk ! 0;1½ Þ is a measurable function.

Problem 2.230 a : a 2 0;1½ Þ; andm fj j�1 a;1ð �ð Þ

¼ 0n o

6¼ ;:

(Solution If not, otherwise, suppose that for every a 2 0;1½ Þ;

m x : a\ f xð Þj jf gð Þ ¼ m fj j�1 a;1ð Þð Þ

¼ m fj j�1 a;1ð �ð Þ

[ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :We have to arrive at a contradiction. Since, for every a 2 0;1½ Þ;

m x : a\ f xð Þj jf gð Þ[ 0; we have, for every a 2 0;1½ Þ; x : a\ f xð Þj jf g 6¼ ;; andhence

supp fð Þ ¼ f�1 C� 0f gð Þ� �� f�1 C� 0f gð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}is unbounded. This shows that supp fð Þ is an unbounded subset of Rk: Since f 2Cc Rk� �

; supp fð Þ is a compact subset of Rk; and hence, supp fð Þ is a bounded subsetof Rk: This is a contradiction. ∎)


It follows that f 2 L1 Rk� �

; and

fk k1¼ min a : a 2 0;1½ Þ; andm fj j�1 a;1ð �ð Þ

¼ 0n o

:

Thus, Cc Rk� � � L1 Rk

� �:

Problem 2.231 sup f xð Þj j : x 2 Rk� �

\1:

(Solution Since f 2 Cc Rk� �

; supp fð Þ is a compact subset of Rk; and f : Rk ! C iscontinuous. Since f : Rk ! C is continuous, fj j : Rk ! 0;1½ Þ is continuous.Since fj j : Rk ! 0;1½ Þ is continuous, and supp fð Þ is a compact subset of Rk;fj j supp fð Þð Þ is a compact subset of 0;1½ Þ; and hence

f xð Þj j : f xð Þ 6¼ 0f g ¼ fj j f�1 C� 0f gð Þ� � � fj j f �1 C� 0f gð Þ� �� ¼ fj j supp fð Þð Þ|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

is bounded. This shows that f xð Þj j : f xð Þ 6¼ 0f g is bounded, and hencef xð Þj j : x 2 Rk

� �is bounded. It follows that sup f xð Þj j : x 2 Rk

� �\1: ∎)

Problem

2.232 sup f xð Þj j : x 2 Rk� ��min a : a 2 0;1½ Þ; andm fj j�1 a;1ð �ð Þ

¼ 0

n o:

(Solution Let us fix any x0 2 Rk: It suffices to show that

f x0ð Þj j �min a : a 2 0;1½ Þ; andm fj j�1 a;1ð �ð Þ

¼ 0n o

:

Let us fix any a0 2 0;1½ Þ such that

m x : a0\ f xð Þj jf gð Þ ¼ m fj j�1 a0;1ð Þð Þ

¼ m fj j�1 a0;1ð �ð Þ

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, m x : a0\ f xð Þj jf gð Þ ¼ 0: It suffices to show that f x0ð Þj j � a0: If not,

otherwise, let a0\ f x0ð Þj j: We have to arrive at a contradiction. Since a0\ f x0ð Þj j;we have x0 2 x : a0\ f xð Þj jf g: Since a0;1ð Þ is an open set and fj j is continuous,fj j�1 a0;1ð Þð Þ ¼ x : a0\ f xð Þj jf gð Þ is open in Rk: Since x : a0\ f xð Þj jf g is open inRk; and x0 2 x : a0\ f xð Þj jf g; x0 is an interior point of x : a0\ f xð Þj jf g; and hencem x : a0\ f xð Þj jf gð Þ[ 0: This is a contradiction. ∎)

Problem

2.233 sup f xð Þj j : x 2 Rk� � ¼ min a : a 2 0;1½ Þ; andm fj j�1 ða;1gð Þ

¼ 0

n o:

374 2 Lp-Spaces

(Solution If not, otherwise, suppose that

sup f xð Þj j : x 2 Rk� �

\min a : a 2 0;1½ Þ; andm fj j�1 a;1ð Þð Þ

¼ 0n o

:

We have to arrive at a contradiction. There exists a positive real number c suchthat

sup f xð Þj j : x 2 Rk� �

\c\min a : a 2 0;1½ Þ; andm fj j�1 a;1ð Þð Þ

¼ 0n o

;

that is

0�ð Þsup f xð Þj j : x 2 Rk� �

\c\min a : a 2 0;1½ Þ; andm x : a\ f xð Þj jf gð Þ ¼ 0f g:

Since sup f xð Þj j : x 2 Rk� �

\c; we have, for every x 2 Rk; f xð Þj j\c: Now,since c 2 0;1½ Þ and

m x : c\ f xð Þj jf gð Þ ¼ m ;ð Þ ¼ 0;

we have

c 2 a : a 2 0;1½ Þ; andm x : a\ f xð Þj jf gð Þ ¼ 0f g:

It follows that

c\ð Þ min a : a 2 0;1½ Þ; andm x : a\ f xð Þj jf gð Þ ¼ 0f gð Þ� c:

Thus, c\c: This is a contradiction. ∎)

Conclusion 2.234 Let k be any positive integer. Then

1. Cc Rk� � � L1 Rk

� �;

2. for every f 2 Cc Rk� �

; sup f xð Þj j : x 2 Rk� � ¼ fk k1:

Note 2.235

Definition Let X be a locally compact Hausdorff space. Let f : X ! C: If, forevery e[ 0; there exists a compact subset K of X such that, for every x 2 Kc;f xð Þj j\e; then we say that f vanishes at infinity. The collection of all continuousfunctions f : X ! C such that f vanishes at infinity is denoted by C0 Xð Þ:Problem 2.236 Cc Xð Þ � C0 Xð Þ:(Solution Let us take any f 2 Cc Xð Þ: We have to show that f 2 C0 Xð Þ: Sincef 2 Cc Xð Þ; f : X ! C is continuous, and

f�1 0ð Þ� �c� ��¼ f�1 0f gð Þ� �c� ��¼ f �1 C� 0f gð Þ� ��¼ supp fð Þ|fflfflfflffl{zfflfflfflffl}


is a compact subset of X: It remains to show that f vanishes at infinity. For thispurpose, let us take any e[ 0: Let

x 2 f�1 0ð Þ� �c� �� c|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f�1 0ð Þ� �0� f�1 0ð Þ:

It suffices to show that f xð Þj j\e: Since x 2 f�1 0ð Þ; we have f xð Þj j ¼ 0j j ¼ 0\e;and hence f xð Þj j\e: ∎)

Conclusion 2.237 Let X be a locally compact Hausdorff space. Then,Cc Xð Þ � C0 Xð Þ:Note 2.238 Let X be a compact Hausdorff space. It follows that X is a locallycompact Hausdorff space.

Problem 2.239 Cc Xð Þ ¼ C0 Xð Þ:(Solution By Conclusion 2.237, it remains to show that C0 Xð Þ � Cc Xð Þ: For thispurpose, let us take any f 2 C0 Xð Þ: We have to show that f 2 Cc Xð Þ: Since f 2C0 Xð Þ; f : X ! C is continuous and f vanishes at infinity. It remains to show that

f�1 0ð Þ� �c� ��¼ f�1 0f gð Þ� �c� ��¼ f �1 C� 0f gð Þ� ��¼ supp fð Þ|fflfflfflffl{zfflfflfflffl}is compact, that is f�1 0ð Þð Þc� ��

is compact. Since f�1 0ð Þð Þc� ��is a closed subset

of the compact space ; f�1 0ð Þð Þc� ��. is compact. ∎)

Notation By C Xð Þ we mean the collection of all continuous functions f : X ! C:If X is a compact set, then C0 Xð Þ ¼ð ÞCc Xð Þ ¼ C Xð Þ: Thus, if X is a compact set,

then C Xð Þ ¼ Cc Xð Þ ¼ C0 Xð Þ:Since z : z 2 C; and zj j ¼ 1f g is a compact set,

C z : z 2 C; and zj j ¼ 1f gð Þ ¼ Cc z : z 2 C; and zj j ¼ 1f gð Þ¼ C0 z : z 2 C; and zj j ¼ 1f gð Þ ð Þ

Conclusion 2.240

1. If X is a compact Hausdorff space, then Cc Xð Þ ¼ C0 Xð Þ;

2: C z : z 2 C; and zj j ¼ 1f gð Þ ¼ Cc z : z 2 C; and zj j ¼ 1f gð Þ¼ C0 z : z 2 C; and zj j ¼ 1f gð Þ:

Note 2.241 Let X be a locally compact Hausdorff space. Let f ; g 2 C0 Xð Þ:It follows that f : X ! C; and g : X ! C are continuous, and hence f þ gð Þ :

X ! C is continuous. We shall show that f þ gð Þ vanishes at infinity.For this purpose, let us take any e[ 0: Since f 2 C0 Xð Þ; there exists a compact

subset K1 of X such that, for every x 2 K1ð Þc; f xð Þj j\ e2 : Similarly, there exists a

376 2 Lp-Spaces

compact subset K2 of X such that, for every x 2 K2ð Þc; g xð Þj j\ e2 : Since K1;K2 are

compact sets, K1 [K2 is a compact set. For every

x 2 K1 [K2ð Þc|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ K1ð Þc \ K2ð Þc� K1ð Þc;

we have f xð Þj j\ e2 : Similarly, for every x 2 K1 [K2ð Þc; we have g xð Þj j\ e

2 : Itfollows that, for every x 2 K1 [K2ð Þc; we have

f þ gð Þ xð Þj j ¼ f xð Þþ g xð Þj j � f xð Þj j þ g xð Þj j\ e2þ g xð Þj j\ e

2þ e

2¼ e;

and hence for every x 2 K1 [K2ð Þc; we have f þ gð Þ xð Þj j\e: Thus, f þ gð Þ van-ishes at infinity. Since f þ gð Þ is continuous and f þ gð Þ vanishes at infinity, wehave f þ gð Þ 2 C0 Xð Þ:

Let a 2 C� 0f g: Since f : X ! C is continuous, af : X ! C is continuous. Weshall show that af vanishes at infinity.

For this purpose, let us take any e[ 0: Since f 2 C0 Xð Þ; there exists a compactsubset K of X such that, for every x 2 Kc; f xð Þj j\ e

aj j ; and hence, for every x 2 Kc;

afð Þ xð Þj j\e: Thus, af vanishes at infinity. Since af is continuous and af vanishes atinfinity, we have af 2 C0 Xð Þ:

This shows that C0 Xð Þ is a complex linear space.For every f 2 C0 Xð Þ; there exists a compact subset K of X such that, for every

x 2 Kc; f xð Þj j\1: Since f : X ! C is continuous and K is compact, f Kð Þ iscompact. Since f Kð Þ is compact, f Kð Þ is bounded, and hence fj j Kð Þ is bounded.Since fj j Kð Þ is bounded, and, for every x 2 Kc; fj j xð Þ\1; fj j Xð Þ is bounded,sup f xð Þj j : x 2 Xf g is hence a nonnegative real number. Let us denote

0�ð Þsup f xð Þj j : x 2 Xf g

by fk k:Since f � gð Þ : X ! C is continuous and K1 [K2 is compact, f � gð Þ K1 [K2ð Þ

is compact. Since f � gð Þ K1 [K2ð Þ is compact, f � gð Þ K1 [K2ð Þ is bounded, andhence f � gj j K1 [K2ð Þ is bounded. Since f � gj j K1 [K2ð Þ is bounded, and, forevery x 2 K1 [K2ð Þc; f � gj j xð Þ\e; f � gj j Xð Þ is bounded, sup f xð Þ � g xð Þjf j :x 2 Xg is hence a nonnegative real number. For every f ; g 2 C0 Xð Þ; put

d f ; gð Þ � sup f xð Þ � g xð Þj j : x 2 Xf g � 0ð Þ:Problem 2.242 C0 Xð Þ; dð Þ is a metric space.


1. for every f 2 C0 Xð Þ; d f ; fð Þ ¼ 0;2. if d f ; gð Þ ¼ 0 then f ¼ g;3. for every f ; g 2 C0 Xð Þ; d f ; gð Þ ¼ d g; fð Þ;4. for every f ; g; h 2 C0 Xð Þ; d f ; gð Þ� d f ; hð Þþ d h; gð Þ:


For 1: Let us take any f 2 C0 Xð Þ: We have to show that d f ; fð Þ ¼ 0:

LHS ¼ d f ; fð Þ ¼ sup f xð Þ � f xð Þj j : x 2 Xf g ¼ sup 0f g ¼ 0 ¼ RHS:

For 2: Let us take any f ; g 2 C0 Xð Þ: Suppose that

sup f xð Þ � g xð Þj j : x 2 Xf g ¼ d f ; gð Þ ¼ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} :We have to show that f ¼ g: Since

sup f xð Þ � g xð Þj j : x 2 Xf g ¼ 0;

for every y 2 X; we have

0� f yð Þ � g yð Þj j � sup f xð Þ � g xð Þj j : x 2 Xf g ¼ 0;

and hence for every y 2 X; f yð Þ � g yð Þj j ¼ 0: It follows that for every y 2 X;f yð Þ ¼ g yð Þ; and hence f ¼ g:

For 3: Let us take any f ; g 2 C0 Xð Þ: We have to show that d f ; gð Þ ¼ d g; fð Þ:

LHS ¼ d f ; gð Þ ¼ sup f xð Þ � g xð Þj j : x 2 Xf g ¼ sup g xð Þ � f xð Þj j : x 2 Xf g¼ d g; fð Þ ¼ RHS:

For 4: Let us take any f ; g; h 2 C0 Xð Þ: We have to show thatd f ; gð Þ� d f ; hð Þþ d h; gð Þ: Since

d f ; gð Þ ¼ sup f xð Þ � g xð Þj j : x 2 Xf g ¼ sup f xð Þ � h xð Þð Þþ h xð Þ � g xð Þð Þj j : x 2 Xf g� sup f xð Þ � h xð Þj j þ h xð Þ � g xð Þj j : x 2 Xf g� sup f xð Þ � h xð Þj j : x 2 Xf gþ sup h xð Þ � g xð Þj j : x 2 Xf g ¼ d f ; hð Þþ d h; gð Þ;

we have d f ; gð Þ� d f ; hð Þþ d h; gð Þ: ∎)Clearly,

1. fk k ¼ 0 , f ¼ 0;

2. for every a 2 C; and for every f 2 C0 Xð Þ; afk k ¼ aj j fk k;3. for every f ; g 2 C0 Xð Þ; f þ gk k� fk kþ gk k:

Thus, C0 Xð Þ; kkð Þ is a normed linear space, and d is the metric over C0 Xð Þinduced by the norm kk: We have seen that Cc Xð Þ � C0 Xð Þ:Problem 2.243 Cc Xð Þ is dense in C0 Xð Þ:(Solution For this purpose, let us take any f 2 C0 Xð Þ; and e[ 0: Since f 2 C0 Xð Þ;there exists a compact subset K of X such that, for every x 2 Kc; f xð Þj j\e: ByConclusion 1.176, there exists g 2 Cc Xð Þ such that

378 2 Lp-Spaces

1. for all x 2 X; 0� g xð Þ� 1;2. for all x 2 K; g xð Þ ¼ 1:

Problem 2:244 The product f � gð Þ 2 Cc Xð Þ:(Solution Since f 2 C0 Xð Þ; f is continuous. Since g 2 Cc Xð Þ; g is continuous andsupp gð Þ is compact. Since, f is continuous and g is continuous, f � gð Þ is continu-ous. It remains to show that

supp gð Þ ¼ g�1 0ð Þ� �c� �� f�1 0ð Þ� �c \ g�1 0ð Þ� �c� ��¼ f�1 0ð Þ [ g�1 0ð Þ� �c� ��¼ f � gð Þ�1 0ð Þ

c �¼ supp f � gð Þ|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}

is compact. Since supp f � gð Þ � supp gð Þ; supp gð Þ is compact and supp f � gð Þ isclosed, supp f � gð Þ is compact. ∎)

Since, for every x 2 Kc;

f � gð Þ xð Þj j ¼ f xð Þj j g xð Þj j � f xð Þj j1 ¼ f xð Þj j\e;

we have f � gð Þ 2 C0 Xð Þ: Thus, Cc Xð Þ is dense in C0 Xð Þ: ∎)

Conclusion 2.245 Let X be a locally compact Hausdorff space. Then Cc Xð Þ is adense subset of the normed linear space C0 Xð Þ:Note 2.246 Let X be a locally compact Hausdorff space. By Problem 2.242,C0 Xð Þ; dð Þ is a metric space. Let fnf g be a Cauchy sequence in C0 Xð Þ:Let us take any e[ 0: There exists a positive integer N such that

m; n�N ) sup fm xð Þ � fn xð Þj j : x 2 Xf g\ e2:

It follows that, for every m; n�N; and for every x 2 X; we havefm xð Þ � fn xð Þj j\ e

2 : Thus, for every x 2 X; fn xð Þf g is a Cauchy sequence in C:Now, since C is complete, for every x 2 X; there exists a unique f xð Þ 2 C such thatlimn!1 fn xð Þ ¼ f xð Þ: Thus, f : X ! C is a function. Let us fix any m0 �N; and anyx0 2 X: We have

fm0 x0ð Þ � fN x0ð Þj j\ e2 ;

fm0 x0ð Þ � fNþ 1 x0ð Þj j\ e2 ;

fm0 x0ð Þ � fNþ 2 x0ð Þj j\ e2 ;

..

.

Now, since limn!1 fn x0ð Þ ¼ f x0ð Þ; we have

e2�

limn!1 fm0 x0ð Þ � fn x0ð Þj j ¼ fm0 x0ð Þ � lim

n!1 fn x0ð Þ�� ¼ fm0 x0ð Þ � f x0ð Þj j:


Thus, for every x0 2 X and for every m0 �N; we have fm0 x0ð Þ � f x0ð Þj j � e2 : It

follows that, for every m0 �N;

sup fm0 xð Þ � f xð Þj j : x 2 Xf g� e2

\eð Þ:

Also, limm!1 fm ¼ f uniformly on X: Since each fm 2 C0 Xð Þ; each fm is con-tinuous. Since, each fm is continuous and limm!1 fm ¼ f uniformly on X; f iscontinuous.

Problem 2.247 f 2 C0 Xð Þ:(Solution It remains to show that f vanishes at infinity. For this purpose, let us taked[ 0:Since limm!1 fm ¼ f uniformly onX; there exists a positive integerN such that

sup fN xð Þ � f xð Þj j : x 2 Xf g\ d2:

Since fN 2 C0 Xð Þ; there exists a compact subset K of X such that for everyx 2 Kc; fN xð Þj j\ d

2 : It follows that, for every x 2 Kc; we have

f xð Þj j ¼ fN xð Þ � fN xð Þ � f xð Þð Þj j� fN xð Þj j þ fN xð Þ � f xð Þj j� fN xð Þj j þ sup fN yð Þ � f yð Þj j : y 2 Xf g\ fN xð Þj j þ d

2\

d2þ d

2¼ d:

Thus, for every x 2 Kc; we have f xð Þj j\d; where K is a compact subset of X:This proves that f 2 C0 Xð Þ: ∎)

Since f 2 C0 Xð Þ; and, for every m0 �N;

d fm0 ; fð Þ ¼ sup fm0 xð Þ � f xð Þj j : x 2 Xf g\e|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};fmf g converges to f 2 C0 Xð Þð Þ in the metric space C0 Xð Þ; dð Þ: Thus, we have

shown that the metric space C0 Xð Þ; dð Þ is complete.

Conclusion 2.248 Let X be a locally compact Hausdorff space. Then C0 Xð Þ is aBanach space, whose norm is defined by

fk k � sup f xð Þj j : x 2 Xf g:Note 2.249 Letf 2 C z : z 2 C; and zj j ¼ 1f gð Þ:

By Problem 2.239,

C z : z 2 C; and zj j ¼ 1f gð Þ ¼ Cc z : z 2 C; and zj j ¼ 1f gð Þ¼ C0 z : z 2 C; and zj j ¼ 1f gð Þ:

380 2 Lp-Spaces

Now, since

f 2 C z : z 2 C; and zj j ¼ 1f gð Þ;

we have

f 2 Cc z : z 2 C; and zj j ¼ 1f gð Þ;

and hence, by Conclusion 2.50, f 2 L1 Rð Þ: It follows that f : R ! C is a Lebesguemeasurable2p-periodic functionsatisfying 1

2p

R p�p f sð Þj jds 2 0;1½ Þ:TheFourier series

X1n¼�1

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1Aof f is a function t 7! limN!1 sN f ; tð Þ from R to C; where

sN f ; tð Þ �XNn¼�N

f nð Þ� �eint

� �;

and

f nð Þ � 12p

Zp�p

f sð Þe�insds:

Thus,

sN f ; tð Þ ¼XNn¼�N

f nð Þ� �eint

� �¼XNn¼�N

12p

Zp�p

f sð Þe�insds

0@ 1Aeint

0@ 1A¼XNn¼�N

12p

Zp�p

f sð Þe�inseintds

0@ 1A0@ 1A¼XNn¼�N

12p

Zp�p

f sð Þein t�sð Þds

0@ 1A¼ 1

2p

Zp�p

XNn¼�N

f sð Þein t�sð Þ !

ds

¼ 12p

Zp�p

f sð ÞXNn¼�N

ein t�sð Þ !

ds

¼ 12p

Zp�p

f sð ÞDN t � sð Þð Þds;


where

DN : u 7!XNn¼�N

einu|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 1þ 2 cos uþ 2 cos 2uþ � � � þ 2 cosNu

from R to R: Clearly, each DN : R ! R is continuous and 2p-periodic, and henceeach

DN 2 C z : z 2 C; and zj j ¼ 1f gð Þ:

Observe that, for every positive integer N; and for every real u; we haveDN �uð Þ ¼ DN uð Þ: By Conclusion 2.173, limN!1 sN � fk k2¼ 0: We want toanswer the following question:

Is it true that, for every f 2 C z : z 2 C; and zj j ¼ 1f gð Þ; and for every t 2 R;

limN!1

sN f ; tð Þ � f tð Þj j ¼ 0?

By Conclusion 2.177, C z : z 2 C; and zj j ¼ 1f gð Þ is a Banach space, whosenorm is defined by

fk k � sup f xð Þj j : x 2 �p;p½ �f g:

For each positive integer N; suppose that

KN : f 7! sN f ; 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 12p

Zp�p

f sð ÞDN �sð Þð Þds

¼ 12p

Zp�p

f sð ÞDN sð Þð Þds

is the mapping from Banach space C z : z 2 C; and zj j ¼ 1f gð Þ to C: Thus, forevery f 2 C z : z 2 C; and zj j ¼ 1f gð Þ; we have

KN fð Þ ¼ 12p

Zp�p

f sð ÞDN sð Þð Þds:

Problem 2.250 Each KN is a linear functional.

(Solution Let us fix a positive integer N. Let a; b 2 C and f ; g 2C z : z 2 C; and zj j ¼ 1f gð Þ: We have to show that

382 2 Lp-Spaces

sN af þ bg; 0ð Þ ¼ a � sN f ; 0ð Þþ b � sN g; 0ð Þ:

LHS ¼ sN af þ bg; 0ð Þ ¼ 12p

Zp�p

af þ bgð Þ sð ÞDN 0� sð Þð Þds

¼ 12p

Zp�p

af sð Þþ bg sð Þð ÞDN 0� sð Þð Þds

¼ 12p

Zp�p

af sð ÞDN 0� sð Þþ bg sð ÞDN 0� sð Þð Þds

¼ 12p

aZp�p

f sð ÞDN 0� sð Þð Þdsþ bZp�p

g sð ÞDN 0� sð Þð Þds0@ 1A

¼ a12p

Zp�p

f sð ÞDN 0� sð Þð Þds0@ 1Aþ b

12p

Zp�p

g sð ÞDN 0� sð Þð Þds0@ 1A

¼ a � sN f ; 0ð Þþ b � sN g; 0ð Þ ¼ RHS:∎)

Problem 2.251 Each KN is bounded and KNk k� 12p

R p�p DN sð Þj jds:

(Solution Let us fix a positive integer N. Let us take any f 2C z : z 2 C; and zj j ¼ 1f gð Þ satisfying

sup f xð Þj j : x 2 �p; p½ �f g� 1:

Now, since

KN fð Þj j ¼ sN f ; 0ð Þj j ¼ 12p

Zp�p

f sð ÞDN 0� sð Þð Þds��

�� ¼ 12p

Zp�p

f sð ÞDN �sð Þð Þds��

��¼ 1

2p

Zp�p

f sð ÞXNn¼�N

ein �sð Þ !

ds

�� 1

2p

Zp�p

f sð ÞXNn¼�N

e�ins

��ds

¼ 12p

Zp�p

f sð Þj jXNn¼�N

e�ins

��ds� 1

2p

Zp�p

1XNn¼�N

e�ins

��ds

¼ 12p

Zp�p

XNn¼�N

eins��

��ds� 12p

Zp�p

XNn¼�N

eins�� ds

¼ 12p

Zp�p

XNn¼�N

1

!ds ¼ 1

2p

Zp�p

2N þ 1ð Þds ¼ 2N þ 1ð Þ;


we have KN fð Þj j � 2Nþ 1ð Þ: It follows that KN is bounded. Since, for every f 2C z : z 2 C; and zj j ¼ 1f gð Þ satisfying sup f xð Þj j : x 2 �p; p½ �f g� 1;

KN fð Þj j � 12p

Zp�p

XNn¼�N

eins��

��ds ¼ 12p

Zp�p

DN sð Þj jds;

we have KNk k� 12p

R p�p DN sð Þds: ∎)

Since each KN is a bounded linear functional from Banach spaceC z : z 2 C; and zj j ¼ 1f gð Þ to C, each KN is continuous. Since, for every positiveinteger N;

Zp�p

DN sð Þj jds ¼Zp�p

XNn¼�N

eins��

��ds¼Zp�p

1þ 2 cos sþ 2 cos 2sþ � � � þ 2 cosNsj jds

¼Zp�p

1þ 2 cos sþ cos 2sþ � � � þ cosNsð Þj jds

¼Zp�p

1þ 2 cossþNs

2

� �sin Ns

2

sin s2

�� ds¼Zp�p

sin s2 þ 2 cos sþNs

2

� �sin Ns

2

sin s2

�� ds¼Zp�p

sin sþNs2 � Ns

2

� �þ 2 cos sþNs2

� �sin Ns

2

sin s2

�� ds¼Zp�p

sin sþNs2 þ Ns

2

� �sin s

2

�� ds ¼ Zp�p

sin Nsþ s2

� �sin s

2

�� ds¼Zp�p

sin Nþ 12

� �s

� �sin s

2

�� ds ¼ Zp�p

sin Nþ 12

� �s

� �� sin s

2

�� ds

¼ 2Zp0

sin Nþ 12

� �s

� �� sin s

2

�� ds ¼ 2Zp0

sin Nþ 12

� �s

� �� sin s

2

ds� 2Zp0

sin Nþ 12

� �s

� �� s2

ds

384 2 Lp-Spaces

¼ 4Zp0

sin Nþ 12

� �s

� �� 1s ds ¼ 4Zu¼ Nþ 1

2ð Þp

u¼0

sin uj jNþ 12

ud

uNþ 1

2

!

¼ 4Zu¼ Nþ 1

2ð Þp

u¼0

sin uj j 1udu

¼ 4Zu¼p

u¼0

sin uj j 1uduþ

Zu¼2p

u¼p

sin uj j 1uduþ � � �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}Nterms

0BBBBB@

1CCCCCAþZu¼ Nþ 1

2ð Þp

u¼Np

sin uj j 1udu

0BBBBB@

1CCCCCA

� 4Zu¼p

u¼0

sin uj j 1pduþ

Zu¼2p

u¼p

sin uj j 12p

duþ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Nterms

0BBBBB@

1CCCCCAþZu¼ Nþ 1

2ð Þp

u¼Np

sin uj j 1udu

0BBBBB@

1CCCCCA

¼ 41p

Zu¼p

u¼0

sin uj jduþ 12

Zu¼2p

u¼p

sin uj jduþ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Nterms

0BBBBB@

1CCCCCAþZu¼ Nþ 1

2ð Þp

u¼Np

sin uj j 1udu

0BBBBB@

1CCCCCA

¼ 41p

Zu¼p

u¼0

sin uj jduþ 12

Zu¼p

u¼0

sin uj jduþ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Nterms

0BBBB@1CCCCAþ

Zu¼ Nþ 12ð Þp

u¼Np

sin uj j 1udu

0BBBBB@

1CCCCCA

¼ 41p

1þ 12þ � � � þ 1

N

� � Zu¼p

u¼0

sinudu

0@ 1AþZu¼ Nþ 1

2ð Þp

u¼Np

sin uj j 1udu

0BB@1CCA

¼ 41p

1þ 12þ � � � þ 1

N

� �� 2

� �þ

Zu¼ Nþ 12ð Þp

u¼Np

sin uj j 1udu

0BB@1CCA

� 41p

1þ 12þ � � � þ 1

N

� �� 2

� �� ¼ 8

p1þ 1

2þ � � � þ 1

N

� �;


we have, for every positive integer N;

4p2

1þ 12þ � � � þ 1

N

� �� 1

2p

Zp�p

DN sð Þj jds;

and hence

1 ¼ð Þ limN!1

4p2

1þ 12þ � � � þ 1

N

� �� lim

N!112p

Zp�p

DN sð Þj jds:

It follows that

limN!1

12p

Zp�p

DN sð Þj jds ¼ 1:

Let us fix a positive integer N: Let us define a function gN : R ! �1; 1f g asfollows: For every t 2 R;

gN tð Þ � 1 if t 2 DNð Þ�1 0;1½ Þð Þ�1 if t 2 DNð Þ�1 �1; 0ð Þð Þ:

�Since DN is 2p-periodic, gN is 2p-periodic. Also, for every t 2 R; we have

gN �tð Þ ¼ gN tð Þ: Since DN : R ! R is continuous, DN : R ! R is measurable, andhence DNð Þ�1 �1; 0ð Þð Þ; DNð Þ�1 0;1½ Þð Þ are measurable sets. SinceDNð Þ�1 �1; 0ð Þð Þ; DNð Þ�1 0;1½ Þð Þ are measurable sets and

gN ¼ v DNð Þ�1 0;1½ Þð Þ � v DNð Þ�1 �1;0ð Þð Þ;

gN : R ! �1; 1f g is a measurable function. It is easy to find conviction thatthere exists a sequence fkf g of functions fk : R ! �1; 1½ � such that1. for every t 2 R; limk!1 fk tð Þ ¼ gN tð Þ;2. each fk is 2p-periodic,3. each fk is continuous.

From 2 and 3, each fk 2 C z : z 2 C; and zj j ¼ 1f gð Þ: By 1, on usingTheorem 1.136,

limk!1

Zp�p

fk tð Þdt0@ 1A ¼

Zp�p

gN tð Þdt:

386 2 Lp-Spaces

For each k ¼ 1; 2; . . .; fk : R ! �1; 1½ �; and fk 2 C z : z 2 C; and zj j ¼ 1f gð Þ; so

fkk k ¼ sup fk tð Þj j : t 2 �p; p½ �f g� 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since

limk!1

KN fkð Þ ¼ limk!1

12p

Zp�p

fk sð ÞDN sð Þð Þds

¼ 12p

limk!1

Zp�p

fk sð ÞDN sð Þð Þds ¼ 12p

Zp�p

limk!1

fk sð ÞDN sð Þð Þ� �

ds

¼ 12p

Zp�p

limk!1

fk sð Þ� �

DN sð Þds ¼ 12p

Zp�p

gN sð ÞDN sð Þds ¼ 12p

Zp�p

DN sð Þj jds;

we have

limk!1

KN fkð Þ ¼ 12p

Zp�p

DN sð Þj jds � 0ð Þ;

and hence

limk!1

KN fkð Þj j ¼ 12p

Zp�p

DN sð Þj jds:

Since for each k ¼ 1; 2; . . .; fk 2 C z : z 2 C; and zj j ¼ 1f gð Þ; fkk k� 1; and eachKN is a bounded linear functional from Banach space C z : z 2 C; and zj j ¼ 1f gð Þto C; we have

KNk k ¼ sup KN fð Þj j : f 2 C z : z 2 C; and zj j ¼ 1f gð Þ; and fk k� 1f gð Þ� limk!1

KN fkð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1

2p

Zp�p

DN sð Þj jds;

and hence 12p

R p�p DN sð Þj jds� KNk k: Now, since KNk k� 1

2p

R p�p DN sð Þj jds; we have

12p

R p�p DN sð Þj jds ¼ KNk k: Since, for every positive integer N;

12p

Zp�p

DN sð Þj jds ¼ KNk k; and limN!1

12p

Zp�p

DN sð Þj jds ¼ 1;


we have limN!1 KNk k ¼ 1: It follows that, for every positive real number M;there exists N 2 N such that M\ KNk k: Hence the statement

‘there exists a positive real number M such that; for every N 2 N; KNk k�M’

is false.Since KN : N 2 Nf g is sequence of bounded linear functional, from Banach

space C z : z 2 C; and zj j ¼ 1f gð Þ to normed linear space C, by Conclusion 2.187,there exists a dense Gd-set V such that for every f 2 V ;

sup KN fð Þj j : N 2 Nf g ¼ 1

or there exists a positive real number M such that for every N 2 N; KNk k�Mð Þ:

Now, since ‘there exists a positive real number M such that for every N 2 N;KNk k�M0 is false, there exists a dense Gd-set

V � C z : z 2 C; and zj j ¼ 1f gð Þð Þ

such that for every f 2 V ;

sup sN f ; 0ð Þj j : N 2 Nf g ¼ sup KN fð Þj j : N 2 Nf g ¼ 1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :In the above discussion, we can take any real t in place of 0 in sN f ; 0ð Þ:

Conclusion 2.252 Corresponding to each real number t; there exists a dense Gd-subset Et of C z : z 2 C; and zj j ¼ 1f gð Þ such that for every f 2 Et;sup sN f ; tð Þj j : N 2 Nf g ¼ 1:

It follows that corresponding to each real number t; there exists a dense Gd-subset Et of C z : z 2 C; and zj j ¼ 1f gð Þ such that for every f 2 Et;s1 f ; tð Þ; s2 f ; tð Þ; � � �f g is an unbounded sequence, and hence for every f 2 Et;s1 f ; tð Þ; s2 f ; tð Þ; � � �f g is not a convergent sequence. Since Et is dense, we have

Et 6¼ ;; and hence there exists g 2 Et: It follows that s1 g; tð Þ; s2 g; tð Þ; � � �f g is not aconvergent sequence, and hence ‘limN!1 sN g; tð Þ � g tð Þj j ¼ 0’ is false.

Thus, the question raised above has a negative answer.

Exercises

2:1 Let u : R ! R be a convex and differentiable function. Show that u0 : R ! Ris an increasing function.

2:2 Let f : R ! 0;1½ �; and g : R ! 0;1½ � be measurable functions. Showthat

388 2 Lp-Spaces

Z1�1

f � gð Þdm0@ 1A�

Z1�1

f 3dm

0@ 1A13 Z1

�1g

32dm

0@ 1A23

:

2:3 Let ℳ be a r-algebra in X: Let l : ℳ ! 0;1½ � be a positive measure on ℳ:Let S be the collection of all measurable simple functions s : X ! C such thatl x : s xð Þ 6¼ 0f gð Þ\1: Show that S is a dense subset of L2 lð Þ:

2:4 Let X; dð Þ; Y ; qð Þ be metric spaces. Let X be complete. Let f : X ! Y becontinuous. Let X0 be a dense subset of X and f X0ð Þ be a dense subset of Y :Suppose that, for every a; b 2 X0; d a; bð Þ ¼ q f að Þ; f bð Þð Þ: Show that for everya; b 2 X; d a; bð Þ ¼ q f að Þ; f bð Þð Þ:

2:5 Suppose that for every integer n; cn is a complex number. Suppose thatP1�1

cnj j2 � 0: Show that there exists f 2 L2 z : z 2 C; and zj j ¼ 1f gð Þ such that

for every integer n;

cn ¼Zp�p

f sð Þe�insds:

2:6 Let X be a real normed linear space. Let M be a subspace of X: Let f : M ! Rbe a bounded real linear functional. Let us fix any real number a: Let

F : xþ tx0ð Þ 7! f xð Þþ tað Þ

be a function from xþ tx0 : t 2 R; and x 2 Mf g to R: Show that

a. for every x 2 M; F xð Þ ¼ f xð Þ;b. F : xþ tx0 : t 2 R; and x 2 Mf g ! R is real linear.

2:7 Let X be a normed linear space. Let x0 be a nonzero member of X: Show thatthere exists a bounded linear functional F : X ! C such that Fk k ¼ 1 andF x0ð Þ ¼ i x0:k k

2:8 Show that there exists a dense Gd-subset A of C z : z 2 C; and zj j ¼ 1f gð Þ suchthat for every f 2 A;

supXNn¼�N

Zp�p

f sð Þe�insds

0@ 1Aeinffiffi2

p0@ 1A��

�� : N 2 N

8<:9=; ¼ 1:


2:9 Let H be a Hilbert space. Let uk : k 2 If g be an orthonormal set in H: Let F bea nonempty finite subset of I: Let MF be the linear span of uk : k 2 Ff g: Letu : I ! C be a function such that for every i 2 I � Fð Þ; u ið Þ ¼ 0: Show thatthere exists y 2 MF such that

a. for every i 2 I; u ið Þ ¼ y; uið Þ;b. yk k2¼P

j2Fy; uj� �� 2:

2:10 Let p 2 1;1½ �: Let X be any nonempty set. Let ℳ be a r-algebra in X: Letl : ℳ ! 0;1½ � be a positive measure on ℳ: Show that if f 2 Lp lð Þ; thenfj j 2 Lp lð Þ and

fj jð Þk kp¼ fk kp:

390 2 Lp-Spaces

Chapter 3Fourier Transforms

In this chapter, we introduce total variation, and prove the Radon-Nikodym theo-rem. The Fubini theorem and change-of-variable theorem are also proved. Finally,we prove the Plancherel theorem on Fourier transforms.

3.1 Total Variations

Note 3.1

Definition Let X be a nonempty set. Let M be a r-algebra in X. Suppose that, forevery positive integer n, En 2 M: Let E 2 M: If

1. for every distinct positive integer m; n, Em \En ¼ ;;2. E ¼ E1 [E2 [E3 [ . . .; then we shall say that the sequence E1;E2;E3; . . .f g is a

partition of E.

Definition Let X be a nonempty set. Let M be a r-algebra in X. Let l : M ! C:If, for every E 2 M; and for every partition E1;E2;E3; . . .f g of E,

1. the series l E1ð Þþ l E2ð Þþ l E3ð Þþ � � � is absolutely convergent(and hence every rearrangement of l E1ð Þþ l E2ð Þþ l E3ð Þþ � � � has the samesum [cf. WR[1], Theorem 3.55]),

2. l Eð Þ ¼ l E1ð Þþ l E2ð Þþ l E3ð Þþ � � � ;

then we say that l is a complex measure on M:

a. Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be acomplex measure on M:


391




Problem 3.2 l ;ð Þ ¼ 0:

(Solution Since l : M ! C is a complex measure on M, ; 2 M; and;; ;; ;; . . .f g is a partition of ;;

l ;ð Þ ¼ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � � :

It follows that l ;ð Þ ¼ 0: ■)

b. Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be acomplex measure onM: Let E1; . . .;Enf g be any finite collection of members inM such that i 6¼ j implies Ei \Ej ¼ ;:

Problem 3.3 l E1 [ � � � [Enð Þ ¼ l E1ð Þþ � � � þ l Enð Þ:

(Solution

LHS ¼ l E1 [ � � � [Enð Þ ¼ l E1 [ � � � [En [;[ ;[;[ � � �ð Þ¼ l E1ð Þþ � � � þ l Enð Þþ l ;ð Þþ l ;ð Þþ l ;ð Þþ � � �¼ l E1ð Þþ � � � þ l Enð Þþ 0þ 0þ 0þ � � � ¼ l E1ð Þþ � � � þ l Enð Þ ¼ RHS:

■)

c. Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be acomplex measure on M: Let A;B 2 M: Let A � B:

Problem 3.4 l B� Að Þ ¼ l Bð Þ � l Að Þ:

(Solution Since B ¼ B� Að Þ [A; and B� Að Þ \A ¼ ;; by b,

l Bð Þ ¼ l B� Að Þ [Að Þ ¼ l B� Að Þþ l Að Þ;

and hence l B� Að Þ ¼ l Bð Þ � l Að Þ: ■)Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be a

complex measure on M: Let E 2 M:It follows that the sequence E; ;; ;; . . .f g is a partition of E, and hence

l Eð Þj j ¼ l Eð Þj j þ 0j j þ 0j j þ � � �¼ l Eð Þj j þ l ;ð Þj j þ l ;ð Þj j þ � � �ð Þ 2 l E1ð Þj j þ l E2ð Þj j þ � � � : E1;E2; . . .f g is a partition of Ef g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

392 3 Fourier Transforms

Thus, by 1,

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Ef g

is a nonempty set of nonnegative real numbers, and hence

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Ef g 2 0;1½ �:

We denote

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Ef g

by lj j Eð Þ:Thus, lj j : M ! 0;1½ �; and for every E 2 M, l Eð Þj j is a real number satis-

fying l Eð Þj j � lj j Eð Þ: Also, lj j Eð Þ ¼ 1 if and only if the set


of nonnegative real numbers is not bounded above.

Problem 3.5 lj j ;ð Þ ¼ 0:

(Solution Observe that ;; ;; ;; . . .f g is the only partition of ;; so

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of ;f g¼ l ;ð Þj j þ l ;ð Þj j þ l ;ð Þj j þ � � �f g ¼ 0j j þ 0j j þ 0j j þ � � �f g ¼ 0f g;

and hence

lj j ;ð Þ ¼ sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of ;f g¼ sup 0f g ¼ 0:

Thus, lj j ;ð Þ ¼ 0 \1ð Þ: ■)Let A1;A2;A3; . . .f g be any sequence of members in M such that i 6¼ j implies

Ai \Aj ¼ ;:

Problem 3.6 lj j A1 [A2 [A3 [ � � �ð Þ� lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � :

(Solution Let E11;E12;E13; . . .f g be a partition of A1, E21;E22;E23; . . .f g be apartition of A2; etc. It suffices to show that

lj j A1 [A2 [A3 [ � � �ð Þ� l E11ð Þj j þ l E12ð Þj j þ l E13ð Þj j þ � � �ð Þþ l E21ð Þj j þ l E22ð Þj j þ l E23ð Þj j þ � � �ð Þþ l E31ð Þj j þ l E32ð Þj j þ l E33ð Þj j þ � � �ð Þþ � � � :

3.1 Total Variations 393

Clearly,

E11;E12;E13; . . .;E21;E22;E23; . . .; . . .f g

is a partition of A1 [A2 [A3 [ � � � : So,

lj j A1 [A2 [A3 [ � � �ð Þ� l E11ð Þj j þ l E12ð Þj j þ l E13ð Þj j þ � � �ð Þþ l E21ð Þj j þ l E22ð Þj j þ l E23ð Þj j þ � � �ð Þþ l E31ð Þj j þ l E32ð Þj j þ l E33ð Þj j þ � � �ð Þþ � � � :

■)Let A;B be members of M: Let A � B:

Problem 3.7 lj j Að Þ� lj j Bð Þ:(Solution If lj j Bð Þ ¼ 1; then lj j Að Þ� lj j Bð Þ is trivially true. So, we consider thecase when lj j Bð Þ\1: Let E1;E2;E3; . . .f g be a partition of A. It suffices to showthat

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � � lj j Bð Þ:

Since E1;E2;E3; . . .f g is a partition of A, and A � B, B� A;E1;E2;E3; . . .f g is apartition of B, and hence

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � � l B� Að Þj j þ l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � � lj j Bð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus,

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � � lj j Bð Þ:

■)Let A;B be members of M: Let A\B ¼ ;:

Problem 3.8 lj j Að Þþ lj j Bð Þ� lj j A[Bð Þ:

(Solution If lj j A[Bð Þ ¼ 1; then lj j Að Þþ lj j Bð Þ� lj j A[Bð Þ is trivially true. So,we consider the case when lj j A[Bð Þ\1:


lj j A[Bð Þ\ lj j Að Þþ lj j Bð Þ:


We have to arrive at a contradiction.Since lj j A[Bð Þ\ lj j Að Þþ lj j Bð Þ; we have

0\ lj j Að Þþ lj j Bð Þ � lj j A[Bð Þ:

Since A � A[B, lj j Að Þ� lj j A[Bð Þ: Since lj j Að Þ� lj j A[Bð Þ; andlj j A[Bð Þ\1;

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Af g ¼ lj j Að Þ\1|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} :Similarly, lj j Bð Þ\1: Let us put e � lj j Að Þþ lj j Bð Þ � lj j A[Bð Þ [ 0ð Þ:Since e[ 0; and

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Af g ¼ lj j Að Þ\1;

there exists a partition E1;E2;E3; . . .f g of A such that

lj j Að Þ � e2\ l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � :

Similarly, there exists a partition F1;F2;F3; . . .f g of B such that

lj j Bð Þ � e2\ l F1ð Þj j þ l F2ð Þj j þ l F3ð Þj j þ � � � :

It follows that

lj j A[Bð Þ ¼ lj j Að Þþ lj j Bð Þ � lj j Að Þþ lj j Bð Þ � lj j A[Bð Þð Þ ¼ lj j Að Þþ lj j Bð Þ � e

¼ lj j Að Þ � e2

� �þ lj j Bð Þ � e

2

� �\ l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � �ð Þþ l F1ð Þj j þ l F2ð Þj j þ l F3ð Þj j þ � � �ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ l E1ð Þj j þ l F1ð Þj j þ l E2ð Þj j þ l F2ð Þj j þ � � � ;

and hence

lj jðA[BÞ\ lðE1Þj j þ lðF1Þj j þ lðE2Þj j þ lðF2Þj j þ � � � :

Clearly,

E1;F1;E2;F2;E3;F3; . . .f g

is a partition of A[B; and hence

lðE1Þj j\ lðF1Þj j þ lðE2Þj j þ lðF2Þj j þ � � � � lj jðA[BÞ:


This is a contradiction. ■)Let A1; . . .;An be members of M: Suppose that, for distinct i; j, Ai \Aj ¼ ;:

Problem 3.9 lj j A1ð Þþ � � � þ lj j Anð Þ� lj j A1 [ � � � [Anð Þ:

(Solution Since

lj j A1ð Þþ � � � þ lj j Anð Þ ¼ lj j A1ð Þþ lj j A2ð Þð Þþ lj j A3ð Þþ � � � þ lj j Anð Þ� lj j A1 [A2ð Þþ lj j A3ð Þþ � � � þ lj j Anð Þ� lj j A1 [A2ð Þ [A3ð Þþ lj j A4ð Þþ � � � þ lj j Anð Þ¼ lj j A1 [A2 [A3ð Þþ lj j A4ð Þþ � � � þ lj j Anð Þ� � � � � lj j A1 [ � � � [Anð Þ;

we have

lj j A1ð Þþ � � � þ lj j Anð Þ� lj j A1 [ � � � [Anð Þ:

■)Let A1;A2;A3; . . .f g be any sequence of members in M such that i 6¼ j implies

Ai \Aj ¼ ;:

Problem 3.10 lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � ¼ lj j A1 [A2 [A3 [ � � �ð Þ:

(Solution Since

lj j A1 [A2 [A3 [ � � �ð Þ� lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � ;

it suffices to show that

lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � � lj j A1 [A2 [A3 [ � � �ð Þ:

If lj j A1 [A2 [A3 [ � � �ð Þ ¼ 1; then

lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � � lj j A1 [A2 [A3 [ � � �ð Þ

is trivially true. So, we consider the case when

lj j A1 [A2 [A3 [ � � �ð Þ\1:

Since

A1 � A1 [A2 [A3 [ � � � ;

we have

lj j A1ð Þ� lj j A1 [A2 [A3 [ � � �ð Þ \1ð Þ;


and hence

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of A1f g ¼ lj j A1ð Þ\1|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} :Similarly, lj j A2ð Þ\1, lj j A3ð Þ\1; etc.If not, otherwise, let

lj j A1 [A2 [A3 [ � � �ð Þ\ lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � :

We have to arrive at a contradiction. Since

lj j A1 [A2 [A3 [ � � �ð Þ\ lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � � ;0\ lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � �ð Þ � lj j A1 [A2 [A3 [ � � �ð Þ:

There exists a positive real number e such that

e\ lj j A1ð Þþ lj j A2ð Þþ lj j A3ð Þþ � � �ð Þ � lj j A1 [A2 [A3 [ � � �ð Þ [ 0ð Þ:

Since A1 � A1 [A2 [A3 [ � � � ; we have

lj j A1ð Þ� lj j A1 [A2 [A3 [ � � �ð Þ \1ð Þ;

and hence

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of A1f g ¼ lj j A1ð Þ\1|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} :Now, since e[ 0; there exists a partition E11;E12;E13; . . .f g of A1 such that

lj j A1ð Þ � e2\ l E11ð Þj j þ l E12ð Þj j þ l E13ð Þj j þ � � � :

Similarly, there exists a partition E21;E22;E23; . . .f g of A2 such that

lj j A2ð Þ � e4\ l E21ð Þj j þ l E22ð Þj j þ l E23ð Þj j þ � � � ;

etc. It follows that

lj j A1 [A2 [A3 [ � � �ð Þ\ lj j A1ð Þþ lj j A2ð Þþ � � �ð Þ � e

¼ lj j A1ð Þ � e2

� �þ lj j A2ð Þ � e

4

� �þ � � �\ l E11ð Þj j þ l E12ð Þj j � � �ð Þ þ l E21ð Þj j þ l E22ð Þj j þ � � �ð Þþ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};


and hence

lj j A1 [A2 [A3 [ � � �ð Þ\ l E11ð Þj j þ l E12ð Þj j þ l E13ð Þj j þ � � �ð Þþ l E21ð Þj j þ l E22ð Þj j þ l E23ð Þj j þ � � �ð Þþ � � � :

Clearly,

E11;E12;E13; . . .;E21;E22;E23; . . .; . . .f g

is a partition of A1 [A2 [A3 [ � � � : It follows that

l E11ð Þj j þ l E12ð Þj j þ l E13ð Þj j þ � � �ð Þþ l E21ð Þj j þ l E22ð Þj j þ l E23ð Þj j þ � � �ð Þþ � � �� lj j A1 [A2 [A3 [ � � �ð Þ:

This is a contradiction. ■)If we recollect the above results, we get the following

Conclusion 3.11 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! C be a complex measure on M: Then lj j : M ! 0;1½ � is a positivemeasure on M:

Definition Here, the positive measure lj j is called the total variation measure of l:

Note 3.12 Let N be a positive integer. Let z1; . . .; zN 2 C:

There exist a1; . . .; aN 2 R such that z1 ¼ z1j jeia1 ; . . .; zN ¼ zNj jeiaN : Let usdefine a function S : �p; p½ � ! P 1; . . .;Nf gð Þ as follows: For every h 2 �p; p½ �

S hð Þ � k : k 2 1; . . .;Nf g and cos ak � hð Þ[ 0f g:For every h 2 �p; p½ �;

z1j j cosð Þþ� �

a1 � hð Þþ � � � þ zNj j cosð Þþ� �

aN � hð Þ ¼X

k2 1;...;Nf gzkj j cosð Þþ� �

ak � hð Þ� �

¼Xk2S hð Þ

zkj j cosð Þþ� �

ak � hð Þ ¼Xk2S hð Þ

zkj j cos ak � hð Þ

¼ ReXk2S hð Þ

zkj j cos ak � hð Þþ i zkj j sin ak � hð Þð Þ

0@ 1A¼ Re

Xk2S hð Þ

zkj jei ak�hð Þ� �0@ 1A ¼ Re

Xk2S hð Þ

zkj jeiak e�ih� �0@ 1A ¼ Re

Xk2S hð Þ

zke�ih

� �0@ 1A�Xk2S hð Þ

zke�ih

� �� ¼ e�ih

Xk2S hð Þ

zk

�� ¼ e�ih

�� Xk2S hð Þ

zk

�� ¼ 1

Xk2S hð Þ

zk

�� ¼

Xk2S hð Þ

zk

��;


and hence for every h 2 �p; p½ �;



aN � hð Þ�Xk2S hð Þ

zk

��:

Since cos and modulus are continuous functions, and, for every real u;

cosð Þþ� �

uð Þ ¼ 12

cosuþ cosuj jð Þ ¼ max cosu; 0f gð Þ;

cosð Þþ is a continuous function,h 7! z1j j cosð Þþ

� �a1 � hð Þþ � � � þ zNj j cosð Þþ

� �aN � hð Þ

is hence a continuous function from compact set �p; p½ � to 0;1½ Þ: It follows thatthere exists h0 2 �p; p½ � such that, for every h 2 �p; p½ �;



aN � hð Þ

� z1j j cosð Þþ� �

a1 � h0ð Þþ � � � þ zNj j cosð Þþ� �

aN � h0ð Þ �X

k2S h0ð Þzk

��

0@ 1A:

Clearly, cosð Þþ is a 2p-periodic function. Since

h 7! z1j j cosð Þþ� �


aN � hð Þ

is a continuous function from compact set �p; p½ � to 0;1½ Þ;

h 7! z1j j cosð Þþ� �


aN � hð Þ

is Riemann integrable over �p; p½ �: Now, by the first mean value theorem, thereexists h� 2 �p; p½ � such that

Zp�p



aN � hð Þ� �

dh

¼ z1j j cosð Þþ� �

a1 � h�ð Þþ � � � þ zNj j cosð Þþ� �

aN � h�ð Þ� �

p� pð Þð Þ:


Observe that

2 z1j j þ � � � þ zNj jð Þ ¼ z1j j þ � � � þ zNj jð Þ sinp2� sin � p

2

� �� ¼ z1j j þ � � � þ zNj jð Þ sinuj

p2�p

2

� �¼ z1j j þ � � � þ zNj jð Þ

Zp2

�p2

cosu du ¼ z1j j þ � � � þ zNj jð ÞZp�p

cosð Þþu� �

du

¼ z1j jZp�p

cosð Þþu� �

duþ � � � þ zNj jZp�p

cosð Þþu� �

du

¼ z1j jZa1 þp

a1�p

cosð Þþu� �

duð Þþ � � � þ zNj jZaN þ p

aN�p

cosð Þþu� �

duð Þ

¼ z1j jZa1�p

a1 þ p

cosð Þþu� �

�duð Þþ � � � þ zNj jZaN�p

aN þ p

cosð Þþu� �

�duð Þ

¼ z1j jZp�p

cosð Þþ� �

a1 � hð Þdhþ � � � þ zNj jZp�p

cosð Þþ� �

aN � hð Þdh

¼Zp�p



aN � hð Þ� �

dh

and


a1 � h�ð Þþ � � � þ zNj j cosð Þþ� �

aN � h�ð Þ� �

p� �pð Þð Þ¼ z1j j cosð Þþ

� �a1 � h�ð Þ þ � � � þ zNj j cosð Þþ

� �aN � h�ð Þ

� �2pð Þ

� z1j j cosð Þþ� �

a1 � h0ð Þþ � � � þ zNj j cosð Þþ� �

aN � h0ð Þ� �

2pð Þ�X

k2S h0ð Þzk

�� 2pð Þ:

Thus,

2 z1j j þ � � � þ zNj jð Þ�X

k2S h0ð Þzk

�� 2pð Þ;

that is

XNk¼1

zkj j � pX

k2S h0ð Þzk

��:


Conclusion 3.13 Let N be a positive integer. Let z1; . . .; zN 2 C: Then there exists anonempty subset S0 of 1; . . .;Nf g such thatXN

k¼1

zkj j � pXk2S0

zk

��:

Note 3.14 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C

be a complex measure on M: Let E 2 M:

Problem 3.15 lj j Eð Þ\1:

(Solution If not, otherwise, let lj j Eð Þ ¼ 1: We have to arrive at a contradiction.Since

sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Ef g ¼ð Þ lj j Eð Þ ¼ 1;


is a nonempty set of nonnegative real numbers, which is not bounded above. Itfollows that there exists a partition E1;E2;E3; . . .f g of E such that

p 1þ l Eð Þj jð Þ\ l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ sup l E1ð Þj j þ � � � þ l Enð Þj j : n 2 Nf g:

It follows that there exists a positive integer N such that

p 1þ l Eð Þj jð Þ\ l E1ð Þj j þ � � � þ l ENð Þj j:

By the conclusion of Note 3.2, there exists a nonempty subset S0 of 1; . . .;Nf gsuch that

p 1þ l Eð Þj jð Þ\ l E1ð Þj j þ � � � þ l ENð Þj j � pXk2S0

l Ekð Þ��

�� ¼ p l [ k2S0Ekð Þj j;

and hence

1�ð Þ1þ l Eð Þj j\ l [ k2S0Ekð Þj j:

It follows that

1\ l [ k2S0Ekð Þj j � l Eð Þj j � l Eð Þj j � l [ k2S0Ekð Þj jj j � l Eð Þ � l [ k2S0Ekð Þj j¼ l [ k2S0Ekð Þ [ E � [ k2S0Ekð Þð Þð Þ � l [ k2S0Ekð Þj j¼ l [ k2S0Ekð Þþ l E � [ k2S0Ekð Þð Þð Þ � l [ k2S0Ekð Þj j ¼ l E � [ k2S0Ekð Þð Þj j;


and hence 1\ l E � [ k2S0Ekð Þð Þj j: Since lj j is a positive measure,

1 ¼ð Þ lj j Eð Þ ¼ lj j [ k2S0Ekð Þþ lj j E � [ k2S0Ekð Þð Þ;

and hence

lj j [ k2S0Ekð Þ ¼ 1 or lj j E � [ k2S0Ekð Þð Þ ¼ 1:

Put A1 � [ k2S0Ek; and B1 � E � [ k2S0Ekð Þ:Thus,

E ¼ A1 [B1;A1 \B1

¼ ;; 1\ l A1ð Þj j; 1\ l B1ð Þj j; and lj j A1ð Þ ¼ 1 or lj j B1ð Þ ¼ 1ð Þ:

For definiteness, let lj j A1ð Þ ¼ 1: As above, there exists A2;B2 such that

A1 ¼ A2 [B2;A2 \B2

¼ ;; 1\ l A2ð Þj j; 1\ l B2ð Þj j; and lj j A2ð Þ ¼ 1 or lj j B2ð Þ ¼ 1ð Þ:

For definiteness, let lj j A2ð Þ ¼ 1: As above, there exists A3;B3 such that

A2 ¼ A3 [B3;A3 \B3

¼ ;; 1\ l A3ð Þj j; 1\ l B3ð Þj j; and lj j A3ð Þ ¼ 1 or lj j B3ð Þ ¼ 1ð Þ; etc:

Thus, we get a countable collection B1;B2;B3; . . .f g of sets in M such thatm 6¼ n implies Bm \Bn ¼ ;: Also, for each positive integer n, 1\ l Bnð Þj j: SinceB1;B2;B3; . . .f g is a countable collection of sets in M such that m 6¼ n implies

Bm \Bn ¼ ;; and l : M ! C is a complex measure on M; the seriesl B1ð Þþ l B2ð Þþ l B3ð Þþ � � � is absolutely convergent, and hence limn!1 l Bnð Þj j ¼0: Since for each positive integer n, 1\ l Bnð Þj j; we have limn!1 l Bnð Þj j 6¼ 0: Thisis a contradiction. ■)

Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be acomplex measure on M:

Problem 3.16 lj j Xð Þ\1: Also, for every E 2 M, l Eð Þ 2 D 0; lj j Xð Þþ 1ð Þ:

(Solution Let us take any E 2 M: We have to show that l Eð Þ � 0j j\ lj j Xð Þþ 1:Here,

l Eð Þ � 0j j ¼ l Eð Þj j � lj j Eð Þ� lj j Xð Þ\ lj j Xð Þþ 1;

so

l Eð Þ � 0j j\ lj j Xð Þþ 1:

■)


Conclusion 3.17 A complex measure is bounded.

Note 3.18 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C

and m : M ! C be complex measures on M:

Problem 3.19 lþ mð Þ : E 7! l Eð Þþ m Eð Þð Þ from M to C is a complex measure.

(Solution Let E 2 M: Let E1;E2;E3; . . .f g be a partition of E. We have to showthat

1. the series lþ mð Þ E1ð Þþ lþ mð Þ E2ð Þþ lþ mð Þ E3ð Þþ � � � is absolutely convergent,2. lþ mð Þ Eð Þ ¼ lþ mð Þ E1ð Þþ lþ mð Þ E2ð Þþ lþ mð Þ E3ð Þþ � � � :

For 1: We have to show that

l E1ð Þþ m E1ð Þj j þ l E2ð Þþ m E2ð Þj j þ l E3ð Þþ m E3ð Þj j þ � � �

is convergent. It suffices to show that

l E1ð Þj j þ m E1ð Þj jð Þþ l E2ð Þj j þ m E2ð Þj jð Þþ l E3ð Þj j þ m E3ð Þj jð Þþ � � �

is convergent. Since E1;E2;E3; . . .f g is a partition of E, and l : M ! C is acomplex measure on M;

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � �

is convergent. Similarly,

m E1ð Þj j þ m E2ð Þj j þ m E3ð Þj j þ � � �

is convergent. Since

l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � ; and m E1ð Þj j þ m E2ð Þj j þ m E3ð Þj j þ � � �

are convergent,

l E1ð Þj j þ m E1ð Þj jð Þþ l E2ð Þj j þ m E2ð Þj jð Þþ l E3ð Þj j þ m E3ð Þj jð Þþ � � �

is convergent.For 2:

LHS ¼ lþ mð Þ Eð Þ ¼ l Eð Þþ m Eð Þ ¼ l E1ð Þþ l E2ð Þþ l E3ð Þþ � � �ð Þþ m E1ð Þþ m E2ð Þþ m E3ð Þþ � � �ð Þ

¼ l E1ð Þþ m E1ð Þð Þþ l E2ð Þþ m E2ð Þð Þþ l E3ð Þþ m E3ð Þð Þþ � � �¼ lþ mð Þ E1ð Þþ lþ mð Þ E2ð Þþ lþ mð Þ E3ð Þþ � � � ¼ RHS:

■)Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be a

complex measures on M: Let c 2 C:


Problem 3.20 clð Þ : E 7! c l Eð Þð Þ from M to C is a complex measure.

(Solution Let E 2 M: Let E1;E2;E3; . . .f g be a partition of E. We have to showthat

1. the series clð Þ E1ð Þþ clð Þ E2ð Þþ clð Þ E3ð Þþ � � � is absolutely convergent,2. clð Þ Eð Þ ¼ clð Þ E1ð Þþ clð Þ E2ð Þþ clð Þ E3ð Þþ � � � :


cj j l E1ð Þj j þ cj j l E2ð Þj j þ cj j l E3ð Þj j þ � � �

is convergent. It suffices to show that


is convergent. Since E1;E2;E3; . . .f g is a partition of E, and l : M ! C is acomplex measure on M;


is convergent.For 2:

LHS ¼ clð Þ Eð Þ ¼ c l Eð Þð Þ ¼ c l E1ð Þþ l E2ð Þþ l E3ð Þþ � � �ð Þ¼ c l E1ð Þð Þþ c l E2ð Þð Þþ c l E3ð Þð Þþ � � �¼ clð Þ E1ð Þþ clð Þ E2ð Þþ clð Þ E3ð Þþ � � � ¼ RHS:

■)Thus, the collection C of all complex measures on M, is a complex linear space.For every l 2 M; put

lk k � lj j Xð Þ 2 0;1½ Þð Þ:

Problem 3.21 C; k kð Þ is a normed linear space.


1. if lk k ¼ 0 then l ¼ 0;2. for every l 2 C and every c 2 C, clk k ¼ cj j lk k;3. for every l; m 2 C; lþ mk k� lk kþ mk k:

For 1: Let lk k ¼ 0; that is lj j Xð Þ ¼ 0: We have to show that l ¼ 0: For thispurpose, let us take any E 2 M: We have to show that l Eð Þ ¼ 0; that is l Eð Þj j ¼0: Since 0� l Eð Þj j � lj j Eð Þ� lj j Xð Þ ¼ 0; we have l Eð Þj j ¼ 0:

For 2: Let l 2 C; and c 2 C: We have to show that clk k ¼ cj j lk k; that is,clj j Xð Þ ¼ cj j lj j Xð Þð Þ: Since


0j j Xð Þ ¼ sup 0 E1ð Þj j þ 0 E2ð Þj j þ 0 E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g¼ sup 0j j þ 0j j þ 0j j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g ¼ sup 0f g ¼ 0;

we have 0lj j Xð Þ ¼ 0j j Xð Þ ¼ 0 ¼ 0j j lj j Xð Þð Þ; and hence 0lj j Xð Þ ¼ 0j j lj j Xð Þð Þ: Itremains to show that, for nonzero complex number c, clj j Xð Þ ¼ cj j lj j Xð Þð Þ:

For this purpose, let c 6¼ 0: We have to show that clj j Xð Þ ¼ cj j lj j Xð Þð Þ:

LHS ¼ clj j Xð Þ ¼ sup clð Þ E1ð Þj j þ clð Þ E2ð Þj j þ clð Þ E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g¼ sup c l E1ð Þð Þj j þ c l E2ð Þð Þj j þ c l E3ð Þð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g¼ sup cj j l E1ð Þj j þ cj j l E2ð Þj j þ cj j l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g¼ cj j � sup l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Xf g¼ cj j lj j Xð Þð Þ ¼ RHS:

For 3: Let l; m 2 C: We have to show that lþ mk k� lk kþ mk k; that islþ mj j Xð Þ� lj j Xð Þþ mj j Xð Þ: For this purpose, let us take a partitionE1;E2;E3; . . .f g is a partition of X: It suffices to show that

lþ mð Þ E1ð Þj j þ lþ mð Þ E2ð Þj j þ lþ mð Þ E3ð Þj j þ � � �� l E1ð Þj j þ l E2ð Þj j þ l E3ð Þj j þ � � �ð Þ þ m E1ð Þj j þ m E2ð Þj j þ m E3ð Þj j þ � � �ð Þ;

that is

l E1ð Þþ m E1ð Þj j þ l E2ð Þþ m E2ð Þj j þ l E3ð Þþ m E3ð Þj j þ � � �� l E1ð Þj j þ m E1ð Þj jð Þþ l E2ð Þj j þ m E2ð Þj jð Þþ l E3ð Þj j þ m E3ð Þj jð Þþ � � � :

Since for every positive integer n,

l Enð Þþ m Enð Þj j � l Enð Þj j þ m Enð Þj j;

we have

l E1ð Þþ m E1ð Þj j þ l E2ð Þþ m E2ð Þj j þ l E3ð Þþ m E3ð Þj j þ � � �� l E1ð Þj j þ m E1ð Þj jð Þþ l E2ð Þj j þ m E2ð Þj jð Þþ l E3ð Þj j þ m E3ð Þj jð Þþ � � � :

■)

Definition Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !R � Cð Þ be a complex measure on M: We know that lj j : M ! 0;1½ Þ � Cð Þ isalso a complex measure on M:

Now, from the above discussion,

12

lj j þ lð Þ : M ! R; and12

lj j � lð Þ : M ! R


are also complex measures on M: Since for every E 2 M;

l Eð Þ�max l Eð Þ;� l Eð Þð Þf g ¼ l Eð Þj j � lj j Eð Þ;

for every E 2 M;

0� lj j Eð Þ � l Eð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ lj j � lð Þ Eð Þ;

and hence

12

lj j � lð Þ : M ! 0;1½ Þ:

Thus,

12

lj j � lð Þ : M ! 0;1½ Þ

is a positive measure on M: Similarly,

12

lj j þ lð Þ : M ! 0;1½ Þ

is a positive measure onM: 12 lj j þ lð Þ is denoted by lþ ; and 12 lj j � lð Þ is denoted

by l�:Thus, lþ and l� are positive measures on M such that l ¼ lþ � l�; and

lj j ¼ lþ þ l�:Here, lþ is called the positive variation of the signed measure l; and l� is

called the negative variation of the signed measure l: Also, the ordered pairlþ ; l�ð Þ is called the Jordan decomposition of the signed measure l:

Definition Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C

be a complex measure onM: Let A 2 M: By k is concentrated on A, we mean thatevery E 2 M, k A\Eð Þ ¼ k Eð Þ:

Problem 3.22 k is concentrated on A if and only if E 2 M satisfying A\E ¼ ;implies k Eð Þ ¼ 0:

(Solution Let k be concentrated on A. Let E 2 M satisfying A\E ¼ ;: We haveto show that k Eð Þ ¼ 0: Since k is concentrated on A, and E 2 M;0 ¼ k ;ð Þ ¼ð Þk A\Eð Þ ¼ k Eð Þ and hence k Eð Þ ¼ 0:Conversely, suppose that for every E 2 M satisfying A\E ¼ ; implies k Eð Þ ¼

0 . We have to show that k is concentrated on A. For this purpose, let us take anyE 2 M: We have to show that

k A\Eð Þ ¼ k Eð Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} ¼ k A\Eð Þþ k Ac \Eð Þ;


that is k Ac \Eð Þ ¼ 0: Since A\ Ac \Eð Þ ¼ ;; and Ac \Eð Þ 2 M; by theassumption, k Ac \Eð Þ ¼ 0: ■)

I. Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C be acomplex measure on M: Let A 2 M: Let k be concentrated on A.

Problem 3.23 The measure kj j : M ! 0;1½ Þ � Cð Þ is concentrated on A.

(Solution For this purpose, let us take any E 2 M satisfying A\E ¼ ;: We haveto show that

sup k E1ð Þj j þ k E2ð Þj j þ k E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of Ef g ¼ kj j Eð Þ ¼ 0|fflfflfflfflfflffl{zfflfflfflfflfflffl} :For this purpose, let us take any partition E1;E2;E3; . . .f g of E: It suffices to

show that

k E1ð Þj j þ k E2ð Þj j þ k E3ð Þj j þ � � � ¼ 0;

that is k E1ð Þ ¼ 0; k E2ð Þ ¼ 0; k E3ð Þ ¼ 0; etc. Since ; � A\E1 �ð ÞA\E ¼ ;; wehave A\E1 ¼ ;: Since k is concentrated on A, E1 2 M; and A\E1 ¼ ;; k E1ð Þ ¼0: Similarly, k E2ð Þ ¼ 0; etc. ■)

Definition Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M !C; and k2 : M ! C be complex measures on M: If there exist A;B 2 M such that

1. A\B ¼ ;;2. k1 is concentrated on A,3. k2 is concentrated on B,

then we write k1 ? k2; and read it as: k1 and k2 are mutually singular.

II. Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M ! C; andk2 : M ! C be complex measures on M: Let k1 ? k2:

Problem 3.24 k1j j ? k2j j:(Solution Since k1 ? k2; there exist A;B 2 M such that

1. A\B ¼ ;;2. k1 is concentrated on A,3. k2 is concentrated on B.

On using I, we have A\B ¼ ;; k1j j is concentrated on A; and k2j j is concen-trated on B. Thus, k1j j ? k2j j: ■)


III. Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M ! C;and k2 : M ! C be complex measures on M: Let l : M ! 0;1½ Þ be apositive measure. It follows that k1 þ k2ð Þ : M ! C is a complex measureon M: Let k1 ? l; and k2 ? l:

Problem 3.25 k1 þ k2ð Þ? l:

(Solution Since k1 ? l; there exist A1;B 2 M such that

1. A1 \B ¼ ;;2. k1 is concentrated on A1;3. l is concentrated on B.

Since k2 ? l; there exist A2;C 2 M such that

a. A2 \C ¼ ;;b. k2 is concentrated on A2;c. l is concentrated on C.


1′. (A1 [A2Þ \ B\Cð Þ ¼ ;;2′. k1 þ k2ð Þ is concentrated on A1 [A2;3′. l is concentrated on B\C:

For 1′: Since

; � ðA1 [A2Þ \ B\Cð Þ ¼ A1 \ B\Cð Þð Þ [ A2 \ B\Cð Þð Þ � A1 \Bð Þ [ A2 \ B\Cð Þð Þ¼ ;[ A2 \ B\Cð Þð Þ ¼ A2 \ B\Cð Þ � A2 \C ¼ ;;

we have A1 [A2ð Þ \ B\Cð Þ ¼ ;:For 2′: Let E 2 M satisfying (A1 [A2Þ \E ¼ ;: We have to show that

k1 þ k2ð Þ Eð Þ ¼ 0; that is k1 Eð Þþ k2 Eð Þ ¼ 0: Here, it suffices to show that k1 Eð Þ ¼0; and k2 Eð Þ ¼ 0: Since ; � A1 \E �ð Þ (A1 [A2Þ \E ¼ ;; we have A1 \E ¼ ;;and hence, by (2), k1 Eð Þ ¼ 0: Similarly, k2 Eð Þ ¼ 0:

For 3′: Let E 2 M satisfying (B\CÞ \E ¼ ;: We have to show that l Eð Þ ¼ 0:

LHS ¼ l Eð Þ ¼ l B� Cð Þ \Eð Þþ l C � Bð Þ \Eð Þþ l B[Cð Þc \Eð Þ¼ l B\E \Ccð Þþ l C \E \Bcð Þþ l E \Bc \Ccð Þ¼ 0þ l C \E\Bcð Þþ l E\Bc \Ccð Þ ¼ 0þ l E \Bc \Ccð Þ ¼ 0 ¼ RHS:

■)Similarly, k1 � k2ð Þ? l; and i k1 þ k2ð Þ? l:


Definition Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C

be any complex measure on M: Let l : M ! 0;1½ � be any positive measure onM: If l Eð Þ ¼ 0 implies k Eð Þ ¼ 0; then we write k l; and we say that k isabsolutely continuous with respect to l:

IV. Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M ! C; andk2 : M ! C be complex measures on M: Let l : M ! 0;1½ Þ be a positivemeasure. It follows that k1 þ k2ð Þ : M ! C is a complex measure on M: Letk1 l; and k2 l:

Problem 3.26 k1 þ k2ð Þ l:

(Solution Let E 2 M satisfying l Eð Þ ¼ 0:We have to show that k1 þ k2ð Þ Eð Þ ¼ 0;that is, k1 Eð Þþ k2 Eð Þ ¼ 0: It suffices to show that k1 Eð Þ ¼ 0; and k2 Eð Þ ¼ 0: SinceE 2 M, l Eð Þ ¼ 0; and k1 l; we have k1 Eð Þ ¼ 0: Similarly, k2 Eð Þ ¼ 0: ■)

Similarly, k1 � k2ð Þ l; and i k1 þ k2ð Þ l:

V. Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C be acomplex measure on M: Let l : M ! 0;1½ Þ be a positive measure. Let k l: Since k : M ! C is a complex measure onM; kj j : M ! 0;1½ Þ � Cð Þ is acomplex measure on M:

Problem 3.27 kj j l:

(Solution Let E 2 M satisfying l Eð Þ ¼ 0: We have to show that ðsupf k E1ð Þj j þk E2ð Þj j þ k E3ð Þj j þ � � � :fE1;E2;E3; . . .g is a partition of Eg ¼Þ kj j Eð Þ ¼ 0: For thispurpose, let us take any partition E1;E2;E3; . . .f g of E: It suffices to show that

k E1ð Þj j þ k E2ð Þj j þ k E3ð Þj j þ � � � ¼ 0;

that is k E1ð Þ ¼ 0, k E2ð Þ ¼ 0, k E3ð Þ ¼ 0; etc.Since E;E1 2 M;E1 � E; l Eð Þ ¼ 0; and l : M ! 0;1½ Þ is a positive mea-

sure, we have l E1ð Þ ¼ 0: Now, since k l, k E1ð Þ ¼ 0: Similarly, k E2ð Þ ¼ 0,k E3ð Þ ¼ 0; etc. ■)

VI. Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M ! C; andk2 : M ! C be complex measures on M: Let l : M ! 0;1½ Þ be a positivemeasure. Let k1 l; and k2 ? l:


Problem 3.28 k2 ? k1:

(Solution Since k2 ? l; there exist A2;B 2 M such that

1. A2 \B ¼ ;;2. k2 is concentrated on A2;3. l is concentrated on B.

It suffices to show that k1 is concentrated on A2ð Þc: For this purpose, let us takeany E 2 M satisfying A2ð Þc \E ¼ ;: We have to show that k1 Eð Þ ¼ 0: SinceA2ð Þc \E ¼ ;; we have E � A2 � Bcð Þ; and hence E \B ¼ ;: Since l is concen-trated on B, and E \B ¼ ;, l Eð Þ ¼ 0: Since l Eð Þ ¼ 0; and k1 l, k1 Eð Þ ¼ 0: ■)

(VII) Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C be acomplex measure on M: Let l : M ! 0;1½ Þ be a positive measure. Letk l; and k? l:

Problem 3.29 k ¼ 0:

(Solution Let E 2 M: We have to show that k Eð Þ ¼ 0: Since k l; and k? l;by (VI), k? k: Since k? k; there exist A;B 2 M such that

1. A\B ¼ ;;2. k is concentrated on A,3. k is concentrated on B.

LHS ¼ k Eð Þ ¼ k B\Eð Þþ k Bc \Eð Þ ¼ k B\Eð Þþ 0 ¼ k B\Eð Þ ¼ 0 ¼ RHS:

■)

Conclusion 3.30 Let X be any nonempty set. Let M be a r-algebra in X. Letk : M ! C be a complex measure on M: Let l : M ! 0;1½ Þ be a positivemeasure. Let k l; and k? l: Then k ¼ 0:

Note 3.31 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !0;1½ � be a positive measure on M: Suppose that X has r-finite measure, that isthere exists a countable collection E1;E2; . . .f g of members in M such that E ¼E1 [E2 [ � � � ; each 0�ð Þl Eið Þ\1; and E1;E2; . . . are pairwise disjoint.

Clearly,

12 1þ l E1ð Þð Þ vE1

þ 14 1þ l E2ð Þð Þ vE2


þ � � ��

: X ! 0; 1ð Þ:

By Lemma 1.89, 12 1þl E1ð Þð Þ vE1


þ 18 1þl E3ð Þð Þ vE3

þ � � � is a mea-

surable function.


Also,ZX

12 1þ l E1ð Þð Þ vE1



þ � � �� dl¼ZX

12 1þl E1ð Þð Þ vE1

þ 14 1þl E2ð Þð Þ vE2


þ � � ��

dl

¼ 12 1þ l E1ð Þð Þ

ZX

vE1dlþ 1

4 1þ l E2ð Þð Þ

ZX

vE2dlþ 1

8 1þl E3ð Þð Þ

ZX

vE3dlþ � � �

¼ 12 1þ l E1ð Þð Þ l E1ð Þþ 1

4 1þ l E2ð Þð Þ l E2ð Þþ 18 1þ l E3ð Þð Þ l E3ð Þþ � � �

� 12þ 1

4þ 1

8þ � � � ¼ 1 \1ð Þ:

Conclusion 3.32 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ � be a positive measure on M: Suppose that X has r-finite measure.Then there exists a measurable function w : X ! 0; 1ð Þ such that

RX wj jdl 2 0; 1½ �:

3.2 Radon–Nikodym Theorem

Note 3.33 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !0;1½ � be a positive measure on M: Suppose that X has r-finite measure. Letk : M ! 0;1½ Þ be a positive measure on M:

By Conclusion 3.32, there exists a measurable function w : X ! 0; 1ð Þ such thatRX w dl 2 0; 1½ �: Thus, w 2 L1 lð Þ:Let u : E 7! k Eð Þþ

RE w dl

� �from M to 0;1½ Þ:

Problem 3.34 u : M ! 0;1½ Þ is a positive measure on M:

(Solution For this purpose, let us take any countable collection A1;A2;A3; . . .f g ofmembers in M such that i 6¼ j implies Ai \Aj ¼ ;: We have to show that

u A1 [A2 [A3 [ � � �ð Þ ¼ u A1ð Þþu A2ð Þþu A3ð Þþ � � � ;


that is

k A1 [A2 [A3 [ � � �ð ÞþZ

A1 [A2 [A3 [ ��ð Þ

wdl ¼ k A1ð ÞþZA1

wdl

0B@1CA

þ k A2ð ÞþZA2

wdl

0B@1CAþ k A3ð Þþ

ZA3

wdl

0B@1CAþ � � � ;

that is

k A1 [A2 [A3 [ � � �ð ÞþZ

A1 [A2 [A3 [ ��ð Þ

wdl ¼ k A1ð Þþ k A2ð Þþ k A3ð Þþ � � �ð Þ

þZA1

wdlþZA2

wdlþZA3

wdlþ � � �

0B@1CA;

that is ZA1 [A2 [A3 [ ��ð Þ

w dl ¼ZA1

w dlþZA2

w dlþZA3

w dlþ � � � :

This is known to be true, by Lemma 1.131. ■)

Problem 3.35 For every measurable function g : X ! 0;1½ Þ satisfying g 2 L1 lð Þ;ZX

g du ¼ZX

g dkþZX

gw dl:

(Solution Case I: when g is a characteristic function, say vE; where E 2 M: Here,

LHS ¼ZX

gdu ¼ZX

vEdu ¼ 1 � u E \Xð Þð Þþ 0 � u Ec \Xð Þð Þ ¼ u Eð Þ

¼ k Eð ÞþZE

w dl;


and

RHS ¼ZX

g dkþZX

gw dl ¼ZX

vEdkþZX

vEw dl ¼ k Eð ÞþZX

vEw dl

¼ k Eð ÞþZE

w dl:

Thus, LHS = RHS.Case II: when g is a simple function, say

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ;

where each ai 2 0;1½ Þ; and each g�1 aið Þ 2 M: On using Case I,

LHS ¼ZX

g du ¼ZX

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ

� �du

¼ a1

ZX

v g�1 a1ð Þð Þduþ � � � þ an

ZX

v g�1 anð Þð Þdu

¼ a1

ZX

v g�1 a1ð Þð ÞdkþZX

v g�1 a1ð Þð Þw dl

0@ 1Aþ � � � þ an

ZX

v g�1 anð Þð ÞdkþZX

v g�1 anð Þð Þw dl

0@ 1A¼ZX

a1v g�1 a1ð Þð Þ þ � � � þ anv g�1 anð Þð Þ

� �dkþ

ZX

a1v g�1 a1ð Þð Þwþ � � � þ anv g�1 anð Þð Þw� �

dl

¼ZX

g dkþZX

a1v g�1 a1ð Þð Þwþ � � � þ anv g�1 anð Þð Þw� �

dl ¼ZX

g dkþZX

gw dl ¼ RHS:

Case III: when g : X ! 0;1½ Þ; and g is not a simple function. By Lemma 1.98,there exists a sequence snf g of simple measurable functions sn : X ! 0;1½ Þ suchthat for every x in X, 0� s1 xð Þ� s2 xð Þ� � � � ; and limn!1 snðxÞ ¼ gðxÞ: ByTheorem 1.125,

RX g du ¼ limn!1

RX sndu

� �; and

RX g dk ¼ limn!1

RX sndk

� �: By

Case II, for each positive integer n,RX sndu ¼

RX sndkþ

RX snw dl. It follows that

ZX

g du ¼ limn!1

ZX

sndu

0@ 1A ¼ limn!1

ZX

sndkþZX

snw dl

0@ 1A|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

:

3.2 Radon–Nikodym Theorem 413

Since for every x in X, limn!1 sn xð Þ ¼ g xð Þ; we have, for every x in X,

limn!1

snwð Þ xð Þð Þ ¼ limn!1

w xð Þð Þ sn xð Þð Þ ¼ w xð Þð Þ g xð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ gwð Þ xð Þ:

Thus, for every x in X, limn!1 snwð Þ xð Þð Þ ¼ gwð Þ xð Þ: Since for every x in X,0� s1 xð Þ� s2 xð Þ� � � � ; and w : X ! 0; 1ð Þ; we have, for every x in X,

0� s1 xð Þð Þ w xð Þð Þ� s2 xð Þð Þ w xð Þð Þ� � � � :

Thus, for every x in X, 0� s1wð Þ xð Þ� s2wð Þ xð Þ� � � � : Further, each snwð Þ :X ! 0;1½ Þ is a measurable function. Now, by Theorem 1.125,

ZX

g du�ZX

g dk ¼ZX

gdu� limn!1

ZX

sndk

0@ 1A ¼ limn!1

ZX

sndu

0@ 1A� limn!1

ZX

sndk

0@ 1A¼ lim

n!1

ZX

sndu�ZX

sndk

0@ 1A ¼ limn!1

ZX

snwð Þdl

0@ 1A ¼ZX

gwð Þdl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence ZX

g du�ZX

g dk ¼ZX

gw dl:

So, ZX

g du ¼ZX

g dkþZX

gwdl:

Thus, in all cases, ZX

gdu ¼ZX

g dkþZX

gw dl:

■)

Conclusion 3.36 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ � be a positive measure on M: Suppose that X has r-finite measure.Let k : M ! 0;1½ Þ be a positive measure on M: Then there exists a measurablefunction w : X ! 0; 1ð Þ such that

RX w dl 2 0; 1½ �; and u : E 7! k Eð Þþ

RE w dl

� �from M to 0;1½ Þ is a positive measure on M: Also, for every measurable functiong : X ! 0;1½ Þ satisfying g 2 L1 lð Þ;

RX g du ¼

RX g dkþ

RX gw dl:


Note 3.37 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !0;1½ � be a positive measure on M: Suppose that X has r-finite measure. Letk : M ! 0;1½ Þ be a positive measure on M:

By Conclusion 3.32, there exists a measurable function w : X ! 0; 1ð Þ such thatw 2 L1 lð Þ; and

RX w dl 2 0; 1½ �: By Conclusion 3.36, u : E 7! k Eð Þþ

RE w dl

� �is

a positive measure from M to 0;1½ Þ: It follows that u Xð Þ is a nonnegative realnumber, and hence the constant function 1 defined on X is a member of L2 uð Þ:Now, by Conclusion 3.36, and Lemma 2.21, for every f 2 L2 uð Þ; we havef ¼ð Þ f � 1ð Þ 2 L1 uð Þ; andZ

X

fj jdk�ZX

fj jdkþZX

fj jwdl ¼ZX

fj jdu ¼ fk k1¼ f � 1k k1 � fk k2 1k k2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ fk k2

ZX

1j j2du

0@ 1A12

¼ fk k2ffiffiffiffiffiffiffiffiffiffiffiu Xð Þ

p\1

and hence for every f 2 L2 uð Þ, f 2 L1 kð Þ: Thus, L2 uð Þ � L1 kð Þ; and L2 uð Þ �L1 uð Þ: Now, by Lemma 1.135, for every f 2 L2 uð Þ;

ZX

f dk

��

ZX

fj jdk \1ð Þ:

Thus, for every f 2 L2 uð Þ,RX f dk 2 C: Let

K : f 7!ZX

f dk

be the mapping from the Hilbert space L2 uð Þ � L1 kð Þð Þ to C: Now, by Lemma1.134, K : L2 uð Þ ! C is a linear functional. Also, since for every f 2 L2 uð Þ;

K fð Þj j ¼ZXf dk

�� \1|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}; K is a bounded linear functional on L2 uð Þ: Since L2 uð Þ is

a Hilbert space, L2 uð Þ is a Banach space. Since L2 uð Þ is a Banach space, and K is abounded linear functional on L2 uð Þ, K is a continuous linear functional on theHilbert space L2 uð Þ; and hence by Conclusion 2.93, there exists g 2 L2 uð Þ suchthat for every f 2 L2 uð Þ;Z

X

f dk ¼ K fð Þ ¼ f ; gð Þ|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ¼ZX

f � �gð Þdu:


Since g 2 L2 uð Þ; �g 2 L2 uð Þ: Thus, we can say that there exists h 2L2 uð Þ � L1 uð Þð Þ such that for every f 2 L2 uð Þ;Z

X

f dk ¼ZX

f � hð Þdu: ð�Þ

(From the definition of L2 uð Þ; h as a point function of X, is determined uniquelyonly a.e. (relative to u).)

Problem 3.38 h xð Þ 2 0; 1½ � a.e. on X with respect to u; that isu h�1 C� 0; 1½ �ð Þð Þ ¼ 0:

(Solution We want to apply Lemma 1.154. Here, u Xð Þ\1, h 2 L1 uð Þ; 0; 1½ � is aclosed subset of C: Let us take any E 2 M satisfying u Eð Þ[ 0: It suffices to showthat

k Eð Þk Eð Þþ

RE w dl

¼ k Eð Þu Eð Þ ¼

RX vEdku Eð Þ ¼

RX vE � h duu Eð Þ ¼

� RE h duu Eð Þ 2 0; 1½ �;

that is

0� k Eð Þk Eð Þþ

RE w dl

� 1:

This is trivially true because k : M ! 0;1½ Þ; l : M ! 0;1½ � is a positivemeasure on M; and w : X ! 0; 1ð Þ: ■)

Since h xð Þ 2 0; 1½ � a.e. on X with respect to u; we can assume that h xð Þ 2 0; 1½ �for every x 2 X:

Let ka : E 7! k E\ h�1 0; 1½ Þð Þð Þ be a function from M to 0;1½ Þ: Now, sincek : M ! 0;1½ Þ is a positive measure on M; ka : M ! 0;1½ Þ is a positivemeasure on M:

Let ks : E 7! k E\ h�1 1Þð Þð Þ be a function from M to 0;1½ Þ: Now, since k :M ! 0;1½ Þ is a positive measure on M; ks : M ! 0;1½ Þ is a positive measureon M: Since h : X ! 0; 1½ �; from the definitions of ka and ks, k ¼ ka þ ks:

Problem 3.39 ks ? l:

(Solution It suffices to show:

1. ks is concentrated on h�1 0; 1½ Þð Þ;2. l is concentrated on h�1 1ð Þ:


For 1: It suffices to show that ks h�1 0; 1½ Þð Þð Þ ¼ 0:

LHS ¼ ks h�1 0; 1½ Þð Þ� �

¼ k h�1 0; 1½ Þð Þ \ h�1 1Þð Þ� �

¼ k ;ð Þ ¼ 0 ¼ RHS:

For 2: It suffices to show that l h�1 1ð Þð Þ ¼ 0: Since vh�1 1ð Þ 2 L2 uð Þ; from �ð Þ;

k h�1 1ð Þ� �

¼ZX

vh�1 1ð Þdk ¼ZX

vh�1 1ð Þ � h� �

du ¼ZX

vh�1 1ð Þ � h� �

dkþZX

vh�1 1ð Þ � h� �

w dl

¼Z

h�1 1ð Þ

h dkþZ

h�1 1ð Þ

hw dl ¼Z

h�1 1ð Þ

1 dkþZ

h�1 1ð Þ

1w dl ¼ k h�1 1ð Þ� �

þZ

h�1 1ð Þ

w dl;

and henceRh�1 1ð Þ w dl ¼ 0: Now, by Lemma 1.151, w ¼ 0 a.e. on h�1 1ð Þ; that is

l h�1 1ð Þ� �

¼ l h�1 1ð Þ \X� �

¼ l h�1 1ð Þ \w�1 ð0; 1ð Þ� �

¼ 0:

Thus, l h�1 1ð Þð Þ ¼ 0: ■)By (*), for every positive integer n,Z

X

vE 1þ hþ � � � þ hnð Þð Þdk

¼ZX

vE 1þ hþ � � � þ hnð Þð Þ � hð Þdu ¼ZX

vE 1þ hþ � � � þ hnð Þ � hð Þdu

¼ZX

vE 1þ hþ � � � þ hnð Þ � hð ÞdkþZX

vE 1þ hþ � � � þ hnð Þ � hð Þw dl;

so for every positive integer n,ZE

1� hnþ 1� �dk ¼

ZX

vE 1� hnþ 1� �� dk ¼

ZX

vE 1þ hþ � � � þ hnð Þ 1� hð Þð Þdk

¼ZX

vE 1þ hþ � � � þ hnð Þ � vE 1þ hþ � � � þ hnð Þ � hð Þdk

¼ZX

vE 1þ hþ � � � þ hnð Þdk�ZX

vE 1þ hþ � � � þ hnð Þ � hð Þdk

¼ZX

vE 1þ hþ � � � þ hnð Þ � hð Þw dl;


and hence on using the fact l h�1 1ð Þð Þ ¼ 0;

ka Eð Þ ¼ k E \ h�1 0; 1½ Þð Þ� �

¼ZE

vh�1 0;1½ Þð Þdk ¼ZE

1� limn!1

hnþ 1� �

dk

¼ZE

limn!1

1� hnþ 1� �

dk ¼ limn!1

ZE

1� hnþ 1� �

dk

¼ limn!1

ZX

vE 1þ hþ � � � þ hnð Þ � hð Þw dl

¼ limn!1

Zh�1 0;1½ Þð Þ


0B@þ

Zh�1 1ð Þ


1CA¼ lim

n!1

Zh�1 0;1½ Þð Þ

vE 1þ hþ � � � þ hnð Þ � hð Þw dlþ 0

0B@1CA

¼ limn!1

Zh�1 0;1½ Þð Þ


¼ limn!1

ZX

vh�1 0;1½ Þð ÞvE 1þ hþ � � � þ hnð Þ � h� �

w dl

¼ limn!1

ZE

vh�1 0;1½ Þð Þ 1þ hþ � � � þ hnð Þ � h� �

w dl

¼ZE

limn!1

vh�1 0;1½ Þð Þ 1þ hþ � � � þ hnð Þ � h� �

w� �

dl:


Thus, for every E 2 M; ka Eð Þ ¼RE H dl; where

H � limn!1

vh�1 0;1½ Þð Þ 1þ hþ � � � þ hnð Þ � h� �

w

¼ limn!1

vh�1 0;1½ Þð Þ1� hnþ 1

1� h

� � h

� w

¼ vh�1 0;1½ Þð Þ1� limn!1 hnþ 1

1� h

� � h

� w

¼ vh�1 0;1½ Þð Þ1� 01� h

� � h

� w ¼ vh�1 0;1½ Þð Þ

hw1� h

:

SinceZX

Hj jdl ¼ZX

vh�1 0;1½ Þð Þhw

1� h

�� dl ¼ZX

vh�1 0;1½ Þð Þhw

1� hdl ¼

ZX

H dl ¼ ka Xð Þ\1;

we haveRX Hj jdl\1; and hence H 2 L1 lð Þ:

Problem 3.40 ka l:

(Solution For this purpose, let us take any E 2 M satisfying l Eð Þ ¼ 0: We haveto show that ka Eð Þ ¼ 0:

LHS ¼ ka Eð Þ ¼REH dl ¼ 0 ¼ RHS: ■)

Conclusion 3.41 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ � be a positive measure on M: Suppose that X has r-finite measure.Let k : M ! 0;1½ Þ be a positive measure on M: Then there exist positive mea-sures ka : M ! 0;1½ Þ; and ks : M ! 0;1½ Þ such that

1. k ¼ ka þ ks;2. ks ? l;3. ka l;4. there exists a function H : X ! 0;1½ Þ such that H 2 L1 lð Þ; and, for every

E 2 M; ka Eð Þ ¼RE H dl:

Theorem 3.42 Let X be any nonempty set. Let M be a r-algebra in X. Let l :M ! 0;1½ � be a positive measure onM: Suppose that X has r-finite measure. Letk : M ! C be a complex measure onM: Then there exists unique pair of complexmeasures ka : M ! C; and ks : M ! C such that

1. k ¼ ka þ ks;2. ks ? l;3. ka l;4. there exists a unique function h : X ! C such that h 2 L1 lð Þ; and for every

E 2 M;


ka Eð Þ ¼ZE

h dl:

Proof Existence: Since k : M ! C is a complex measure onM; Re kð Þ : M ! R;

and Im kð Þ : M ! R are signed measures. It follows that Re kð Þð Þþ ; Re kð Þð Þ�;Im kð Þð Þþ ; Im kð Þð Þ� are positive measures on M satisfying

Re kð Þ ¼ Re kð Þð Þþ� Re kð Þð Þ�; and Im kð Þ ¼ Im kð Þð Þþ� Im kð Þð Þ�:

By Conclusion 3.41,

Re kð Þð Þþ ¼ Re kð Þð Þþ� �

a þ Re kð Þð Þþ� �

s; Re kð Þð Þþ� �

s ? l; Re kð Þð Þþ� �

a l;

Re kð Þð Þ� ¼ Re kð Þð Þ�ð Þa þ Re kð Þð Þ�ð Þs; Re kð Þð Þ�ð Þs ? l; Re kð Þð Þ�ð Þa l;

Im kð Þð Þþ ¼ Im kð Þð Þþ� �

a þ Im kð Þð Þþ� �

s; Im kð Þð Þþ� �

s ? l; Im kð Þð Þþ� �

a l;

and

Im kð Þð Þ�¼ Im kð Þð Þ�ð Þa þ Im kð Þð Þ�ð Þs; Im kð Þð Þ�ð Þs ? l; Im kð Þð Þ�ð Þa l:

Put

ka ¼ Re kð Þð Þþ� �

a� Re kð Þð Þ�ð Þa� �

þ i Im kð Þð Þþ� �

a� Im kð Þð Þ�ð Þa� �

;

and

ks ¼ Re kð Þð Þþ� �

s� Re kð Þð Þ�ð Þs� �


s� Im kð Þð Þ�ð Þs� �

:

Problem 3:43 ka : M ! C; and ks : M ! C are complex measures on M:

(Solution Its proof is clear. ■)For 1:

LHS ¼ k ¼ Re kð Þþ i Im kð Þð Þ ¼ Re kð Þð Þþ� Re kð Þð Þ��

þ i Im kð Þð Þþ� Im kð Þð Þ��

¼ Re kð Þð Þþ� �

a þ Re kð Þð Þþ� �

s� Re kð Þð Þ�ð Þa þ Re kð Þð Þ�ð Þs� ��


a þ Im kð Þð Þþ� �

s� Im kð Þð Þ�ð Þa þ Im kð Þð Þ�ð Þs� ��




a� Im kð Þð Þ�ð Þa� ��

þ Re kð Þð Þþ� �



s� Im kð Þð Þ�ð Þs� ��

¼ ka þ ks ¼ RHS:



Re kð Þð Þþ� �



s� Im kð Þð Þ�ð Þs� ��

? l:

This is clearly true, from Problem 3.25.For 3: We have to show that





l:

This is clearly true, by Problem 3.26.For 4: There exists a function H1 : X ! 0;1½ Þ such that H1 2 L1 lð Þ; and for

every E 2 M;


a Eð Þ ¼ZE

H1dl:

There exists a function H2 : X ! 0;1½ Þ such that H2 2 L1 lð Þ; and for everyE 2 M;

Re kð Þð Þ�ð Þa Eð Þ ¼ZE

H2dl:


Im kð Þð Þþ� �

a Eð Þ ¼ZE

H3dl:


Im kð Þð Þ�ð Þa Eð Þ ¼ZE

H4dl:

Put h ¼ H1 � H2ð Þþ i H3 � H4ð Þ:Clearly, h : X ! C is a function such that h 2 L1 lð Þ: For every E 2 M;


ka Eð Þ ¼ Re kð Þð Þþ� �




Eð Þ


a Eð Þ � Re kð Þð Þ�ð Þa Eð Þþ i Im kð Þð Þþ� �

a Eð Þ � Im kð Þð Þ�ð Þa Eð Þ� �

¼ZE

H1dl�ZE

H2dlþ iZE

H3dl�ZE

H4dl

0@ 1A¼ZE

H1 � H2ð Þþ i H3 � H4ð Þð Þdl ¼ZE

h dl:

Thus, ka Eð Þ ¼RE h dl:

Uniqueness: Let ka : M ! C; and ks : M ! C be complex measures such that

k ¼ ka þ ks; ks ? l; ka l:

Let k0a : M ! C; and k0s : M ! C be complex measures such that

k ¼ k0a þ k0s; k0s ? l; k0a l:

We have to show that ka ¼ k0a; and ks ¼ k0s: Since

k0a þ k0s ¼ k ¼ ka þ ks;

k0a � ka ¼ ks � k0s:

Since ka l; and k0a l; k0a � ka� �

l: Similarly, ks � k0s� �

? l: It followsthat k0a � ka

� �? l: Since k0a � ka

� �? l; and k0a � ka

� � l; by Problem 3.29,

ks � k0s ¼� �

k0a � ka ¼ 0; and hence ka ¼ k0a: Similarly, ks ¼ k0s:Next, let h : X ! C be a function such that h 2 L1 lð Þ; and for every E 2 M;

ka Eð Þ ¼RE h dl: Let h1 : X ! C be a function such that h1 2 L1 lð Þ; and for every

E 2 M; ka Eð Þ ¼RE h1dl: We have to show that h ¼ h1 a.e. on X. Since for every

E 2 M; ZE

h1dl ¼ ka Eð Þ ¼ZE

h dl;

for every E 2 M; ZE

h1 � hð Þdl ¼ZE

h1dl�ZE

h dl ¼ 0

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:


Since for every E 2 M;RE h1 � hð Þdl; and h1 � hð Þ 2 L1 lð Þ; by Lemma 1.95,

h1 � hð Þ ¼ 0 a.e. on X, and hence h ¼ h1 a.e. on X. ■The result (4) of Theorem 3.42, known as the Radon–Nikodym theorem, is due

to J. Radon (16.12.1887–25.05.1956), and O. M. Nikodym (13.08.1887–04.05.1974). The idea of the proof given here is due to von Neumann (28.12.1903–08.02.1957).

Radon’s doctoral dissertation was on the calculus of variations.Nikodym’s father died in an accident, and soon after his mother also died. He

was brought up by his mother’s grandparents. He wrote a book entitled Theory ofTensors. He suffered an electric shock in 1971 in the USA.

Here, the ordered pair ka; ksð Þ is called the Lebesgue decomposition of k relativeto l. The function h : X ! C is called the Radon–Nikodym derivative of ka withrespect to l; and we write

dka ¼ h dl ordkadl

¼ h:

In short, we write

dk ¼ h dlþ d ksð Þ;

and call it the Lebesgue decomposition of k with respect to l:

Note 3.44 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !0;1½ � be a positive measure on M: Let k : M ! C be a complex measure on M:Let k l:

Problem 3.45 For every e[ 0; there exists d[ 0 such that l Eð Þ\d impliesk Eð Þj j\e:

(Solution If not, otherwise, suppose that there exists e[ 0 such that correspondingto each d[ 0; there exists E 2 M satisfying l Eð Þ\d and e� k Eð Þj j: We have toarrive at a contradiction.

It follows that there exist e[ 0 such that corresponding to each positive integern, there exists En 2 M satisfying l Enð Þ\ 1

2n and e� k Enð Þj j � kj j Enð Þð Þ: Sincel : M ! 0;1½ � is a positive measure on M; and for every positive integer n,En 2 M; and l Enð Þ\ 1

2n ; we have for every positive integer n,

l E1 [E2 [E3 [ � � �ð Þ �P1n¼1

l Enð Þ \P1n¼1

12n ¼ 1

� ;

l E2 [E3 [E4 [ � � �ð Þ�P1n¼2

l Enð Þ \P1n¼2

12n ¼ 1

2

� ;

l E3 [E4 [E5 [ � � �ð Þ�P1n¼3

l Enð Þ \P1n¼3

12n ¼ 1

4

� ;

..

.:


Clearly,

E1 [E2 [E3 [ � � �ð Þ E2 [E3 [E4 [ � � �ð Þ E3 [E4 [E5 [ � � �ð Þ � � � :

Now, by Lemma 1.99 (5),

0 ¼ limn!1

12n�1 �

� limn!1

l En [Enþ 1 [Enþ 2 [ � � �ð Þ ¼ l Að Þ � 0ð Þ;

where

A � E1 [E2 [E3 [ � � �ð Þ \ E2 [E3 [E4 [ � � �ð Þ \ E3 [E4 [E5 [ � � �ð Þ \ � � � 2 Mð Þ:

Since k l; by Note 3.4 (V), kj j l: Since A 2 M; l Að Þ ¼ 0; and kj j l;kj j Að Þ ¼ 0; and hence

0� limn!1

kj j Enð Þ� limn!1

kj j En [Enþ 1 [Enþ 2 [ � � �ð Þ

¼ kj j E1 [E2 [E3 [ � � �ð Þ \ E2 [E3 [E4 [ � � �ð Þ \ E3 [E4 [E5 [ � � �ð Þ \ � � �ð Þ ¼ 0:

It follows that limn!1 kj j Enð Þ ¼ 0: Since for every positive integern; e� kj j Enð Þ; e� limn!1 kj j Enð Þ ¼ 0ð Þ; and hence e� 0: This is a contradiction. ■)

Conclusion 3.46 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ � be a positive measure on M: Let k : M ! C be a complexmeasure on M: Then k l if and only if for every e[ 0; there exists d[ 0 suchthat l Eð Þ\d implies k Eð Þj j\e:

Proof of the remaining part Suppose that, for every e[ 0; there exists d[ 0 suchthat l Eð Þ\d implies k Eð Þj j\e: We have to show that k l:

For this purpose, let us take any E 2 M satisfying l Eð Þ ¼ 0: We have to showthat k Eð Þ ¼ 0: If not, otherwise, let k Eð Þ 6¼ 0: We have to arrive at a contradiction.Since k Eð Þ 6¼ 0; k Eð Þj j[ 0: Now, by the assumption, there exists d[ 0 such thatl Fð Þ\d implies k Fð Þj j\ k Eð Þj j: Now, since l Eð Þ ¼ 0 \dð Þ; we havek Eð Þj j\ k Eð Þj j: This is a contradiction. ■

Note 3.47 Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C

be a complex measure on M:Since for every E 2 M; 0�ð Þ k Eð Þj j � kj j Eð Þ; we have k kj j:

Problem 3.48 0? kj j:

(Solution It suffices to show: 1. 0 is concentrated on ;; 2. kj j is concentrated on X.For 1: Let E 2 M satisfying ;\E ¼ ;: We have to show that 0 Eð Þ ¼ 0: This is

true, by the definition of zero measure.For 2: Let E 2 M satisfying X \E ¼ ;:We have to show that kj j Eð Þ ¼ 0: Since

E ¼ð ÞX \E ¼ ;; kj j Eð Þ ¼ kj j ;ð Þ ¼ 0: ■)


Since k ¼ kþ 0; k kj j; and 0? kj j; by Theorem 3.42, the ordered pair k; 0ð Þis the Lebesgue decomposition of k relative to kj j, and there exists a functionh : X ! C such that h 2 L1 kj jð Þ; and for every E 2 M;

k Eð Þ ¼ZE

h d kj j:

Problem 3.49 1� hj j a.e. on X with respect to kj j:(Solution It suffices to show that for every r 2 0; 1ð Þ;

kj j x : h xð Þj j\rf gð Þ ¼ 0:

For this purpose, let us fix any r 2 0; 1ð Þ: We have to show that

sup k E1ð Þj j þ k E2ð Þj j þ k E3ð Þj j þ � � � : E1;E2;E3; . . .f g is a partition of x : h xð Þj j\rf gf g ¼ 0:

Let us take any partition E1;E2;E3; . . .f g of x : h xð Þj j\rf g: Since for everypositive integer n,

k Enð Þj j ¼ZEn

h d kj j

��

ZEn

hj jd kj j �ZEn

r d kj j ¼ r kj j Enð Þð Þ;

we haveX1n¼1

k Enð Þj j �X1n¼1

r kj j Enð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ rX1n¼1

kj j Enð Þ ¼ r kj j x : h xð Þj j\rf gð Þð Þ;

and hence

kj j x : h xð Þj j\rf gð Þ� r kj j x : h xð Þj j\rf gð Þð Þ:

It follows that

0� kj j x : h xð Þj j\rf gð Þð Þ 1� rð Þ ¼ kj j x : h xð Þj j\rf gð Þ � r kj j x : h xð Þj j\rf gð Þð Þ� 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence

kj j x : h xð Þj j\rf gð Þð Þ 1� rð Þ ¼ 0:

Now, since r 2 0; 1ð Þ; kj j x : h xð Þj j\rf gð Þ ¼ 0: ■)


Problem 3.50 hj j � 1 a.e. on X with respect to kj j; that iskj j h�1 z : z 2 C and 1\ zj jf gð Þð Þ ¼ 0:

(Solution We want to apply Lemma 1.154. Here, kj j Xð Þ\1; h 2 L1 kj jð Þ;z : z 2 C and zj j � 1f g is a closed subset of C: Let us take any E 2 M satisfyingkj j Eð Þ[ 0: It suffices to show that

k Eð Þkj j Eð Þ ¼

� RE h d kj jkj j Eð Þ 2 z : z 2 C and zj j � 1f g;

that is

k Eð Þj jkj j Eð Þ ¼

k Eð Þkj j Eð Þ

�� 1|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} :This is known to be true. ■)Since hj j � 1 a.e. on X with respect to kj j; and 1� hj j a.e. on X with respect to kj j;

kj j h�1 z : z 2 C and zj j 6¼ 1f gð Þ� �

¼ 0:

Since kj j h�1 z : z 2 C and zj j 6¼ 1f gð Þð Þ ¼ 0; we can redefine h onh�1 z : z 2 C and zj j 6¼ 1f gð Þ so that h xð Þ ¼ 1 for every x 2h�1 z : z 2 C and zj j 6¼ 1f gð Þ; and hence for every x 2 X; h xð Þj j ¼ 1:

Conclusion 3.51 Let X be any nonempty set. Let M be a r-algebra in X. Letk : M ! C be a complex measure on M: Then there exists a function h : X !z : z 2 C and zj j ¼ 1f g such that h 2 L1 kj jð Þ; and for every E 2 M;

k Eð Þ ¼ZE

h d kj j:

In short, we say that there exists a measurable function h : X !z : z 2 C and zj j ¼ 1f g such that

dk ¼ h � d kj jð Þ:

Here, dk ¼ h � d kj jð Þ is called the polar representation of k:

Note 3.52 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M !0;1½ � be a positive measure on M: Let g : X ! C be a measurable function suchthat g 2 L1 lð Þ:


Problem 3.53 k : E 7!RE g dl from M to C is a complex measure on M:

(Solution For this purpose, let us take any E 2 M; and a partition E1;E2;E3; . . .f gof E. We have to show that

1. the seriesRE1g dl

�� þ RE2g dl

�� þ RE3g dl

�� þ � � � is convergent,2.RE g dl ¼

RE1g dlþ

RE2g dlþ

RE3g dlþ � � � :


ZE1

Re gð Þð Þþ dl�ZE1

Re gð Þð Þ�dlþ iZE1

Im gð Þð Þþ dl�ZE1

Im gð Þð Þ�dl

0@ 1A��

þZE2




Im gð Þð Þ�dl

0@ 1A��

þZE3




Im gð Þð Þ�dl

0B@1CA

��þ � � �

is convergent. Now, since for every positive integer n,ZEn

Re gð Þð Þþ dl�ZEn

Re gð Þð Þ�dlþ iZEn

Im gð Þð Þþ dl�ZEn

Im gð Þð Þ�dl

0@ 1A��

�ZEn

Re gð Þð Þþ dl

��þ

ZEn

Re gð Þð Þ�dl

��þ

ZEn

Im gð Þð Þþ dl

��þ

ZEn

Im gð Þð Þ�dl

��

¼ZEn

Re gð Þð Þþ dlþZEn

Re gð Þð Þ�dlþZEn

Im gð Þð Þþ dlþZEn

Im gð Þð Þ�dl;


X1n¼1

ZEn

Re gð Þð Þþ dlþZEn

Re gð Þð Þ�dlþZEn

Im gð Þð Þþ dlþZEn

Im gð Þð Þ�dl

0@ 1Ais convergent, that is

X1n¼1

ZEn

Re gð Þð Þþ dl;X1n¼1

ZEn

Re gð Þð Þ�dl;X1n¼1

ZEn

Im gð Þð Þþ dl;X1n¼1

ZEn

Im gð Þð Þ�dl


are convergent. Since g 2 L1 lð Þ; we have

X1n¼1

ZEn

Re gð Þð Þþ dl�ZE

Re gð Þð Þþ dl�ZX

Re gð Þð Þþ dl�ZX

gj jdl\1;

and henceP1

n¼1

REn

Re gð Þð Þþ dl is convergent. Similarly,

X1n¼1

ZEn

Re gð Þð Þ�dl;X1n¼1

ZEn

Im gð Þð Þþ dl;X1n¼1

ZEn

Im gð Þð Þ�dl

are convergent.For 2:

LHS ¼ZE

gdl ¼ZE

Re gð Þð Þþ dl�ZE

Re gð Þð Þ�dlþ iZE

Im gð Þð Þþ dl�ZE

Im gð Þð Þ�dl

0@ 1A¼X1n¼1

ZEn

Re gð Þð Þþ dl�X1n¼1

ZEn

Re gð Þð Þ�dlþ iX1n¼1

ZEn

Im gð Þð Þþ dl�X1n¼1

ZEn

Im gð Þð Þ�dl

0@ 1A¼X1n¼1

ZEn

Re gð Þð Þþ dl�ZEn

Re gð Þð Þ�dlþ iZEn

Im gð Þð Þþ dl�ZEn

Im gð Þð Þ�dl

0@ 1A0@ 1A¼X1n¼1

ZEn

g dl ¼ RHS:

■)Now, by Conclusion 3.51, there exists a function h : X ! z : z 2 C and zj j ¼ 1f g

such that h 2 L1 kj jð Þ; and for every E 2 M;RE g dl ¼

� �k Eð Þ ¼

RE h d kj j: It follows

that for every E 2 M;

ZE

h d kj j ¼ZE

g dl ¼ZE

h � 1hg

� dl ¼

ZE

h � hj j2

hg

!dl ¼

ZE

h � �hgð Þdl;

and hence for every E 2 M;ZE

h d kj j ¼ZE

h � �hgð Þdl:


In short, d kj j ¼ �hgð Þdl: Now, since kj j and l are positive measures on M;�hg� 0 a.e. on X with respect to l: Since �hg� 0 a.e. on X with respect to l;

�hg ¼ �hg�� ¼ �h

�� gj j ¼ hj j gj j ¼ 1 gj j ¼ gj j a:e:

on X with respect to l: Since �hg ¼ gj j a.e. on X with respect to l; for every E 2 M;ZE

h d kj j ¼ZE

h � �hgð Þdl ¼ZE

h � gj jdl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence for every E 2 M;

ZE

h d kj j ¼ZE

h � gj jdl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZE

h � d F 7!ZF

gj jdl

0@ 1A; by Lemma 1:132

Thus, for every E 2 M; kj j Eð Þ ¼RE gj jdl:

Conclusion 3.54 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ � be a positive measure on M: Let g : X ! C be a measurablefunction such that g 2 L1 lð Þ: Let k : E 7!

RE g dl from M to C be a complex

measure on M: Then, for every E 2 M;

kj j Eð Þ ¼ZE

gj jdl:

Note 3.55 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! R

be a complex measure on M:By Conclusion 3.51, there exists a function h : X ! z : z 2 C and zj j ¼ 1f g such

that h 2 L1 lj jð Þ; and for every E 2 M;

l Eð Þþ i0 ¼ l Eð Þ ¼ZE

h d lj j

|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}¼ZE

Re hð Þð Þd lj j þ iZE

Im hð Þð Þd lj j:

It follows that for every E 2 M;RE Im hð Þð Þd lj j ¼ 0; and hence Im hð Þ ¼ 0 a.e.

on X with respect to lj j: So, h is real a.e. on X with respect to lj j; and hence, we canassume that h : X ! R: Now since h : X ! z : z 2 C and zj j ¼ 1f g; for every x 2X; h xð Þ ¼ þ 1 or � 1: It follows that h�1 1ð Þ \ h�1 �1ð Þ ¼ ;; and h�1 1ð Þ [ h�1

�1ð Þ ¼ X: For every E 2 M;


lþ Eð Þ ¼ 12

lj j þ lð Þ�

Eð Þ ¼ 12

lj j Eð Þþ l Eð Þð Þ ¼ 12

lj j Eð ÞþZE

h d lj j

0@ 1A¼ 1

2lj j Eð Þþ

ZE \ h�1 1ð Þ

h d lj j þZ

E\ h�1 �1ð Þ

h d lj j

0B@1CA

0B@1CA

¼ 12

lj j Eð ÞþZ

E \ h�1 1ð Þ

1 d lj j þZ

E\ h�1 �1ð Þ

�1ð Þd lj j

0B@1CA

0B@1CA

¼ 12

lj j Eð Þþ lj j E \ h�1 1ð Þ� �

� lj j E \ h�1 �1ð Þ� ��

¼ 12

lj j E \ h�1 1ð Þ� �

þ lj j E \ h�1 �1ð Þ� ��

þ lj j E \ h�1 1ð Þ� �

� lj j E \ h�1 �1ð Þ� ��

¼ lj j E \ h�1 1ð Þ� �

¼Z

E \ h�1 1ð Þ

1 d lj j ¼Z

E \ h�1 1ð Þ

h d lj j ¼ l E \ h�1 1ð Þ� �

:

Thus, for every E 2 M; lþ Eð Þ ¼ l E \ h�1 1ð Þð Þ: Also, for every E 2 M;

l� Eð Þ ¼ lj j � lþð Þ Eð Þ ¼ lj j Eð Þ � lþ Eð Þ ¼ lj j Eð Þ � lj j E \ h�1 1ð Þ� �

¼ lj j E \ h�1 1ð Þ� �

þ lj j E \ h�1 �1ð Þ� ��

� lj j E\ h�1 1ð Þ� �

¼ lj j E\ h�1 �1ð Þ� �

¼ �Z

E\ h�1 �1ð Þ

�1ð Þd lj j ¼ �Z

E \ h�1 �1ð Þ

h d lj j ¼ �l E\ h�1 �1ð Þ� �

:

Thus, for every E 2 M; l� Eð Þ ¼ �l E \ h�1 �1ð Þð Þ:

Conclusion 3.56 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! R be a complex measure on M: Then there exist A;B 2 M such that

1. A\B ¼ ;;A[B ¼ X;2. for every E 2 M; lþ Eð Þ ¼ l E \Að Þ; and l� Eð Þ ¼ �l E\Bð Þ:

Definition Here, the ordered pair A;Bð Þ is called the Hahn decomposition of X,induced by l:

Lemma 3.57 Let X be any nonempty set. Let M be a r-algebra in X. Let l :M ! R be a complex measure on M: Let k1 : M ! 0;1½ Þ; and k2 : M !0;1½ Þ be positive measures on M: Let l ¼ k1 � k2: Then lþ � k1; and l� � k2:

Proof Let us take any E 2 M: We have to show that lþ Eð Þ� k1 Eð Þ;l� Eð Þ� k2 Eð Þ:


By Conclusion 3.56, there exist A;B 2 M such that

1. A\B ¼ ;;A[B ¼ X;2. lþ Eð Þ ¼ l E \Að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ k1 � k2ð Þ E\Að Þ ¼ k1 E \Að Þ � k2 E \Að Þ� k1 E \Að Þ� k1 Eð Þ;

and

l� Eð Þ ¼ �l E\Bð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ � k1 � k2ð Þ E \Bð Þ

¼ k2 E \Bð Þ � k1 E \Bð Þ� k2 E \Bð Þ� k2 Eð Þ:

Thus, lþ Eð Þ� k1 Eð Þ; and l� Eð Þ� k2 Eð Þ: ■

3.3 Bounded Linear Functionals on Lp

Note 3.58 Let p 2 1;1½ Þ: Let q 2 1;1ð �ð Þ be the exponent conjugate to p. LetX be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ � be apositive measure on M: Let g 2 Lq lð Þ:

Problem 3.59 Ug : f 7!RX f � gð Þdl is a function from Lp lð Þ to C:

(Solution Let us take any f 2 Lp lð Þ: We have to show thatRX f � gð Þdl 2 C: By

Lemma 2.24, f � gð Þ 2 L1 lð Þ and f � gk k1 � fk kp gk kq: Since f � gð Þ 2 L1 lð Þ;RX f � gð Þdl 2 C: ■)

Problem 3.60 The function

Ug : f 7!ZX

f � gð Þdl

from Lp lð Þ to C is a bounded linear functional on the normed linear space Lp lð Þ:(Solution Let f1; f2 2 Lp lð Þ; and a; b 2 C: Since Lp lð Þ is a complex linear space,af1 þ bf2ð Þ 2 Lp lð Þ: Since af1 þ bf2ð Þ 2 Lp lð Þ; and g 2 Lq lð Þ; by Lemma 2.24

a f1 � gð Þþ b f2 � gð Þ ¼ af1 þ bf2ð Þ � gð Þ 2 L1 lð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :


Similarly, f1 � gð Þ; f2 � gð Þ 2 L1 lð Þ: Now, by Lemma 1.134,

Ug af1 þ bf2ð Þ ¼ZX

af1 þ bf2ð Þ � gð Þdl ¼ZX

a f1 � gð Þþ b f2 � gð Þð Þdl

¼ aZX

f1 � gð Þdlþ bZX

f2 � gð Þdl ¼ a Ug f1ð Þ� �

þ b Ug f2ð Þ� �

;

and hence

Ug af1 þ bf2ð Þ ¼ a Ug f1ð Þ� �

þ b Ug f2ð Þ� �

:

This shows that Ug is a linear functional on the normed linear space Lp lð Þ: Now,it remains to show that

Ug fð Þ�� : f 2 Lp lð Þ and fk kp � 1n o

is bounded above, that is

ZX

f � gð Þdl

�� : f 2 Lp lð Þ and fk kp � 1

8<:9=;

is bounded above. For this purpose, let us take any f 2 Lp lð Þ satisfying fk kp � 1:Here, by Lemma 1.135,

ZX

f � gð Þdl

��

ZX

f � gj jdl ¼ f � gk k1 � fk kp gk kq � 1 gk kq¼ gk kq\1;

so

ZX

f � gð Þdl

�� gk kq\1:

It follows that

ZX

f � gð Þdl

�� : f 2 Lp lð Þ and fk kp � 1

8<:9=;

is bounded above. ■)


We further see that, for every g 2 Lq lð Þ;

Ug

�� ¼� �

sup Ug fð Þ�� : f 2 Lp lð Þ and fk kp � 1n o� �

� gk kq;

that is for every g 2 Lq lð Þ, Uk kg � gk kq:

Conclusion 3.61 Let p 2 1;1½ Þ: Let q 2 1;1ð �ð Þ be the exponent conjugate top. Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ � be apositive measure on M: Then, for every g 2 Lq lð Þ; Ug : f 7!

RX f � gð Þdl is a

bounded linear functional on the normed linear space Lp lð Þ; and

Uk kg � gk kq:

Note 3.62 Let p 2 1;1½ �: Let X be any nonempty set. Let M be a r-algebra inX. Let l : M ! 0;1½ Þ be a positive measure on M:

Since l : M ! 0;1½ Þ is a positive measure on M; for every E 2 M and forevery r 2 1;1½ Þ; Z

X

vEj jrdl ¼ZX

vEdl ¼ l Eð Þ 2 0;1½ Þ;

and hence for every E 2 M and for every r 2 1;1½ Þ,RX vEj jrdl\1: This shows

that for every E 2 M; and for every r 2 1;1½ Þ; vE 2 Lr lð Þ: Now, since for everyr 2 1;1½ Þ; Lr lð Þ is a complex linear space, we find that, for every r 2 1;1½ Þ; eachmeasurable simple function s : X ! C is a member of Lr lð Þ:

Problem 3.63 If s : X ! C is a measurable simple function, then, s 2 L1 lð Þ:

(Solution Let s : X ! C be a measurable simple function. We have to show thats 2 L1 lð Þ: Since L1 lð Þ is a linear space, it suffices to show that, for every E 2 M;vEj j ¼ð ÞvE 2 L1 lð Þ; that is the essential supremum of vE is different from1; that is

a : a 2 0;1½ Þ; and l v�1E a;1ð �ð Þ

� �¼ 0

� 6¼ ;:

Since v�1E 1;1ð �ð Þ ¼ ;;

1 2 a : a 2 0;1½ Þ; and l v�1E a;1ð �ð Þ

� �¼ 0

� ;

and hence

a : a 2 0;1½ Þ; and l v�1E a;1ð �ð Þ

� �¼ 0

� 6¼ ;:

■)

3.3 Bounded Linear Functionals on Lp 433

Conclusion 3.64 Let p 2 1;1½ �: Let X be any nonempty set. LetM be a r-algebrain X. Let l : M ! 0;1½ Þ be a positive measure on M: If s : X ! C is a mea-surable simple function, then s 2 Lp lð Þ:

Problem 3.65 For every r 2 0;1½ Þ; L1 lð Þ � Lr lð Þ:(Solution Let us fix any f 2 L1 lð Þ; and r 2 0;1½ Þ: We have to show that f 2Lr lð Þ; that is

RX fj jrdl\1: Since f 2 L1 lð Þ; by Conclusion 2.13 fj j xð Þ� fk k1 a.

e. on X, and henceZX

fj jrdl�ZX

fk k1� �r

dl ¼ fk k1� �r

l Xð Þ\1;

and henceRX fj jrdl\1: ■)

Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ Þ bea positive measure on M: Let f : X ! C be a measurable function. Let f 2 L1 lð Þ:

Since f 2 L1 lð Þ; we have fk k1\1; and l fj j�1 fk k1;1� ��

¼ 0:

Put E � fj j�1 fk k1;1� �� c

: Observe that

fj j�1 fk k1;1� �� c

¼ fj j�1 fk k1;1� �c� �

¼ fj j�1 0; fk k1� ��

¼ x : fj j xð Þ� fk k1�

¼ x : f xð Þj j � fk k1�

:

Thus,

E ¼ x : f xð Þj j � fk k1�

;

and hence f is bounded on E. Since f is bounded on E, Re fð Þð Þþ ; Re fð Þð Þ�;Im fð Þð Þþ ; Im fð Þð Þ� are bounded on E. Also, l X � Eð Þ ¼ 0: Since Re fð Þð Þþ :X ! 0;1½ Þ is a measurable function, and Re fð Þð Þþ is bounded on E, by Lemma1.98 there exists a sequence snf g of simple measurable functions sn : X ! 0;1½ Þsuch that for every x in X,

limn!1

sn xð Þ ¼ Re fð Þð Þþ� �

xð Þ;

and

limn!1

sn ¼ Re fð Þð Þþ� �

uniformly on E. Similarly, there exists a sequence tnf g of simple measurablefunctions tn : X ! 0;1½ Þ such that for every x in X,


limn!1

tn xð Þ ¼ Re fð Þð Þ�ð Þ xð Þ; and limn!1

tn ¼ Re fð Þð Þ�ð Þ uniformly onE:

Also, there exists a sequence unf g of simple measurable functions un : X !0;1½ Þ such that for every x in X,

limn!1

un xð Þ ¼ Im fð Þð Þþ� �

xð Þ; and limn!1

un ¼ Im fð Þð Þþ� �

uniformly onE:

Further, there exists a sequence vnf g of simple measurable functions vn : X !0;1½ Þ such that for every x in X,

limn!1

vn xð Þ ¼ Im fð Þð Þ�ð Þ xð Þ; and limn!1

vn ¼ Im fð Þð Þ�ð Þ uniformly onE:

For every positive integer n, put

Sn � sn � tnð Þþ i un � vnð Þ:

Now, since sn; tn; un; vn are simple measurable functions on X, Sn is a simplemeasurable function on X, and hence each Sn 2 L1 lð Þ:

Since limn!1 sn ¼ Re fð Þð Þþ� �

uniformly on E, limn!1 tn ¼ Re fð Þð Þ�ð Þ uni-

formly on E, limn!1 un ¼ Im fð Þð Þþ� �

uniformly on E, and limn!1 vn ¼Im fð Þð Þ�ð Þ uniformly on E, we have

limn!1

Sn ¼ limn!1

sn � tnð Þþ i un � vnð Þð Þ

¼ Re fð Þð Þþ� Re fð Þð Þ��

þ i Im fð Þð Þþ� Im fð Þð Þ��

¼ f

uniformly on E, and hence, limn!1 Sn ¼ f uniformly on E. Also, for every x in X,limn!1 Sn xð Þ ¼ f xð Þ:

Conclusion 3.66 Let X be any nonempty set. Let M be a r-algebra in X. Letl : M ! 0;1½ Þ be a positive measure on M: Let f : X ! C be a measurablefunction. Let f 2 L1 lð Þ: Then there exists a sequence Snf g of simple measurablefunctions Sn : X ! C; and a set E 2 M such that each Sn 2 L1 lð Þ; l X � Eð Þ ¼ 0;limn!1 Sn ¼ f uniformly on E and, for every x in X, limn!1 Sn xð Þ ¼ f xð Þ:

Note 3.67 Let p 2 1;1½ Þ: Let X be any nonempty set. Let M be a r-algebra inX. Let l : M ! 0;1½ Þ be a positive measure on M: Let U : Lp lð Þ ! C be abounded linear functional on Lp lð Þ:

Since U : Lp lð Þ ! C; in view of Conclusion 3.64 for every E 2 M, U vEð Þ 2 C:Let

k : E 7! U vEð Þ

be the function from M to C:


Problem 3.68 k is a complex measure on M:

(Solution For this purpose, let us take any E 2 M; and any partitionE1;E2;E3; . . .f g of E. We have to show that1. the series k E1ð Þþ k E2ð Þþ k E3ð Þþ � � � converges absolutely, 2.

k Eð Þ ¼ k E1ð Þþ k E2ð Þþ k E3ð Þþ � � � :For 2: We have to show that

U vEð Þ ¼ U vE1

� �þU vE2

� �þU vE3

� �þ � � � :

Observe that

U vEð Þ ¼ U vE1 [E2 [E3 [ ��

and

U vE1

� �þU vE2

� �þU vE3

� �þ � � � ¼ lim

n!1U vE1

� �þ � � � þU vEn

� �� ¼ lim

n!1U vE1

þ � � � þ vEn

� �� ¼ lim

n!1U vE1 [ �� [En

� �� ;

so it suffices to show that

limn!1

U vE1 [ �� [En

� �� ¼ U vE1 [E2 [E3 [ ��

� �:

Since U : Lp lð Þ ! C is a bounded linear functional on Lp lð Þ; by Conclusion2.184, U : Lp lð Þ ! C is continuous on Lp lð Þ; and hence it suffices to show that

limn!1

vE1 [ �� [En¼ vE1 [E2 [E3 [ ��

in Lp lð Þ; that is

limn!1

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� p¼ 0:

Observe that

limn!1

ZX

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdl0@ 1A0@ 1A1

p

¼ limn!1

ZX

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdl0@ 1A1

p

0B@1CA

¼ limn!1

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� p

so it suffices to show that


limn!1

ZX

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdl0@ 1A ¼ 0:

LHS ¼ limn!1

ZX

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdl0@ 1A

¼ limn!1

ZE1

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdlþ � � � þZEn

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdl0@ 1A0@

þZ

Enþ 1

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdlþ ZEnþ 2

vE1 [E2 [E3 [ �� vE1 [ �� [En

�� pdlþ � � �

1CA¼ lim

n!1

ZE1

1� 1j jpdlþ � � � þZEn

1� 1j jpdl

0@ 1AþZ

Enþ 1

1� 0j jpdlþZ

Enþ 2

1� 0j jpdlþ � � �

0B@1CA

¼ limn!1

ZE1

0 dlþ � � � þZEn

0 dl

0@ 1AþZ

Enþ 1

1 dlþZ

Enþ 2

1 dlþ � � �

0B@1CA

¼ limn!1

l Enþ 1ð Þþ l Enþ 2ð Þþ � � �ð Þ ¼ limn!1

l Enþ 1 [Enþ 2 [ � � �ð Þ

¼ l \1n¼1 Enþ 1 [Enþ 2 [ � � �ð Þ

� �¼ l ;ð Þ ¼ 0 ¼ RHS:

For 1: Since every ‘rearrangement’ of E1;E2;E3; . . .f g gives the same union E,by (1), every rearrangement of k E1ð Þþ k E2ð Þþ k E3ð Þþ � � � gives the same sum,and hence k E1ð Þþ k E2ð Þþ k E3ð Þþ � � � converges absolutely [see, WR[1],Theorem 3.54]. ■)

Problem 3.69 k l:

(Solution Let us take any E 2 M such that l Eð Þ ¼ 0: We have to show thatU vEð Þ ¼ð Þk Eð Þ ¼ 0; that is U vEð Þ ¼ 0: It suffices to show that vE ¼ 0: Since

vEk kp¼ZX

vEj jpdl

0@ 1A1p

¼ZX

vEdl

0@ 1A1p

¼ l Eð Þð Þ1p¼ 0ð Þ

1p¼ 0;

we have vEk kp¼ 0; and hence vE ¼ 0: ■)Clearly, 0? l: Since k ¼ kþ 0; k l; and 0? l; by Theorem 3.42 the ordered

pair k; 0ð Þ is the Lebesgue decomposition of k relative to l. There exists a functiong : X ! C such that g 2 L1 lð Þ; and for every E 2 M;

U vEð Þ ¼ð Þk Eð Þ ¼ZE

g dl ¼ZX

vE � g dl

0@ 1A:


Thus, for every E 2 M;

U vEð Þ ¼ZX

vE � g dl:

Problem 3.70 For every measurable simple function s : X ! C;U sð Þ ¼RX s � g dl:

(Solution Let

s ¼ a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ;

where each ai 2 C; and each s�1 aið Þ 2 M:

RHS ¼ U sð Þ ¼ U a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ

� �¼ a1U v s�1 a1ð Þð Þ

� �þ � � � þ anU v s�1 anð Þð Þ

� �¼ a1

ZX

v s�1 a1ð Þð Þ � g dlþ � � � þ an

ZX

v s�1 anð Þð Þ � g dl

¼ZX

a1v s�1 a1ð Þð Þ � gþ � � � þ anv s�1 anð Þð Þ � g� �

dl

¼ZX

a1v s�1 a1ð Þð Þ þ � � � þ anv s�1 anð Þð Þ

� �� g dl ¼

ZX

s � g dl ¼ LHS:

■)By Conclusion 3.64, for every r 2 0;1½ Þ; L1 lð Þ � Lr lð Þ:Let us take any f 2 L1 lð Þ:We shall try to show that f 2 Lp lð Þ; and U fð Þ ¼

RX f � g dl:

Since f 2 L1 lð Þ � Lp lð Þð Þ; f 2 Lp lð Þ: Now, since f 2 Lp lð Þ; and U : Lp lð Þ !C; U fð Þ 2 C: Since f 2 L1 lð Þ; and g 2 L1 lð Þ; f � gð Þ 2 L1 lð Þ; and henceR

X f � g dl� �

2 C: It remains to show that

U fð Þ ¼ZX

f � g dl:

By Conclusion 3.66, there exists a sequence snf g of simple measurable functionssn : X ! C; and a set E 2 M such that each sn 2 L1 lð Þ; l X � Eð Þ ¼ 0; andlimn!1 sn ¼ f uniformly on E, and for every x in X, limn!1 sn xð Þ ¼ f xð Þ: Sinceeach sn is a simple measurable function, each sn 2 Lp lð Þ: Sincef 2 L1 lð Þ � Lp lð Þð Þ, f 2 Lp lð Þ:


Problem 3.71 limn!1 sn ¼ f in Lp lð Þ:

(Solution We have to show that U fð Þ ¼RX f � g dl; that is

limn!1

ZX

sn � fj jpdl

0@ 1A1p

¼ limn!1

sn � fk kp¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl};that is

limn!1

ZX

sn � fj jpdl

0@ 1A ¼ 0:

For this purpose, let us take any e[ 0: Since limn!1 sn ¼ f uniformly on E,there exists a positive integer N such that for every positive integer n�N; and, forevery x 2 E;

sn xð Þj j � f xð Þj j � sn � fj j xð Þ ¼ sn xð Þ � f xð Þj j\e1p|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus, for every positive integer n�N; sn � fj jp\e on E. Now, for every pos-itive integer n�N;

ZX

sn � fj jpdl� 0

�� ¼

ZX

sn � fj jpdl ¼ZE

sn � fj jpdlþZ

X�E

sn � fj jpdl

¼ZE

sn � fj jpdlþ 0 ¼ZE

sn � fj jpdl�ZE

edl ¼ l Eð Þð Þe;

so,

limn!1

ZX

sn � fj jpdl

0@ 1A ¼ 0:

■)Since U : Lp lð Þ ! C is a bounded linear functional on the normed linear space

Lp lð Þ; U : Lp lð Þ ! C is continuous. Since U : Lp lð Þ ! C is continuous, andlimn!1 sn ¼ f in Lp lð Þ; limn!1 U snð Þ ¼ U fð Þ: Since each sn : X ! C is mea-surable simple function, U snð Þ ¼

RX sn � g dl: Now, since


limn!1

ZX

sn � g dl ¼ limn!1

U snð Þ ¼ U fð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};we have

limn!1

ZX

sn � gð Þdl ¼ U fð Þ:


limn!1

ZX

sn � gð Þdl ¼ZX

f � gð Þdl:

For this purpose, we shall apply Theorem 1.136.Since limn!1 sn ¼ f uniformly on E, there exists a positive integer N such that

for every positive integer n�N;

snj j � fj j �ð Þ sn � fj j\1

on E, and hence sNj j � 1þ fj j on E, sNþ 1j j � 1þ fj j on E, sNþ 2j j � 1þ fj j on E,etc. Thus, sN � gj j � 1þ fj jð Þ gj j on E, sNþ 1 � gj j � 1þ fj jð Þ gj j on E,

sNþ 2 � gj j � 1þ fj jð Þ gj j

on E, etc. Since f 2 L1 lð Þ, fj j 2 L1 lð Þ: Now, since 1 2 L1 lð Þ; and L1 lð Þ is acomplex linear space, 1þ fj jð Þ 2 L1 lð Þ: Since g 2 L1 lð Þ, gj j 2 L1 lð Þ: Since1þ fj jð Þ 2 L1 lð Þ; and gj j 2 L1 lð Þ; 1þ fj jð Þ gj j 2 L1 lð Þ:Also, for every x in X,

limn!1

sn � gð Þ xð Þ ¼ limn!1

sn xð Þ� �

g xð Þð Þ ¼ f xð Þð Þ g xð Þð Þ ¼ f � gð Þ xð Þ;

and hence for every x in x 2 X;

limn!1

sn � gð Þ xð Þ ¼ f � gð Þ xð Þ:

Now, by applying Theorem1.136 on the sequence sN � g; sNþ 1 � g; sNþ 2 � g; . . .f g;we get

limn!1

ZX

sn � gð Þdl ¼ZX

f � gð Þdl:


Thus, for every f 2 L1 lð Þ;

U fð Þ ¼ZX

f � gð Þdl:

Conclusion 3.72 Let p 2 1;1½ Þ: Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ Þ be a positive measure onM: Let U : Lp lð Þ ! C

be a bounded linear functional on Lp lð Þ: Then there exists a function g : X ! C

such that g 2 L1 lð Þ; and for every f 2 L1 lð Þ;

U fð Þ ¼ZX

f � gð Þdl:

Lemma 3.73 Let X be any nonempty set. Let M be a r-algebra in X. Let l :

M ! 0;1½ Þ be a positive measure on M: Let U : L1 lð Þ ! C be a bounded linearfunctional on L1 lð Þ: Then there exists a function g : X ! C such that g 2 L1 lð Þ;and for every f 2 L1 lð Þ � L1 lð Þð Þ;

U fð Þ ¼ZX

f � gð Þdl:

Also, gk k1 � Uk k:

Proof Since U : L1 lð Þ ! C is a bounded linear functional on the normed linearspace L1 lð Þ, Uk k is a nonnegative real number. If U ¼ 0; then 0 serves the purposeof g.

So, we consider the case when U 6¼ 0:Since U 6¼ 0; and U : L1 lð Þ ! C is a bounded linear functional on the normed

linear space L1 lð Þ; Uk k[ 0:By Conclusion 3.72, there exists a function g : X ! C such that g 2 L1 lð Þ; and

for every f 2 L1 lð Þ; U fð Þ ¼RX f � gð Þdl: It suffices to show that g 2 L1 lð Þ; and

gk k1 � Uk k: By Conclusion 2.20, it suffices to show that g xð Þj j � Uk k holds a.e.on X, that is g xð Þ 2 D 0; Uk k½ � holds a.e. on X, where D 0; Uk k½ � denotes the closeddisk with center 0 and radius Uk k: Now, by Lemma 1.154, it suffices to show thatfor every E 2 M satisfying l Eð Þ[ 0;R

E g dll Eð Þ 2 D 0; Uk k½ �;


that is for every E 2 M satisfying l Eð Þ[ 0;RE g dl

�� l Eð Þ � Uk k:

For this purpose, let us fix any E 2 M satisfying l Eð Þ[ 0: Since E 2 M;vE 2 L1 lð Þ; and hence

U vEð Þ ¼ZX

vE � gð Þdl

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZE

g dl:

It follows that

ZE

g dl

�� ¼ U vEð Þj j � Uk k vEk k1¼ Uk k

ZX

vEj jdl ¼ Uk kZX

vEdl ¼ Uk kl Eð Þ;

and hence RE g dl

�� l Eð Þ � Uk k 2 0;1ð Þð Þ:

■

Lemma 3.74 Let X be any nonempty set. Let M be a r-algebra in X. Let l :

M ! 0;1½ Þ be a positive measure on M: Let U : L1 lð Þ ! C be a bounded linearfunctional on L1 lð Þ: Then there exists a function g : X ! C such that

1. g 2 L1 lð Þ;2. for every f 2 L1 lð Þ; U fð Þ ¼

RX f � gð Þdl;

3. Uk k ¼ gk k1:

Proof By Lemma 3.73, there exists a function g : X ! C such that g 2 L1 lð Þ; forevery f 2 L1 lð Þ � L1 lð Þð Þ;

U fð Þ ¼ZX

f � gð Þdl; and gk k1 � Uk k:


a. for every f 2 L1 lð Þ; U fð Þ ¼RX f � gð Þdl;

b. Uk k� gk k1:


For a: Let us take any f 2 L1 lð Þ: We have to show that

U fð Þ ¼ZX

f � gð Þdl:

Since L1 lð Þ contains all measurable simple functions on X, and byTheorem 2.47 the collection of all measurable simple functions on X is dense inL1 lð Þ; we find that L1 lð Þ is a dense subset of L1 lð Þ: Now, since f 2 L1 lð Þ; thereexists a sequence fnf g in L1 lð Þ such that limn!1 fn ¼ f in L1 lð Þ; and hence

U fð Þ ¼ U limn!1

fn� �

¼ limn!1

U fnð Þ ¼ limn!1

ZX

fn � gð Þdl:


limn!1RX fn � gð Þdl ¼

RX f � gð Þdl; that is, limn!1

RX fn � gð Þdl�

RX f � gð Þdl

� �¼ 0;

that is, limn!1RX fn � gð Þ � f � gð Þð Þdl

� �¼ 0; that is, limn!1

RX fn � fð Þ � g dl

� �¼

0; that is, limn!1RX fn � fð Þ � g dl

�� ¼ 0:It is enough to show that

limn!1

ZX

fn � fð Þ � gj jdl

0@ 1A ¼ 0:

SinceZX

fn � fð Þ � gj jdl ¼ fn � fð Þ � gk k1 � fn � fk k1 gk k1; and limn!1

fn � fk k1¼ 0;

we have

limn!1

ZX


0@ 1A ¼ 0:

For b: By a, for every f 2 L1 lð Þ;

U fð Þj j ¼ZX

f � gð Þdl

��

ZX

f � gj jdl ¼ f � gk k1 � fk k1 gk k1;


so, for every f 2 L1 lð Þ; U fð Þj j � gk k1 fk k1: Now, since U : L1 lð Þ ! C is abounded linear functional on L1 lð Þ;

Uk k� gk k1:

■

Lemma 3.75 Let p 2 1;1ð Þ: Let q 2 1;1ð Þð Þ be the exponent conjugate to p. LetX be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ Þ be apositive measure on M: Let U : Lp lð Þ ! C be a bounded linear functional onLp lð Þ: Then, there exists a function g : X ! C such that g 2 Lq lð Þ; and for everyf 2 L1 lð Þ � Lp lð Þð Þ;

U fð Þ ¼ZX

f � gð Þdl:

Also, gk kq � Uk k:

Proof Since U : Lp lð Þ ! C is a bounded linear functional on the normed linearspace Lp lð Þ; Uk k is a nonnegative real number. By Conclusion 3.72, there exists afunction g : X ! C such that g 2 L1 lð Þ; and for every f 2 L1 lð Þ;

U fð Þ ¼ZX

f � gð Þdl:

It suffices to show that g 2 Lq lð Þ:By Lemma 1.68, there exists a measurable function b : X ! C such that bj j ¼ 1;

and g ¼ b � gj j: Put a � 1b : Clearly, a : X ! C is a measurable function, aj j ¼ 1;

and gj j ¼ a � g:

Problem 3:76 For every positive integer n;

v x: gj j xð Þ� nf g gj jq�1a� �

2 L1 lð Þ:

(Solution It suffices to show that for every positive integer n,

t : t 2 0;1½ Þ; and l v x: gj j xð Þ� nf g gj jq�1a�� 1

t;1ð �ð Þ�

¼ 0� �

6¼ ;:


Since

v x: gj j xð Þ� nf g gj jq�1a�� 1

n q�1ð Þ;1� i� �

¼ y : v x: gj j xð Þ� nf g gj jq�1a�� yð Þ 2 n q�1ð Þ;1

� in o¼ y : v x: gj j xð Þ� nf g yð Þ g yð Þj jq�1a yð Þ

�� 2 n q�1ð Þ;1� in o

¼ y : n q�1ð Þ\ v x: gj j xð Þ� nf g yð Þ g yð Þj jq�1a yð Þ�� n o

¼ y : n q�1ð Þ\ v x: gj j xð Þ� nf g yð Þ� �

g yð Þj jq�1 a yð Þj jn o

¼ y : n q�1ð Þ\ v x: gj j xð Þ� nf g yð Þ� �

g yð Þj jq�11n o

¼ y : n q�1ð Þ\ v x: g xð Þj j � nf g yð Þ� �

g yð Þj jq�1n o

¼ ;;

we have

n q�1ð Þ 2 t : t 2 0;1½ Þ; and l v x: gj j xð Þ� nf g gj jq�1a�� 1

t;1ð �ð Þ�

¼ 0� �

;

and hence each

t : t 2 0;1½ Þ; and l v x: gj j xð Þ� nf g gj jq�1a�� 1

t;1ð �ð Þ�

¼ 0� �

6¼ ;:

■)

Problem 3:77 For every positive integer n; v x: gj j xð Þ� nf g gj jq�1a�� p¼ gj jq on

x : g xð Þj j � nf g:

(Solution Let us fix any positive integer n, and any y 2 X satisfying g yð Þj j � n: Wehave to show that

v x: gj j xð Þ� nf g gj jq�1a�� p� �

yð Þ ¼ gj jqð Þ yð Þ;

that is

v x: gj j xð Þ� nf g yð Þ g yð Þj jq�1a yð Þ�� p¼ g yð Þj jq:

LHS ¼ v x: gj j xð Þ� nf g yð Þ g yð Þj jq�1a yð Þ�� p¼ 1 � g yð Þj jq�1a yð Þ

�� p¼ g yð Þj jq�1� �p

a yð Þj jp¼ g yð Þj jpq�p a yð Þj jp¼ g yð Þj jq a yð Þj jp

¼ g yð Þj jq 1ð Þp¼ g yð Þj jq¼ RHS:

■)


It follows that, for every positive integer n,

0�ð ÞZ

x: gj j xð Þ� nf g

gj jqdl ¼ZX

v x: gj j xð Þ� nf g gj jqdl ¼ZX

v x: gj j xð Þ� nf g gj jq�1 gj jdl

¼ZX

v x: gj j xð Þ � nf g gj jq�1 a � gð Þdl ¼ZX

v x: gj j xð Þ� nf g gj jq�1a� �

g dl ¼ U v x: gj j xð Þ� nf g gj jq�1a� �

;

and hence for every positive integer n, U v x: gj j xð Þ� nf g gj jq�1a� �

is a nonnegative real,

and

Zx: gj j xð Þ� nf g

gj jqdl¼ U v x: gj j xð Þ� nf g gj jq�1a� �

¼ U v x: gj j xð Þ � nf g gj jq�1a� ��

� Uk k v x: gj j xð Þ� nf g gj jq�1a��

p¼ Uk k

ZX

v x: gj j xð Þ� nf g gj jq�1a�� pdl

0@ 1A1p

¼ Uk kZ


v x: gj j xð Þ � nf g gj jq�1a�� pdlþ Z

x:n\ gj j xð Þf g

v x: gj j xð Þ� nf g gj jq�1a�� pdl

0B@1CA

1p

¼ Uk kZ


v x: gj j xð Þ � nf g gj jq�1a�� pdlþ Z

x:n\ gj j xð Þf g

0 � gj jq�1a�� pdl

0B@1CA

1p

¼ Uk kZ


v x: gj j xð Þ � nf g gj jq�1a�� pdlþ 0

0B@1CA

1p

¼ Uk kZ

x: gj j xð Þ � nf g

gj jqdl

0B@1CA

1p

:

Thus, for every positive integer n,


gj jqdl

0B@1CA

1p Z


gj jqdl

0B@1CA

1q

¼Z


gj jqdl

0B@1CA

1pþ 1

q

¼Z


gj jqdl

0B@1CA

1

¼Z


gj jqdl� Uk kZ


gj jqdl

0B@1CA

1p

:




gj jqdl

0B@1CA

1p Z


gj jqdl

0B@1CA

1q

� Uk k

0BB@1CCA� 0;

that is


gj jqdl

0B@1CA

1q

0BB@1CCA

qp Z


gj jqdl

0B@1CA

1q

� Uk k

0BB@1CCA� 0;

and hence


gj jqdl

0B@1CA

1q

¼ 0; orZ


gj jqdl

0B@1CA

1q

� Uk k:

Now, since 0� Uk k; in all cases, for every positive integer n,

ZX

v x: gj j xð Þ� nf g gj jqdl

0@ 1A1q

¼Z


gj jqdl

0B@1CA

1q

� Uk k;

that is for every positive integer n,ZX

v x: gj j xð Þ� nf g gj jqdl� Uk kq \1ð Þ:

Now, since

v x: gj j xð Þ� 1f g gj jq � v x: gj j xð Þ� 2f g gj jq � v x: gj j xð Þ� 3f g gj jq � � � � ;

by Theorem 1.125,

Uk kq �ð Þ limn!1

ZX

v x: gj j xð Þ� nf g gj jqdl ¼ZX

limn!1

v x: gj j xð Þ� nf g gj jq� �

dl

¼ZX

limn!1

v x: gj j xð Þ� nf g

� �gj jqdl ¼

ZX

vX gj jqdl ¼ZX

gj jqdl;


and hence

RX gj jqdl

� �1q � Uk k\1: This shows that g 2 Lq lð Þ; and gk kq � Uk k: ■

Note 3.78 Let p 2 1;1ð Þ: Let q 2 1;1ð Þð Þ be the exponent conjugate to p. LetX be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ Þ be apositive measure on M: Let U : Lp lð Þ ! C be a bounded linear functional onLp lð Þ:

Problem 3.79 There exists a function g : X ! C such that1. g 2 Lq lð Þ; 2. for every f 2 Lp lð Þ; U fð Þ ¼

RX f � gð Þdl; 3. Uk k ¼ gk kq:

(Solution By Lemma 3.75, there exists a function g : X ! C such that g 2 Lq lð Þ;for every f 2 L1 lð Þ � Lp lð Þð Þ; U fð Þ ¼

RX f � gð Þdl; and gk kq � Uk k: It suffices to

show that

a. for every f 2 Lp lð Þ; U fð Þ ¼RX f � gð Þdl;

b. Uk k� gk kq;

For a: Let us take any f 2 Lp lð Þ: We have to show that U fð Þ ¼RX f � gð Þdl:

Since L1 lð Þ contains all measurable simple functions on X, and by Theorem 2.47,the collection of all measurable simple functions on X is dense in Lp lð Þ; we findthat L1 lð Þ is a dense subset of L1 lð Þ: Now, since f 2 Lp lð Þ; there exists asequence fnf g in L1 lð Þ such that limn!1 fn ¼ f in Lp lð Þ; and hence

U fð Þ ¼ U limn!1

fn� �

¼ limn!1

U fnð Þ ¼ limn!1

ZX

fn � gð Þdl:

It suffices to show thatlimn!1

RX fn � gð Þdl ¼

RX f � gð Þdl; that is, limn!1

RX fn � gð Þdl�

RX f � gð Þdl

� �¼ 0;

that is limn!1RX fn � gð Þ � f � gð Þð Þdl

� �¼ 0; that is, limn!1

RX fn � fð Þ � g dl

� �¼

0; that is limn!1RX fn � fð Þ � g dl

�� ¼ 0:It suffices to show that

limn!1

ZX


0@ 1A ¼ 0:

SinceZX

fn � fð Þ � gj jdl ¼ fn � fð Þ � gk k1 � fn � fk kp gk kq; and limn!1

fn � fk kp¼ 0;


we have

limn!1

ZX


0@ 1A ¼ 0:

For b: By a, for every f 2 Lp lð Þ;

U fð Þj j ¼ZX

f � gð Þdl

��

ZX

f � gj jdl ¼ f � gk k1 � fk kp gk kq;

so, for every f 2 Lp lð Þ; U fð Þj j � gk kq fk kp: Now, since U : Lp lð Þ ! C is abounded linear functional on Lp lð Þ; Uk k� gk kq: ■)

If we recollect the results of Lemma 3.74, and Conclusion 3.80, we get thefollowing

Conclusion 3.80 Let p 2 1;1½ Þ: Let q 2 1;1ð �ð Þ be the exponent conjugate top. Let X be any nonempty set. LetM be a r-algebra in X. Let l : M ! 0;1½ Þ be apositive measure on M: Let U : Lp lð Þ ! C be a bounded linear functional onLp lð Þ: Then there exists a function g : X ! C such that

1. g 2 Lq lð Þ;2. for every f 2 Lp lð Þ; U fð Þ ¼

RX f � gð Þdl;

3. Uk k ¼ gk kq:

Theorem 3.81 Let p 2 1;1ð Þ: Let q 2 1;1ð Þð Þ be the exponent conjugate top. Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ � be apositive measure on M: Suppose that X has r-finite measure. Let U : Lp lð Þ ! C

be a bounded linear functional on Lp lð Þ: The, there exists a function g : X ! C

such that

1. g 2 Lq lð Þ;2. for every f 2 Lp lð Þ; U fð Þ ¼

RX f � gð Þdl;

3. Uk k ¼ gk kq:

Proof Case I: when l Xð Þ\1: Conclusion 3.80 proves the theorem.

Case II: l Xð Þ ¼ 1: By Conclusion 3.32, there exists a measurable functionw : X ! 0; 1ð Þ such that

RX w dl ¼

� � RX wj jdl 2 0; 1½ �: Thus, for every E 2 M;R

E w dl�� R

X w dl 2 0; 1½ �; and w 2 L1 lð Þ: This shows that ~l : E 7!RE w dl is a

mapping from M to 0; 1½ �: By Lemma 1.131, ~l : M ! 0; 1½ � is a positive measure.Let us take any F 2 Lp ~lð Þ: It follows that

RX Fj jpd~l\1: We shall try to show

that w1pF 2 Lp lð Þ; that is


ZX

w Fj jpdl ¼ZX

w1pF

�� pdl\1

|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl};

that isRX w Fj jpdl\1: By Lemma 1.132,

1[ð ÞZX

Fj jpd~l ¼ZX

Fj jp�wð Þdl;

and henceRX w Fj jpdl\1: Thus, Kp : F 7!w

1pF is a mapping from normed linear

space Lp ~lð Þ to normed linear space Lp lð Þ:

Problem 3:82 Kp is linear.

(Solution Let us take any F;G 2 Lp ~lð Þ; and a; b 2 C: We have to show that

w1p � aFþ bGð Þ ¼ a w

1pF

� �þ b w

1pG

� �: This is clearly true. ■)

Problem 3:83 Kp : Lp ~lð Þ ! Lp lð Þ is 1-1, and onto.

(Solution 1-1 ness: Let w1pF ¼ w

1pG; where F;G 2 Lp ~lð Þ: We have to show that

F ¼ G: Since w1pF ¼ w

1pG, w

1p F � Gð Þ ¼ 0. Since w : X ! 0; 1ð Þ; for every x 2 X,

w1p

� �xð Þ 6¼ 0: Now, since w

1p F � Gð Þ ¼ 0, F � Gð Þ ¼ 0; and hence F ¼ G:

Onto-ness: Let us take any G 2 Lp lð Þ: It suffices to show that w�1pG 2 Lp ~lð Þ;

that isZX

Gj jpdl ¼ZX

1w

Gj jp�

� w dl ¼ZX

1w

Gj jp�

d~l ¼ZX

w�1pG

�� pd~l\1

|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that isRX Gj jpdl\1: This is true, because G 2 Lp lð Þ: ■)

Let F 2 Lp ~lð Þ: Clearly,

ZX

Fj jp�w dl

0@ 1A1p

¼ZX

Fj jpd~l

0@ 1A1p

¼ Fk kp¼ w1pF

�� p¼

ZX

w1pF

�� pdl0@ 1A1

p

¼ZX

w � Fj jpdl

0@ 1A1p

:


Thus, Kp : F 7!w1pF is an isometry from normed linear space Lp ~lð Þ onto normed

linear space Lp lð Þ: Similarly, Kq : F 7!w1qF is an isometry from normed linear

space Lq ~lð Þ onto normed linear space Lq lð Þ: Now, since U : Lp lð Þ ! C;

U � Kp� �

: F 7! U w1pF

� �is a mapping from Lp ~lð Þ to C: Next, since Kp : Lp ~lð Þ ! Lp lð Þ is an isometry, andU : Lp lð Þ ! C is a bounded linear functional, U � Kp

� �: Lp ~lð Þ ! C is a bounded

linear functional. Clearly, U � Kp

�� ¼ Uk k:Since ~l : M ! 0;1½ Þ; and U � Kp

� �: Lp ~lð Þ ! C is a bounded linear func-

tional, by Case I, there exists a function ~g : X ! C such that

a. ~g 2 Lq ~lð Þ;b. for every f 2 Lp ~lð Þ; U � Kp

� �fð Þ ¼

RX f � ~gð Þd~l;

c. Uk k ¼ð Þ U � Kp� �� ¼ ~gk kq:

Since ~g 2 Lq ~lð Þ; w1q~g 2 Lq lð Þ: Put g � w

1q~g ¼ Kq ~gð Þ� �

: Thus, g 2 Lq lð Þ: Next,let us take any f 2 Lp lð Þ: We have to show that U fð Þ ¼

RX f � gð Þdl: Since f 2

Lp lð Þ; w�1pf ¼

� �Kp� ��1

fð Þ 2 Lp ~lð Þ; and hence, by b,

U fð Þ ¼ U � Kp� �

Kp� ��1

fð Þ� �

¼ U � Kp� �

w�1pf

� �¼ZX

w�1pf � ~g

� �d~l

¼ZX

w�1pf � ~g

� �� w dl ¼

ZX

w1qf � ~g

� �dl ¼

ZX

f � w1q~g

� �� dl ¼

ZX

f � gð Þdl:

Thus,U fð Þ ¼RX f � gð Þdl: Since gk kq¼ Kq ~gð Þ

�� q¼ ~gk kq; by (c), gk kq¼ Uk k:■

Theorem 3.84 Let X be any nonempty set. Let M be a r-algebra in X. Let l :M ! 0;1½ � be a positive measure onM: Suppose that X has r-finite measure. LetU : L1 lð Þ ! C be a bounded linear functional on L1 lð Þ: Then, there exists afunction g : X ! C such that

1. g 2 L1 lð Þ;2. for every f 2 L1 lð Þ; U fð Þ ¼

RX f � gð Þdl;

3. Uk k ¼ gk k1:

Proof Case I: when l Xð Þ\1: Conclusion 3.80 proves the theorem.

Case II: l Xð Þ ¼ 1: By Conclusion 3.32, there exists a measurable functionw : X ! 0; 1ð Þ such that

RX w dl ¼

� � RX wj jdl 2 0; 1½ �: Thus, for every E 2 M;R

E w dl�� R

X w dl 2 0; 1½ �; and w 2 L1 lð Þ: This shows that ~l : E 7!RE w dl is a


mapping from M to 0; 1½ �: By Lemma 1.131, l : M ! 0; 1½ � is a positive measure.Since w : X ! 0; 1ð Þ; for every E 2 M; ~l Eð Þ� l Eð Þ: It follows thatL1 lð Þ � L1 ~lð Þ:

Problem 3:85 L1 ~lð Þ � L1 lð Þ:(Solution Let f 2 L1 ~lð Þ: We have to show that f 2 L1 lð Þ: Since f 2 L1 ~lð Þ;there exists k 2 0;1ð Þ such thatZ

fj j�1 k;1ð �ð Þ

w dl ¼ ~l fj j�1 k;1ð �ð Þ� �

¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since w : X ! 0; 1ð Þ is a measurable function, and l Xð Þ 6¼ 0, w ¼ 0 a.e. on X is

false. Since w ¼ 0 a.e. on X is false, andR

fj j�1 k;1ð �ð Þ w dl ¼ 0; we have

l fj j�1 k;1ð �ð Þ� �

¼ 0; and hence f 2 L1 lð Þ: ■)

Thus, L1 ~lð Þ ¼ L1 lð Þ:Let us take any F 2 L1 ~lð Þ: It follows that

RX Fj j1d~l\1: We shall try to show

that wF 2 L1 lð Þ; that is ZX

w Fj jdl ¼ZX

wFj jdl\1;

|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}that is

RX w Fj jdl\1: By Lemma 1.132, 1[ð Þ

RX Fj jd~l ¼

RX Fj j � wð Þdl; and

henceRX w Fj jdl\1: Thus, K1 : F 7!wF is a mapping from normed linear space

L1 ~lð Þ to normed linear space L1 lð Þ:

Problem 3:86 K1 is linear.

(Solution Let us take any F;G 2 L1 ~lð Þ; and a; b 2 C: We have to show thatw � aFþ bGð Þ ¼ a wFð Þþ b wGð Þ: This is clearly true. ■)

Problem 3:87 K1 : L1 ~lð Þ ! L1 lð Þ is 1-1, and onto.

(Solution 1-1 ness: Let wF ¼ wG; where F;G 2 L1 ~lð Þ: We have to show thatF ¼ G: Since wF ¼ wG, w F � Gð Þ ¼ 0. Since w : X ! 0; 1ð Þ; for every x 2 X;w xð Þ 6¼ 0: Now, since w F � Gð Þ ¼ 0, F � Gð Þ ¼ 0; and hence F ¼ G:

Onto-ness: Let us take any G 2 L1 lð Þ: It suffices to show that 1wG 2 L1 ~lð Þ; that is

ZX

Gj jdl ¼ZX

1w

Gj j�

� w dl ¼ZX

1w

Gj j�

d~l ¼ZX

1wG

�� d~l\1

|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl};


that isRX Gj jdl\1: This is true, because G 2 L1 lð Þ: ■)

Let F 2 L1 ~lð Þ: Clearly,ZX

Fj j � w dl ¼ZX

Fj jd~l ¼ Fk k1¼ wFk k1¼ZX

wFj jdl ¼ZX

w � Fj jdl:

Thus, K1 : F 7!wF is an isometry from normed linear space L1 ~lð Þ onto normedlinear space L1 lð Þ:

Now, since U : L1 lð Þ ! C, U � K1ð Þ : F 7!U wFð Þ is a mapping from L1 ~lð Þ toC: Now, since K1 : L1 ~lð Þ ! L1 lð Þ is an isometry, and U : L1 lð Þ ! C is a boundedlinear functional, U � K1ð Þ : L1 ~lð Þ ! C is a bounded linear functional. Clearly,U � K1k k ¼ Uk k:Since ~l : M ! 0;1½ Þ; and U � K1ð Þ : L1 ~lð Þ ! C is a bounded linear func-

tional, by Case I, there exists a function ~g : X ! C such that

a. ~g 2 L1 ~lð Þ ¼ L1 lð Þð Þ;b. for every f 2 L1 ~lð Þ, U � K1ð Þ fð Þ ¼

RX f � ~gð Þd~l;

c. Uk k ¼ð Þ U � K1ð Þk k ¼ ~gk k1:

Put g ¼ ~g: We get

I. g 2 L1 lð Þ;II. for every f 2 L1 ~lð Þ, U � K1ð Þ fð Þ ¼

RX f � gð Þd~l;

III. Uk k ¼ð Þ U � K1ð Þk k ¼ gk k1:

Thus, it suffices to show that for every f 2 L1 lð Þ;

U fð Þ ¼ZX

f � gð Þdl:

For this purpose, let us take any f 2 L1 lð Þ: We have to show that U fð Þ ¼RX f � gð Þdl: Since f 2 L1 lð Þ, 1

w f 2 L1 ~lð Þ; and hence, by II,

U fð Þ ¼ U w1wf

� � ¼ U K1

1wf

� � ¼ U � K1ð Þ 1

wf

� ¼ZX

1wf � g

� d~l

¼ZX

1wf � g

� � w dl ¼

ZX

f � gð Þdl:

Thus, U fð Þ ¼RX f � gð Þdl: ■

If we recollect the results of Theorem 3.81, and Theorem 3.84, we get thefollowing


Theorem 3.88 Let p 2 1;1½ Þ: Let q 2 1;1ð �ð Þ be the exponent conjugate to p. LetX be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ � be apositive measure on M: Suppose that X has r-finite measure. Let U : Lp lð Þ ! C

be a bounded linear functional on Lp lð Þ: Then there exists a unique functiong : X ! C such that

1. g 2 Lq lð Þ;2. for every f 2 Lp lð Þ, U fð Þ ¼

RX f � gð Þdl;

3. Uk k ¼ gk kq:

Proof of the remaining part Uniqueness: If not, otherwise, let g : X ! C andh : X ! C be functions satisfying

1. g; h 2 Lq lð Þ;2. for every f 2 Lp lð Þ, U fð Þ ¼

RX f � gð Þdl; and U fð Þ ¼

RX f � hð Þdl

3. Uk k ¼ gk kq; and Uk k ¼ hk kq:4. g ¼ h a.e. on X is false.

We have to arrive at a contradiction. From 2, for every f 2 Lp lð Þ,RX f �

g� hð Þdl ¼ 0: It follows that, for every E 2 M;ZE

g� hð Þdl ¼ZX

vE � g� hð Þdl ¼ 0;

and hence for every E 2 M,RE g� hð Þdl ¼ 0: It follows that g� h ¼ 0 a.e. on X,

that is, g ¼ h a.e. on X. This contradicts 4. ■

Theorem 3.89 Let p 2 1;1½ Þ: Let q 2 1;1ð �ð Þ be the exponent conjugate to p. LetX be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ � be apositive measure on M: Suppose that X has r-finite measure. Then the dual spaceLp lð Þð Þ� of the normed linear space Lp lð Þ is isometrically isomorphic to thenormed linear space Lq lð Þ:

In short, we say that the dual space of Lp lð Þ is Lq lð Þ:

Proof Let us take any U 2 Lp lð Þð Þ�: It follows that U : Lp lð Þ ! C is a boundedlinear functional on Lp lð Þ: By Theorem 3.88, there exists a unique function gU :X ! C such that

1. gU 2 Lq lð Þ;2. for every f 2 Lp lð Þ, U fð Þ ¼

RX f � gUð Þdl;

3. Uk k ¼ gUk kq:

It follows that g : U 7! gU from Lp lð Þð Þ� to Lq lð Þ is a mapping that preservesnorms.


Problem 3:90 g : Lp lð Þð Þ�! Lq lð Þ is linear.

(Solution Let U;W 2 Lp lð Þð Þ�; and a; b 2 C: We have to show that g aUþ bWð Þ ¼agU þ bgW a.e. on X, that is

gðaUþ bWÞ � agU � bgW ¼ 0 a:e: onX:

For this purpose, it suffices to show that for every E 2 M;ZE

g aUþbWð Þ � agU � bgW� �

dl ¼ 0;

that is for every E 2 M;ZE

g aUþ bWð Þ� �

dl ¼ aZE

gUdlþ bZE

gWdl:

From 2, for every E 2 M, U vEð Þ ¼RE gUdl, W vEð Þ ¼

RE gWdl; and

aUþ bWð Þ vEð Þ ¼RE g aUþbWð Þdl: Thus, it suffices to show that, for every E 2 M;

aUþ bWð Þ vEð Þ ¼ aU vEð Þþ bW vEð Þ: This is trivially true. ■)It is clear that g : Lp lð Þð Þ�! Lq lð Þ is 1-1.

(Reason Let gU ¼ gW; where U;W 2 Lp lð Þð Þ�: We have to show that U ¼ W: Forthis purpose, let us take any f 2 Lp lð Þ: We have to show thatZ

X

f � gUð Þdl ¼ U fð Þ ¼ W fð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ZX

f � gUð Þdl;

that is,RX f � gUð Þdl ¼

RX f � gUð Þdl: This is true, because gU ¼ gW:)

Problem 3:91 g : Lp lð Þð Þ�! Lq lð Þ is onto.

(Solution Let us take any F 2 Lq lð Þ: Let U : g 7!RX g � Fð Þdl be the function

from Lp lð Þ to C:

Problem 3.92 U : Lp lð Þ ! C is linear.

(Solution Let g; h 2 Lp lð Þ; and a; b 2 C: We have to show thatZX

agþ bhð Þ � Fð Þdl ¼ aZX

g � Fð Þdlþ bZX

h � Fð Þdl:


LHS ¼ZX

agþ bhð Þ � Fð Þdl ¼ZX

a g � Fð Þþ b h � Fð Þð Þdl

¼ aZX

g � Fð Þdlþ bZX

h � Fð Þdl ¼ RHS:

■)Let g 2 Lp lð Þ; where gk kp � 1: Now

U gð Þj j ¼ZX

g � Fð Þdl

��

ZX

g � Fj jdl ¼ g � Fk k1 � gk kp Fk kq � 1 Fk kq

¼ Fk kq \1ð Þ:

Thus, for every g 2 Lp lð Þ satisfying gk kp � 1; U gð Þj j � Fk kq \1ð Þ; and henceU : Lp lð Þ ! C is bounded. Thus, U 2 Lp lð Þð Þ�: It suffices to show that gU ¼ F a.e.on X.

By the definition of gU; for every f 2 Lp lð Þ,RX f � Fð Þdl ¼

� �U fð Þ ¼R

X f � gUð Þdl; and hence for every f 2 Lp lð Þ;

ZX

f � F � gUð Þð Þdl ¼ZX

f � Fð Þ � f � gUð Þð Þdl ¼ 0

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:

Thus, for every E 2 M,RE F � gUð Þdl ¼

ZXvE � F � gUð Þð Þdl ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}; and hence

for every E 2 M;RE F � gUð Þdl ¼ 0: It follows that F � gUð Þ ¼ 0 a.e. on X, that is

gU ¼ F a.e. on X. ■)Thus, g : U 7! gU from Lp lð Þð Þ� to Lq lð Þ is a 1-1, onto mapping that preserves

norms. Hence

g : Lp lð Þð Þ�! Lq lð Þ

is an isometric isomorphic. It follows that Lp lð Þð Þ� is isometrically isomorphic toLq lð Þ: ■


3.4 Lebesgue Points

Lemma 3.93 Let X be any nonempty set. Let M be a r-algebra in X. Let l :M ! C be a complex measure on M: Let A1;A2;A3; . . .f g be a countable col-lection of members in M such that A1 � A2 � A3 � � � � : Then,limn!1 l Anð Þ ¼ l A1 [A2 [A3 [ � � �ð Þ:

Proof

Problem 3:94 Re lð Þ : M ! R is a signed measure on M:

(Solution For this purpose, let us take any E 2 M; and a partition E1;E2;E3; . . .f gof E. We have to show that

Re lð Þð Þ Eð Þ ¼ Re lð Þð Þ E1ð Þþ Re lð Þð Þ E2ð Þþ Re lð Þð Þ E3ð Þþ � � � :

Since l : M ! C is a complex measure on M, E 2 M; and E1;E2;E3; . . .f g isa partition of E, we have

Re lð Þð Þ Eð Þþ i Im lð Þð Þ Eð Þð Þ ¼ l Eð Þ ¼ l E1ð Þþ l E2ð Þþ l E3ð Þþ � � �¼ Re lð Þð Þ E1ð Þþ i Im lð Þð Þ E1ð Þð Þð Þþ Re lð Þð Þ E2ð Þþ i Im lð Þð Þ E2ð Þð Þð Þþ Re lð Þð Þ E3ð Þþ i Im lð Þð Þ E3ð Þð Þð Þþ � � � ¼ Re lð Þð Þ E1ð Þþ Re lð Þð Þ E2ð Þþ Re lð Þð Þ E3ð Þþ � � �ð Þþ i Im lð Þð Þ E1ð Þþ Im lð Þð Þ E2ð Þþ Im lð Þð Þ E3ð Þþ � � �ð Þ;

and hence

Re lð Þð Þ Eð Þ ¼ Re lð Þð Þ E1ð Þþ Re lð Þð Þ E2ð Þþ Re lð Þð Þ E3ð Þþ � � � :

■)Similarly, Im lð Þ : M ! R is a signed measure on M: It follows that

Re lð Þ ¼ Re lð Þð Þþ� Re lð Þð Þ�;

where Re lð Þð Þþ ; Re lð Þð Þ� are positive measures on M:Also,

Im lð Þ ¼ Im lð Þð Þþ� Im lð Þð Þ�;

where Im lð Þð Þþ ; Im lð Þð Þ� are positive measures on M:Thus,

l ¼ Re lð Þð Þþ� Re lð Þð Þ��

þ i Im lð Þð Þþ� Im lð Þð Þ��

:

3.4 Lebesgue Points 457

LHS ¼ limn!1

l Anð Þ ¼ limn!1

Re lð Þð Þþ Anð Þ � Re lð Þð Þ� Anð Þ� ��

þ i Im lð Þð Þþ Anð Þ � Im lð Þð Þ� Anð Þ� ��

¼ limn!1

Re lð Þð Þþ Anð Þ� ��

� limn!1

Re lð Þð Þ� Anð Þð Þ�þ i lim

n!1Im lð Þð Þþ Anð Þ

� �� lim

n!1Im lð Þð Þ� Anð Þð Þ

� �¼ Re lð Þð Þþ A1 [A2 [A3 [ � � �ð Þ � Re lð Þð Þ� A1 [A2 [A3 [ � � �ð Þ� �þ i Im lð Þð Þþ A1 [A2 [A3 [ � � �ð Þ � Im lð Þð Þ� A1 [A2 [A3 [ � � �ð Þ� �

¼ l A1 [A2 [A3 [ � � �ð Þ ¼ RHS:

■

Lemma 3.95 Let X be any nonempty set. Let M be a r-algebra in X. Let l :M ! C be a complex measure on M: Let A1;A2;A3; . . .f g be a countable col-lection of members in M such that A1 A2 A3 � � � : Thenlimn!1 l Anð Þ ¼ l A1 \A2 \A3 \ � � �ð Þ:

Proof Since l : M ! C is a complex measure on M, Re lð Þ : M ! R; andIm lð Þ : M ! R are signed measures on M:

It follows that Re lð Þ ¼ Re lð Þð Þþ� Re lð Þð Þ�; where Re lð Þð Þþ ; Re lð Þð Þ� arepositive measures on M: Also, Im lð Þ ¼ Im lð Þð Þþ� Im lð Þð Þ�; whereIm lð Þð Þþ ; Im lð Þð Þ� are positive measures on M: Thus,

l ¼ Re lð Þð Þþ� Re lð Þð Þ��

þ i Im lð Þð Þþ� Im lð Þð Þ��

:

LHS ¼ limn!1

l Anð Þ ¼ limn!1

Re lð Þð Þþ Anð Þ � Re lð Þð Þ� Anð Þ� ��

þ i Im lð Þð Þþ Anð Þ � Im lð Þð Þ� Anð Þ� ��¼ lim

n!1Re lð Þð Þþ Anð Þ

� �� lim

n!1Re lð Þð Þ� Anð Þð Þ

� �þ i lim

n!1Im lð Þð Þþ Anð Þ

� �� lim

n!1Im lð Þð Þ� Anð Þð Þ

� �¼ Re lð Þð Þþ A1 \A2 \A3 \ � � �ð Þ�

� Re lð Þð Þ� A1 \A2 \A3 \ � � �ð ÞÞþ i Im lð Þð Þþ A1 \A2 \A3 \ � � �ð Þ�

� Im lð Þð Þ� A1 \A2 \A3 \ � � �ð ÞÞ ¼ l A1 \A2 \A3 \ � � �ð Þ ¼ RHS:

■

Note 3.96 Let M be the r-algebra of Borel sets in R: Let l : M ! C be acomplex measure on M:

(For every x 2 R; �1; xð Þ is an open set in R; and hence �1; xð Þ is a Borel setin R: Now, since M is the r-algebra of Borel sets in R; and l : M ! C; for everyx 2 R; l �1; xð Þð Þ 2 C:Þ


Let f : x 7! l �1; xð Þð Þ be the function from R to C: Let a 2 R; and A 2 C: Let

f 0 að Þ ¼ A (that is limx!af xð Þ�f að Þ

x�a ¼ 0; that is for every e[ 0; there exists d[ 0such that

0\ x� aj j\d ) f xð Þ � f að Þx� a

� A

�� \e;

that is, for every e[ 0; there exists d[ 0 such that

0\ x� aj j\d ) f xð Þ � f að Þ � A x� að Þj j\e x� aj jÞ:

Let us fix any e[ 0:Since f 0 að Þ ¼ A; there exists d[ 0 such that

x� aj j\d ) f xð Þ � f að Þ � A x� að Þj j � e2x� aj j:

Let s; tð Þ be an open interval that contains a, and m s; tð Þð Þ ¼ð Þ t � sð Þ\d: Thuss\a\t:

Problem 3.97 l s;tð Þð Þm s;tð Þð Þ � A�� \e; that is, l s; tð Þð Þ � A t � sð Þj j\e t � sð Þ:

(Solution We can find a sequence snf g such that s\ � � �\s2\s1\a \tð Þ; andlimn!1 sn ¼ s: It follows that s1; t½ Þ � s2; t½ Þ � s3; t½ Þ � � � � : Since each sn; t½ Þ is aBorel set in R; and M is the r-algebra of Borel sets in R; each sn; t½ Þ 2 M: Now,since l : M ! C is a complex measure on M; by Lemma 3.93

limn!1

l sn; t½ Þð Þ ¼ l s1; t½ Þ [ s2; t½ Þ [ s3; t½ Þ [ � � �ð Þ ¼ l s; tð Þð Þð Þ;

and hencelimn!1 l sn; t½ Þð Þ ¼ l s; tð Þð Þ: Similarly, limn!1 m sn; t½ Þð Þ ¼ m s; tð Þð Þ: It follows

that

l s; tð Þð Þ � A t � sð Þ ¼ l s; tð Þð Þ � A � m s; tð Þð Þ ¼ limn!1

l sn; t½ Þð Þ � A � m sn; t½ Þð Þð Þ

¼ limn!1

l sn; t½ Þð Þ � A � t � snð Þð Þ ¼ limn!1

l sn; t½ Þð Þ � A t � limn!1

sn� �

¼ limn!1

l sn; t½ Þð Þ � A t � sð Þ ¼ limn!1

l �1; tð Þ � �1; snð Þð Þ � A t � sð Þ

¼ limn!1

l �1; tð Þð Þ � l �1; snð Þð Þð Þ � A t � sð Þ ¼ limn!1

f tð Þ � f snð Þð Þ � A t � sð Þ

¼ f tð Þ � f að Þ � A t � að Þð Þ � limn!1

f snð Þ � f að Þ � A sn � að Þð Þ;


and hence

l s; tð Þð Þ � A t � sð Þj j ¼ f tð Þ � f að Þ � A t � að Þð Þ � limn!1

f snð Þ � f að Þ � A sn � að Þð Þ��

� f tð Þ � f að Þ � A t � að Þj j þ limn!1

f snð Þ � f að Þ � A sn � að Þð Þ��

� e2t � aj j þ lim

n!1f snð Þ � f að Þ � A sn � að Þð Þ

�� ¼ e2t � aj j

þ limn!1

f snð Þ � f að Þ � A sn � að Þj j � e2t � aj j þ lim

n!1

e2sn � aj j

¼ e2t � aj j þ e

2limn!1

sn � a�� ¼ e

2t � aj j þ e

2s� aj j

¼ e2

t � aj j þ s� aj jð Þ ¼ e2

t � að Þþ a� sð Þð Þ ¼ e2

t � sð Þ\e t � sð Þ:

Thus, l s; tð Þð Þ � A t � sð Þj j\e t � sð Þ: ■)

Conclusion 3.98 Let M be the r-algebra of Borel sets in R: Let l : M ! C be acomplex measure on M: Let f : x 7! l �1; xð Þð Þ be the function from R to C: Leta 2 R; and A 2 C: Let f 0 að Þ ¼ A Then for every e[ 0; there exists d[ 0 such thatfor every open interval I that contains a; and mðIÞ\d;

l Ið Þm Ið Þ � A

�� \e:

Note 3.99 Let M be the r-algebra of Borel sets in R: Let l : M ! C be acomplex measure on M: Let f : x 7! l �1; xð Þð Þ be the function from R to C: Leta 2 R; and A 2 C: Suppose that, for every e[ 0; there exists d[ 0 such that forevery open interval I that contains a, and m Ið Þ\d;


�� \e:

Problem 3.100 f 0 að Þ ¼ A:

(Solution We first try to show that l af gð Þ ¼ 0: If not, otherwise, let l af gð Þ 6¼ 0:We have to arrive at a contradiction. Since l af gð Þ 6¼ 0, l af gð Þj j[ 0: By theassumption, there exists d[ 0 such that for every open interval I that contains a,

and m Ið Þ\d; l Ið Þm Ið Þ � A�� \ l af gð Þj j: It follows that, for every positive integer n[ 1;

l a� d2n

; aþ d2n

� � � A � m a� d

2n; aþ d

2n

� � �� \ l af gð Þj j � m a� d

2n; aþ d

2n

� � ¼ l af gð Þj j 2d

2n

� ;


and hence

0� l af gð Þj j ¼ l af gð Þ � A � 0j j ¼ l af gð Þ � A � m af gð Þj j

¼ l \1n¼1 a� d

2n; aþ d

2n

� � � A � m \1

n¼1 a� d2n

; aþ d2n

� � �� ¼ l \1

n¼1 a� d2n

; aþ d2n

� � � A lim

n!1m a� d

2n; aþ d

2n

� � �� ¼ lim

n!1l a� d

2n; aþ d

2n

� � � A lim

n!1m a� d

2n; aþ d

2n

� � �� ¼ lim

n!1l a� d

2n; aþ d

2n

� � � A � m a� d

2n; aþ d

2n

� � � �� ¼ lim

n!1l a� d

2n; aþ d

2n

� � � A � m a� d

2n; aþ d

2n

� � �� limn!1

l af gð Þj j 2d2n

¼ 0ð Þ:

Thus, l af gð Þj j ¼ 0; that is l af gð Þ ¼ 0: This is a contradiction.Now, we shall try to show that f 0 að Þ ¼ A: For this purpose, let us take any e[ 0:

By the assumption, there exists d[ 0 such that for every open interval I that

contains a, and m Ið Þ\d; l Ið Þm Ið Þ � A�� \e: Let us take any x 2 a� d

2 ; aþ d2

� �where

x 6¼ a: It suffices to show that f xð Þ�f að Þx�a � A

�� \e:

Case I: when x 2 a� d2 ; a

� �: Here, for every positive integer n,

l x� x� a�d2ð Þ

2n ; aþ d2nþ 1

� � m x� x� a�d

2ð Þ2n ; aþ d

2nþ 1

� � � A

��\

e2;

that is for every positive integer n;

l x�x� a� d

2

� �2n

; aþ d2nþ 1

� � � A � m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � �� \

e2� m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � ;


and hence

f að Þ � f xð Þ � A a� xð Þj j ¼ l �1; að Þð Þ � l �1; xð Þð Þ � A a� xð Þj j¼ l �1; að Þ � �1; xð Þð Þ � A a� xð Þj j ¼ l x; a½ Þð Þ � A a� xð Þj j ¼ l x; a½ Þð Þþ 0� A � m x; a½ �ð Þj j¼ l x; a½ Þð Þþ l af gð Þ � A � m x; a½ �ð Þj j ¼ l x; a½ Þ [ af gð Þ � A � m x; a½ �ð Þj j ¼ l x; a½ �ð Þ � A � m x; a½ �ð Þj j

¼ l \1n¼1 x�

x� a� d2

� �2n

; aþ d2nþ 1

� � � A � m \1

n¼1 x�x� a� d

2

� �2n

; aþ d2nþ 1

� � �� ¼ lim

n!1l x�

x� a� d2

� �2n

; aþ d2nþ 1

� � � A lim

n!1m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � �� ¼ lim

n!1l x�

x� a� d2

� �2n

; aþ d2nþ 1

� � � A � m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � � �� ¼ lim

n!1l x�

x� a� d2

� �2n

; aþ d2nþ 1

� � � A � m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � �� lim

n!1

e2� m x�

x� a� d2

� �2n

; aþ d2nþ 1

� � ¼ e

2� limn!1

m x�x� a� d

2

� �2n

; aþ d2nþ 1

� � ¼ e

2� m \1

n¼1 x�x� a� d

2

� �2n

; aþ d2nþ 1

� � ¼ e

2� m x; a½ �ð Þ ¼ e

2� a� xð Þ ¼ e

2� a� xj j\e a� xj j:

Thus,

f xð Þ � f að Þx� a

� A

�� ¼ f að Þ � f xð Þ � A a� xð Þa� x

�� \e; and hence;f xð Þ � f að Þ

x� a� A

�� \e:

Case II: when x 2 a; aþ d2

� �: Here, for every positive integer n,

l a� d2nþ 1 ; x

� �� m a� d

2nþ 1 ; x� �� A

��\ e

2;

that is for every positive integer n,

l a� d2nþ 1 ; x

� � � A � m a� d

2nþ 1 ; x

� � �� \ e2� m a� d

2nþ 1 ; x

� � ;

and hence

f xð Þ � f að Þ � A x� að Þj j ¼ f xð Þ � f að Þ � A � m a; x½ Þð Þj j¼ l �1; xð Þð Þ � l �1; að Þð Þ � A � m a; x½ Þð Þj j¼ l �1; xð Þ � �1; að Þð Þ � A � m a; x½ Þð Þj j ¼ l a; x½ Þð Þ � A � m a; x½ Þð Þj j


¼ l \1n¼1 a� d

2nþ 1; x

� � � A � m \1

n¼1 a� d2nþ 1

; x

� � �� ¼ lim

n!1l a� d

2nþ 1 ; x

� � � A � lim

n!1m a� d

2nþ 1 ; x

� � �� ¼ lim

n!1l a� d

2nþ 1 ; x

� � � A � m a� d

2nþ 1 ; x

� � � �� ¼ lim

n!1l a� d

2nþ 1; x

� � � A � m a� d

2nþ 1; x

� � �� lim

n!1

e2� m a� d

2nþ 1 ; x

� � ¼ e

2� limn!1

m a� d2nþ 1 ; x

� � ¼ e

2� m \1

n¼1 a� d2nþ 1 ; x

� � ¼ e � m a; x½ Þð Þ ¼ e � x� aj j:

Thus, f xð Þ�f að Þx�a � A

�� \e: So, in all cases, f xð Þ�f að Þx�a � A

�� \e: This shows that

f 0 að Þ ¼ A: ■)If we combine this result with Conclusion 3.98, we get the following

Conclusion 3.101 Let M be the r-algebra of Borel sets in R: Let l : M ! C be acomplex measure on M: Let f : x 7! l �1; xð Þð Þ be the function from R to C: Leta 2 R; and A 2 C: Then f 0 að Þ ¼ A if and only if for every e[ 0; there exists d[ 0such that for every open interval I that contains a, and m Ið Þ\d;


�� \e:

Note 3.102

Definition Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let a 2 Rk; and r be a positive real number. The setx : x 2 Rk and x� aj j\r�

is denoted by B a; rð Þ; and is called an open ball withcenter a and radius r.

a. Let 0\r1 � r2: Let the intersection of two open balls B a1; r1ð Þ and B a2; r2ð Þ benonempty.

Problem 3.103 B a1; r1ð Þ � B a2; 3r2ð Þ:

(Solution Let us take any x 2 B a1; r1ð Þ: We have to show that x 2 B a2; 3r2ð Þ; thatis x� a2j j\3r2: Since x 2 B a1; r1ð Þ, x� a1j j\r1: From the assumption, thereexists y 2 Rk such that y� a1j j\r1; and y� a2j j\r2: Hence

x� a2j j � x� a1j j þ a1 � yj j þ y� a2j j\r1 þ y� a1j j þ y� a2j j\r1 þ r1 þ y� a2j j ¼ 2r1 þ y� a2j j\2r1 þ r2 � 2r2 þ r2 ¼ 3r2;

and hence x� a2j j\3r2: ■)


b. Let 0\r; and a 2 C: Let m be the Lebesgue measure on Rk:

Problem 3.104 m B a; 3rð Þð Þ ¼ 3k � m B a; rð Þð Þ:

(Solution We know that the formula for m B a; rð Þð Þ in R2 is pr2; the formula form B a; rð Þð Þ in R3 is 4

3 pr3: Similarly, we can suppose that the formula for m B a; rð Þð Þ

in Rk is krk; where k is a constant.

LHS ¼ m B a; 3rð Þð Þ ¼ k 3rð Þk¼ 3k � krk� �

¼ 3k � m B a; rð Þð Þ ¼ RHS:

■)

c. Let B a1; r1ð Þ; . . .;B aN ; rNð Þ be N open balls in Rk; where 0\rN � � � � � r1:

Put B1 � B a1; r1ð Þ; . . .;BN � B aN ; rNð Þ: Now, we construct a subset S of1; . . .;Nf g as follows:In order to aid comprehension, here we present a concrete example of N ¼ 6:Here we have 6 balls B1;B2;B3;B4;B5;B6; and B1 is a ball of ‘largest size’. B1

definitely has nonempty intersection with B1: Let B1;B2;B4f g be the collection ofall balls that have nonempty intersection with B1: Thus, B3;B5;B6f g is the col-lection of all balls that have empty intersection with B1: Here B3 is a ball of ‘largestsize’ that has empty intersection with B1: Next, let B3;B5f g be the collection of allballs that have nonempty intersection with B3: Thus, B6f g is the collection of allballs that have empty intersection with B3 [B1: We shall take 1; 3; 6f g for S.

Since B1;B2;B4f g is the collection of all balls that have nonempty intersectionwith B1; by a, B1 [B2 [B4 � B a1; 3r1ð Þ: Since B3;B5f g is the collection of all ballsthat have nonempty intersection with B3, B3 [B5 � B a3; 3r3ð Þ: SinceB1 [B2 [B4 � B a1; 3r1ð Þ; and B3 [B5 � B a3; 3r3ð Þ,B1 [B2 [B3 [B4 [B5 [B6 � B a1; 3r1ð Þ [B a3; 3r3ð Þ [B a6; 3r6ð Þ¼ [ n2SB an; 3rnð Þð Þ:With similar construction demonstrated above, we can find a nonempty subset

S of 1; . . .;Nf g such that

1. all pair of balls in Bn : n 2 Sf g are disjoint,2. [ N

n¼1B an; rnð Þ � [ n2SB an; 3rnð Þ:Also, from 2,

m [ Nn¼1B an; rnð Þ

� ��m [ n2SB an; 3rnð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � X

n2Sm B an; 3rnð Þð Þ

¼Xn2S

3k � m B an; rnð Þð Þ� �

¼ 3kXn2S

m B an; rnð Þð Þ:

Hence,3. m [ N

n¼1B an; rnð Þ� �

� 3kP

n2S m B an; rnð Þð Þ:


Conclusion 3.105 Let B a1; r1ð Þ; . . .;B aN ; rNð Þ be N open balls in Rk: Then thereexists a nonempty subset S of 1; . . .;Nf g such that

1. all pair of balls in B an; rnð Þ : n 2 Sf g are disjoint,2. [ N

n¼1B an; rnð Þ � [ n2SB an; 3rnð Þ;3. m [ N

n¼1Bðan; rnÞ� �

� 3kP

n2S mðBðan; rnÞÞ.

Note 3.106 Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let l : M ! C be a complex measure on M: Let a 2 Rk:

For every r 2 0;1ð Þ; B a; rð Þ is an open set in Rk; and hence, for every r 20;1ð Þ; B a; rð Þ is a Borel set in Rk: Thus, for every r 2 0;1ð Þ; B a; rð Þ 2 M; andm B a; rð Þð Þ 2 0;1ð Þ: By Conclusion 3.17, lj j : M ! 0;1½ Þ is a positive measure,and hence for every r 2 0;1ð Þ; lj j B a; rð Þð Þ 2 0;1½ Þ. It follows that

lj j B a; rð Þð Þm B a; rð Þð Þ : r 2 0;1ð Þ

� �is a nonempty set of nonnegative real numbers, and hence

suplj j B a; rð Þð Þm B a; rð Þð Þ : r 2 0;1ð Þ

� �2 0;1½ �:

Thus,

Ml : x 7! suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �is a function from Rk to 0;1½ �:

Problem 3.107 Ml : Rk ! 0;1½ � is lower semicontinuous.

(Solution For this purpose, let us take any a 2 0;1ð Þ: It suffices to show thatMlð Þ�1 a;1ð �ð Þ is open in Rk: Let us take any a 2 Mlð Þ�1 a;1ð �ð Þ: Observe that

Mlð Þ�1 a;1ð �ð Þ ¼ x : Mlð Þ xð Þ 2 a;1ð �f g ¼ x : a\ Mlð Þ xð Þf g

¼ x : a\suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �� :

It follows that

a\suplj j B a; rð Þð Þm B a; rð Þð Þ : r 2 0;1ð Þ

� �;


and hence there exists r0 2 0;1ð Þ such that

a\lj j B a; r0ð Þð Þm B a; r0ð Þð Þ :

We have to show that a is an interior point of Mlð Þ�1 a;1ð �ð Þ:Since limt!0þ 1þ tð Þk¼ 1; and 1\ 1

a �lj j B a;r0ð Þð Þm B a;r0ð Þð Þ ; there exists t0 [ 0 such that

1þ t0ð Þk\ 1a� lj j B a; r0ð Þð Þm B a; r0ð Þð Þ :

It suffices to show that B a; r0t0ð Þ � Mlð Þ�1 a;1ð �ð Þ:For this purpose, let us take any x 2 B a; r0t0ð Þ; that is x� aj j\r0t0: We have to

show that x 2 Mlð Þ�1 a;1ð �ð Þ; that is

a\suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �:

Clearly, B a; r0ð Þ � B x; r0 þ r0t0ð Þ: (Reason: If y� aj j\r0; theny� xj j � y� aj j þ a� xj j\r0 þ x� aj j\r0 þ r0t0:Þ It follows that

a � m B x; r0 þ r0t0ð Þð Þ ¼ 1þ t0ð Þka � r0ð Þk

r0 þ r0t0ð Þkm B x; r0 þ r0t0ð Þð Þ

¼ 1þ t0ð Þka � m B a; r0ð Þð Þ\ lj j B a; r0ð Þð Þ� lj j B x; r0 þ r0t0ð Þð Þ;

and hence

a\lj j B x; r0 þ r0t0ð Þð Þm B x; r0 þ r0t0ð Þð Þ � sup

lj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �:

Thus,

a\suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �:

■)

Conclusion 3.108 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let l : M ! C be a complex measure on M: Then

Ml : x 7! suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �


is a lower semicontinuous function from Rk to 0;1½ �: Also, Ml : Rk ! 0;1½ � is ameasurable function.

Here the function Ml : Rk ! 0;1½ � is called the maximal function of l:

Proof of the remaining part It suffices to show that for every open interval a; bð Þ;Mlð Þ�1 a; bð Þð Þ 2 M: Since

a; bð Þ ¼ a;1ð � � b;1½ � ¼ a;1ð � � \1n¼1 b� 1

n;1

� �;

we have

Mlð Þ�1 a; bð Þð Þ ¼ Mlð Þ�1 a;1ð �ð Þ � \1n¼1 Mlð Þ�1 b� 1

n;1

� �� :

Since M l is lower semicontinuous, Mlð Þ�1 a;1ð �ð Þ 2 M; and eachMlð Þ�1 b� 1

n ;1� ��

2 M: Now, since M is a r-algebra,

Mlð Þ�1 a; bð Þð Þ ¼ Mlð Þ�1 a;1ð �ð Þ � \1n¼1 Mlð Þ�1 b� 1

n;1

� �� 2 M|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence Mlð Þ�1 a; bð Þð Þ 2 M: ■

Note 3.109 Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let l : M ! C be a complex measure on M: Let a be a positive realnumber.

Since M l : Rk ! 0;1½ � is a lower semicontinuous function, Mlð Þ�1 a;1ð �ð Þ isan open set in Rk; and hence Mlð Þ�1 a;1ð �ð Þ 2 M:

Problem 3.110 m Mlð Þ�1 a;1ð �ð Þ� �

� 3klj j Rkð Þ

a :

(Solution By Conclusion 1.51,

m Mlð Þ�1 a;1ð �ð Þ� �

¼ sup m Kð Þ : K � Mlð Þ�1 a;1ð �ð Þ; andK is a compact setn o

:

Let us take a compact set K such that K � Mlð Þ�1 a;1ð �ð Þ: It suffices to show

that m Kð Þ� 3klj j Rkð Þ

a :


Let us take any a 2 K � Mlð Þ�1 a;1ð �ð Þ� �

: Observe that

Mlð Þ�1 a;1ð �ð Þ ¼ x : Mlð Þ xð Þ 2 a;1ð �f g ¼ x : a\ Mlð Þ xð Þf g

¼ x : a\suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �� :

It follows that

a\suplj j B a; rð Þð Þm B a; rð Þð Þ : r 2 0;1ð Þ

� �;

and hence there exists ra 2 0;1ð Þ such that

a\lj j B a; rað Þð Þm B a; rað Þð Þ :

Thus, for every a 2 K; there exists ra 2 0;1ð Þ such that

a\lj j B a; rað Þð Þm B a; rað Þð Þ :

It follows that B a; rað Þ : a 2 Kf g is an open cover of the compact set K, andhence there exist finite-many a1; . . .; aN in K such that

K � B a1; ra1ð Þ [ � � � [B aN ; raNð Þ:

By Conclusion 3.105, there exists a nonempty subset S of 1; . . .;Nf g such that

1. all pair of balls in B an; ranð Þ : n 2 Sf g are disjoint,2. K �ð Þ[ N

n¼1B an; ranð Þ � [ n2SB an; 3ranð Þ:

From 2,

m Kð Þ�m [ n2SB an; 3ranð Þð Þ�Xn2S

m B an; 3ranð Þð Þ

¼Xn2S

3k � m B an; ranð Þð Þ ¼ 3kXn2S

m B an; ranð Þð Þ

¼ 3kXn2S

lj j B an; ranð Þð Þa

¼ 3k

a

Xn2S

lj j B an; ranð Þð Þ

¼ 3k

alj j [ n2SB an; ranð Þð Þ� 3k

alj j Rk� �

;

and hence, m Kð Þ� 3klj j Rkð Þ

a : ■)


Conclusion 3.111 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let l : M ! C be a complex measure on M: Let a be apositive real number. Then

m Mlð Þ�1 a;1ð �ð Þ� �

� 3klj j Rk� �a

:

Note 3.112 Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let f 2 L1 Rk

� �: Let a be a positive real number.

Since f 2 L1 Rk� �

; f : Rk ! C is a Lebesgue measurable function, fj j : Rk !0;1½ Þ is a Lebesgue measurable function, and

RRk fj jdm 2 0;1½ Þ: SinceR

Rk fj jdm 2 0;1½ Þ; and a is a positive real number,

RRk

fj jdma 2 0;1½ Þ: Since fj j :

Rk ! 0;1½ Þ is a Lebesgue measurable function, and a;1ð Þ is open in 0;1½ Þ;fj j�1 a;1ð Þð Þ 2 M; and hence m fj j�1 a;1ð Þð Þ

� �2 0;1½ �:

Problem 3.113 m fj j�1 a;1ð Þð Þ� �

�RRk

fj jdma 2 0;1½ Þð Þ:

(Solution Since

fj j�1 a;1ð Þð Þ ¼ x : a\ fj j xð Þf g;

0� a � m fj j�1 a;1ð Þð Þ� �

¼Z

fj j�1 a;1ð Þð Þ

adm�Z

fj j�1 a;1ð Þð Þ

fj jdm�ZRk

fj jdm\1;

and hence

m fj j�1 a;1ð Þð Þ� �

�RRk fj jdm

a:

■)Thus, for every f 2 L1 Rk

� �; and for every positive real number a,

m fj j�1 a;1ð Þð Þ� �

is a nonnegative real number. Also,

a � m fj j�1 a;1ð Þð Þ� �

: a 2 0;1ð Þn o

is a nonempty set of nonnegative real numbers, which hasRRk fj jdm 2 0;1½ Þð Þ as

an upper bound, and 0 as a lower bound, and hence

a � m fj j�1 a;1ð Þð Þ� �

: a 2 0;1ð Þn o


is a bounded set of nonnegative real numbers. Thus, the function

a 7! a � m fj j�1 a;1ð Þð Þ� �

is a bounded function from 0;1ð Þ to 0;1½ Þ:

Definition Let f : Rk ! C be a Lebesgue measurable function.It follows that fj j : Rk ! 0;1½ Þ is a Lebesgue measurable function. Since fj j :

Rk ! 0;1½ Þ is a Lebesgue measurable function, and for every positive real a;a;1ð Þ is open in 0;1½ Þ; we have, for every positive real a; fj j�1 a;1ð Þð Þ 2 M;

and hence for every positive real a; m fj j�1 a;1ð Þð Þ� �

2 0;1½ �:If the function

a 7! a � m fj j�1 a;1ð Þð Þ� �

is a bounded function from 0;1ð Þ to 0;1½ Þ; then we say that f belongs to weak L1:From the above discussion we find that if f 2 L1 Rk

� �; then f belongs to weak L1:

Definition Let f be any Lebesgue measurable function defined on Rk: If thefunction

a 7! a � m fj j�1 a;1ð �ð Þ� �

is a bounded function over 0;1ð Þ; then we say that f belongs to weak L1:

Definition Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let f 2 L1 Rk

� �:

Since f 2 L1 Rk� �

; f : Rk ! C is a Lebesgue measurable function. It follows thatfj j : Rk ! 0;1½ Þ is a Lebesgue measurable function, and hence

l : E 7!ZE

fj jdm

from M to 0;1½ Þ is a positive measure. It follows that Ml : Rk ! 0;1½ � is ameasurable function. Here M l is denoted by Mf : Thus,

Mf : x 7! suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �:


Observe that

suplj j B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �¼ sup

l B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ� �

¼ sup

RB x;rð Þ fj jdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )

is a measurable function from Rk to 0;1½ �:

Problem 3.114 Mf belongs to weak L1:

(Solution For this purpose, let us take any a 2 0;1ð Þ: Now, on using Conclusion3.111, we get

0� a � m Mfj j�1 a;1ð �ð Þ� �

¼ a � m Mfð Þ�1 a;1ð �ð Þ� �

¼ a � m Mlð Þ�1 a;1ð �ð Þ� �

� a � 3klj j Rk� �a

¼ 3k lj j Rk� �

\1ð Þ;

and hence

0� a � m Mfj j�1 a;1ð �ð Þ� �

� 3k lj j Rk� �

¼ 3kl Rk� �

¼ 3kZRk

fj jdm ¼ 3k fk k1\1:

This shows that the function

a 7! a � m Mfj j�1 a;1ð �ð Þ� �

is a bounded function over 0;1ð Þ; and hence Mf belongs to weak L1: ■)

Conclusion 3.115 M can be thought of a mapping that sends members of L1 tomembers of weak L1: Also, if f 2 L1 Rk

� �; and a 2 0;1ð Þ; then

m Mfð Þ�1 a;1ð �ð Þ� �

� 3k fk k1a :


� �: Let a 2 Rk:

Observe that, for positive real r, m B a; rð Þð Þ is a positive real number. Sincef 2 L1 Rk

� �; f : Rk ! C is a measurable function, and hence x 7! f xð Þ � f að Þj j is a

measurable function. It follows thatRB a;rð Þ f � f að Þj jdm exists. Further,


ZB a;rð Þ

f � f að Þj jdm�Z

B a;rð Þ

fj j þ f að Þj jð Þdm ¼Z

B a;rð Þ

fj jdmþ f að Þj jZ

B a;rð Þ

1 dm

¼Z

B a;rð Þ

fj jdmþ f að Þj j � m B a; rð Þð Þ�ZRk

fj jdmþ f að Þj j � m B a; rð Þð Þ\1:

ThusRB a;rð Þ f � f að Þj jdm is a nonnegative real number. It follows thatR

B a;rð Þ f � f að Þj jdmm B a; rð Þð Þ

is a nonnegative real number. We further assume that f : Rk ! C is continuous ata.

Problem 3.117 limr!0þ

RB a;rð Þ

f�f að Þj jdmm B a;rð Þð Þ ¼ 0:

(Solution Let us take any e[ 0: Since f : Rk ! C is continuous at a, there existsd[ 0 such that for every x 2 B a; dð Þ; f xð Þ � f að Þj j\e; and hence for everyr 2 0; dð Þ; R

B a;rð Þ f � f að Þj jdmm B a; rð Þð Þ �

RB a;rð Þ edm

m B a; rð Þð Þ ¼e � m B a; rð Þð Þm B a; rð Þð Þ ¼ e:

It follows that

limr!0þ

RB a;rð Þ f � f að Þj jdm

m B a; rð Þð Þ ¼ 0:

■)

Definition Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let f 2 L1 Rk

� �: Let a 2 Rk: If

limr!0þ


m B a; rð Þð Þ ¼ 0 or equivalently; limn!1

sup


m B a; rð Þð Þ : r 2 0;1n

� ( ) !¼ 0

!;

then we say that a is a Lebesgue point of f.


Conclusion 3.118 If f : Rk ! C is continuous at a, and f 2 L1 Rk� �

; then a is aLebesgue point of f.


� �:

We shall try to show that

limn!1

sup

RB x;rð Þ f � f xð Þj jdm

m B x; rð Þð Þ : r 2 0;1n

� ( ) !¼ 0 holds a:e: onRk;

that is,

m x : 0\ limn!1

sup



� ( ) !( ) !¼ 0:

Let

Tfð Þ : x 7! limn!1

sup



� ( ) !

be the function from Rk to 0;1½ �: We have to show that m x : 0\ Tfð Þ xð Þf gð Þ ¼ 0:For this purpose, let us take any positive integer p. Since f 2 L1 Rk

� �; by

Conclusion 2.50, there exists gp 2 Cc Rk� �

� L1 Rk� ��

such that gp � f��

1\1p2 :

Since gp 2 Cc Rk� �

; gp : Rk ! C is continuous, and hence gp : Rk ! C is con-tinuous at every point of Rk: Since gp : Rk ! C is continuous at every point of Rk;

and gp 2 L1 Rk� �

; every point of Rk is a Lebesgue point of gp: It follows that forevery x 2 Rk;

limn!1

sup

RB x;rð Þ gp � gp xð Þ

�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( ) !¼ 0:

Let

Tgp� �

: x 7! limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( ) !

be a function from Rk to 0;1½ �:We have, for every x 2 Rk; Tgp� �

xð Þ ¼ 0: Observethat, for every x 2 Rk;


Tfð Þ xð Þ ¼ limn!1

sup



� ( ) !

¼ limn!1

sup


� �þ f � gp� �

� f � gp� �

xð Þ� �� dm


� ( ) !

� limn!1

sup


�� þ f � gp� �

� f � gp� �

xð Þ��

dm


� ( ) !

¼ limn!1

sup


�� dmþRB x;rð Þ f � gp

� �� f � gp� �

xð Þ�� dm


� ( ) !

¼ limn!1

sup


�� dmm B x; rð Þð Þ þ

RB x;rð Þ f � gp

� �� f � gp� �

xð Þ�� dm


� ( ) !

� limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( )þ sup


� �� f � gp� �

xð Þ�� dm


� ( ) !

¼ limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( ) !

þ limn!1

sup


� �� f � gp� �

xð Þ�� dm


� ( ) !

¼ Tgp� �

xð Þþ limn!1

sup


� �� f � gp� �

xð Þ�� dm


� ( ) !

¼ 0þ limn!1

sup


� �� f � gp� �

xð Þ�� dm


� ( ) !

� limn!1

sup


�� þ f xð Þ � gp xð Þ�� dm


� ( ) !

¼ limn!1

sup


�� dmþ f xð Þ � gp xð Þ�� R

B x;rð Þ 1 dm


� ( ) !

¼ limn!1

sup


�� dmþ f xð Þ � gp xð Þ�� m B x; rð Þð Þ


� ( ) !

¼ limn!1

sup


�� dmm B x; rð Þð Þ þ f xð Þ � gp xð Þ

�� : r 2 0;1n

� ( ) !

¼ limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( )þ f xð Þ � gp xð Þ�� !

¼ limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;

1n

� ( ) !þ f xð Þ � gp xð Þ��

� limn!1

sup


�� dmm B x; rð Þð Þ : r 2 0;1ð Þ

( ) !þ f xð Þ � gp xð Þ��

¼ limn!1

M f � gp� ��

xð Þþ f � gp�� xð Þ ¼ M f � gp

� �� xð Þþ f � gp

�� xð Þ;

so

Tfð Þ xð Þ� M f � gp� ��

xð Þþ f � gp�� xð Þ:


It follows that, for every positive integer p,

x :1p\ Tfð Þ xð Þ

� �� x :

12p

\ M f � gp� ��

xð Þ� �

[ x :12p

\ f � gp�� xð Þ

� �¼ M f � gp

� �� 1 12p

;1� ��

[ f � gp�� 1 1

2p;1

� �� ;

and hence for every positive integer p,

m x :1p\ Tfð Þ xð Þ

� �� m M f � gp

� �� 1 12p

;1� ��

[ f � gp�� 1 1

2p;1

� �� m M f � gp

� �� 1 12p

;1� ��

þm f � gp�� 1 1

2p;1

� �� 3k

RRk f � gp�� dm

12p

þm f � gp�� 1 1

2p;1

� �� 3k

RRk f � gp�� dm

12p

þRRk f � gp�� dm

12p

¼ 3k þ 1� �

2pZRk

f � gp�� dm ¼ 3k þ 1

� �2p f � gp��

1\ 3k þ 1� �

2p � 1p2

¼ 2 � 3k þ 1� � 1

p:

Thus, for every positive integer p,

m x :1p\ Tfð Þ xð Þ

� �� \2 � 3k þ 1

� � 1p:

It follows that

0�m x : 0\ Tfð Þ xð Þf gð Þ ¼ m [1p¼1 x :

1p\ Tfð Þ xð Þ

� �� ¼ lim

p!1m x :

1p\ Tfð Þ xð Þ

� �� lim

p!12 � 3k þ 1� � 1

p¼ 0;

and hence m x : 0\ Tfð Þ xð Þf gð Þ ¼ 0:

Conclusion 3.120 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let f 2 L1 Rk

� �: Then almost every point of Rk is a Lebesgue

point of f.

Note 3.121

Definition Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let l : M ! C be a complex measure on M: Let a 2 Rk: If

limr!0þ

l B a; rð Þð Þm B a; rð Þð Þ exists;


then

limr!0þ

l B a; rð Þð Þm B a; rð Þð Þ

is denoted by Dlð Þ að Þ; and is called the symmetric derivative of l at a.

a. Let M be the r-algebra of Borel sets in the k-dimensional Euclidean space Rk:

Let f 2 L1 Rk� �

: Let a 2 Rk� �

be a Lebesgue point of f.

Problem 3.122 f að Þ ¼ limr!0þ

RB a;rð Þ

f dmm B a;rð Þð Þ :

(Solution Since a 2 Rk� �

is a Lebesgue point of f, we have

0� limr!0þ

RB a;rð Þ Re fð Þ � Re fð Þð Þ að Þð Þdm

m B a; rð Þð Þ

�� ¼ lim

r!0þ


m B a; rð Þð Þ

��

¼ limr!0þ


�� m B a; rð Þð Þ � lim

r!0þ

RB a;rð Þ Re fð Þ � Re fð Þð Þ að Þj jdm

m B a; rð Þð Þ

� limr!0þ


m B a; rð Þð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

limr!0þ


m B a; rð Þð Þ

�� ¼ 0:

It follows that

limr!0þ



Similarly,

limr!0þ

RB a;rð Þ Im fð Þ � Im fð Þð Þ að Þð Þdm



Thus,

limr!0þ

RB a;rð Þ f dm

m B a; rð Þð Þ � f að Þ ¼ limr!0þ

RB a;rð Þ f dm� f að Þm B a; rð Þð Þ

m B a; rð Þð Þ

¼ limr!0þ

RB a;rð Þ f dm� f að Þ

RB a;rð Þ 1 dm

m B a; rð Þð Þ ¼ limr!0þ

RB a;rð Þ f � f að Þð Þ dm

m B a; rð Þð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Hence,

f að Þ ¼ limr!0þ

RB a;rð Þ f dm

m B a; rð Þð Þ :

■)

b. Let M be the r-algebra of Borel sets in the k-dimensional Euclidean space Rk:Let l : M ! C be a complex measure on M: Let l m:

Since m : M ! 0;1½ � is a positive measure on M; Rk has r-finite measure,l m; 0?m; and l ¼ lþ 0; by Theorem 3.42 there exists a function f : X ! C

such that f 2 L1 lð Þ; and, for every E 2 M; l Eð Þ ¼RE f dm: Thus, l; 0ð Þ is the

Lebesgue decomposition of l relative to m, and the function f : X ! C is the

Radon–Nikodym derivative of l with respect to m. In short, dldm ¼ f :

Let a 2 Rk� �

be a Lebesgue point of f. It follows from a, that

f að Þ ¼ limr!0þ

RB a;rð Þ f dm

m B a; rð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ limr!0þ

l B a; rð Þð Þm B a; rð Þð Þ ¼ Dlð Þ að Þ;

and hence f and Dl coincide at all Lebesgue points of f. By Conclusion 3.120,almost every point of Rk is a Lebesgue point of f, so f ¼ Dl a.e. on Rk: It followsthat for every E 2 M;

l Eð Þ ¼ð ÞZE

f dm ¼ZE

Dlð Þdm:

Hence, for every E 2 M; l Eð Þ ¼RE Dlð Þdm:

Conclusion 3.123 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let l : M ! C be a complex measure on M: Let l m: Letf : X ! C be the Radon–Nikodym derivative of l with respect to m. Then f ¼ Dla.e. on Rk; and for every E 2 M;


l Eð Þ ¼ZE

Dlð Þdm:

Note 3.124

Definition Let M be the r-algebra of Borel sets in the k-dimensional Euclideanspace Rk: Let a 2 Rk: Suppose that for every n ¼ 1; 2; 3; . . .;En 2 M: If there exista sequence rnf g of positive real numbers, and a real number a[ 0 such thatlimn!1 rn ¼ 0; and for every n ¼ 1; 2; 3; . . .;

1. En � B a; rnð Þ; (and hence m Enð Þ�m B a; rnð Þð Þ; that is m Enð Þm B a;rnð Þð Þ 2 0; 1½ �:)

2. a� m Enð Þm B a;rnð Þð Þ ;

then we say that the sequence Enf g shrinks to a nicely.Let M be the r-algebra of Borel sets in the k-dimensional Euclidean space Rk:

Suppose that for every x 2 Rk; En xð Þf g shrinks to x nicely. Let f 2 L1 Rk� �

:

Let a 2 Rk� �

be a Lebesgue point of f. It follows that

limr!0þ



From the assumption, En að Þf g shrinks to a nicely, and hence there exist asequence rnf g of positive real numbers, and a real number a[ 0 such thatlimn!1 rn ¼ 0; and for every n ¼ 1; 2; 3; . . .;

1. En að Þ � B a; rnð Þ;2. a � m B a; rnð Þð Þ�m En að Þð Þ:

Now, since limn!1 rn ¼ 0;

0 ¼ limr!0þ


m B a; rð Þð Þ ¼ limn!1

RB a;rnð Þ f � f að Þj jdm

m B a; rnð Þð Þ � limn!1

RB a;rnð Þ f � f að Þj jdm

m En að Þð Þa

� limn!1

REn að Þ f � f að Þj jdm

m En að Þð Þa

¼ a limn!1


m En að Þð Þ � 0;

hence

a limn!1


m En að Þð Þ ¼ 0:

It follows that limn!1

REn að Þ

f�f að Þj jdmm En að Þð Þ ¼ 0; because a[ 0: Thus,


0� limn!1

REn að Þ Re fð Þ � Re fð Þð Þ að Þð Þdm

m En að Þð Þ

�� ¼ lim

n!1


m En að Þð Þ

��

¼ limn!1


�� m En að Þð Þ � lim

n!1

REn að Þ Re fð Þ � Re fð Þð Þ að Þj jdm

m En að Þð Þ

� limn!1


m En að Þð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

limn!1


m En að Þð Þ

�� ¼ 0:

It follows that

limn!1



Similarly,

limn!1

REn að Þ Im fð Þ � Im fð Þð Þ að Þð Þdm


Thus,

limn!1

REn að Þ f dm

m En að Þð Þ � f að Þ ¼ limn!1

REn að Þ f dm� f að Þm En að Þð Þ

m En að Þð Þ

¼ limn!1

REn að Þ f dm� f að Þ

REn að Þ 1 dm

m En að Þð Þ ¼ limn!1

REn að Þ f � f að Þð Þdm

m En að Þð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Hence,

f að Þ ¼ limn!1

REn að Þ f dm

m En að Þð Þ :


Thus, f and

x 7! limn!1

REn xð Þ f dm

m En xð Þð Þ

coincide at all Lebesgue points of f. By Conclusion 3.120, almost every point of Rk

is a Lebesgue point of f, so

f xð Þ ¼ limn!1

REn xð Þ f dm

m En xð Þð Þ a:e: onRk:

Conclusion 3.125 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk. Suppose that for every x 2 Rk; En xð Þf g shrinks to x nicely. Letf 2 L1 Rk

� �: Then at every Lebesgue point a of f,

f að Þ ¼ limn!1

REn að Þ f dm

m En að Þð Þ ; and f xð Þ ¼ limn!1

REn xð Þ f dm

m En xð Þð Þ a:e:onRk:

Note 3.126 Let M be the r-algebra of Borel sets in R. Let f 2 L1 Rð Þ: Let F :

x 7!R

�1;xð Þ f dm be the function from R to C: Then, clearly, at every Lebesgue

point a of f, F0 að Þ ¼ f að Þ:Problem 3.127 F0 xð Þ ¼ f xð Þ a.e. on R:

(Solution By Conclusion 3.120, it suffices to show that F0 and f coincide at allLebesgue points of f. For this purpose, let us take any Lebesgue point a of f. Wehave to show that

limh!0

R�1;aþ hð Þð Þ f dm�

R�1;að Þð Þ f dm

h¼ lim

h!0

F aþ hð Þ � F að Þh

¼ F0 að Þ ¼ f að Þ|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl};that is for every sequence dnf g of positive real numbers satisfying lim

n!1dn ¼ 0;

limn!1

R�1;aþ dnð Þð Þ f dm�


dn¼ f að Þ; and lim

n!1

R�1;a�dnð Þð Þ f dm�


�dn¼ f að Þ:

For this purpose, let us take any sequence dnf g of positive real numbers satis-fying limn!1 dn ¼ 0: We have to show that


limn!1

R½a;aþ dn½ � f dm

m a; aþ dn½ �ð Þ ¼ limn!1

R½a;aþ dnð Þ f dm

m a; aþ dn½ �ð Þ ¼ limn!1



m a; aþ dn½ �ð Þ

¼ limn!1



dn¼ f að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and

limn!1

R½a�dn;a½ � f dm

m a� dn; a½ �ð Þ ¼ limn!1

Ra�dn;a½ Þ f dm

m a� dn; a½ �ð Þ ¼ limn!1

�Ra�dn;a½ Þ f dm

�m a� dn; a½ �ð Þ

¼ limn!1



�m a� dn; a½ �ð Þ ¼ limn!1



�dn¼ f að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

limn!1

R½a;aþ dn½ � f dm

m a; aþ dn½ �ð Þ ¼ f að Þ; and limn!1

R½a�dn;a½ � f dm

m a� dn; a½ �ð Þ ¼ f að Þ:

Problem 3:128 For every x 2 R; x; xþ d1½ �; x; xþ d2½ �; x; xþ d3½ �; . . .f g shrink atx nicely.

(Solution We know that each x; xþ dn½ � 2 M: Put, for every n ¼ 1; 2; 3; . . .; rn �32 dn; and a � 1

3 : It is easy to see that all the conditions for ‘shrinking at x nicely’ aresatisfied. ■)

Now, by Conclusion 3.125,

f að Þ ¼ limn!1

Ra;aþ dn½ � f dm

m a; aþ dn½ �ð Þ :

Problem 3:129 For every x 2 R; x� d1; x½ �; x� d2; x½ �; x� d3; x½ �; . . .f g shrink atx nicely.

(Solution We know that each x� dn; x½ � 2 M: Put, for every n ¼ 1; 2; 3; . . ., rn �32 dn; and a � 1

3 : It is easy to see that all the conditions for ‘shrinking at x nicely’ aresatisfied. ■)

Now, by Conclusion 3.125,

f að Þ ¼ limn!1

Ra�dn;a½ � f dm

m a� dn; a½ �ð Þ :

■)


Conclusion 3.130 Let M be the r-algebra of Borel sets in R: Let f 2 L1 Rð Þ: LetF : x 7!

R�1;xð Þ f dm be the function from R to C: Then, at every Lebesgue point

a of f, F0 að Þ ¼ f að Þ: Also, F0 xð Þ ¼ f xð Þ a.e. on R:

3.5 Metric Density

Note 3.131 Let M be the r-algebra of all Lebesgue measurable sets in the k-dimensional Euclidean space Rk: Let E 2 M: Let a 2 Rk:

For every real r[ 0; B a; rð Þ 2 M: Since E 2 M; for every real r[ 0;B a; rð Þ 2 M; and M is a r-algebra, we have, for every real r[ 0;B a; rð Þ ð ÞE\B a; rð Þ 2 M; and hence, for every real r[ 0;

0�ð Þm E\B a; rð Þð Þ�m B a; rð Þð Þ 2 0;1ð Þð Þ:

This shows that, for every real r[ 0;

m E \B a; rð Þð Þm B a; rð Þð Þ 2 0; 1½ �:

Definition If

limr!0þ

m E \B a; rð Þð Þm B a; rð Þð Þ

exists, then

limr!0þ

m E \B a; rð Þð Þm B a; rð Þð Þ 2 0; 1½ �ð Þ

is called the metric density of E at a.Let M be the r-algebra of all Lebesgue measurable sets in the k-dimensional

Euclidean space Rk: Let E 2 M: Let m Eð Þ\1:We shall try to show that

limr!0þ

m E \B x; rð Þð Þm B x; rð Þð Þ ¼ 1

holds a.e in E, and

limr!0þ

m E \B x; rð Þð Þm B x; rð Þð Þ ¼ 0


holds a.e in Ec; that is for every sequence dnf g of positive real numbers satisfyinglimn!1 dn ¼ 0;

limn!1

m E \B x; dnð Þð Þm B x; dnð Þð Þ ¼ 1

holds a.e in E, and

limn!1

m E \B x; dnð Þð Þm B x; dnð Þð Þ ¼ 0

holds a.e in Ec: So for every sequence dnf g of positive real numbers satisfyinglimn!1 dn ¼ 0;

limn!1

m E\B x; dnð Þð Þm B x; dnð Þð Þ ¼ vE xð Þ

holds a.e. on Rk: Since ZRk

vEj jdm ¼ m Eð Þ \1ð Þ;

vE 2 L1 Rk� �

: Clearly, for every x 2 Rk; B x; dnð Þf g shrinks to x nicely. Now, byConclusion 3.125,

vE xð Þ ¼ limn!1

RB x;dnð Þ vEdm

m B x; dnð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ limn!1

m E \B x; dnð Þð Þm B x; dnð Þð Þ

holds a.e. on Rk .

Conclusion 3.132 Let M be the r-algebra of all Lebesgue measurable sets in thek-dimensional Euclidean space Rk: Let E 2 M: Let m Eð Þ\1: Then the metricdensity of E at x is 1 a.e. in E, and the metric density of E at x is 0 a.e. in Ec:

Note 3.133 Let M be the r-algebra of all Lebesgue measurable sets in R: LetE 2 M: Let m Eð Þ\1:

We claim that there does not exist e 2 0; 12� �

such that for every open interval I,

e\m E \ Ið Þm Ið Þ \ 1� eð Þ:

3.5 Metric Density 483

If not, otherwise, let there exist e 2 0; 12� �

such that for every open interval I,

e\m E \ Ið Þm Ið Þ \ 1� eð Þ:

We have to arrive at a contradiction. It follows that, for every x 2 R; and forevery r[ 0;

e\m E \ x� r; xþ rð Þð Þm x� r; xþ rð Þð Þ \ 1� eð Þ;

and hence for every x 2 R;

0\ð Þe� limr!0þ

m E \ x� r; xþ rð Þð Þm x� r; xþ rð Þð Þ � 1� eð Þ \1ð Þ:

Thus,

x : limr!0þ

m E \ x� r; xþ rð Þð Þm x� r; xþ rð Þð Þ 62 0; 1f g

� �¼ R:

By Conclusion 3.132,

1 ¼ m Rð Þ ¼ð Þm x : limr!0þ

m E \ x� r; xþ rð Þð Þm x� r; xþ rð Þð Þ 62 0; 1f g

� �� ¼ 0;

and hence 1 ¼ 0: This is a contradiction.

Conclusion 3.134 Let E be a Lebesgue measurable set in R with finite Lebesguemeasure. Then e 2 0; 12

� �does not exist, such that for every open interval I,

e\ m E \ Ið Þm Ið Þ \ 1� eð Þ:

Theorem 3.135 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Suppose that, for every x 2 Rk; En xð Þf g shrinks to x nicely.Let l : M ! C be a complex measure on M: Let l?m: Then

limn!1

l En xð Þð Þm En xð Þð Þ ¼ 0

holds a.e. on Rk .

Proof Since l : M ! C is a complex measure on M; Re lð Þ : M ! R; andIm lð Þ : M ! R are signed measures. It follows that Re lð Þð Þþ ; Re lð Þð Þ�;Im lð Þð Þþ ; Im lð Þð Þ� are positive measures on M satisfying Re lð Þ ¼ Re lð Þð Þþ

� Re lð Þð Þ�; and Im lð Þ ¼ Im lð Þð Þþ� Im lð Þð Þ�: Now, since


limn!1

l En xð Þð Þm En xð Þð Þ

¼ limn!1

Re lð Þð Þþ En xð Þð Þ � Re lð Þð Þ� En xð Þð Þþ i Im lð Þð Þþ En xð Þð Þ � Im lð Þð Þ� En xð Þð Þ� �

m En xð Þð Þ

¼ limn!1

Re lð Þð Þþ En xð Þð Þm En xð Þð Þ � lim

n!1

Re lð Þð Þ� En xð Þð Þm En xð Þð Þ

� þ i lim

n!1

Im lð Þð Þþ En xð Þð Þm En xð Þð Þ � lim

n!1

Im lð Þð Þ� En xð Þð Þm En xð Þð Þ

� ;


limn!1

Re lð Þð Þþ En xð Þð Þm En xð Þð Þ � lim

n!1

Re lð Þð Þ� En xð Þð Þm En xð Þð Þ

� þ i lim

n!1

Im lð Þð Þþ En xð Þð Þm En xð Þð Þ � lim

n!1

Im lð Þð Þ� En xð Þð Þm En xð Þð Þ

� ¼ 0

holds a.e. on Rk: It suffices to show that

limn!1

Re lð Þð Þþ En xð Þð Þm En xð Þð Þ ¼ 0 holds a:e: onRk;

limn!1

Re lð Þð Þ� En xð Þð Þm En xð Þð Þ ¼ 0 holds a:e: onRk;

limn!1

Im lð Þð Þþ En xð Þð Þm En xð Þð Þ ¼ 0 holds a:e: onRk; and

limn!1

Im lð Þð Þ� En xð Þð Þm En xð Þð Þ ¼ 0 holds a:e: onRk:

This shows that it is enough to prove limn!1l En xð Þð Þm En xð Þð Þ ¼ 0 a.e. with the additional

condition that l : M ! 0;1½ Þ:From assumption, for every x 2 Rk; En xð Þf g shrinks to x nicely, and hence for

every x 2 Rk; there exist a sequence rn xð Þf g of positive real numbers, and a realnumber a xð Þ[ 0 such that limn!1 rn xð Þ ¼ 0; and, for every n ¼ 1; 2; 3; . . .; 1.En xð Þ � B x; rn xð Þð Þ; 2. a xð Þ � m B x; rn xð Þð Þð Þ�m En xð Þð Þ:

Now, for every x 2 Rk;

limn!1

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ � lim

n!1

l B x; rn xð Þð Þð Þm En xð Þð Þ

a xð Þ� lim

n!1

l En xð Þð Þm En xð Þð Þ

a xð Þ¼ a xð Þ lim

n!1

l En xð Þð Þm En xð Þð Þ ;


so for every x 2 Rk;

0�ð Þ limn!1

l En xð Þð Þm En xð Þð Þ �

1a xð Þ lim

n!1

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ ;

and hence it suffices to show that limr!0þl B x;rð Þð Þm B x;rð Þð Þ ¼ 0 holds a.e., that is

limn!1

supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ 0 holds a:e:;

that is, limn!1 fn xð Þ ¼ 0 holds a.e., where, for every positive integer n,

fn : x 7! supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �is a function from Rk to 0;1½ �; that is �Dl ¼ 0 holds a:e:, where

Dl : x 7! limn!1

fn xð Þ limn!1

supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� :

Let us fix any positive integer n.

Problem 3:136 fn : Rk ! 0;1½ � is lower semicontinuous.

(Solution For this purpose, let us take any a 2 0;1ð Þ: It suffices to show thatfnð Þ�1 a;1ð �ð Þ is open in Rk: Let us take any

a 2 fnð Þ�1 a;1ð �ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ x : fn xð Þ 2 a;1ð �f g ¼ x : a\fn xð Þf g

¼ x : a\supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� :

It follows that

a\supl B a; rð Þð Þm B a; rð Þð Þ : r 2 0;

1n

� � �;

and hence there exists r0 2 0; 1n� �

such that a\ l B a;r0ð Þð Þm B a;r0ð Þð Þ : We have to show that a is

an interior point of fnð Þ�1 a;1ð �ð Þ: Since r0 2 0; 1n� �

; limt!0þ 1þ tð Þk¼ 1; and

1\ 1a �

l B a;r0ð Þð Þm B a;r0ð Þð Þ ; there exists t0 [ 0 such that

r0 þ r0t0 ¼ð Þr0 1þ t0ð Þ\ 1n; and 1þ t0ð Þk\ 1

a� l B a; r0ð Þð Þm B a; r0ð Þð Þ :



B a; r0t0ð Þ � fnð Þ�1 a;1ð �ð Þ:

For this purpose, let us take any x 2 B a; r0t0ð Þ; that is x� aj j\r0t0: We have toshow that x 2 fnð Þ�1 a;1ð �ð Þ; that is

a\supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �:

Clearly, B a; r0ð Þ � B x; r0 þ r0t0ð Þ: (Reason: If y� aj j\r0; theny� xj j � y� aj j þ a� xj j\r0 þ x� aj j\r0 þ r0t0:Þ It follows that

a � m B x; r0 þ r0t0ð Þð Þ ¼ 1þ t0ð Þka � r0ð Þk

r0 þ r0t0ð Þkm B x; r0 þ r0t0ð Þð Þ

¼ 1þ t0ð Þka � m B a; r0ð Þð Þ\ l B a; r0ð Þð Þ� l B x; r0 þ r0t0ð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence

a\l B x; r0 þ r0t0ð Þð Þm B x; r0 þ r0t0ð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � sup

l B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �:

Thus,

a\supl B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �:

■)

Problem 3:137 fn : Rk ! 0;1½ � is a Borel measurable function.

(Solution It suffices to show that for every open interval a; bð Þ; fnð Þ�1 a; bð Þð Þ 2M: Since

a; bð Þ ¼ a;1ð � � b;1½ � ¼ a;1ð � � \1n¼1 b� 1

n;1

� �;

we have

fnð Þ�1 a; bð Þð Þ ¼ fnð Þ�1 a;1ð �ð Þ � \1n¼1 fnð Þ�1 b� 1

n;1

� �� :


Since fn is lower semicontinuous, fnð Þ�1 a;1ð �ð Þ 2 M; and eachfnð Þ�1 b� 1

n ;1� ��

2 M: Now, since M is a r-algebra,

fnð Þ�1 a; bð Þð Þ ¼ fnð Þ�1 a;1ð �ð Þ � \1n¼1 fnð Þ�1 b� 1

n ;1� ��

2 M; and hence

fnð Þ�1 a; bð Þð Þ 2 M: ■)Since each fn is a Borel measurable function, and Dl : x 7! limn!1 fn xð Þ; �Dl is

a Borel measurable function. We have to show that �Dl ¼ 0 holds a:e:, that is

limn!1

m x :1n\ �Dlð Þ xð Þ

� �� ¼ m [1

n¼1 x :1n\ �Dlð Þ xð Þ

� �� ¼ m x : 0\ �Dlð Þ xð Þf gð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

limn!1

m x :1n\ �Dlð Þ xð Þ

� �� ¼ 0:

For this purpose, let us take any positive integer N. It suffices to show that

m x :1N\ �Dlð Þ xð Þ

� �� \3k � 1

N:

Since l?m; there exist A;B 2 M such that A\B ¼ ;; l is concentrated on A,and m is concentrated on B. By Conclusion 1.234, l : M ! 0;1½ Þ is regular, andhence

l Að Þ ¼ sup l Kð Þ : K � A; andK is a compact setf g:

It follows that there exists a compact subset K of Rk such that K � A; and

l Rk� �

� 1N2 ¼ l A[Acð Þ � 1

N2 ¼ lðAÞþ l Acð Þð Þ � 1N2 ¼ lðAÞþ 0ð Þ � 1

N2

¼ lðAÞ � 1N2 \lðKÞ|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence K � A; and l Rk� �

� 1N2 \l Kð Þ: Since K � A � Bc; and m is concen-

trated on B, m Kð Þ ¼ 0: Since l Rk� �

� 1N2 \l Kð Þ; we have

l Kcð Þ ¼ Re lð Þð Þþ Rk � K� �

¼ l Rk� �

� l Kð Þ\ 1N2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};


and hence l Kcð Þ\ 1N2 : Let l1 : E 7! l E \Kð Þ � l Eð Þð Þ be the function from M to

0;1½ Þ: Clearly, l1 is a complex measure. Now, since l : M ! 0;1½ Þ is a com-plex measure, l� l1ð Þ : M ! 0;1½ Þ is a complex measure on M; and hencel� l1j j ¼ l� l1ð Þ: Now,

l� l1j j Rk� �

¼ l� l1ð Þ Rk� �

¼ l Rk� �

� l1 Rk� �

¼ l Rk� �

� l Rk \K� �

¼ l Rk� �

� l Kð Þ ¼ l Rk � K� �

¼ l Kcð Þ\ 1N2

;

so l� l1j j Rk� �

\ 1N2 : Since K is a compact subset of Rk; K is closed, and hence Kc

is open. Let x 2 Kc: Now, since Kc is open, there exists r0 [ 0 such that 0\r\r0implies B x; rð Þ � Kc; and hence 0\r\r0 implies B x; rð Þ \Kc ¼ B x; rð Þ:

Problem 3:138 For every x 2 Kc; �Dlð Þ xð Þ� M l� l1ð Þð Þ xð Þ:

(Solution Let x 2 Kc: Here,

�D l� l1ð Þð Þ xð Þ ¼ limn!1

supl� l1ð Þ B x; rð Þð Þ


� � �� ¼ lim

n!1sup

l B x; rð Þð Þ � l1 B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ lim

n!1sup

l B x; rð Þð Þ � l B x; rð Þ \Kð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ lim

n!1sup

l B x; rð Þ � B x; rð Þ \Kð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ lim

n!1sup

l B x; rð Þ \ Kcð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ lim

n!1sup

l B x; rð Þð Þm B x; rð Þð Þ : r 2 0;

1n

� � �� ¼ �Dlð Þ xð Þ; so �Dlð Þ xð Þ ¼ �D l� l1ð Þð Þ xð Þ:

Since



� � �� sup

l� l1ð Þ B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �;

we have

�Dlð Þ xð Þ ¼ �D l� l1ð Þð Þ xð Þ ¼ limn!1



� � �� sup

l� l1ð Þ B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �¼ M l� l1ð Þð Þ xð Þð Þ;


and hence

�Dlð Þ xð Þ� M l� l1ð Þð Þ xð Þ:

■)It follows that, for every x 2 Rk;

x :1N\ �Dlð Þ xð Þ

� �� K [ x :

1N\ M l� l1ð Þð Þ xð Þ

� �;

and hence

m x :1N\ �Dlð Þ xð Þ

� �� m K [ x :


� �� m Kð Þþm x :


� �� ¼ 0þm x :


� �� ¼ m x :


� �� ¼ m M l� l1ð Þð Þ�1 1

N;1

� �� 3k

l� l1j j Rk� �

1N

¼ 3kN l� l1j j Rk� ��

\3kN � 1N2 ¼ 3k � 1

N:

Thus, for every x 2 Rk; m x : 1N\ �Dlð Þ xð Þ

� � �\3k � 1

N : ■

Theorem 3.139 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Suppose that, for every x 2 Rk; En xð Þf g shrink to x nicely. Letl : M ! C be a complex measure on M: Let dl ¼ h dmþ d lsð Þ be the Lebesguedecomposition of l with respect to m; where h 2 L1 mð Þ; ls ?m; l� lsð Þ m;and for every E 2 M; l� lsð Þ Eð Þ ¼

RE h dm (see Theorem 3.42). Then

limn!1

l En xð Þð Þm En xð Þð Þ ¼ h xð Þ holds a:e: onRk:

Proof Here,

limn!1

l En xð Þð Þm En xð Þð Þ ¼ lim

n!1

REn xð Þ h dmþ ls En xð Þð Þ

m En xð Þð Þ

¼ limn!1

REn xð Þ h dm

m En xð Þð Þ þ ls En xð Þð Þm En xð Þð Þ

!

¼ limn!1

REn xð Þ h dm

m En xð Þð Þ þ limn!1

ls En xð Þð Þm En xð Þð Þ ;


so

limn!1


n!1

REn xð Þ h dm

m En xð Þð Þ þ limn!1

ls En xð Þð Þm En xð Þð Þ :

Since ls ?m; by Theorem 3.135,

limn!1

ls En xð Þð Þm En xð Þð Þ ¼ 0 a:e: onRk;

and hence

limn!1


n!1

REn xð Þ h dm


Since h 2 L1 mð Þ; by Conclusion 3.125

h xð Þ ¼ limn!1

REn xð Þ h dm


Now, since

limn!1


n!1

REn xð Þ h dm

m En xð Þð Þ a:e: onRk;

we have

limn!1

l En xð Þð Þm En xð Þð Þ ¼ h xð Þ a:e: onRk:

■)

Lemma 3.140 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let l : M ! C be a complex measure on M: Let l?m:Then Dlð Þ xð Þ ¼ 0 a.e. on Rk:

Proof We have to show that

limr!0þ

l B x; rð Þð Þm B x; rð Þð Þ ¼ 0 a:e: onRk;

that is, for almost every x 2 Rk; and for every sequence rn xð Þf g of positive realnumbers satisfying limn!1 rn xð Þ ¼ 0;


limn!1

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ ¼ 0:

It is clear that, for every x 2 Rk; B x; rn xð Þð Þf g shrinks to x nicely. Since l?m;dl ¼ 0 dmþ dl is the Lebesgue decomposition of l; by Theorem 3.139

limn!1

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ ¼ 0 xð Þ ¼ 0ð Þ

holds a.e. on Rk: ■

Theorem 3.141 Let M be the r-algebra of Borel sets in the k-dimensionalEuclidean space Rk: Let l : M ! 0;1½ � be a positive measure on M: Let l?m:Then

limr!0þ

l B x; rð Þð Þm B x; rð Þð Þ ¼

� Dlð Þ xð Þ ¼ 1 a:e: onRk

with respect to l:

Proof We have to show that

l x : limr!0þ

l B x; rð Þð Þm B x; rð Þð Þ 6¼ 1

� �� ¼ 0:

Since l?m; there exist A;B 2 M such that A\B ¼ ;; l is concentrated on A,and m is concentrated on B. Since l is concentrated on A, l Acð Þ ¼ 0: Since l Acð Þ ¼0; it suffices to show that

l A\ x : limr!0þ


� �� ¼ 0:

Here,

A\ x : limr!0þ


� �¼ x : x 2 A; and lim

r!0þ

l B x; rð Þð Þm B x; rð Þð Þ ¼ 1 is false

� �¼ x : x 2 A; and for every sequence r1 xð Þ; r2 xð Þ; . . .f g of positive real numbers satisfying lim

n!1rn xð Þ

�n¼ 0;

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ

� �is not bounded above

is false

�¼ x : x 2 A; and there exists a sequence r1 xð Þ; r2 xð Þ; . . .f g of positive real numbers satisfying lim

n!1rn xð Þ

�n¼ 0 such that

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ

� �is bounded above

�


¼ x : x 2 A; and there exists a sequence r1 xð Þ; r2 xð Þ; . . .f g of positive real numbers satisfying limn!1

rn xð Þ�n

¼ 0 such that for every positive integer n;l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ\N for some positive integerN

�¼ [1

N¼1 x : x 2 A; and there exists a sequence r1 xð Þ; r2 xð Þ; . . .f g of positive real numbers satisfying limn!1

rn xð Þ�n

¼ 0 such that for every positive integer n;l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ\N

�¼ [1

N¼1EN ;

where, for every positive integer N; EN � x : x 2 Af , and there exists a sequencer1 xð Þ; r2 xð Þ; . . .f g of positive real numbers satisfying limn!1 rn xð Þ ¼ 0 such that

for every positive integer n; l B x;rn xð Þð Þð Þm B x;rn xð Þð Þð Þ\N

o,

so

A\ x : limr!0þ


� �¼ [1

N¼1EN :

Clearly,

E1 � E2 � E3 � � � � � A; so l A\ x : limr!0þ


� �� ¼ l [1

N¼1EN� �

¼ limn!1

l Enð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence, it suffices to show that for every positive integer N, l ENð Þ ¼ 0:

For this purpose, let us fix any positive integer N. Next let us fix any positiveinteger j. Since m is concentrated on B, and A\B ¼ ;;

inf m Vð Þ : A � V ; andV is openf g ¼ m Að Þ ¼ 0|fflfflfflfflfflffl{zfflfflfflfflfflffl};and hence

inf m Vð Þ : A � V ; andV is openf g ¼ 0:

Now, for every positive integer j, there exists an open set Vj such that A � Vj;

and m Vj� �

\0þ 1j : Thus, A � V1;A � V2;A � V3; etc. Also, m V1ð Þ\ 1

1 ;m V2ð Þ\ 1

2 ;m V3ð Þ\ 13 ; etc.


Problem 3:142 For every x 2 EN ; there exists a positive real r xð Þ such thatB x; r xð Þð Þ � Vj; and

l B x; r xð Þð Þð Þm B x; r xð Þð Þð Þ\N:

(Solution Let x 2 EN � A � Vj� �

: Now, since Vj is open, there exists a real s[ 0such that B x; sð Þ � Vj: Since x 2 EN ; there exists a sequence r1 xð Þ; r2 xð Þ; . . .f g ofpositive real numbers satisfying limn!1 rn xð Þ ¼ 0; and for every positive integer n,

l B x; rn xð Þð Þð Þm B x; rn xð Þð Þð Þ\N:

Since limn!1 rn xð Þ ¼ 0 \sð Þ; there exists a positive integer l such that rl xð Þ\s;and hence

B x; rl xð Þð Þ � B x; sð Þ � Vj� �

:

Thus, B x; rl xð Þð Þ � Vj; and

l B x; rl xð Þð Þð Þm B x; rl xð Þð Þð Þ\N;

that is, B x; r xð Þð Þ � Vj; and

l B x; r xð Þð Þð Þ\N � m B x; r xð Þð Þð Þ;

where r xð Þ � rl xð Þ: ■)Note that each r xð Þ depends on j and N. It is clear that

EN � [ x2ENB x;13r xð Þ

� � :

Since for every x 2 EN ;

B x;13r xð Þ

� �

� B x; r xð Þð Þ � Vj;

Also, [ x2ENB x; 13 r xð Þ� �

is an open set. Thus, for every positive integer N; j; wehave

EN � [ x2ENB x;13r xð Þ

� � Vj:


Problem 3:143 l [ x2ENB x; 13 r xð Þ� ��

� 3k Nj :

(Solution Here, [ x2ENB x; 13 r xð Þ� �

is an open subset of Rk; so there exist ‘closedballs’ B1;B2; . . . such that

[ x2ENB x;13r xð Þ

� ¼ B1 [B2 [B3 [ � � � :

It suffices to show that, for every positive integer n,

l B1 [ � � � [Bnð Þ� 3kNj:

For this purpose, let us fix a positive integer n, We have to show thatl Knð Þ� 3k N

j ; where Kn � B1 [ � � � [Bn: Since

[ x2ENB x;13r xð Þ

� ¼ B1 [B2 [B3 [ � � � B1 [ � � � [Bn ¼ Knð Þ;

B x; 13 r xð Þ� �

: x 2 EN�

is an open cover of the compact set Kn: So, there existfinite-many x1; . . .; xl 2 EN such that

Kn � B x1;13r x1ð Þ

� [ � � � [B xl;

13r xlð Þ

� :

Now, by Conclusion 3.105, there exists a nonempty subset S of 1; . . .; lf g suchthat

1. all pair of balls in B xn; 13 r xnð Þ� �

: n 2 S�

are disjoint,2. Kn �ð Þ[ l

n¼1B xn; 13 r xnð Þ� �

� [ n2SB xn; r xnð Þð Þ:Since

l Knð Þ� l [ n2SB xn; r xnð Þð Þð Þ�Xn2S

l B xn; r xnð Þð Þð Þ\Xn2S

N � m B xn; r xnð Þð Þð Þ

¼ NXn2S

m B xn; r xnð Þð Þð Þ ¼ NXn2S

3k � m B xn;13r xnð Þ

� � ¼ N � 3k

Xn2S

m B xn;13r xnð Þ

� � ¼ N � 3km [ n2SB xn;

13r xnð Þ

� � �N � 3km [ x2ENB x;

13r xð Þ

� � �N � 3k � m Vj

� �\N � 3k � 1

j;

we have l Knð Þ� 3k Nj : ■)


Since for every positive integer N; j; EN � [ x2ENB x; 13 r xð Þ� �

; we have, for everypositive integer N, EN � \1

j¼1 [ x2ENB x; 13 r xð Þ� ��

:

Now, since for every positive integer N; j;

l [ x2ENB x;13r xð Þ

� � � 3k

Nj; l ENð Þ� 3k

Nj;

it follows that 0�ð Þl ENð Þ� limj!1 3k Nj ¼ 0ð Þ: Thus l ENð Þ ¼ 0: ■

Note 3.144

Definition Let a\b: Let f : a; b½ � ! C be any function. By f is absolutely con-tinuous on a; b½ �; we mean:

for every e[ 0; there exists d[ 0 such that for every positive integer n, and forall open intervals a1; b1ð Þ; . . .; an; bnð Þ contained in a; b½ � satisfying

i 6¼ j ) ai; bið Þ \ aj; bj� �

¼ ;;

and

b1 � a1ð Þþ � � � þ bn � anð Þ\d; f b1ð Þ � f a1ð Þj j þ � � � þ f bnð Þ � f anð Þj j\e:

Clearly, if f : a; b½ � ! C is absolutely continuous, then f is uniformly continuous(take n ¼ 1), and hence f is continuous.

Let a\b: Let f : a; b½ � ! R be any function. Let f be monotonically increasing.Let f be absolutely continuous on a; b½ �: Let M be the set of all Lebesgue mea-surable subsets of R: Let E be a subset of a; bð Þ such that E 2 M; and m Eð Þ ¼ 0:

Problem 3.145 f Eð Þ 2 M; and m f Eð Þð Þ ¼ 0:

(Solution By Conclusion 1.258(5), it suffices to find F 2 M such that m Fð Þ ¼ 0;and f Eð Þ � F; and hence it suffices to show that for every positive integer n, thereexists Fn 2 M such that m Fnð Þ� 1

n ; and f Eð Þ � Fn:

For this purpose, let us fix any positive integer n0:Since f : a; b½ � ! R is absolutely continuous, there exists d[ 0 such that for

every positive integer n, and for all open intervals a1; b1ð Þ; . . .; an; bnð Þ contained ina; b½ � satisfying


¼ ;

and

b1 � a1ð Þþ � � � þ bn � anð Þ\d; f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ

¼ f b1ð Þ � f a1ð Þj j þ � � � þ f bnð Þ � f anð Þj j\ 1n0

:


Since

inf m Vð Þ : E � V ; andV is openf g ¼ m Eð Þ ¼ 0\d;

inf m Vð Þ : E � V ; andV is openf g\d;

there therefore exists an open set V1 such that E � V1; and m V1ð Þ\d:Put V � V1 \ a; bð Þ � a; bð Þ � a; b½ �ð Þ:Clearly, V is open. Since E � V1; and E � a; bð Þ; E � V1 \ a; bð Þ ¼ Vð Þ; and

hence E � V � a; b½ �ð Þ: Thus f Eð Þ � f Vð Þ:Since

V ¼ð ÞV1 \ a; bð Þ � V1;m Vð Þ�m V1ð Þ \dð Þ;

hence m Vð Þ\d: Since V is an open subset of R; there exists a sequence an; bnð Þf gof open intervals such that


¼ ;;

and

a; b½ � V ¼ a1; b1ð Þ [ a2; b2ð Þ [ a3; b3ð Þ [ � � � :


d[ð Þm Vð Þ�m a1; b1ð Þ [ � � � [ an; bnð Þð Þ¼ m a1; b1ð Þð Þþ � � � þm an; bnð Þð Þ ¼ b1 � a1ð Þþ � � � þ bn � anð Þ;

and hence for every positive integer n, b1 � a1ð Þþ � � � þ bn � anð Þ\d: Now, forevery positive integer n,

m f a1ð Þ; f b1ð Þ½ � [ � � � [ f anð Þ; f bnð Þ½ �ð Þ¼ m f a1ð Þ; f b1ð Þ½ �ð Þþ � � � þm f anð Þ; f bnð Þ½ �ð Þ

¼ f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ\ 1n0

;

and hence for every positive integer n,

m f a1ð Þ; f b1ð Þ½ � [ � � � [ f anð Þ; f bnð Þ½ �ð Þ\ 1n0

:


It follows that

m f Vð Þð Þ ¼ m f a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þð Þþ 0

¼ m f a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þð Þð Þþm f a1ð Þ; f b1ð Þ; f a2ð Þ; f b2ð Þ; � � �f gð Þ¼ m f a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þð Þ [ f a1ð Þ; f b1ð Þ; f a2ð Þ; f b2ð Þ; � � �f gð Þ¼ m f a1; b1½ � [ a2; b2½ �ð Þ [ � � �ð Þ ¼ m f a1; b1½ �ð Þ [ f an; bn½ �ð Þ [ � � �ð Þ¼ m [1

n¼1 f a1; b1½ �ð Þ [ � � � [ f an; bn½ �ð Þð Þ� �

¼ m [1n¼1 f a1ð Þ; f b1ð Þ½ � [ � � � [ f anð Þ; f bnð Þ½ �ð Þ

� �¼ lim

n!1m f a1ð Þ; f b1ð Þ½ � [ � � � [ f anð Þ; f bnð Þ½ �ð Þ� 1

n0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence m f Vð Þð Þ� 1

n0: Since

f Vð Þ ¼ f a1; b1ð Þ [ a2; b2ð Þ [ � � �ð Þ ¼ f a1; b1ð Þð Þ [ f a2; b2ð Þð Þ [ � � �¼ f a1ð Þ; f b1ð Þð Þ [ f a2ð Þ; f b2ð Þð Þ [ � � � 2 M;

we have f Vð Þ 2 M: Thus, f Vð Þ 2 M; m f Vð Þð Þ� 1n0; and f Eð Þ � f Vð Þ: ■)

Conclusion 3.146 Let a\b: Let f : a; b½ � ! R be any function. Let f be mono-tonically increasing. Let f be absolutely continuous on a; b½ �: Let M be the set of allLebesgue measurable subsets of R: Let E be a subset of a; b½ � such that E 2 M; andm Eð Þ ¼ 0: Then f Eð Þ 2 M; and m f Eð Þð Þ ¼ 0:

Proof Case I: when a 62 E; and b 62 E: In this case, E � a; bð Þ: Now, by the abovediscussion, f Eð Þ 2 M; and m f Eð Þð Þ ¼ 0:Case II: when a 2 E; and b 62 E: Here, E � af gð Þ � a; bð Þ: Since af g is a closedset, af g is a Borel set, and hence af g 2 M: Since af g 2 M;E 2 M; and M is ar-algebra, E � af gð Þ 2 M: Since

m E � af gð Þ ¼ m Eð Þ � m af gð Þ ¼ m Eð Þ � 0 ¼ m Eð Þ ¼ 0;

we have m E � af gð Þ ¼ 0: Now, by Case I, f E � af gð Þ 2 M; andm f E � af gð Þð Þ ¼ 0: Since f að Þf g is a closed set, f að Þf g 2 M: Since f að Þf g 2 M;f E � af gð Þ 2 M; and M is a r-algebra,

f Eð Þ ¼ f E � af gð Þ [ af gð Þ¼ f E � af gð Þ [ f af gð Þ ¼ f E � af gð Þ [ f að Þf g 2 M|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence f Eð Þ 2 M: Next,


0�m f Eð Þð Þ ¼ m f E � af gð Þ [ f að Þf gð Þ�m f E � af gð Þð Þþm f að Þf gð Þ ¼ 0þm f að Þf gð Þ ¼ m f að Þf gð Þ ¼ 0;

so m f Eð Þð Þ ¼ 0:Case III: when b 2 E; and a 62 E: This case is similar to Case II.Case IV: when a 2 E; and b 2 E: Here, E � a; bf gð Þ � a; bð Þ: Since a; bf g is aclosed set, a; bf g is a Borel set, and hence a; bf g 2 M: Since a; bf g 2 M;E 2 M;and M is a r-algebra, E � a; bf gð Þ 2 M: Since

m E � a; bf gð Þ ¼ m Eð Þ � m a; bf gð Þ ¼ m Eð Þ � 0 ¼ m Eð Þ ¼ 0;

we have m E � a; bf gð Þ ¼ 0: Now, by Case I, f E � a; bf gð Þ 2 M; andm f E � a; bf gð Þð Þ ¼ 0: Since f a; bf gð Þ ¼ð Þ f að Þ; f bð Þf g is a closed set,f a; bf gð Þf g 2 M: Since f a; bf gð Þ 2 M; f E � a; bf gð Þ 2 M; and M is a r-

algebra,

f Eð Þ ¼ f E � a; bf gð Þ [ a; bf gð Þ ¼ f E � a; bf gð Þ [ f a; bf gð Þ 2 M|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence f Eð Þ 2 M: Next,

0�m f Eð Þð Þ ¼ m f E � a; bf gð Þ [ f a; bf gð Þð Þ�m f E � a; bf gð Þð Þþm f a; bf gð Þð Þ¼ 0þm f a; bf gð Þð Þ ¼ m f að Þ; f bð Þf gð Þ ¼ 0; som f Eð Þð Þ ¼ 0:

■

Note 3.147 Let a\b: Let f : a; b½ � ! R be any continuous function. Let f bemonotonically increasing (that is, x\y ) f xð Þ� f yð ÞÞ: Let M be the set of allLebesgue measurable subsets of R: Suppose that, if E is a subset of a; b½ � satisfyingE 2 M and m Eð Þ ¼ 0; then f Eð Þ 2 M and m f Eð Þð Þ ¼ 0:

Problem 3.148 Iþ fð Þ : a; b½ � ! R is 1-1, where I : x 7! x is the mapping froma; b½ � to R:

(Solution Let xþ f xð Þ ¼ I xð Þþ f xð Þ ¼ Iþ fð Þ xð Þ Iþ fð Þ yð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ I yð Þþ f yð Þ ¼ yþ f yð Þ:

We have to show that x ¼ y: If not, otherwise, let x 6¼ y: We have to arrive at acontradiction. Since x 6¼ y; either x\y or y\x: For definiteness, let x\y: Now,since f is monotonically increasing, f xð Þ� f yð Þ: Since xþ f xð Þ ¼ yþ f yð Þ; andx\y; we have f xð Þ[ f yð Þ; a contradiction. ■)


Problem 3.149 Iþ fð Þ : a; b½ � ! R is strictly increasing.

(Solution Let x\y: Let x; y 2 a; b½ �: We have to show that

xþ f xð Þ ¼ Iþ fð Þ xð Þ\ Iþ fð Þ yð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ yþ f yð Þ;

that is xþ f xð Þ\yþ f yð Þ: Since x\y; and f is monotonically increasing,f xð Þ� f yð Þ: Since x\y; and f xð Þ� f yð Þ; xþ f xð Þ\yþ f yð Þ: ■)

Since I and f are continuous, Iþ f is continuous. Since Iþ fð Þ : a; b½ � ! R isstrictly increasing, and continuous, Iþ fð Þ : a; b½ � ! aþ f að Þ; bþ f bð Þ½ � is ahomeomorphism.

Observe that, if x; yð Þ is an open interval contained in a; b½ �;

Iþ fð Þ x; yð Þð Þ ¼ xþ f xð Þ; yþ f yð Þð Þ;

and hence

m Iþ fð Þ x; yð Þð Þð Þ ¼ m xþ f xð Þ; yþ f yð Þð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ yþ f yð Þð Þ � xþ f xð Þð Þ

¼ y� xð Þþ f yð Þ � f xð Þð Þ ¼ m x; yð Þð Þþ f yð Þ � f xð Þð Þ ¼ m x; yð Þð Þþm f x; yð Þð Þð Þ:

Thus, if J is an interval contained in a; b½ �; then m Iþ fð Þ Jð Þð Þ ¼ m Jð Þþm f Jð Þð Þ:

Let E be a subset of a; b½ � such that E 2 M; and inf m Vð Þ : E � V ; andfV is openg ¼ m Eð Þ ¼ 0|fflfflfflfflfflffl{zfflfflfflfflfflffl} :Problem 3.150 Iþ fð Þ Eð Þ 2 M; and m Iþ fð Þ Eð Þð Þ ¼ 0:

(Solution It suffices to show that

inf m Vð Þ : Iþ fð Þ Eð Þ � V ; andV is openf g ¼ 0;

that is for every e[ 0; inf m Vð Þ : Iþ fð Þ Eð Þ � V ; andV is openf g\e: For thispurpose, let us take any e[ 0:

Since

inf m Vð Þ : E � V ; andV is openf g ¼ 0 \eð Þ;

there exists an open set V1 such that E � V1; and m V1ð Þ\ e2 : Since m Eð Þ ¼ 0; by

assumption, f Eð Þ 2 M and

inf m Wð Þ : f Eð Þ � W ; andW is openf g ¼ð Þm f Eð Þð Þ ¼ 0 \eð Þ;


and hence there exists an open set W1 such that f Eð Þ � W1; and m W1ð Þ\ e2 : Since

f is continuous, and W1 is open, E �ð Þf�1 W1ð Þ is open. Since f�1 W1ð Þ is open, andV1 is open, V1 \ f�1 W1ð Þ is open. Since E � V1; and E � f�1 W1ð Þ,E � V1 \ f�1 W1ð Þð Þ:

Since V1 \ f�1 W1ð Þ is open, there exist countable-many disjoint intervalsJ1; J2; J3; . . . such that

V1 \ f�1 W1ð Þ ¼ J1 [ J2 [ J3 [ � � � ;

and hence

Iþ fð Þ V1 \ f�1 W1ð Þ� �

¼ I þ fð Þ J1 [ J2 [ J3 [ � � �ð Þ ¼ Iþ fð Þ J1ð Þ [ Iþ fð Þ J2ð Þ [ � � �ð Þ:

It follows that

m Iþ fð Þ V1 \ f�1 W1ð Þ� ��

¼ m Iþ fð Þ J1ð Þ [ Iþ fð Þ J2ð Þ [ � � �ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ m Iþ fð Þ J1ð Þð Þþm Iþ fð Þ J2ð Þð Þþ � � �¼ m J1ð Þþm f J1ð Þð Þð Þþ m J2ð Þþm f J2ð Þð Þð Þþ � � �¼ m J1ð Þþm J2ð Þþ � � �ð Þþ m f J1ð Þð Þþm f J2ð Þð Þþ � � �ð Þ¼ m J1 [ J2 [ J3 [ � � �ð Þþ m f J1ð Þð Þþm f J2ð Þð Þþ � � �ð Þ¼ m J1 [ J2 [ J3 [ � � �ð Þþm f J1ð Þ [ f J2ð Þ [ � � �ð Þ¼ m J1 [ J2 [ J3 [ � � �ð Þþm f J1 [ J2 [ J3 [ � � �ð Þð Þ¼ m V1 \ f�1 W1ð Þ

� �þm f V1 \ f�1 W1ð Þ

� �� ;

and hence

m Iþ fð Þ V1 \ f�1 W1ð Þ� ��

¼ m V1 \ f�1 W1ð Þ� �

þm f V1 \ f�1 W1ð Þ� ��

:

Since

m Iþ fð Þ V1 \ f�1 W1ð Þ� ��

¼ m V1 \ f�1 W1ð Þ� �

þm f V1 \ f�1 W1ð Þ� ��

�m V1ð Þþm f V1 \ f�1 W1ð Þ� ��

\e2þm f V1 \ f�1 W1ð Þ

� �� e

2þm f f�1 W1ð Þ

� �� e

2þm W1ð Þ\ e

2þ e

2¼ e;

we have

m Iþ fð Þ V1 \ f�1 W1ð Þ� ��

\e:

Since E � V1 \ f�1 W1ð Þð Þ; we have Iþ fð Þ Eð Þ � Iþ fð Þ V1 \ f�1 W1ð Þð Þ: SinceIþ fð Þ : a; b½ � ! aþ f að Þ; bþ f bð Þ½ � is a strictly increasing homeomorphism, and


V1 \ f�1 W1ð Þ is an open set contained in a; b½ �, Iþ fð Þ V1 \ f�1 W1ð Þð Þ is an openset. Since Iþ fð Þ V1 \ f�1 W1ð Þð Þ is an open set,

Iþ fð Þ Eð Þ � Iþ fð Þ V1 \ f�1 W1ð Þ� �

;

and

m Iþ fð Þ V1 \ f�1 W1ð Þ� ��

\e;

we have inf m Vð Þ : Iþ fð Þ Eð Þ � V ; andV is openf g\e: ■)

Conclusion 3.151 Let a\b: Let f : a; b½ � ! R be any continuous function. Let fbe monotonically increasing. Let M be the set of all Lebesgue measurable subsetsof R: Suppose that, if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; thenf Eð Þ 2 M and m f Eð Þð Þ ¼ 0: Then

1. ðIþ f Þ : ½a; b� ! ½aþ f ðaÞ; bþ f ðbÞ� is a strictly increasing homeomorphism,2. if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; then Iþ fð Þ Eð Þ 2 M

and m Iþ fð Þ Eð Þð Þ ¼ 0:

Note 3.152 Let a\b: Let f : a; b½ � ! R be any continuous function. Let f bemonotonically increasing. Let M be the set of all Lebesgue measurable subsets ofR: Suppose that, if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; thenf Eð Þ 2 M and m f Eð Þð Þ ¼ 0:

By Conclusion 3.151

1. ðIþ f Þ : ½a; b� ! ½aþ f ðaÞ; bþ f ðbÞ� is a strictly increasing homeomorphism,2. if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; then Iþ fð Þ Eð Þ 2 M

and m Iþ fð Þ Eð Þð Þ ¼ 0:

Let l : E 7!m Iþ fð Þ Eð Þð Þ be a mapping from E : E 2 M; andE � a; b½ �f g to0;1½ �:Clearly, l is a positive measure on the r-algebra E : E 2 M; andE � a; b½ �f g:

Since

l a; b½ �ð Þ ¼ m Iþ fð Þ a; b½ �ð Þð Þ ¼ m aþ f að Þ; bþ f bð Þð Þð Þ¼ bþ f bð Þð Þ � aþ f að Þð Þ\1;

we have

l : fE : E 2 M; andE � ½a; b�g ! 0;1½ Þ:

From 2, if E is a subset of a; b½ � satisfying m Eð Þ ¼ 0; thenl Eð Þ ¼ð Þm Iþ fð Þ Eð Þð Þ ¼ 0; and hence l m: Now, since 0?m; the ordered pairl; 0ð Þ is the Lebesgue decomposition of l relative to m. By Theorem 3.42, thereexists a function h : a; b½ � ! C such that h 2 L1 mð Þ; and for every E 2 M satis-fying E � a; b½ �; m Iþ fð Þ Eð Þð Þ ¼ð Þl Eð Þ ¼

RE h dm: Hence, for every x 2 a; bð �;


Za;x½ �

1 dm� f xð Þ � f að Þð Þ ¼ x� að Þþ f xð Þ � f að Þð Þ ¼ xþ f xð Þð Þ � aþ f að Þð Þ

¼ m aþ f að Þ; xþ f xð Þ½ �ð Þ ¼ m Iþ fð Þ a; x½ �ð Þð Þ ¼Za;x½ �

h dm:

Thus, for every x 2 a; bð �;

f xð Þ � f að Þ ¼Za;x½ �

1 dm�Za;x½ �

h dm ¼Za;x½ �

1� hð Þdm ¼Za;x½ Þ

1� hð Þdm;

and hence for every x 2 a; b½ �;

f xð Þ � f að Þ ¼Za;x½ Þ

1� hð Þdm:

Now, by the arguments similar to Note 3.126,

f 0 xð Þ ¼ f 0 xð Þ � 0 ¼ f � f að Þð Þ0 xð Þ ¼ 1� hð Þ xð Þ a:e: on a; b½ �:

Since f 0 xð Þ ¼ 1� hð Þ xð Þ a.e. on a; b½ �; f is differentiable a.e. on a; b½ �:Since m : fE : E 2 M; andE � ½a; b�g ! 0;1½ Þ we have 1 2 L1 mð Þ: Since 1 2

L1 mð Þ; and h 2 L1 mð Þ; 1� hð Þ 2 L1 mð Þ: Since 1� hð Þ 2 L1 mð Þ; and f 0 xð Þ ¼1� hð Þ xð Þ a.e. on a; b½ �; f 0 2 L1 mð Þ:Since f 0 xð Þ ¼ 1� hð Þ xð Þ a.e. on a; b½ �; for every x 2 a; bð �;

f xð Þ � f að Þ ¼Za;x½ Þ

1� hð Þdm ¼Za;x½ Þ

f 0ð Þdm ¼Zxa

f 0 tð Þdt;

and hence for every x 2 a; b½ �;

Zxa

f 0 tð Þdt ¼ f xð Þ � f að Þ:

Conclusion 3.153 Let a\b: Let f : a; b½ � ! R be any continuous function. Letf be monotonically increasing. Let M be the set of all Lebesgue measurable subsetsof R: Suppose that, if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; thenf Eð Þ 2 M and m f Eð Þð Þ ¼ 0: Then


1. f is differentiable a.e. on a; b½ �;2. f 0 2 L1 mð Þ;3. for every x 2 a; b½ �,

R xa f

0 tð Þdt ¼ f xð Þ � f að Þ:

Note 3.154 Let a\b: Let f : a; b½ � ! R be any continuous function. Let f bemonotonically increasing. Let M be the set of all Lebesgue measurable subsets ofR: Let f be differentiable a.e. on a; b½ �; f 0 2 L1 mð Þ; and, for every x 2 a; b½ �,R xa f

0 tð Þdt ¼ f xð Þ � f að Þ:

Problem 3.155 f is absolutely continuous on a; b½ �:

(Solution For this purpose, let us take any e[ 0: Since f 0 2 L1 mð Þ; for everyE 2 M;

RE f

0dm� �

2 R: Let

l : E 7!ZE

f 0dm ¼ZE

f 0ð Þþ dm�ZE

f 0ð Þ�dm

0@ 1Abe a mapping from M to R: Since

E 7!ZE

f 0ð Þþ dm

is a measure, and

E 7!ZE

f 0ð Þ�dm

is a measure,

l : E 7!ZE

f 0ð Þþ dm�ZE

f 0ð Þ�dm

0@ 1A ¼ZE

f 0dm

0@ 1Ais a signed measure on M: Let E 2 M; and m Eð Þ ¼ 0: It follows thatRE f 0ð Þþ dm ¼ 0; and

RE f 0ð Þ�dm ¼ 0; and hence

l Eð Þ ¼ZE

f 0dm ¼ZE

f 0ð Þþ dm�ZE

f 0ð Þ�dm ¼ 0

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}:


This shows that l m: Now, by Conclusion 3.46, there exists d[ 0 such thatm Eð Þ\d implies l Eð Þj j\e: Let n be a positive integer. Let a1; b1ð Þ; . . .; an; bnð Þ beopen intervals contained in a; b½ � satisfying


¼ ;;

and

b1 � a1ð Þþ � � � þ bn � anð Þ\d:


f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ ¼ f b1ð Þ � f a1ð Þj j þ � � � þ f bnð Þ � f anð Þj j\e;

that is

f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ\e:

Here,

l a1;b1ð Þð Þ ¼ l a;b1ð Þ � a; a1ð �ð Þ ¼ l a; b1ð Þð Þ � l a; a1ð �ð Þ ¼ l a;b1ð Þð Þ � l a; a1ð Þð Þ

¼Za;b1ð Þ

f 0dm�Za;a1ð Þ

f 0dm ¼Zb1a

f 0 tð Þdt �Za1a

f 0 tð Þdt ¼ f b1ð Þ � f að Þð Þ � f a1ð Þ � f að Þð Þ

¼ f b1ð Þ � f a1ð Þ; so l a1;b1ð Þð Þ ¼ f b1ð Þ � f a1ð Þ:

Similarly, l a2; b2ð Þð Þ ¼ f b2ð Þ � f a2ð Þ; etc. Since

m a1; b1ð Þ [ � � � [ an; bnð Þð Þ ¼ m a1; b1ð Þð Þþ � � � þm an; bnð Þð Þ¼ b1 � a1ð Þþ � � � þ bn � anð Þ\d;

we have

m a1; b1ð Þ [ � � � [ an; bnð Þð Þ\d;

and hence

f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ ¼ l a1; b1ð Þð Þþ � � � þ l an; bnð Þð Þ� l a1; b1ð Þð Þþ � � � þ l an; bnð Þð Þj j ¼ l a1; b1ð Þ [ � � � [ an; bnð Þð Þj j\e|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :


Thus,

f b1ð Þ � f a1ð Þð Þþ � � � þ f bnð Þ � f anð Þð Þ\e:Þ

Now, if we combine this result with Conclusions 3.146 and 3.153, we get thefollowing

Conclusion 3.156 Let a\b: Let f : a; b½ � ! R be any continuous function. Letf be monotonically increasing. Let M be the set of all Lebesgue measurable subsetsof R: Then the following statements are equivalent:

a. f is absolutely continuous on a; b½ �;b. if E is a subset of a; b½ � satisfying E 2 M and m Eð Þ ¼ 0; then f Eð Þ 2 M and

m f Eð Þð Þ ¼ 0;c. f is differentiable a.e. on a; b½ �; f 0 2 L1 mð Þ; and for every x 2 a; b½ �,R x

a f0 tð Þdt ¼ f xð Þ � f að Þ:

Note 3.157 Let a\b: Let f : a; b½ � ! R be any absolutely continuous function. Let

F : x 7! sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j : a ¼ t0\t1\ � � �\tN ¼ xf g is a partition of a; x½ �f g

be a function from a; b½ � to 0;1½ �:

Problem 3.158 F is monotonically increasing.

(Solution Let x\y; where x; y 2 a; b½ �: We have to show that

Sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j : a ¼ t0\t1\ � � �\tN ¼ xf g is a partition of a; x½ �f g¼ F xð Þ�F yð Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

¼ sup f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj j : a ¼ s0\s1\ � � �\sM ¼ yf g is a partition of a; y½ �f g:

Let us take any partition a ¼ t0\t1\ � � �\tN ¼ xf g of a; x½ �: It follows that

a ¼ t0\t1\ � � �\tN ¼ x\yf g

is a partition of a; y½ �; and hence

f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j � f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jþ f yð Þ � f xð Þj j �F yð Þ:

Thus,

f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j �F yð Þ:


This shows that

F xð Þ ¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j :fa ¼ t0\t1\ � � �\tN ¼ xf g is a partition of a; x½ � g�F yð Þ;

and hence F xð Þ�F yð Þ: ■)

Problem 3.159 F � f ;Fþ f are monotonically increasing.

(Solution Let x\y; where x; y 2 a; b½ �: Let us take any partition

a ¼ t0\t1\ � � �\tN ¼ xf g

of a; x½ �: It follows that

a ¼ t0\t1\ � � �\tN ¼ x\yf g

is a partition of a; y½ �; and hence

f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j þ f yð Þ � f xð Þj j �F yð Þ:

Thus,

f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j �F yð Þ � f yð Þ � f xð Þj j:

This shows that

F xð Þ ¼ð Þsup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j :fa ¼ t0\t1\ � � �\tN ¼ xf g is a partition of a; x½ � g�F yð Þ � f yð Þ � f xð Þj j;

and hence,

f yð Þ � f xð Þ�ð Þ f yð Þ � f xð Þj j �F yð Þ � F xð Þ:

It follows that

F xð Þ � f xð Þ�F yð Þ � f yð Þ;

and hence, F � f is a monotonically increasing function. Since

f xð Þ � f yð Þ�ð Þ f yð Þ � f xð Þj j �F yð Þ � F xð Þ;

it follows that F xð Þþ f xð Þ�F yð Þþ f yð Þ; and hence Fþ f is a monotonicallyincreasing function. ■)


Problem 3.160 F is an absolutely continuous function on a; b½ �:

(Solution Let us take any e[ 0: Since f : a; b½ � ! R is an absolutely continuousfunction, there exists d[ 0 such that for every positive integer n, and for all openintervals a1; b1ð Þ; . . .; an; bnð Þ contained in a; b½ � satisfying


¼ ;� �

;

and

b1 � a1ð Þþ � � � þ bn � anð Þ\d; f b1ð Þ � f a1ð Þj j þ � � � þ f bnð Þ � f anð Þj j\ e2:

Let us take any positive integer n. Let a1; b1ð Þ; . . .; an; bnð Þ be open intervalscontained in a; b½ � satisfying


¼ ;� �

; and b1 � a1ð Þþ � � � þ bn � anð Þ\d:


F b1ð Þ � F a1ð Þð Þþ � � � þ F bnð Þ � F anð Þð Þ¼ F b1ð Þ � F a1ð Þj j þ � � � þ F bnð Þ � F anð Þj j\e|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

F b1ð Þ � F a1ð Þð Þþ � � � þ F bnð Þ � F anð Þð Þ\e:

Since

F b1ð Þ � F a1ð Þ¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j : a ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a; b1½ �f g� sup f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj j : a ¼ s0\s1\ � � �\sM ¼ a1f g is a partition of a; a1½ �f g¼ sup f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj jð Þþ f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jð Þf: a ¼ s0\s1\ � � �\sM ¼ a1f g is a partition of a; a1½ �; and a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �g� sup f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj j : a ¼ s0\s1\ � � �\sM ¼ a1f g is a partition of a; a1½ �f g¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j : a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �f g;

we have

F b1ð Þ � F a1ð Þ ¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jf: a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �g:


Similarly,

F b2ð Þ � F a2ð Þ ¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jf: a2 ¼ t0\t1\ � � �\tN ¼ b2f g is a partition of a2; b2½ �g; etc:

Now,

F b1ð Þ � F a1ð Þð Þþ � � � þ F bnð Þ � F anð Þð Þ¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j : a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �f gþ � � �þ sup f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj j : an ¼ s0\s1\ � � �\sM ¼ bnf g is a partition of an; bn½ �f g¼ sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jð Þþ � � � þ f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj jð Þf: a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �; . . .; an ¼ s0\s1\ � � �\sM ¼ bnf g is a partition of an; bn½ �g

Let a1 ¼ t0\t1\ � � �\tN ¼ b1f g be a partition of a1; b1½ �; . . .; an ¼ s0f\s1\ � � �\sM ¼ bng be a partition of an; bn½ �:

It follows that t0; t1ð Þ; . . .; tN�1; tNð Þ; . . .; s0; s1ð Þ; . . .; sM�1; sMð Þ are disjoint openintervals contained in a; b½ �; and

t1 � t0ð Þþ � � � þ tN � tN�1ð Þð Þþ � � � þ s1 � s0ð Þþ � � �ðþ sM � sM�1ð ÞÞ\ b1 � a1ð Þþ � � � þ bn � anð Þ\d;

and hence

f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jð Þþ � � � þ f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj jð Þ\ e2:

It follows that

F b1ð Þ � F a1ð Þð Þþ � � � þ F bnð Þ � F anð Þð Þ ¼ð Þsup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jð Þþ � � � þ f s1ð Þ � f s0ð Þj j þ � � � þ f sMð Þ � f sM�1ð Þj jð Þf: a1 ¼ t0\t1\ � � �\tN ¼ b1f g is a partition of a1; b1½ �; . . .; an ¼ s0\s1\ � � �\sM ¼ bnf g is a partition of an; bn½ �g

� e2

\eð Þ;

and hence

F b1ð Þ � F a1ð Þð Þþ � � � þ F bnð Þ � F anð Þð Þ\e:

■)Since F is an absolutely continuous function on a; b½ �, and f : a; b½ � ! R is an

absolutely continuous function, F � f ;Fþ f are absolutely continuous functions ona; b½ �:

Conclusion 3.161 Let a\b: Let f : a; b½ � ! R be any absolutely continuousfunction. Let F : x 7! sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj jf : a ¼ t0\t1f\ � � �\tN ¼ xg is a partition of a; x½ �g be a function from a; b½ � to 0;1½ �: Then F,F � f ;Fþ f are monotonically increasing, and F, F � f ;Fþ f are absolutelycontinuous functions on a; b½ �:


3.6 Vitali–Caratheodory Theorem

Note 3.162 Let a\b: Let f : a; b½ � ! R be any absolutely continuous function.Then, clearly, f is differentiable a.e. on a; b½ �; f 0 2 L1 mð Þ; and for every x 2 a; b½ �;

Zxa


(Solution Let

F : x 7! sup f t1ð Þ � f t0ð Þj j þ � � � þ f tNð Þ � f tN�1ð Þj j :fa ¼ t0\t1\ � � �\tN ¼ xf g is a partition of a; x½ �g

be the function from a; b½ � to 0;1½ �: By Conclusion 3.161, F � f ;Fþ f aremonotonically increasing, and F � f ;Fþ f are absolutely continuous functions ona; b½ �: It follows that 1

2 F � fð Þ; 12 Fþ fð Þ are monotonically increasing, and12 F � fð Þ; 12 Fþ fð Þ are absolutely continuous functions on a; b½ �: Now, byConclusion 3.156, 12 F � fð Þ is differentiable a.e. on a; b½ �; 1

2 F0 � f 0ð Þ 2 L1; and forevery x 2 a; b½ �;

Zxa

12

F0 tð Þ � f 0 tð Þð Þ�

dt ¼ 12

F xð Þ � f xð Þð Þ � 12

F að Þ � f að Þð Þ:

Also, 12 Fþ fð Þ is differentiable a.e. on a; b½ �; 1

2 F0 þ f 0ð Þ 2 L1; and, for everyx 2 a; b½ �;

Zxa

12

F0 tð Þþ f 0 tð Þð Þ�

dt ¼ 12

F xð Þþ f xð Þð Þ � 12

F að Þþ f að Þð Þ:

Since 12 Fþ fð Þ is differentiable a.e. on a; b½ �; and 1

2 F � fð Þ is differentiable a.e.on a; b½ �; f ¼ð Þ 12 Fþ fð Þ � 1

2 F � fð Þ is differentiable a.e. on a; b½ �; and hence f isdifferentiable a.e. on a; b½ �: Since 1

2 F0 þ f 0ð Þ 2 L1 mð Þ, 12 F0 � f 0ð Þ 2 L1 mð Þ; and

L1 mð Þ is a linear space,

f 0 ¼ð Þ 12

F0 þ f 0ð Þ � 12

F0 � f 0ð Þ 2 L1 mð Þ;

and hence f 0 2 L1 mð Þ: Next, for every x 2 a; b½ �;


Zxa

f 0 tð Þdt ¼Zxa

12

F0 tð Þþ f 0 tð Þð Þ � 12


dt

¼Zxa

12

F0 tð Þþ f 0 tð Þð Þ�

dt �Zxa

12


dt

¼ 12

FðxÞþ f ðxÞð Þ � 12

FðaÞþ f ðaÞð Þ�

� 12

FðxÞ � f ðxÞð Þ � 12

FðaÞ � f ðaÞð Þ�

¼ f ðxÞ � f ðaÞ;

so for every x 2 a; b½ �; Zxa


■)

Conclusion 3.163 Let a\b: Let f : a; b½ � ! R be any absolutely continuousfunction. Then, f is differentiable a.e. on a; b½ �; f 0 2 L1 mð Þ; and, for every x 2 a; b½ �;

Zxa


Note 3.164 Let a\b: Let f : a; b½ � ! 0;1½ Þ be a measurable function. Let f 2L1 mð Þ; and f 6¼ 0:

By Lemma 1.98, there exists a sequence snf g of simple measurable functionssn : a; b½ � ! 0;1½ Þ such that for every x in a; b½ �;

0� s1 xð Þ� s2 xð Þ� s3 xð Þ� � � � ; and limn!1

sn xð Þ ¼ f xð Þ:

Put t1 � s1 � 0ð Þ; t2 � s2 � s1 � 0ð Þ, t3 � s3 � s2 � 0ð Þ; etc.Since each sn is a simple function, each sn a; b½ �ð Þ is a finite set, and hence

tn a; b½ �ð Þ ¼ sn � sn�1ð Þ a; b½ �ð Þ � sn a; b½ �ð Þ � sn�1 a; b½ �ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}is a finite set. It follows that each tn can be expressed as

c1vE1þ � � � þ cnvEn

;

where each ci is positive, and each Ei is a Lebesgue measurable set. So, we cansuppose

t1 þ t2 þ t3 þ � � � ¼ a1vA1þ a2vA2

þ � � � ;

3.6 Vitali–Caratheodory Theorem 511

where each ai [ 0; and each Ai is a Lebesgue measurable set. Since f 2 L1 mð Þ; andf : a; b½ � ! 0;1½ Þ; Z

a;b½ �

f dm ¼Za;b½ �

fj j dm\1

|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}:

Clearly,

a1vA1þ a2vA2

þ � � � ¼ t1 þ t2 þ t3 þ � � � ¼ limn!1

sn|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f ;

so

1[ð ÞZa;b½ �

f dm ¼ a1m A1ð Þþ a2m A2ð Þþ � � � ;

and hence

a1m A1ð Þþ a2m A2ð Þþ � � �

is a convergent series of nonnegative real numbers.Let us take any e[ 0:Since each An is a Lebesgue measurable set, by Conclusion 1.258(7), there exist

a closed set Fn; and an open set Vn such that Fn � An � Vn; andm Vn � Fnð Þ\ e

an2nþ 1 :

Let

v : x 7! a1vV1xð Þþ a2vV2

xð Þþ � � ��

be the function from a; b½ � to 0;1½ �: Observe that

a1vV1xð Þþ a2vV2

xð Þþ � � � ¼ sup a1vV1þ � � � þ anvVn

: n is a positive integer� � �

ðxÞ

Problem 3.165 v : a; b½ � ! 0;1½ � is lower semicontinuous.

(Solution Since V1 is open, by Lemma 1.166, vV1: a; b½ � ! 0;1½ � is lower

semicontinuous. Similarly, vV2: a; b½ � ! 0;1½ � is lower semicontinuous, etc. It

follows that, for every positive integer n,

a1vV1þ � � � þ anvVn

� �: a; b½ � ! 0;1½ �

is lower semicontinuous, and hence by Lemma 1.168


v ¼ð Þsup a1vV1þ � � � þ anvVn

: n is a positive integer�

is lower semicontinuous. Thus, v : a; b½ � ! 0;1½ � is lower semicontinuous. ■)Since for every x 2 a; b½ �;

0� a1vV1xð Þþ a2vV2

xð Þþ � � ��

¼ v xð Þð Þ;

v is bounded below. Since a1m A1ð Þþ a2m A2ð Þþ � � � is a convergent series ofnonnegative real numbers, there exists a positive integer N such that

aNþ 1m ANþ 1ð Þþ aN þ 2m ANþ 2ð Þþ � � �ð Þ\ e2:

Let

u : x 7! a1vF1xð Þþ � � � þ aNvFN

xð Þ� �

be the function from a; b½ � to 0;1½ �:

Problem 3.166 u : a; b½ � ! 0;1½ � is upper semicontinuous.

(Solution Since F1 is closed, by Lemma 1.167, vF1: a; b½ � ! 0;1½ � is upper

semicontinuous. Similarly, vF2: a; b½ � ! 0;1½ � is upper semicontinuous, etc. It

follows that, for every positive integer n,

u ¼ð Þ a1vF1þ � � � þ aNvFN

� �: a; b½ � ! 0;1½ �

is upper semicontinuous. ■)Since

u a; b½ �ð Þ ¼ð Þ a1vF1þ � � � þ aNvFN

� �a; b½ �ð Þ

is a finite set of nonnegative real numbers, u is bounded above. Thus,u : a; b½ � ! 0;1½ Þ:

Problem 3.167 u� f � v:

(Solution Since for every positive integer n, Fn � An; we have

u ¼ a1vF1þ � � � þ aNvFN

� a1vA1þ � � � þ aNvAN|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} � a1vA1

þ a2vA2þ � � � ¼ f ;

and hence u� f : Since for every positive integer n, An � Vn; we have

f ¼ð Þa1vA1þ a2vA2

þ � � � � a1vV1þ a2vV2

þ � � � ¼ vð Þ;

and hence f � v: ■)


Since

u : x 7! a1vF1xð Þþ � � � þ aNvFN

xð Þ� �

;

and each Fn is Lebesgue measurable, u : a; b½ � ! 0;1½ Þ is a measurable function.Since

v : x 7! a1vV1xð Þþ a2vV2

xð Þþ � � ��

;

and each Vn is Lebesgue measurable, v : a; b½ � ! 0;1½ Þ is a measurable function.Since u and v are measurable functions, v� uð Þ is a measurable function. SinceZa;b½ �

v� uð Þdm ¼Za;b½ �

a1vV1þ a2vV2

þ � � ��

� a1vF1þ � � � þ aNvFN

� �� dm

¼Za;b½ �

aN þ 1vVN þ 1þ aNþ 2vVN þ 2

þ � � ��

þ a1 vV1� vF1

� �þ � � � þ aN vVN

� vFN

� �� dm

¼Za;b½ �

aN þ 1vVN þ 1þ aNþ 2vVN þ 2

þ � � ��

þ a1 vV1� vF1

� �þ a2 vV2

� vF2

� �þ � � �

� �� aNþ 1 vVNþ 1

� vFN þ 1

� �þ aN þ 2 vVN þ 2

� vFN þ 2

� �þ � � �

� ��dm

¼Za;b½ �

aN þ 1vFN þ 1þ aNþ 2vFN þ 2

þ � � ��

þ a1 vV1� vF1

� �þ a2 vV2

� vF2

� �þ � � �

� �� dm

¼Za;b½ �


þ � � ��

þ a1 vV1�F1

� �þ a2 vV2�F2

� �þ � � �

� �� dm

¼Za;b½ �


þ � � ��

dmþZa;b½ �

a1 vV1�F1

� �þ a2 vV2�F2

� �þ � � �

� �dm

¼Za;b½ �


þ � � ��

dmþ a1m V1 � F1ð Þþ a2m V2 � F2ð Þþ � � �ð Þ

¼Za;b½ �


þ � � ��

dmþ a1e

a121þ 1 þ a2e

a222þ 1 þ � � ��

¼Za;b½ �


þ � � ��

dmþ e2

�Za;b½ �

aNþ 1vAN þ 1þ aNþ 2vANþ 2

þ � � ��

dmþ e2

¼ aNþ 1m ANþ 1ð Þþ aNþ 2m ANþ 2ð Þþ � � �ð Þ þ e2\

e2þ e

2¼ e;

we haveRa;b½ � v� uð Þdm\e:


Conclusion 3.168 Let a\b: Let f : a; b½ � ! 0;1½ Þ be a measurable function. Letf 2 L1 mð Þ; and f 6¼ 0: Let e[ 0: Then there exist functions u : a; b½ � ! 0;1½ Þ; andv : a; b½ � ! 0;1½ Þ such that

1. u� f � v;2. u : a; b½ � ! 0;1½ Þ is upper semicontinuous and bounded above,3. v : a; b½ � ! 0;1½ � is lower semicontinuous and bounded below,

4.R ba v� uð Þdm\e:

Theorem 3.169 Let a\b: Let f : a; b½ � ! R be a measurable function. Let f 2L1 mð Þ: Let e[ 0: Then there exist functions u : a; b½ � ! 0;1½ Þ; and v : a; b½ � !0;1½ Þ such that



Proof Since f : a; b½ � ! R is a measurable function, f þ : a; b½ � ! 0;1½ Þ is ameasurable function, f� : a; b½ � ! 0;1½ Þ is a measurable function, and f ¼f þ � f�: Since f 2 L1 mð Þ; we have f þ �ð Þ fj j 2 L1 mð Þ; and hence f þ 2 L1 mð Þ:Similarly, f� 2 L1 mð Þ:

By Conclusion 3.168, there exist functions u1 : a; b½ � ! 0;1½ Þ; and v1 :a; b½ � ! 0;1½ Þ such that

1. u1 � f þ � v1;2. u1 : a; b½ � ! 0;1½ Þ is upper semicontinuous and bounded above,3. v1 : a; b½ � ! 0;1½ � is lower semicontinuous and bounded below,

4.R ba v1 � u1ð Þdm\ e

2 :

Also, there exist functions u2 : a; b½ � ! 0;1½ Þ; and v2 : a; b½ � ! 0;1½ Þ such that

I. u2 � f� � v2;II. u2 : a; b½ � ! 0;1½ Þ is upper semicontinuous, and bounded above,III. v2 : a; b½ � ! 0;1½ � is lower semicontinuous, and bounded below,

IV.R ba v2 � u2ð Þdm\ e

2 :

From 1, and I,

u1 � v2 � f þ � f� � v1 � u2;

that is

u1 � v2 � f � v1 � u2:

Since v2 is lower semicontinuous, �v2 is upper semicontinuous. Now, since u1 isupper semicontinuous, u1 � v2 is upper semicontinuous. Similarly, v1 � u2 is lower


semicontinuous. Clearly, u1 � v2 is bounded above, and v1 � u2 is bounded below.Also,

Zba

v1 � u2ð Þ � u1 � v2ð Þð Þdm ¼Zba

v1 � u1ð Þþ v2 � u2ð Þð Þdm

¼Zba

v1 � u1ð ÞdmþZba

v2 � u2ð Þdm\ e2þ e

2¼ e

Let us put u � u1 � v2; and v � v1 � u2: We have



Theorem 3.169, known as Vitali–Caratheodory theorem, is due to G. Vitali(23.08.1875–29.02.1932, Italian), and C. Caratheodory (13.09.1873–02.02.1950,Greek). Vitali was the first to give an example of a non-measurable set of realnumbers. His covering theorem is a fundamental result. Caratheodory made sig-nificant contributions to the theory of a real variable, calculus of variations, andmeasure theory.

Note 3.170 Let a\b: Let f : a; b½ � ! R be a function differentiable at every pointof a; b½ �. Let f 0 2 L1 mð Þ:

Let us take any e[ 0:Since f 0 2 L1 mð Þ, f 0 : a; b½ � ! R is a measurable function. Now, by

Theorem 3.169, there exist functions u : a; b½ � ! 0;1½ Þ; and v : a; b½ � ! 0;1½ Þsuch that

1. u� f 0\v;2. u : a; b½ � ! 0;1½ Þ is upper semicontinuous and bounded above,3. v : a; b½ � ! 0;1½ � is lower semicontinuous and bounded below,


From 1, v� f 0ð Þ � v� uð Þ; and hence

Zba

v dm�Zba

f 0dm ¼Zba

v� f 0ð Þdm�Zba

v� uð Þdm

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\e:


Thus,R ba v dm\

R ba f

0dmþ e: Let us take any g[ 0: Let

Fg : x 7!Zxa

v dm� f xð Þþ f að Þþ g � x� að Þ

0@ 1A|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼Zxa

v tð Þdt � f xð Þþ f að Þþ g � x� að Þ

be the function from a; b½ � to R: Clearly, Fg is continuous.Let us fix any x 2 a; b½ Þ:Since v : a; b½ � ! 0;1½ � is lower semicontinuous, v�1 f 0 xð Þ;1ð �ð Þ is open in

a; b½ �: Since f 0\v; we have x 2 v�1 f 0 xð Þ;1ð �ð Þ: Since x 2 v�1 f 0 xð Þ;1ð �ð Þ; andv�1 f 0 xð Þ;1ð �ð Þ is open in a; b½ �; there exists lx [ 0 such that x; xþ lxð Þ �v�1 f 0 xð Þ;1ð �ð Þ: Since f 0 xð Þ exists, there exists mx [ 0 such that for everyt 2 x; xþ mxð Þ;

f tð Þ � f xð Þt � x

\f 0 xð Þþ g:

Put dx � min lx; mxf g [ 0ð Þ:It follows that for every t 2 x; xþ dxð Þ;

f tð Þ � f xð Þt � x

\f 0 xð Þþ g; and f 0 xð Þ\v tð Þ:

Now, for every t 2 x; xþ dxð Þ; we have

Fg tð Þ � Fg xð Þ ¼Z t

a

v dm� f tð Þþ f að Þþ g � t � að Þ

0@ 1A�

Zxa


0@ 1A¼

Zxa

v dmþZ t

x

v dm

0@ 1A� f tð Þþ f að Þþ g � t � að Þ

0@ 1A�

Zxa


0@ 1A


¼Z t

x

v dm� f tð Þ � f xð Þð Þþ g � t � xð Þ�Z t

x

f 0 xð Þdm

� f tð Þ � f xð Þð Þþ g � t � xð Þ¼ f 0 xð Þð Þ t � xð Þ � f tð Þ � f xð Þð Þþ g � t � xð Þ[ f 0 xð Þð Þ t � xð Þ � f 0 xð Þþ gð Þ � t � xð Þþ g � t � xð Þ ¼ 0:

Thus, for every x 2 a; b½ Þ; and for every t 2 x; xþ dxð Þ, Fg xð Þ\Fg tð Þ:

Problem 3.171 0�Fg bð Þ:

(Solution If not, otherwise, let Fg bð Þ\0: We have to arrive at a contradiction.Since

Fg að Þ ¼Zaa

v dm� f að Þþ f að Þþ g � a� að Þ

0@ 1A ¼ 0;

and Fg : a; b½ � ! R is continuous, a 2ð Þ Fg� ��1 0ð Þ is a closed subset of a; b½ �; and

hence sup Fg� ��1 0ð Þ� �

exists, and

sup Fg� ��1

0ð Þ� ��

2 Fg� ��1

0ð Þ� �

:

This shows that

sup Fg� ��1

0ð Þ� ��

¼ max Fg� ��1

0ð Þ� ��

;

and hence

Fg max Fg� ��1

0ð Þ� ��

¼ 0:

Case I: when max Fg� ��1

0ð Þ� �

¼ b: Here, Fg bð Þ ¼ 0: This is a contradiction.

Case II: when max Fg� ��1

0ð Þ� �

\b: In this case, for every

t 2 max Fg� ��1

0ð Þ� �

; max Fg� ��1

0ð Þ� �

þ dmax Fgð Þ�1

0ð Þ� ��

� max Fg� ��1

0ð Þ� �

; b� ��

;

0 ¼ð ÞFg max Fg� ��1

0ð Þ� ��

\Fg tð Þ:

Thus, there exists t0 2 max Fg� ��1

0ð Þ� �

; b� �

such that Fg t0ð Þ is positive. SinceFg t0ð Þ is positive, Fg bð Þ\0; t0\b; and Fg : a; b½ � ! R is continuous, there exists


x0 2 t0; bð Þ such that Fg x0ð Þ ¼ 0: Since x0 2 t0; bð Þ, t0\x0: Since Fg x0ð Þ ¼ 0, x0 2Fg� ��1

0ð Þ; and hence t0\ð Þx0 �max Fg� ��1

0ð Þ� �

: This contradicts

t0 2 max Fg� ��1

0ð Þ� �

; b� �

:

So, in all cases, we arrive at a contradiction. ■)Since

0�Fg bð Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl} ¼Zba

v dm� f bð Þþ f að Þþ g � b� að Þ;

and g is any positive real number, we get

0�Zba

v dm� f bð Þþ f að Þ\Zba

f 0dmþ e

0@ 1A� f bð Þþ f að Þ:

Now, since e is any positive real number, we get

0�Zba

f 0dm

0@ 1A� f bð Þþ f að Þ;

that is

f bð Þ � f að Þ�Zba

f 0dm:

Conclusion 3.172 Let a\b: Let f : a; b½ � ! R be a function differentiable at everypoint of a; b½ �. Let f 0 2 L1 mð Þ: Then

f bð Þ � f að Þ�Zba

f 0dm:

Theorem 3.173 Let a\b: Let f : a; b½ � ! R be a function differentiable at everypoint of a; b½ �. Let f 0 2 L1 mð Þ: Then

f bð Þ � f að Þ ¼Zba

f 0dm:


Proof In view of Conclusion 3.172, it suffices to show that

Zba

f 0dm� f bð Þ � f að Þ:

Since f : a; b½ � ! R is a function differentiable at every point of a; b½ �, �fð Þ :a; b½ � ! R is a function differentiable at every point of a; b½ �: Since f 0 2 L1 mð Þ,�fð Þ0¼

� �� f 0ð Þ 2 L1 mð Þ; and hence �fð Þ02 L1 mð Þ: Since �fð Þ : a; b½ � ! R is a

function differentiable at every point of a; b½ �; and �fð Þ02 L1 mð Þ; by Conclusion3.172,

f að Þ � f bð Þ ¼ � f bð Þð Þ � f að Þð Þ ¼ �fð Þ bð Þ � f|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} að Þ�Zba

�fð Þ0dm

¼Zba

� f 0ð Þð Þdm ¼ �Zba

f 0dm;

and hence

Zba

f 0dm� f bð Þ � f að Þ:

■Theorem 3.173 is known as the fundamental theorem of calculus.

Note 3.174 Let r be any positive real number. Let F : x : x 2 Rk and xj j � r�

!Rk be a continuous mapping. Let e 2 0; rð Þ: Suppose that for every y 2x : x 2 Rk and xj j ¼ r�

; F yð Þ 2 z : z 2 Rk and z� yj j\e�

: Let a 2 Rk such thataj j\ r � eð Þ:

Problem 3.175 a 2 F x : x 2 Rk and xj j\r� � �

:

(Solution If not, otherwise, let a 62 F x : x 2 Rk and xj j\r� � �

; that is, for everyx 2 Rk satisfying xj j\r, a 6¼ F xð Þ: We have to arrive at a contradiction.

Problem 3.176 a 62 F x : x 2 Rk and xj j ¼ r� � �

:

(Solution Let us take any y 2 x : x 2 Rk and xj j ¼ r�

: We have to show thata 6¼ F yð Þ: Since y 2 x : x 2 Rk and xj j ¼ r

� , yj j ¼ r: Now, by the supposition,

F yð Þ 2 z : z 2 Rk and z� yj j\e�

; and hence


r � F yð Þj j ¼ yj j � F yð Þj j � F yð Þ � yj j\e|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} :It follows that r � eð Þ\ F yð Þj j; and hence F yð Þj j¥ r � eð Þ: Now, since

aj j\ r � eð Þ; we have a 6¼ F yð Þ: ■)Since

a 62 F x : x 2 Rk and xj j ¼ r� � �

;

and

a 62 F x : x 2 Rk and xj j\r� � �

;

we have

a 62 F x : x 2 Rk and xj j ¼ r� � �

[F x : x 2 Rk and xj j\r� � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ F x : x 2 Rk and xj j ¼ r�

[ x : x 2 Rk and xj j\r� � �

¼ F x : x 2 Rk and xj j � r� � �

:

Thus, a 62 F x : x 2 Rk and xj j � r� � �

:

Now, we can define a mapping

G : y 7! ra� F yð Þj j a� F yð Þð Þ

from x : x 2 Rk and xj j � r�

to x : x 2 Rk and xj j ¼ r�

� x : x 2 Rk��

and xj j � rgÞ:Since F : x : x 2 Rk and xj j � r

� ! Rk is a continuous mapping, G :

y 7! ra�F yð Þj j a� F yð Þð Þ is a continuous map from x : x 2 Rk and xj j � r

� to

x : x 2 Rk and xj j � r�

: Now, by the Brouwer’s fixed point theorem in ‘algebraictopology’, there exists b 2 Rk such that bj j � r; and r

a�F bð Þj j a� F bð Þð Þ ¼ b: It

follows that bj j ¼ r; and b � a� F bð Þð Þ ¼ r a� F bð Þj j:Since a 62 F x : x 2 Rk and xj j � r

� � �; and b 2 x : x 2 Rk and xj j � r

� ; we

have a 6¼ F bð Þ; and hence

r F bð Þ � bj j � eð Þ ¼ r b� F bð Þj j � eð Þ ¼ r b� F bð Þj j � re ¼ bj j b� F bð Þj j � re

� b � b� F bð Þð Þ � re ¼ r r � eð Þþ b � b� F bð Þð Þ � r2 [ r aj j þ b � b� F bð Þð Þ � r2

¼ bj j aj j þ b � b� F bð Þð Þ � r2 � b � aþ b � b� F bð Þð Þ � r2 ¼ b � aþ b � b� F bð Þð Þ � bj j2

¼ b � aþ b � b� F bð Þð Þ � b � b ¼ b � a� F bð Þð Þ ¼ r a� F bð Þj j[ 0|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} :


Thus e\ F bð Þ � bj j: Since bj j ¼ r; by the supposition, F bð Þ � bj j\e: This is acontradiction. ■)

Conclusion 3.177 Let r be any positive real number. Let F : B 0; r½ � ! Rk be acontinuous mapping. Let e 2 0; rð Þ: Suppose that for every y 2 S 0; rð Þ, F yð Þ 2B y; eð Þ: Then B 0; r � eð Þ � F B 0; rð Þð Þ:

Note 3.178 Let V be an open subset of Rk: Let T : V ! Rk be a continuousmap. Let 0 2 V ; and T 0ð Þ ¼ 0: Let T be differentiable at 0. Let T 0 0ð Þ be a 1-1operator.

Since 0 2 V ; and V is open, there exists r0 [ 0 such that B 0; r0ð Þ � V :

Problem 3.179 For every r 2 0; r0ð Þ, T B 0; rð Þð Þ is Lebesgue measurable.

(Solution For this purpose, let us fix any r 2 0; r0ð Þ: We have to show thatT B 0; rð Þð Þ is Lebesgue measurable. Here,

T B 0; rð Þð Þ ¼ T [1n¼2B 0; 1� 1

n

� r

� �� ¼ [1

n¼2T B 0; 1� 1n

� r

� �� :

Since each B 0; 1� 1n

� �r

� �is compact, and T : V ! Rk is continuous, each

T B 0; 1� 1n

� �r

� �� is compact, and hence each T B 0; 1� 1

n

� �r

� �� is closed. It fol-

lows that each T B 0; 1� 1n

� �r

� �� is Lebesgue measurable, and hence

T B 0; rð Þð Þ ¼ð Þ [1n¼2T B 0; 1� 1

n

� r

� �� is Lebesgue measurable. Thus, T B 0; rð Þð Þ is a Lebesgue measurable subset ofRk:■)

Since T is differentiable at 0, T 0 0ð Þð Þ : Rk ! Rk is a linear operator. Here,T 0 0ð Þð Þ : Rk ! Rk is 1-1, so T 0 0ð Þð Þ�1: Rk ! Rk exists, and is a linear operator.Since T 0 0ð Þð Þ�1: Rk ! Rk is a linear operator, T 0 0ð Þð Þ�1: Rk ! Rk is continuous.

Now, since T : V ! Rk is a continuous map, their composite T 0 0ð Þð Þ�1�T� �

:

V ! Rk is continuous. It follows, as above, that for every r 2 0; r0ð Þ;T 0 0ð Þð Þ�1�T

� �B 0; rð Þð Þ is Lebesgue measurable.

Problem 3.180 limr!0m T 0 0ð Þð Þ�1�Tð Þ B 0;rð Þð Þð Þ

m B 0;rð Þð Þ ¼ 1:

(Solution For this purpose, let us take any e 2 0; 1ð Þ: Since

T 0 0ð Þð Þ�1�T� �0

0ð Þ ¼ T 0 0ð Þð Þ�1� �0

T 0ð Þð Þ� �

� T 0 0ð Þð Þ

¼ T 0 0ð Þð Þ�1� �0

0ð Þ� �

� T 0 0ð Þð Þ ¼ T 0 0ð Þð Þ�1� �

� T 0 0ð Þð Þ ¼ I;


where I denotes the identity operator on Rk; we have

T 0 0ð Þð Þ�1�T� �0

0ð Þ ¼ I;

and hence

limh!0

T 0 0ð Þð Þ�1 T hð Þð Þ � h��

hj j ¼ limh!0

T 0 0ð Þð Þ�1 T hð Þð Þ � 0� h��

hj j

¼ limh!0

T 0 0ð Þð Þ�1 T hð Þð Þ � T 0 0ð Þð Þ�1 0ð Þ � h��

hj j ¼ limh!0

T 0 0ð Þð Þ�1 T hð Þð Þ � T 0 0ð Þð Þ�1 T 0ð Þð Þ � h��

hj j

¼ limh!0

T 0 0ð Þð Þ�1�T� �

0þ hð Þ � T 0 0ð Þð Þ�1�T� �

0ð Þ � I hð Þ��

hj j ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus,

limh!0


hj j ¼ 0:

It follows that there exists d[ 0 such that d\min 1; r0f g; and for every h sat-isfying 0\ h� 0j j\d;


hj j � 0

��\e;

that is, d\min 1; r0f g; and for every nonzero h satisfying hj j\d;

T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \e hj j:

Problem 3:181 For every r 2 0; dð Þ, B 0; r � reð Þ � T 0 0ð Þð Þ�1�T� �

B 0; rð Þð Þ:

(Solution Let us fix any r 2 0; dð Þ: Now, since e 2 0; 1ð Þ; er 2 0; rð Þ: By,Conclusion 3.177, it suffices to show that for every h 2 S 0; rð Þ;T 0 0ð Þð Þ�1�T

� �hð Þ 2 B h; erð Þ. That is, it suffices to show that for every h satisfying

hj j ¼ r; T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \er: For this purpose, let us take any h satisfying

hj j ¼ r: We have to show that

T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \er:


Since hj j ¼ r; and r 2 0; dð Þ; 0\ hj j\d; and hence

T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \e hj j ¼ erð Þ:

Thus, T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \er: ■)

Problem 3:182 For every r 2 0; dð Þ; T 0 0ð Þð Þ�1�T� �

B 0; rð Þð Þ � B 0; rþ erð Þ:

(Solution Let us fix any r 2 0; dð Þ: Let h 2 Rk such that hj j\r: We have to show

that T 0 0ð Þð Þ�1�T� �

hð Þ�� \rþ er: Since hj j\r; and r 2 0; dð Þ; hj j\d: If h ¼ 0;

then

T 0 0ð Þð Þ�1�T� �

hð Þ�� ¼ T 0 0ð Þð Þ�1�T

� �0ð Þ

�� ¼ T 0 0ð Þð Þ�1 T 0ð Þð Þ�� ¼ T 0 0ð Þð Þ�1 0ð Þ

�� ¼ 0j j ¼ 0\rþ er;

so we only consider the case when h is nonzero. Since h is nonzero, and hj j\d;

T 0 0ð Þð Þ�1�T� �

hð Þ�� r\ T 0 0ð Þð Þ�1�T

� �hð Þ

�� hj j

� T 0 0ð Þð Þ�1�T� �

hð Þ � h�� \e hj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}\er;

and hence

T 0 0ð Þð Þ�1�T� �

hð Þ�� \rþ er:

■)Since for every r 2 0; dð Þ;

B 0; r � reð Þ � T 0 0ð Þð Þ�1�T� �

B 0; rð Þð Þ � B 0; rþ erð Þ;

we have, for every r 2 0; dð Þ;

m B 0; r � reð Þð Þ�m T 0 0ð Þð Þ�1�T� �

B 0; rð Þð Þ� �

�m B 0; rþ erð Þð Þ;

and hence for every r 2 0; dð Þ;


1� eð Þk ¼ r � erð Þk

rk¼ m B 0; r � reð Þð Þ

m m B 0; rð Þð Þð Þ �m T 0 0ð Þð Þ�1�T� �


m m B 0; rð Þð Þð Þ � m B 0; rþ erð Þð Þm m B 0; rð Þð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ rþ erð Þk

rk¼ 1þ eð Þk:

Since for every r 2 0; dð Þ;

1� eð Þk �m T 0 0ð Þð Þ�1�T� �


m m B 0; rð Þð Þð Þ � 1þ eð Þk;


m T 0 0ð Þð Þ�1�T� �


m m B 0; rð Þð Þð Þ

0@ 1A1k

�1

��\e:

This shows that

limr!0

m T 0 0ð Þð Þ�1�T� �


m m B 0; rð Þð Þð Þ

0@ 1A1k

¼ 1;

and hence

limr!0

m T 0 0ð Þð Þ�1�T� �


m m B 0; rð Þð Þð Þ ¼ 1:

■)Now, on using Conclusion 1.258(14),

1 ¼ limr!0

m T 0 0ð Þð Þ�1�T� �


m B 0; rð Þð Þ ¼ limr!0

det T 0 0ð Þð Þ�1� ��

� m T B 0; rð Þð Þð Þm B 0; rð Þð Þ

¼ limr!0

1det T 0 0ð Þð Þj j � m T B 0; rð Þð Þð Þ

m B 0; rð Þð Þ ¼ 1det T 0 0ð Þð Þj j limr!0

m T B 0; rð Þð Þð Þm B 0; rð Þð Þ :


Thus,

limr!0

m T B 0; rð Þð Þð Þm B 0; rð Þð Þ ¼ det T 0 0ð Þð Þj j:

Conclusion 3.183 Let V be an open subset of Rk: Let T : V ! Rk be a continuousmap. Let 0 2 V ; and T 0ð Þ ¼ 0: Let T be differentiable at 0. Let T 0 0ð Þ be a 1-1operator. Then

limr!0


Theorem 3.184 Let V be an open subset of Rk: Let T : V ! Rk be a continuousmap. Let 0 2 V ; and T 0ð Þ ¼ 0: Let T be differentiable at 0. Then

limr!0


Proof Case I: when T 0 0ð Þ is 1-1. By Conclusion 3.183,

limr!0


Case II: when T 0 0ð Þ : Rk ! Rk is not 1-1.In this case, dim T 0 0ð Þð Þ Rk

� �� \dim Rk

� �¼ kð Þ; and hence m T 0 0ð Þð Þ Rk

� �� ¼

0: Since T 0 0ð Þ is not 1-1, det T 0 0ð Þð Þ ¼ 0:

Problem 3:185 limr!0m T B 0;rð Þð Þð Þm B 0;rð Þð Þ ¼ 0:

(Solution For this purpose, let us take any e[ 0:Since for every u 2 B 0; 1ð Þ;

T 0 0ð Þð Þ uð Þj j � T 0 0ð Þk k uj j � T 0 0ð Þk k1 ¼ T 0 0ð Þk k \1ð Þ;

T 0 0ð Þð Þ B 0; 1ð Þð Þ is a bounded subset of the linear space T 0 0ð Þð Þ Rk� �

: SinceT 0 0ð Þð Þ B 0; 1ð Þð Þ is a bounded subset of the linear space T 0 0ð Þð Þ Rk

� �;

dim T 0 0ð Þð Þ Rk� ��

\k; and

x 7! distance of x from T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ

is continuous, there exists g[ 0 such that


m x : distance of x from T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ\gf gð Þ\e:

Since T is differentiable at 0,

limh!0

T hð Þ � T 0 0ð Þð Þ hð Þj jhj j ¼ lim

h!0

T hð Þ � 0� T 0 0ð Þð Þ hð Þj jhj j

¼ limh!0

T 0þ hð Þ � T 0ð Þ � T 0 0ð Þð Þ hð Þj jhj j ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

limh!0

T hð Þ � T 0 0ð Þð Þ hð Þj jhj j ¼ 0:

It follows that there exists d[ 0 such that for every h 2 Rk satisfying 0\ hj j\d;we have h 2 V ; and

T hð Þ � T 0 0ð Þð Þ hð Þj jhj j � 0

�� \g;

That is, for every h 2 Rk satisfying 0\ hj j\d; we have h 2 V ; and

T hð Þ � T 0 0ð Þð Þ hð Þj j\g hj j:

Problem 3:186 For every r 2 0; dð Þ; T B 0; rð Þð Þ � x : distance of from T 0 0ð Þð ÞðfB 0; rð Þð ÞÞ\grg:

(Solution Let us take any h 2 Rk satisfying hj j\r: We have to show thatdistance of T hð Þ from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\gr: It suffices to show thatT hð Þ � T 0 0ð Þð Þ hð Þj j\gr: Since hj j\r; and r 2 0; dð Þ, hj j\d:If h ¼ 0; then

T hð Þ � T 0 0ð Þð Þ hð Þj j ¼ T 0ð Þ � T 0 0ð Þð Þ 0ð Þj j ¼ 0� T 0 0ð Þð Þ 0ð Þj j¼ T 0 0ð Þð Þ 0ð Þj j ¼ 0j j ¼ 0\gr;

so we only consider the case when h is nonzero. Since h is nonzero and hj j\d;

T hð Þ � T 0 0ð Þð Þ hð Þj j\g hj j \grð Þ;

and hence T hð Þ � T 0 0ð Þð Þ hð Þj j\gr. ■)



T B 0; rð Þð Þ � x : distance of x from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\grf g;


m T B 0; rð Þð Þð Þ�m x : distance of x from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\grf gð Þ:


T 0 0ð Þð Þ B 0; rð Þð Þ ¼ T 0 0ð Þð Þ r B 0; 1ð Þð Þð Þ ¼ r T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ;


x : distance of x from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\grf g ¼ x : distance of x from r T 0 0ð Þð Þ B 0; 1ð Þð Þð Þð Þ\grf g¼ rx : distance of x from T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ\gf g;

and hence for every r 2 0; dð Þ;

m x : distance of x from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\grf gð Þ¼ m rx : distance of x from T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ\gf gð Þ¼ rk � m x : distance of x from T 0 0ð Þð Þ B 0; 1ð Þð Þð Þ\gf gð Þ\rk � e

¼ m B 0; rð Þð Þm B 0; 1ð Þð Þ � e:

Thus, for every r 2 0; dð Þ;

m T B 0; rð Þð Þð Þ�m x : distance of x from T 0 0ð Þð Þ B 0; rð Þð Þð Þ\grf gð Þ\m B 0; rð Þð Þm B 0; 1ð Þð Þ e:

Hence, for every r 2 0; dð Þ;

m T B 0; rð Þð Þð Þm B 0; rð Þð Þ � 0

�� \ em B 0; 1ð Þð Þ :

This shows that limr!0m T B 0;rð Þð Þð Þm B 0;rð Þð Þ ¼ 0: ■)

Since limr!0m T B 0;rð Þð Þð Þm B 0;rð Þð Þ ¼ 0 ¼ det T 0 0ð Þð Þj jð Þ; we have

limr!0



Thus, in all cases,

limr!0


■)

Theorem 3.187 Let V be an open subset of Rk: Let T : V ! Rk be a continuousmap. Let a 2 V : Let T be differentiable at a. Then

limr!0

m T B a; rð Þð Þð Þm B a; rð Þð Þ ¼ det T 0 að Þð Þj j:

Proof Since V is an open subset of Rk; and a 2 V ; V � að Þ is an open subset of Rk;and 0 2 V � að Þ: Let S : y 7! T yþ að Þ � T að Þð Þ be the mapping from V � að Þ toRk: Since T : V ! Rk is continuous, S : V � að Þ ! Rk is a continuousmap. Clearly, S 0ð Þ ¼ 0: Since T is differentiable at a, S : y 7! T yþ að Þ � T að Þð Þ isdifferentiable at 0, and S0 0ð Þ ¼ T 0 0þ að Þð Þ � I ¼ T 0 að Þ: Now, by Theorem 3.184,

limr!0

m T B a; rð Þð Þð Þm B a; rð Þð Þ ¼ lim

r!0

m T B a; rð Þð Þð Þm B 0; rð Þþ að Þ ¼ lim

r!0

m T B a; rð Þð Þð Þm B 0; rð Þð Þ

¼ limr!0

m T B 0; rð Þþ að Þð Þm B 0; rð Þð Þ ¼ lim

r!0

m T B 0; rð Þþ að Þ � T að Þð Þm B 0; rð Þð Þ

¼ limr!0

m S B 0; rð Þð Þð Þm B 0; rð Þð Þ ¼ det S0 0ð Þð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ det T 0 að Þð Þj j;

so

limr!0

m T B a; rð Þð Þð Þm B a; rð Þð Þ ¼ det T 0 að Þð Þj j:

■

Note 3.188 Let E be a nonempty subset of Rk: Let T : E ! Rk be any mapping.Let a 2 E:

Observe that, for every positive integer p,

E B a;1p

� \E B a;

1pþ 1

� \ E 3að Þ:


Now, since

T : E ! Rk;T xð Þ � T að Þj j

x� aj j : x 2 B a;1p

� \E; and x 6¼ a

� �is either an empty set or a nonempty set of nonnegative real numbers. It follows that

supT xð Þ � T að Þj j


� \E; and x 6¼ a

� �is either �1 or a member of 0;1½ �:

Clearly, for every positive integer p,


x� aj j : x 2 B a;1

pþ 1

� \E; and x 6¼ a

� �� sup

T xð Þ � T að Þj jx� aj j : x 2 B a;

1p

� \E; and x 6¼ a

� �:

It follows that

inf supT xð Þ � T að Þj j


� \E; and x 6¼ a

� �: p ¼ 1; 2; . . .

� �is either �1 or a member of 0;1½ �: Here

inf supT xð Þ � T að Þj j


� \E; and x 6¼ a

� �: p ¼ 1; 2; . . .

� �is denoted by

lim supcx ! ax 2 E

T xð Þ � T að Þj jx� aj j :

Let us assume that

lim supcx ! ax 2 E

T xð Þ � T að Þj jx� aj j 6¼ 1:

It follows that there exists a positive integer n0 such that

lim supcx ! ax 2 E

T xð Þ � T að Þj jx� aj j \n0:


Here, there exists a positive integer p0 such that


x� aj j : x 2 B a;1p0

� \E; and x 6¼ a

� �\n0;

and hence for every x 2 B a; 1p0

� �\E satisfying x 6¼ a, T xð Þ�T að Þj j

x�aj j \n0:

Thus, for every y 2 B a; 1p0

� �\E;

T yð Þ � T að Þj j � n0 y� aj j:

Put

R1 � x; yð Þ : x; yð Þ 2 E E and T xð Þ � T yð Þj j � 1 x� yj jf g;R2 � x; yð Þ : x; yð Þ 2 E E and T xð Þ � T yð Þj j � 2 x� yj jf g; etc:

Clearly, R1 � R2 � R3 � � � � : For every positive integers p; n; put

Fn;p � x : x 2 E and y 2 B x;1p

� \E ) x; yð Þ 2 Rn

� � �:

Thus,

Fn0;p0 ¼ x : x 2 E and y 2 B x;1p0

� \E ) T yð Þ � T xð Þj j � n0 y� xj j

� � �:

It follows that a 2 Fn0;p0 :

Conclusion 3.189 Let E be a nonempty subset of Rk: Let T : E ! Rk be anymapping. Suppose that for every x 2 E;

lim supcy ! xy 2 E

T yð Þ � T xð Þj jy� xj j 6¼ 1:

For every positive integers p; n; put


� \E ) T yð Þ � T xð Þj j � n y� xj j

� � �:


Then

E ¼ [ Fn;p : n ¼ 1; 2; . . .; and p ¼ 1; 2; . . .�

:

Lemma 3.190 Let E be a nonempty Lebesgue measurable subset of Rk: Letm Eð Þ ¼ 0: Let T : E ! Rk be any mapping. Suppose that for every x 2 E;

lim supcy ! xy 2 E


Then m T Eð Þð Þ ¼ 0:

Proof For every positive integer p; n; put


� \E ) T yð Þ � T xð Þj j � n y� xj j

� � �:

By Conclusion 3.189, E ¼ [ Fn;p : n ¼ 1; 2; . . .; and p ¼ 1; 2; . . .�

: It followsthat

T Eð Þ ¼ [ T Fn;p� �

: n ¼ 1; 2; . . .; and p ¼ 1; 2; . . .�

;

and hence

m T Eð Þð Þ�X

m T Fn;p� ��

: n ¼ 1; 2; . . .; and p ¼ 1; 2; . . .�

:

It suffices to show that each m T Fn;p� ��

¼ 0:For this purpose, let us fix any positive integers n and p. Next, let us take any

e[ 0:Since Fn;p � E, m Eð Þ ¼ 0; and m is complete, Fn;p is Lebesgue measurable, and

m Fn;p� �

¼ 0: Since m Fn;p� �

¼ 0; there exists an open set G such that Fn;p � G; andm Gð Þ\e: Now, there exists a sequence x1; x2; . . .f g in Fn;p; and a sequence

r1; r2; . . .f g in 0; 1p

� �such that

Fn;p � B x1; r1ð Þ [B x2; r2ð Þ [ � � � � G; andm B x1; r1ð Þð Þþm B x2; r2ð Þð Þþ � � �\e:

Problem 3:191 T Fn;p \B x1; r1ð Þ� �

� B T x1ð Þ; nr1ð Þ:

(Solution Let us take any x 2 Fn;p \B x1; r1ð Þ � B x1; r1ð Þð Þ: We have to show that

T xð Þ � T x1ð Þj j\nr1: Since x 2 B x1; r1ð Þ � B x1; 1p

� �� ; we have x 2 B x1; 1p

� �:

Since x 2 Fn;p \B x1; r1ð Þ, x 2 Fn;p � Eð Þ: Thus, x 2 B x1; 1p

� �\E: Since x 2


B x1; 1p

� �\E; and x1 2 Fn;p; by the definition of Fn;p;

T xð Þ � T x1ð Þj j � n x� x1j j \nr1ð Þ;

and hence T xð Þ � T x1ð Þj j\nr1: ■)Similarly, T Fn;p \B x2; r2ð Þ

� �� B T x2ð Þ; nr2ð Þ; etc. Thus,

B T x1ð Þ; nr1ð Þ [B T x2ð Þ; nr2ð Þ [ � � � T Fn;p \B x1; r1ð Þ� �

[ T Fn;p \B x2; r2ð Þ� �

[ � � �¼ T Fn;p \B x1; r1ð Þ

� �[ Fn;p \B x2; r2ð Þ� �

[ � � ��

¼ T Fn;p \ B x1; r1ð Þ [B x2; r2ð Þ [ � � �ð Þ� �

¼ T Fn;p� �

;

and hence

m T Fn;p� ��

�m B T x1ð Þ; nr1ð Þð Þþm B T x2ð Þ; nr2ð Þð Þþ � � �¼ nkm B T x1ð Þ; r1ð Þð Þþ nkm B T x2ð Þ; r2ð Þð Þþ � � �¼ nk m B T x1ð Þ; r1ð Þð Þþm B T x2ð Þ; r2ð Þð Þþ � � �ð Þ¼ nk m B x1; r1ð Þð Þþm B x2; r2ð Þð Þþ � � �ð Þ\nke:

Since 0�m T Fn;p� ��

\nke; and e is arbitrary, m T Fn;p� ��

¼ 0: ■

Lemma 3.192 Let V be a nonempty open subset of Rk: Let T : V ! Rk be anymapping. Let E be a nonempty Lebesgue measurable subset of V. Suppose thatT 0 xð Þ exists for every x 2 E: Let m Eð Þ ¼ 0: Then m T Eð Þð Þ ¼ 0:

Proof Let us fix any x 2 E: By Lemma 3.190, it suffices to show that

lim supcy ! xy 2 E


Since T 0 xð Þ exists,

limh!0

T yð Þ � T xð Þ � T 0 xð Þð Þ y� xð Þj jy� xj j ¼ 0:

It follows that there exists d 2 0; 1ð Þ such that 0\ y� xj j\d \1ð Þ impliesy 2 V ; and


T yð Þ � T xð Þj jy� xj j � T 0 xð Þk k ¼ T yð Þ � T xð Þj j

y� xj j � 1y� xj j T 0 xð Þk k y� xj jð Þ

� 1y� xj j T yð Þ � T xð Þj j � 1

y� xj j T 0 xð Þð Þ y� xð Þj j

� 1y� xj j T yð Þ � T xð Þð Þ � 1

y� xj j T 0 xð Þð Þ y� xð Þð Þ��

¼ T yð Þ � T xð Þ � T 0 xð Þð Þ y� xð Þj jy� xj j \1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus, 0\ y� xj j\d implies that

T yð Þ � T xð Þj jy� xj j \ 1þ T 0 xð Þk kð Þ \1ð Þ:

Hence, for every x 2 E, lim sup cy ! xy 2 E

T yð Þ�T xð Þj jy�xj j 6¼ 1: ■

Lemma 3.193 Let V be a nonempty open subset of Rk: Let X be nonemptyLebesgue measurable subset of V. Let T : V ! Rk be any continuous mapping.Suppose that T 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0: Let E be anonempty Lebesgue measurable subset of V. Let m Eð Þ ¼ 0: Then m T Eð Þð Þ ¼ 0:

Proof Since

m T Eð Þð Þ ¼ m T E \Xð Þ [ E � Xð Þð Þð Þ ¼ m T E \Xð Þ [ T E � Xð Þð Þ�m T E \Xð Þð Þþm T E � Xð Þð Þ;

we have

0�ð Þm T Eð Þð Þ�m T E \Xð Þð Þþm T E � Xð Þð Þ:

It suffices to show that m T E \Xð Þð Þ ¼ 0; and m T E � Xð Þð Þ ¼ 0: Since E � V ;we have E � Xð Þ � V � Xð Þ; and hence T E � Xð Þ � T V � Xð Þ: It follows that

0�ð Þm T E � Xð Þð Þ�m T V � Xð Þð Þ ¼ 0ð Þ;

and hence m T E � Xð Þð Þ ¼ 0:It remains to show that m T E \Xð Þð Þ ¼ 0: On applying Lemma 3.192, we have

m T E \Xð Þð Þ ¼ 0: ■

Lemma 3.194 Let V be a nonempty open subset of Rk: Let X be nonemptyLebesgue measurable subset of V. Let T : V ! Rk be a continuous mapping.


Suppose that T 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0: Let E be anonempty Lebesgue measurable subset of V.

Then T Eð Þ is Lebesgue measurable.

Proof Since E is a nonempty Lebesgue measurable, by Conclusion 1.258(9), thereexist Lebesgue measurable sets A and B such that E ¼ A[B;A\B ¼ ;;m Bð Þ ¼ 0;and A is an Fr: Now, T Eð Þ ¼ T Að Þ [ T Bð Þ: It suffices to show that both T Að Þ; T Bð Þare Lebesgue measurable sets.

Since Rk is locally compact Hausdorff space, and Rk is r-compact, every closedset is r-compact. Now, since A is an Fr; A is r-compact. Since A is r-compact,T : V ! Rk is continuous and A � V ; T Að Þ is r-compact, hence T Að Þ is Lebesguemeasurable.

It remains to show that T Bð Þ is Lebesgue measurable. By Lemma 3.193, T Bð Þ isLebesgue measurable, and m T Bð Þð Þ ¼ 0: ■

3.7 Change-of-Variables Theorem

Note 3.195 Let V be a nonempty open subset of Rk: Let X be a nonempty Lebesguemeasurable subset of V. Let T : V ! Rk be a continuous mapping. Suppose thatT 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0:Let n be a positive real number. LetM be the collection of all Lebesgue measurablesubsets of Rk: Let

JT : x 7! det T 0 xð Þð Þ

be the mapping from X to R.

ðJT : X ! R is generally called the Jacobian of T :Þ

Let E 2 M: Since T : V ! Rk is continuous, we have Tj j : V ! 0;1½ Þ; andT�1 B 0; nð Þð Þ � Vð Þ is open, hence T�1 B 0; nð Þð Þ 2 M: Since E;X; T�1 B 0; nð Þð Þ 2M; andM is a r-algebra, E \X \ T�1 B 0; nð Þð Þ 2 M; and hence by Lemma 3.194,

T E \X \ T�1 B 0; nð Þð Þ� �

2 M:

It follows that

m T E \X \ T�1 B 0; nð Þð Þ� ��

2 0;1½ �:

Since E \X \ T�1 B 0; nð Þð Þ � T�1 B 0; nð Þð Þ; we have T E \X \ T�1ðB 0; nð Þð ÞÞ � B 0; nð Þ; and hence


m T E \X \ T�1 B 0; nð Þð Þ� ��

�m B 0; nð Þð Þ \1ð Þ:

Thus, for every positive integer n,

ln : E 7! m T E\X \ T�1 B 0; nð Þð Þ� ��

is a map from M to 0;1½ Þ:

Problem 3.196 For every positive integer n, ln : M ! 0;1½ Þ is a measure.

(Solution Let us fix a positive integer n. Here,

ln ;ð Þ ¼ m T ;\X \T�1 B 0; nð Þð Þ� ��

¼ m T ;ð Þð Þ ¼ m ;ð Þ ¼ 0;

so ln ;ð Þ ¼ 0: Next, let E1;E2; . . .f g be any disjoint collection of sets in M: Wehave to show that

m T E1 [E2 [ � � �ð Þ \X \ T�1 B 0; nð Þð Þ� ��

¼ m T E1 \X \ T�1 B 0; nð Þð Þ� ��

þm T E2 \X \ T�1 B 0; nð Þð Þ� ��

þ � � � :

LHS ¼ m T E1 [E2 [ � � �ð Þ \X \ T�1 B 0; nð Þð Þ� ��

¼ m T E1 \X \ T�1 B 0; nð Þð Þ� �

[ E2 \X \ T�1 B 0; nð Þð Þ� �

[ � � ��

¼ m T E1 \X \ T�1 B 0; nð Þð Þ� �

[ T E2 \X \ T�1 B 0; nð Þð Þ� �

[ � � ��

¼ m T E1 \X \ T�1 B 0; nð Þð Þ� ��

þm T E2 \X \ T�1 B 0; nð Þð Þ� ��

þ � � � ¼ RHS:

■)

Problem 3.197 For every E 2 M; ln Eð Þ ¼ð Þm T E \X \T�1 B 0; nð Þð Þð Þð Þ¼ m T E \ T�1 B 0; nð Þð Þð Þð Þ:

(Solution Let us fix any E 2 M: It suffices to show

m T E \T�1 B 0; nð Þð Þ� �

� T E\X \ T�1 B 0; nð Þð Þ� ��

¼ 0:

Since

T E \ T�1 B 0; nð Þð Þ� �

� T E \X \ T�1 B 0; nð Þð Þ� �

� T E \ T�1 B 0; nð Þð Þ� �

� E\X \T�1 B 0; nð Þð Þ� ��

¼ T E \ T�1 B 0; nð Þð Þ� �

� X� �

� T V � Xð Þ;


we have

0�ð Þm T E\ T�1 B 0; nð Þð Þ� �

� T E \X \T�1 B 0; nð Þð Þ� ��

�m T V � Xð Þð Þ ¼ 0ð Þ;

and hence

m T E \T�1 B 0; nð Þð Þ� �

� T E\X \ T�1 B 0; nð Þð Þ� ��

¼ 0:

■)

Problem 3.198 For every positive integer n, ln m:

(Solution Let m Eð Þ ¼ 0; where E 2 M: We have to show that ln Eð Þ ¼ 0; that is

m T E \X \ T�1 B 0; nð Þð Þ� ��

¼ 0:

Since m Eð Þ ¼ 0; we have m E \Xð Þ ¼ 0: Now, by Lemma 3.193,m T E \Xð Þð Þ ¼ 0: Since

E \X \ T�1 B 0; nð Þð Þ� �

� E \X;

we have

T E \X \ T�1 B 0; nð Þð Þ� �

� T E \Xð Þ:

Now, since

m T E \Xð Þð Þ ¼ 0;

we have

m T E \X \ T�1 B 0; nð Þð Þ� ��

¼ 0:

■)

Problem 3.199 For every positive integer n, and for every x 2 X \T�1 B 0; nð Þð Þ;

limr!0þ

ln B x; rð Þð Þm B x; rð Þð Þ ¼ Dlnð Þ xð Þ ¼ det T 0 xð Þð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ JT xð Þj j:

(Solution For this purpose, let us fix any positive integer n, and x 2X \ T�1 B 0; nð Þð Þ: We have to show that

3.7 Change-of-Variables Theorem 537

limr!0þ

ln B x; rð Þð Þm B x; rð Þð Þ ¼ det T 0 xð Þð Þj j:

Since x 2 X \ T�1 B 0; nð Þð Þ; we have x 2 T�1 B 0; nð Þð Þ: Now, since T�1 B 0; nð Þð Þis open, there exists e[ 0; such that B x; eð Þ � T�1 B 0; nð Þð Þ:

It follows that, for every r 2 0; eð Þ, B x; rð Þ � T�1 B 0; nð Þð Þ � Vð Þ; and hence

ln B x; rð Þð Þ ¼ m T B x; rð Þ \X \ T�1 B 0; nð Þð Þ� ��

¼ m T B x; rð Þ \Xð Þð Þ¼ m T B x; rð Þð Þð Þ � m T B x; rð Þ � Xð Þð Þ ¼ m T B x; rð Þð Þð Þ � 0 ¼ m T B x; rð Þð Þð Þ:

Hence, for every r 2 0; eð Þ

ln B x; rð Þð Þ ¼ m T B x; rð Þð Þð Þ:

By Theorem 3.187,

limr!0

ln B x; rð Þð Þm B x; rð Þð Þ ¼ lim

r!0

m T B x; rð Þð Þð Þm B x; rð Þð Þ ¼ det T 0 xð Þð Þj j|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

so

limr!0þ

ln B x; rð Þð Þm B x; rð Þð Þ ¼ det T 0 xð Þð Þj j:

■)Now, by Conclusion 3.123, for every positive integer n, fn ¼ Dln a.e. on Rk;

where fn : X ! C is the Radon–Nikodym derivative of ln with respect to m. Also,for every E 2 M, ln Eð Þ ¼

RE Dlnð Þdm: It follows that for every positive integer n,

and for every E 2 M;

ZX \ T�1 B 0;nð Þð Þ

vE JTj jdm¼Z

E \X \ T�1 B 0;nð Þð Þ

JTj jdm ¼Z

E \X \ T�1 B 0;nð Þð Þ

Dlnð Þdm

¼ ln E \X \ T�1 B 0; nð Þð Þ� �

¼ m T E \X \ T�1 B 0; nð Þð Þ� �

\X \ T�1 B 0; nð Þð Þ� ��

¼ m T E \X \ T�1 B 0; nð Þð Þ� ��

;

and hence for every positive integer n, and for every E 2 M;

m T E \X \ T�1 B 0; nð Þð Þ� ��

¼Z

X \T�1 B 0;nð Þð Þ

vE JTj jdm:


It follows that for every E 2 M;

m T E \Xð Þð Þ ¼ m T E \X \Vð Þð Þ ¼ m T E \X \ T�1 Rk� ��

¼ m T E \X \ T�1 [1n¼1B 0; nð Þ

� �� ¼ m T E\X \ [1

n¼1T�1 B 0; nð Þð Þ

� �� ¼ m T [1

n¼1 E \X \ T�1 B 0; nð Þð Þ� ��

¼ m [1n¼1T E \X \ T�1 B 0; nð Þð Þ

� �� ¼ lim

n!1m T E\X \ T�1 B 0; nð Þð Þ

� �� ¼ lim

n!1

ZX \ T�1 B 0;nð Þð Þ

vE JTj jdm

0B@1CA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼

Z[1

n¼1 X \ T�1 B 0;nð Þð Þð Þ

vE JTj jdm ¼Z

X \ [1n¼1 T�1 B 0;nð Þð Þð Þð Þ

vE JTj jdm ¼Z

X \ T�1 [1n¼1B 0;nð Þð Þ

vE JTj jdm

¼Z

X \ T�1 Rkð ÞvE JTj jdm ¼

ZX \V

vE JTj jdm ¼ZX

vE JTj jdm;

and hence for every E 2 M;

m T E \Xð Þð Þ ¼ZX

vE JTj jdm:

Conclusion 3.200 Let V be a nonempty open subset of Rk: Let X be a nonemptyLebesgue measurable subset of V. Let T : V ! Rk be a continuous mapping.Suppose that T 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0: Then, for everyLebesgue measurable set E in Rk;

m T E \Xð Þð Þ ¼ZX

vE JTj jdm:

Lemma 3.201 Let V be a nonempty open subset of Rk: Let X be a nonemptyLebesgue measurable subset of V. Let T : V ! Rk be a continuous mapping.Suppose that T 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0: Let T be 1-1 onX. Let E be a Lebesgue measurable set in Rk: Then,Z

T Xð Þ

vEdm ¼ZX

vE � Tð Þ JTj jdm:


Proof Case I: when E is a Borel set.

Problem 3:202 T�1 Eð Þ 2 M:

(Solution Since T : V ! Rk is continuous, T : V ! Rk is Lebesgue measurable,and hence the collection fF : T�1 Fð Þis a Lebesgue measurable set inRkg is a r-algebra containing all open sets in Rk: It follows that fF : T�1 Fð Þis a Lebesgue measurable set inRkg contains all Borel sets. Now, since E is aBorel set, T�1 Eð Þ is a Lebesgue measurable set inRk: ■)

Problem 3:203 T T�1 Eð Þ \Xð Þ ¼ E \T Xð Þ:

(Solution Let T xð Þ 2 T T�1 Eð Þ \Xð Þ; where x 2 T�1 Eð Þ \X � T�1 Eð Þð Þ: It fol-lows that T xð Þ 2 E: Since x 2 T�1 Eð Þ \X � Xð Þ; x 2 X; and hence T xð Þ 2 T Xð Þ:Thus, T xð Þ 2 E \ T Xð Þ: Hence, T T�1 Eð Þ \Xð Þ � E \ T Xð Þ: It remains to showthat E \ T Xð Þ � T T�1 Eð Þ \Xð Þ: For this purpose, let us take any T xð Þ 2 E \T Xð Þ;where x 2 X: We have to show that T xð Þ 2 T T�1 Eð Þ \Xð Þ: It suffices to show thatx 2 T�1 Eð Þ \X: Since T xð Þ 2 E \T Xð Þ � Eð Þ, x 2 T�1 Eð Þ: Now, since x 2 X; wehave x 2 T�1 Eð Þ \X: ■)

Problem 3:204 For every x 2 X, vT�1 Eð Þ xð Þ ¼ vE � Tð Þ xð Þ:

(Solution Let us take any x 2 X:Case I: when x 2 T�1 Eð Þ: It follows that T xð Þ 2 E; and hence

vE � Tð Þ xð Þ ¼ vE T xð Þð Þ ¼ 1|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ¼ vT�1 Eð Þ xð Þ:

Thus, vT�1 Eð Þ xð Þ ¼ vE � Tð Þ xð Þ:Case II: when x 62 T�1 Eð Þ: Now, since x 2 X, T xð Þ 62 E; and hencevE � Tð Þ xð Þ ¼ð ÞvE T xð Þð Þ ¼ 0: Since x 62 T�1 Eð Þ, vT�1 Eð Þ xð Þ ¼ 0 ¼ vE � Tð Þ xð Þð Þ:

Thus, vT�1 Eð Þ xð Þ ¼ vE � Tð Þ xð Þ: ■)Now, by Conclusion 3.200,Z

T Xð Þ

vEdm¼ m E \ T Xð Þð Þ ¼ m T T�1 Eð Þ \X� ��

¼ZX

vT�1 Eð Þ JTj jdm

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZX

vE � Tð Þ JTj jdm;


so ZT Xð Þ

vEdm ¼ZX


Case II: when E is a Lebesgue measurable set satisfying m Eð Þ ¼ 0: Here, for everypositive integer n, there exists an open set Gn such that E � Gn; and m Gnð Þ\ 1

n : Itfollows that E � G1 [G2 [ � � � ;m G1 [G2 [ � � �ð Þ ¼ 0; and G1 [G2 [ � � � is aBorel set. Now, by Case I,

0 ¼ m G1 [G2 [ � � �ð Þ \ T Xð Þð Þ

¼Z

T Xð Þ

v G1 [G2 [ ��ð Þdm ¼ZX

v G1 [G2 [ ��ð Þ � T� �

JTj jdm

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

so v G1 [G2 [ ��ð Þ � T� �

JTj j ¼ 0 a.e. Since E � G1 [G2 [ � � � ; we have

0� vE � v G1 [G2 [ ��ð Þ;

and hence

0�ð ÞZX

vE � Tð Þ JTj jdm�ZX

v G1 [G2 [ ��ð Þ � T� �

JTj jdm ¼ 0ð Þ:

This shows thatRX vE � Tð Þ JTj jdm ¼ 0: Since m Eð Þ ¼ 0;

ZT Xð Þ

vEdm ¼ m E \ T Xð Þð Þ ¼ 0 ¼ZX

vE � Tð Þ JTj jdm;

we have ZT Xð Þ

vEdm ¼ZX



Case III: when E is a Lebesgue measurable set. Now, by Conclusion 1.258(9), thereexist Lebesgue measurable sets A and B such that E ¼ A[B;A\B ¼ ;;m Bð Þ ¼ 0;and A is an Fr: Since m Bð Þ ¼ 0; by Case II,

RT Xð Þ vBdm ¼

RX vB � Tð Þ JTj jdm: Since

A is an Fr; A is a Borel set, and hence by Case I,ZT Xð Þ

vAdm ¼ZX

vA � Tð Þ JTj jdm:

LHS ¼Z

T Xð Þ

vEdm ¼Z

T Xð Þ

vA þ vBð Þdm ¼Z

T Xð Þ

vAdmþZ

T Xð Þ

vBdm

¼ZX

vA � Tð Þ JTj jdmþZX

vB � Tð Þ JTj jdm ¼ZX

vA � Tð Þ JTj j þ vB � Tð Þ JTj jð Þdm

¼ZX

vA þ vBð Þ � Tð Þ JTj jdm ¼ZX

vA[B � Tð Þ JTj jdm ¼ZX

vE � Tð Þ JTj jdm ¼ RHS:

■

Theorem 3.205 Let V be a nonempty open subset of Rk: Let X be a nonemptyLebesgue measurable subset of V. Let T : V ! Rk be a continuous mapping.Suppose that T 0 xð Þ exists for every x 2 X: Let m T V � Xð Þð Þ ¼ 0: Let T be 1-1 onX. Let f : Rk ! 0;1½ � be any Lebesgue measurable function. Then,Z

T Xð Þ

f dm ¼ZX

f � Tð Þ JTj jdm:

Proof Case I: when f : Rk ! 0;1½ Þ is a simple function. It follows that there existdistinct real numbers a1; . . .; an such that f�1 a1ð Þ; . . .; f�1 anð Þ are members of M;and

f ¼ a1v f�1 a1ð Þð Þ þ � � � þ anv f�1 anð Þð Þ:


Hence, on using Lemma 3.201,

LHS ¼Z

T Xð Þ

f dm ¼Z

T Xð Þ

a1v f�1 a1ð Þð Þ þ � � � þ anv f�1 anð Þð Þ

� �dm

¼ a1

ZT Xð Þ

v f�1 a1ð Þð Þdmþ � � � þ an

ZT Xð Þ

v f�1 anð Þð Þdm

¼ a1

ZX

v f�1 a1ð Þð Þ � T� �

JTj jdmþ � � � þ an

ZT Xð Þ

v f�1 anð Þð Þ � T� �

JTj jdm

¼ZX

a1 v f�1 a1ð Þð Þ � T� �

JTj j þ � � � þ an v f�1 anð Þð Þ � T� �

JTj j� �

dm

¼ZX

a1 v f�1 a1ð Þð Þ � T� �

þ � � � þ an v f�1 anð Þð Þ � T� ��

JTj jdm

¼ZX


� �� T

� �JTj jdm ¼

ZX

f � Tð Þ JTj jdm ¼ RHS:

Case II: when f : Rk ! 0;1½ Þ is a measurable function. By Lemma 1.98, thereexists a sequence snf g of simple measurable functions sn : Rk ! 0;1½ Þ such thatfor every x in Rk; 0� s1 xð Þ� s2 xð Þ� � � � ; and limn!1 sn xð Þ ¼ f xð Þ: Now, byTheorem 1.125,

ZT Xð Þ

f dl ¼ limn!1

ZT Xð Þ

sn dl

0B@1CA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ lim

n!1

ZX

sn � Tð Þ JTj jdl

0@ 1A:

Since 0� s1 � s2 � � � � ; we have 0� s1 � Tð Þ� s2 � Tð Þ� � � � ; and hence0� s1 � Tð Þ JTj j � s2 � Tð Þ JTj j � � � � : Here, T is continuous and each sn is mea-surable, each sn � Tð Þ JTj j is measurable. Also, since T is continuous, andlimn!1 sn ¼ f ; we have

limn!1

sn � Tð Þ JTj j ¼ f � Tð Þ JTj j:


ZT Xð Þ

f dl ¼ limn!1

ZX

sn � Tð Þ JTj jdl

0@ 1A ¼ZX

f � Tð Þ JTj jdl:


Thus, ZT Xð Þ

f dl ¼ZX

f � Tð Þ JTj jdl:

■Theorem 3.205 is known as the change-of-variables theorem.

3.8 Fubini Theorem

Note 3.206

Definition Let X be any nonempty set. Let Y be any nonempty set. If A � X; andB � Y ; then A B � X Yð Þ is called a rectangle in X Y :

Definition Let X be any nonempty set. Let S be a r-algebra in X. Let Y be anynonempty set. Let T be a r-algebra in Y. If A 2 S; and B 2 T ; then A B � X Yð Þ is called a measurable rectangle in X Y :

Let R be the collection of all measurable rectangles in X Y : The smallest r-algebra containing R is denoted by S T :

Let E 2 S T : Let a 2 X:

Problem 3.207 y : a; yð Þ 2 Ef g 2 T :

(Solution Put

X � F : F 2 S T and x 2 X ) y : x; yð Þ 2 Ff g 2 Tð Þf g:

It suffices to show that S Tð Þ � X: Since S T is the smallest r-algebracontaining R; it suffices to show that

a. X is a r-algebra in X Y ;b. R � X:

For a:

1. Since X 2 S; and Y 2 T , X Y 2 R � S Tð Þ; and hence X Yð Þ 2S Tð Þ: Let us fix any x 2 X: Here y : x; yð Þ 2 X Yf g ¼ Y 2 Tð Þ; so by thedefinition of X, X Yð Þ 2 X:

2. Let F1;F2; . . .f g be any countable collection of members in X:We have to showthat F1 [F2 [F3 [ � � � is a member of X; that is

(i) F1 [F2 [F3 [ � � �ð Þ 2 S T ;(ii) x 2 X ) y : x; yð Þ 2 F1 [F2 [F3 [ � � �ð Þf g 2 T :


For (i): Here, each Fn 2 X; so each Fn 2 S T : Now, since S T is a r-algebra, F1 [F2 [F3 [ � � �ð Þ 2 S T :

For (ii): Let us fix any a 2 X: We have to show that y : a; yð Þ 2fF1 [F2 [F3 [ � � �ð Þg 2 T : Clearly,

y : a; yð Þ 2 F1 [F2 [F3 [ � � �ð Þf g ¼ y : a; yð Þ 2 F1f g[ y : a; yð Þ 2 F2f g[ � � � :

Since each Fn 2 X; and a 2 X; each y : a; yð Þ 2 Fnf g 2 T : Now, since T is a r-algebra,

y : a; yð Þ 2 F1 [F2 [F3 [ � � �ð Þf g ¼ y : a; yð Þ 2 F1f g[ y : a; yð Þ 2 F2f g[ � � �ð Þ 2 T ;

and hence

y : a; yð Þ 2 F1 [F2 [F3 [ � � �ð Þf g 2 T :

3. Let F 2 X; that is F 2 S T and x 2 X ) y : x; yð Þ 2 Ff g 2 Tð Þ: We have toshow that F0 2 X; that is

(i) F0 2 S T ;(ii) x 2 X ) y : x; yð Þ 2 F0f g 2 T :

For (i): Since F 2 S T ; and S T is a r-algebra, F0 2 S T :For (ii): Let us fix any x 2 X: We have to show that y : x; yð Þ 2 F0f g 2 T : Here,

y : x; yð Þ 2 F0f g ¼ y : x; yð Þ 2 Ff g0: Since y : x; yð Þ 2 Ff g 2 T ; and T is a r-algebra, y : x; yð Þ 2 F0f g ¼ð Þ y : x; yð Þ 2 Ff g02 T ; and hence y : x; yð Þf 2 F0g 2 T :

Thus, X is a r-algebra in X Y :For b: Let us take any A Bð Þ 2 R; where A 2 S; and B 2 T : We have to show

that A Bð Þ 2 X; that is

(i) A Bð Þ 2 S T ;(ii) x 2 X ) y : x; yð Þ 2 A Bð Þf g 2 T :

For (i): Since A Bð Þ 2 R; and R � S Tð Þ; we have A Bð Þ 2 S T :For (ii): Let us fix any a 2 X: We have to show that y : a; yð Þ 2 A Bð Þf g 2 T :

Since y : a; yð Þ 2 A Bð Þf g ¼ B 2 Tð Þ; we have y : a; yð Þ 2 A Bð Þf g 2 T : ■)Thus, R � X:

Conclusion 3.208 Let X be any nonempty set. Let S be a r-algebra in X. Let Y beany nonempty set. Let T be a r-algebra in Y. Let E 2 S T : Let a 2 X: Theny : a; yð Þ 2 Ef g 2 T :Here y : a; yð Þ 2 Ef g is denoted by Ea; and is called an x-section of E. Also, if

b 2 Y ; then x : x; bð Þ 2 Ef g is denoted by Eb; and is called a y-section of E.We have seen that each x-section of members in S T is a T -measurable set.

Similarly, each y-section of members in S T is an S-measurable set.

3.8 Fubini Theorem 545

Note 3.209

Definition Let X be any nonempty set. Let M be a collection of subsets of X. If,

A1;A2;A3; � � � 2 M; andA1 � A2 � A3 � � � �ð Þ ) [1n¼1An

� �2 M;

and

B1;B2;B3; � � � 2 M; andB1 B2 B3 � � �ð Þ ) \1n¼1Bn

� �2 M;

then we say that M is a monotone class.

Definition Let X be any nonempty set. Let S be a r-algebra in X. Let Y be anynonempty set. Let T be a r-algebra in Y : Let R be the collection of all measurablerectangles in X Y : If n is a positive integer, R1; . . .;Rn 2 R; andi 6¼ j ) Ri \Rj ¼ ;� �

; then we say that R1 [ � � � [Rnð Þ is an elementary set inX Y :

Let X be any nonempty set. Let S be a r-algebra in X. Let Y be any nonemptyset. Let T be a r-algebra in Y. Let E be the collection of all elementary sets inX Y :

a: Problem 3.210 S T is a monotone class containing E:

(Solution Since S T is a r-algebra,

A1;A2;A3; � � � 2 S T ; andA1 � A2 � A3 � � � �ð Þ ) [1n¼1An

� �2 S T

� �;

and

B1;B2;B3; � � � 2 S T ; andB1 B2 B3 � � �ð Þ ) \1n¼1Bn

� �2 S T

� �;

and hence S T is a monotone class. Since S T is a r-algebra containing allmeasurable rectangles in X Y ; S T contains E: ■)

b. Let A B and C D be measurable rectangles in X Y ; where A;C 2 S; andB;D 2 T :

Problem 3.211 A Bð Þ \ C Dð Þ is a measurable rectangle in X Y :

(Solution Since A;C 2 S; and S is a r-algebra, A\C 2 S: Similarly, B\D 2 T :It follows that A Bð Þ \ C Dð Þ ¼ð Þ A\Cð Þ B\Dð Þ is a measurable rectangle,and henc, A Bð Þ \ C Dð Þ is a measurable rectangle. ■)

c. Let A B and C D be measurable rectangles in X Y ; where A;C 2 S; andB;D 2 T :


Problem 3.212 A Bð Þ � C Dð Þ can be expressed as a union of two disjointmeasurable rectangles in X Y :

(Solution Clearly,

A Bð Þ � C Dð Þ ¼ A� Cð Þ Bð Þ [ A\Cð Þ B� Dð Þð Þ:

Since A;C 2 S; and S is a r-algebra, A� Cð Þ 2 S: Now, since B 2 T ;A� Cð Þ B is a measurable rectangle in X Y : Similarly, A\Cð Þ B� Dð Þ is ameasurable rectangle in X Y : It is clear that

A� Cð Þ Bð Þ \ A\Cð Þ B� Dð Þð Þ ¼ ;:

■)

d. Let P;Q 2 E:

Problem 3.213 P\Q 2 E:

(Solution Since P 2 E; there exist finite-many measurable rectangles R1; . . .;Rn

such that i 6¼ j ) Ri \Rj ¼ ;� �

; and P ¼ R1 [ � � � [Rn: Similarly, there existfinite-many measurable rectangles S1; . . .; Sm such that i 6¼ j ) Si \ Sj ¼ ;

� �; and

Q ¼ S1 [ � � � [ Sm: Here,

P\Q ¼ R1 [ � � � [Rnð Þ \ S1 [ � � � [ Smð Þ ¼ [ i;jð Þ2 1;...;nf g 1;...;mf g Ri \ Sj� �

:

Since each Ri is a measurable rectangle, and each Sj is a measurable rectangle, by(b), each Ri \ Sj

� �is a measurable rectangle in X Y : Let i; jð Þ; k; lð Þ be distinct

members of 1; . . .; nf g 1; . . .;mf g: It suffices to show thatRi \Rkð Þ \ Sj \ Sl

� �Ri \ Sj� �

\ Rk \ Slð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ;: that is Ri \Rkð Þ \ Sj \ Sl� �

¼ ;:

Since i; jð Þ; k; lð Þ are distinct, either i 6¼ k or j 6¼ l: For definiteness, let i 6¼ k: Itfollows that Ri \Rk ¼ ;; and hence Ri \Rkð Þ \ Sj \ Sl

� �¼ ;: ■)

e. Let P;Q 2 E: Let P\Q ¼ ;: Then, clearly, P[Q 2 E:f. Let P;Q 2 E:

Problem 3.214 P� Q 2 E:

(Solution Since P 2 E; there exist finite-many measurable rectangles R1; . . .;Rn

such that i 6¼ j ) Ri \Rj ¼ ;� �

; and P ¼ R1 [ � � � [Rn: Similarly, there existfinite-many measurable rectangles S1; . . .; Sm such that i 6¼ j ) Si \ Sj ¼ ;

� �; and

Q ¼ S1 [ � � � [ Sm: Here,


P� Q ¼ R1 [ � � � [Rnð Þ � S1 [ � � � [ Smð Þ¼ \ j2 1;...;mf g R1 [ � � � [Rnð Þ � Sj

� �¼ \ j2 1;...;mf g [ i2 1;...;nf g Ri � Sj

� �� ¼ [ i1;...;imð Þ2 1;...;nf gm Ri1 � S1ð Þ \ � � � \ Rim � Smð Þ:

Since each Ri1 is a measurable rectangle, and S1 is a measurable rectangle, by (c),each Ri1 � S1ð Þ 2 E: Similarly, Ri2 � S2ð Þ 2 E; etc. Now, by (d), eachRi1 � S1ð Þ \ � � � \ Rim � Smð Þ 2 E: Let i1; . . .; imð Þ and j1; . . .; jmð Þ be distinctmembers of 1; . . .; nf gm: By (e), it suffices to show that

Ri1 � S1ð Þ \ Rj1 � S1� ��

\ � � � \ Rim � Smð Þ \ Rjm � Sm� ��

¼ Ri1 � S1ð Þ \ � � � \ Rim � Smð Þð Þ \ Rj1 � S1� �

\ � � � \ Rjm � Sm� ��

¼ ;|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is

Ri1 � S1ð Þ \ Rj1 � S1� ��

\ � � � \ Rim � Smð Þ \ Rjm � Sm� ��

¼ ;:

Since i1; . . .; imð Þ and j1; . . .; jmð Þ are distinct, we have i1 6¼ j1 or i2 6¼ j2 or � � � :For definiteness, let i1 6¼ j1: It follows that Ri1 \Rj1 ¼ ;; and henceRi1 � S1ð Þ \ Rj1 � S1

� �¼ ;: This shows that

Ri1 � S1ð Þ \ Rj1 � S1� ��

\ � � � \ Rim � Smð Þ \ Rjm � Sm� ��

¼ ;:

■)

g. Let P;Q 2 E:

Problem 3.215 P[Q 2 E:

(Solution Since P;Q 2 E; we have, by (f), P� Q 2 E: Also, by (d), P\Q 2 E:Now, since P� Qð Þ \ P\Qð Þ ¼ ;; by (e), P[Q ¼ð Þ P� Qð Þ [ P\Qð Þ 2 E:Hence, P[Q 2 E: ■)

Let M be the smallest monotone class containing E: Let P � X Y : Put

X Pð Þ � Q : Q � X Y ; P� Qð Þ 2 M; Q� Pð Þ 2 M; P[Qð Þ 2 Mf g:

h. Let P � X Y ; and Q 2 X Pð Þ:


Problem 3.216 P 2 X Qð Þ:

(Solution Since Q 2 X Pð Þ; we have Q � X Y , P� Qð Þ 2 M, Q� Pð Þ 2 M,P[Qð Þ 2 M: We have to show that P 2 X Qð Þ; that is P � X Y , Q� Pð Þ 2 M,P� Qð Þ 2 M, Q[Pð Þ 2 M: These are clearly true. ■)

i. Let P � X Y :

Problem 3.217 X Pð Þ is a monotone class.

(Solution Let A1;A2;A3; � � � 2 X Pð Þ; andA1 � A2 � A3 � � � � : We have to showthat [1

n¼1An� �

2 X Pð Þ; that is

1. [1n¼1An

� �� X Y ;

2. P� [1n¼1An

� �� 2 M;

3. [1n¼1An

� �� P

� �2 M;

4. P[ [1n¼1An

� �� 2 M:

For 1: Since each An 2 X Pð Þ; each An � X Y ; and hence [1n¼1An

� �� X Y :

For 2: Since each An 2 X Pð Þ; each P� Anð Þ 2 M: Since A1 � A2 � A3 � � � � ;P� A1ð Þ P� A2ð Þ P� A3ð Þ � � � : Now, since M is a monotone class,P� [1

n¼1An� ��

¼� �

\1n¼1 P� Anð Þ 2 M; and hence P� [1

n¼1An� ��

2 M:

For 3: Since each An 2 X Pð Þ; each An � Pð Þ 2 M: Since A1 � A2 � A3 � � � � ;

A1 � Pð Þ � A2 � Pð Þ � A3 � Pð Þ � � � � :

Now, since M is a monotone class,

[1n¼1An

� �� P

� �¼ [1

n¼1 An � Pð Þ 2 M|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence [1

n¼1An� �

� P� �

2 M:

For 4: Since each An 2 X Pð Þ; each P[Anð Þ 2 M: Since A1 � A2 � A3 � � � � ;P[A1ð Þ � P[A2ð Þ � P[A3ð Þ � � � � : Now, since M is a monotone class,

P[ [1n¼1An

� �¼ [1

n¼1 P[Anð Þ 2 M|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence P[ [1

n¼1An� ��

2 M:

Next, let B1;B2;B3; � � � 2 X Pð Þ; and B1 B2 B3 � � � : As above,\1

n¼1Bn� �

2 X Pð Þ: ■)


j. Let P 2 E:

Problem 3.218 M � X Pð Þ:

(Solution Let us take any A 2 E: We shall show that A 2 X Pð Þ; that is

1. A � X Y ;2. P� Að Þ 2 M;3. A� Pð Þ 2 M;4. P[Að Þ 2 M:

For 1: Since A 2 M; and M is a monotone class in X Y ; A � X Y :For 2: Since P;A 2 E; by (f), P� Að Þ 2 E:For 3: Since P;A 2 E; by (f), A� Pð Þ 2 E:For 4: Since P;A 2 E; by (g), P[Að Þ 2 E:Thus, E � X Pð Þ: Now, by (i), X Pð Þ is a monotone class containing E: Thus, by

the definition of M; M � X Pð Þ: ■)

k. Let P 2 M:

Problem 3.219 M � X Pð Þ:

(Solution Take any A 2 E: Now, since P 2 M; by (j), P 2 X Að Þ; and hence by(h), A 2 X Pð Þ: Thus, E � X Pð Þ: Now, by (i), X Pð Þ is a monotone class containingE: Thus, by the definition of M; M � X Pð Þ: ■)

l. Let P;Q 2 M:

Problem 3.220 P� Qð Þ 2 M; and P[Qð Þ 2 M:

(Solution Since P;Q 2 M; by (k), Q 2 X Pð Þ; and hence P� Qð Þ 2 M; andP[Qð Þ 2 M: ■)

m: Problem 3.221 M is a r-algebra in X Y :

(Solution

1. Since S is a r-algebra in X, X 2 S: Similarly, Y 2 T : It follows that X Y is ameasurable rectangle, and hence X Yð Þ 2 E � Mð Þ: Thus, X Yð Þ 2 M:

2. Let P 2 M: We have to show that P0 2 M: Since P 2 M; and X Yð Þ 2 M;by (l), P0 ¼ð Þ X Yð Þ � Pð Þ 2 M; and hence P0 2 M:

3. Let P1;P2;P3; � � � 2 M: We have to show that P1 [P2 [P3 [ � � � 2 M: By (l),P1; P1 [P2ð Þ; P1 [P2 [P3ð Þ; � � � 2 M: Also,

P1 � P1 [P2ð Þ � P1 [P2 [P3ð Þ � � � � :

Now, since M a monotone class, P1 [P2 [P3 [ � � � ¼ P1 [ P1 [P2ð Þ [P1 [P2 [P3ð Þ [ � � � 2 M; and hence, P1 [P2 [P3 [ � � � 2 M: ■)


n: Problem 3.222 S T is the smallest monotone class containing E:

(Solution By (m), M is a r-algebra in X Y containing E: Now, since E containsall measurable rectangles, M is a r-algebra in X Y containing all measurablerectangles. It follows, by the definition of S T ; that S T � M: Since M is thesmallest monotone class containing E; by (a), M � S T : Since M � S T ;and S T � M; we have S T ¼ M: Now, since M is the smallest monotoneclass containing E; S T is the smallest monotone class containing E: ■)

Conclusion 3.223 Let X be any nonempty set. Let S be a r-algebra in X. Let Y beany nonempty set. Let T be a r-algebra in Y. Let E be the collection of allelementary sets in X Y : Then, S T is the smallest monotone class containing E:

Definition Let X; Y ; Z be any nonempty sets. Let f : X Y ! Z: Let a 2 X; andb 2 Y : The function from Y to Z that sends each y 2 Y to f a; yð Þ; is denoted by fa:The function from X to Z that sends each x 2 X to f x; bð Þ; is denoted by f b:

Lemma 3.224 Let X be a nonempty set. Let S be a r-algebra in X. Let Y be anonempty set. Let T be a r-algebra in Y. Let Z be a topological space. Let f :X Y ! Z be a S Tð Þ-measurable function. Let a 2 X; and b 2 Y : Then,

1. fa : Y ! Z is a T -measurable function,2. f b : X ! Z is an S-measurable function.

Proof Let V be an open subset of Z. We shall show that fað Þ�1 Vð Þ 2 T : Sincef : X Y ! Z is a S Tð Þ-measurable function, and V is an open subset of Z, wehave f�1 Vð Þ 2 S Tð Þ; and hence by Conclusion 3.208,

fað Þ�1 Vð Þ ¼ y : fað Þ yð Þ 2 Vf g ¼ y : f a; yð Þ 2 Vf g ¼ y : a; yð Þ 2 f�1 Vð Þ�

2 T|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Thus, fað Þ�1 Vð Þ 2 T : This proves Lemma 3.224(1). The proof of Lemma 3.224

(2) is similar. ■

Note 3.225 Let X be a nonempty set. Let S be a r-algebra in X. Let l : S ! 0;1½ �be a measure. Suppose that X has r-finite measure. In short, we say that X;S; lð Þ isa r-finite measure space. Let Y ; T ; kð Þ be another r-finite measure space. LetQ 2 S Tð Þ:

By Conclusion 3.208, for every x 2 X, Qx 2 T ; and hence for every x 2 X,k Qxð Þ 2 0;1½ �: By uQ; we mean the mapping

uQ : x 7! k Qxð Þ

from X to 0;1½ �: Similarly, by wQ; we mean the mapping


wQ : y 7! l Qyð Þ

from Y to 0;1½ �: Let

X � P : P 2 S Tð Þ;uP isS-measurable; wP is T -measurable; andZX

uPdl ¼ZY

wPdk

8<:9=;:

a. Let A B be a measurable rectangle in X Y ; where A 2 S; and B 2 T :

Problem 3.226 A Bð Þ 2 X; that is

1. A Bð Þ 2 S Tð Þ;2. uðA BÞ isS-measurable,3. w A Bð Þ is T -measurable,4.RX u A Bð Þdl ¼

RY w A Bð Þdk:

(Solution For 1: Since A 2 S; and B 2 T ; and A B is a measurable rectangle inX Y ; hence A Bð Þ 2 S Tð Þ:

For 2: Observe that, for every x 2 X;

u A Bð Þ xð Þ ¼ k A Bð Þx� �

¼k Bð Þ if x 2 A

k ;ð Þ if x 62 A

�¼

k Bð Þ if x 2 A

0 if x 62 A

�¼ k Bð Þð Þ vAð Þð Þ xð Þ;

so, u A Bð Þ ¼ k Bð Þð Þ vAð Þ: Similarly, w A Bð Þ ¼ l Að Þð Þ vBð Þ:Next, let V be any nonempty open subset of 0;1½ �: We have to show that

X or Ac or A ¼ð Þ�u A Bð Þ

��1ðVÞ 2 S: Now, since X;Ac;Af g � S,�u A Bð Þ

��1Vð Þ 2 S:

For 3: Similar to 2.For 4:

LHS ¼ZX

u A Bð Þdl ¼ZX

k Bð Þð Þ vAð Þdl ¼ k Bð Þð ÞZX

vAdl ¼ k Bð Þð Þ l Að Þð Þ

¼ l Að Þð Þ k Bð Þð Þ ¼ l Að Þð ÞZY

vBdk ¼ZY

l Að Þð Þ vBð Þdk ¼ZY

w A Bð Þdk ¼ RHS:

■)

b. Let Q1;Q2;Q3; � � � 2 X: Let Q1 � Q2 � Q3 � � � � :


Problem 3.227 Q1 [Q2 [Q3 [ � � �ð Þ 2 X; that is

1. Q1 [Q2 [Q3 [ � � �ð Þ � S Tð Þ;2. u Q1 [Q2 [Q3 [ ��ð Þ isS-measurable,3. w Q1 [Q2 [Q3 [ ��ð Þ is T -measurable,4.RX u Q1 [Q2 [Q3 [ ��ð Þdl ¼

RY w Q1 [Q2 [Q3 [ ��ð Þdk:

(Solution For every positive integer n, Qn 2 X; so for every positive integer n,Qn 2 S Tð Þ, u Qnð Þ isS-measurable, w Qnð Þ is T -measurable, and

RX u Qnð Þdl ¼R

Y w Qnð Þdk:For 1: Since each Qn 2 S Tð Þ; and S T is a r-algebra in X Y ;

Q1 [Q2 [Q3 [ � � �ð Þ 2 S Tð Þ:For 2: Since Q1 � Q2 � Q3 � � � � ; for every x 2 X; we have Q1ð Þx� Q2ð Þx�

Q3ð Þx� � � � : Since each Qn 2 S Tð Þ; for every x 2 X; we have Q1ð Þx;Q2ð Þx; Q3ð Þx; � � � 2 T : It follows that, for every x 2 X;

u Q1 [Q2 [Q3 [ ��ð Þ xð Þ ¼ k Q1 [Q2 [Q3 [ � � �ð Þx� �

¼ k Q1ð Þx [ Q2ð Þx [ Q3ð Þx [ � � ��

¼ limn!1

k Qnð Þx� �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ lim

n!1uQn

xð Þ;

and hence

u Q1 [Q2 [Q3 [ ��ð Þ ¼ limn!1

uQn:

Since each u Qnð Þ isS-measurable, u Q1 [Q2 [Q3 [ ��ð Þ ¼� �

limn!1 uQnis S-mea-

surable, and hence u Q1 [Q2 [Q3 [ ��ð Þ isS-measurable.For 3: Similar to (2).For 4: Since for every x 2 X, Q1ð Þx� Q2ð Þx� Q3ð Þx� � � � ; and

Q1ð Þx; Q2ð Þx; Q3ð Þx; � � � 2 T ; we have

0�uQ1�uQ2

�uQ3� � � � :

Also, each u Qnð Þ isS-measurable. On using Theorem 1.125, we get

LHS ¼ZX

u Q1 [Q2 [Q3 [ ��ð Þdl ¼ZX

limn!1

uQn

� �dl ¼ lim

n!1

ZX

uQndl

0@ 1A¼ lim

n!1

ZY

w Qnð Þdk

0@ 1A ¼ZY

limn!1

w Qnð Þ

� �dk ¼

ZY

w Q1 [Q2 [Q3 [ ��ð Þdk ¼ RHS:

■)c1. Let Q1;Q2 2 X: Suppose that Q1 \Q2 ¼ ;:


Problem 3.228 Q1 [Q2ð Þ 2 X; that is

1. Q1 [Q2ð Þ � S Tð Þ;2. u Q1 [Q2ð Þ isS-measurable,3. w Q1 [Q2ð Þ is T -measurable,4.RX u Q1 [Q2ð Þdl ¼

RY w Q1 [Q2ð Þdk:

(Solution Since Q1 2 X; we have Q1 2 S Tð Þ, u Q1ð Þ isS-measurable,w Q1ð Þ is T -measurable, and

RX u Q1ð Þdl ¼

RY w Q1ð Þdk: Similarly, Q2 2 S Tð Þ,

u Q2ð Þ isS-measurable, w Q2ð Þ is T -measurable, andRX u Q2ð Þdl ¼

RY w Q2ð Þdk:

For 1: Since Q1;Q2 2 S Tð Þ; and S T is a r-algebra in X Y ,Q1 [Q2ð Þ 2 S Tð Þ:For 2: Since Q1;Q2 2 S Tð Þ; for every x 2 X; we have Q1ð Þx; Q2ð Þx2 T :

Since Q1 \Q2 ¼ ;; for every x 2 X; we have Q1ð Þx \ Q2ð Þx¼ ;:It follows that, for every x 2 X;

u Q1 [Q2ð Þ xð Þ ¼ k Q1 [Q2ð Þx� �

¼ k Q1ð Þx [ Q2ð Þx� �

¼ k Q1ð Þx� �

þ k Q2ð Þx� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ uQ1xð ÞþuQ2

xð Þ ¼ uQ1þuQ2

� �xð Þ;

and hence

u Q1 [Q2ð Þ ¼ uQ1þuQ2

:

Since uQ1;uQ2

areS-measurable, u Q1 [Q2ð Þ ¼� �

uQ1þuQ2

� �is S-measurable,

and hence u Q1 [Q2ð Þ isS-measurable.For 3: Similar to 2.For 4: Since u Q1 [Q2ð Þ ¼ uQ1

þuQ2; and uQ1

;uQ2are nonnegative and S-mea-

surable, we get

LHS ¼ZX

u Q1 [Q2ð Þdl ¼ZX

uQ1þuQ2

� �dl ¼

ZX

uQ1dlþ

ZX

uQ2dl

¼ZY

wQ1dkþ

ZY

wQ2dk ¼

ZY

wQ1þwQ2

� �dk ¼

ZY

w Q1 [Q2ð Þdk ¼ RHS:

■)c2. Let Q1;Q2;Q3; � � � 2 X: Suppose that i 6¼ j\Qi \Qj ¼ ;:


Problem 3.229 Q1 [Q2 [Q3 [ � � �ð Þ 2 X:

(Solution Put P1 � Q1, P2 � Q1 [Q2, P3 � Q1 [Q2 [Q3; etc. By (c1), each Pn 2X: Also, P1 � P2 � P3 � � � � : Now, by (b),

Q1 [Q2 [Q3 [ � � �ð Þ ¼ P1 [P2 [P3 [ � � �ð Þ 2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence

Q1 [Q2 [Q3 [ � � �ð Þ 2 X:

■)

d. Let A 2 S; and l Að Þ\1: Let B 2 T ; and k Bð Þ\1: Let Q1;Q2;Q3; � � � 2 X:Suppose that A Bð Þ Q1 Q2 Q3 � � � :

Problem 3.230 Q1 \Q2 \Q3 \ � � �ð Þ 2 X; that is

1. Q1 \Q2 \Q3 \ � � �ð Þ � S Tð Þ;2. u Q1 \Q2 \Q3 \ ��ð Þ isS-measurable,3. w Q1 \Q2 \Q3 \ ��ð Þ is T -measurable,4.RX u Q1 \Q2 \Q3 \ ��ð Þdl ¼

RY w Q1 \Q2 \Q3 \ ��ð Þdk:

(Solution For every positive integer n, Qn 2 X; so for every positive integer n,Qn 2 S Tð Þ; u Qnð Þ isS-measurable, w Qnð Þ is T -measurable, and

RX u Qnð Þdl ¼R

Y w Qnð Þdk:For 1: Since each Qn 2 S Tð Þ; and S T is a r-algebra in X Y ;

Q1 \Q2 \Q3 \ � � �ð Þ 2 S Tð Þ:For 2: Since Q1 Q2 Q3 � � � ; for every x 2 X we have Q1ð Þx Q2ð Þx

Q3ð Þx � � � : Since each Qn 2 S Tð Þ; for every x 2 X we have,Q1ð Þx; Q2ð Þx; Q3ð Þx; � � � 2 T : Since A Bð Þ Q1; we have, for every x 2 A; B ¼A Bð Þx Q1ð Þx; and for every x 62 A, A Bð Þx¼ ;: Now, since k Bð Þ\1; byLemma 1.99(5),

uQ1 \Q2 \Q3 \ �� xð Þ ¼ k Q1 \Q2 \Q3 \ � � �ð Þx� �

¼ k Q1ð Þx \ Q2ð Þx \ Q3ð Þx \ � � ��

¼ limn!1

k Qnð Þx� �

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ lim

n!1uQn

xð Þ� �

¼ limn!1

uQn

� �xð Þ;


and hence

uQ1 \Q2 \Q3 \ �� ¼ limn!1

uQn:

Since each u Qnð Þ isS-measurable,�u Q1 \Q2 \Q3 \ ��ð Þ ¼

�limn!1 uQn

is S-mea-surable, and hence u Q1 \Q2 \Q3 \ ��ð Þ isS-measurable.

For 3: Similar to 2.For 4: Since for every x 2 X; Q1ð Þx Q2ð Þx Q3ð Þx � � � ; and

Q1ð Þx; Q2ð Þx; Q3ð Þx; � � � 2 T ; we have uQ1�uQ2

�uQ3� � � � : Also, for x 2 A;

Q1ð Þx� A Bð Þx ¼ Bð Þ; and hence for x 2 A; uQ1xð Þ ¼

� �k Q1ð Þx� �

� k Bð Þ\1:

Thus, for every for x 2 X;

uQ1xð Þ

�� k Bð Þð Þ vAð Þ xð Þ:

Also, k Bð Þð Þ vAð Þ 2 L1 lð Þ; and each u Qnð Þ isS-measurable. On usingTheorem 1.136, we get

LHS ¼ZX

u Q1 [Q2 [Q3 [ ��ð Þdl ¼ZX

limn!1

uQn

� �dl ¼ lim

n!1

ZX

uQndl

0@ 1A¼ lim

n!1

ZY

w Qnð Þdk

0@ 1A ¼ZY

limn!1

w Qnð Þ

� �dk ¼

ZY

w Q1 [Q2 [Q3 [ ��ð Þdk ¼ RHS:

■)Since X;S; lð Þ is a r-finite measure space, X has r-finite measure, and hence

there exists a countable collection X1;X2; . . .f g of members in S such that X ¼X1 [X2 [ � � � ; each l Xið Þ\1; and X1;X2; . . . are pairwise disjoint. Similarly, thereexists a countable collection Y1; Y2; . . .f g of members in T such that Y ¼Y1 [ Y2 [ � � � ; each k Yið Þ\1; and Y1; Y2; . . . are pairwise disjoint. Put

M � Q : Q 2 S Tð Þ; and for all positive integers n;m;Q\ Xn Ymð Þ 2 Xf g

e: Problem 3.231 M is a monotone class.

(Solution I. Let Q1;Q2;Q3; � � � 2 M; andQ1 � Q2 � Q3 � � � � : We have to showthat Q1 [Q2 [Q3 [ � � �ð Þ 2 M; that is

1. Q1 [Q2 [Q3 [ � � �ð Þ 2 S Tð Þ;2. for all positive integers n;m; Q1 [Q2 [Q3 [ � � �ð Þ \ Xn Ymð Þ 2 X.


For 1: Since each Qk 2 M; each Qk 2 S Tð Þ; and eachQk \ Xn Ymð Þ 2 X:Since each Qk 2 S Tð Þ; and S Tð Þ is a r-algebra, Q1 [Q2 [Q3 [ � � �ð Þ2 S Tð Þ:

For 2: Let us fix any positive integers m, n. We have to show that

Q1 \ Xn Ymð Þð Þ [ Q2 \ Xn Ymð Þð Þ [ � � � ¼ Q1 [Q2 [Q3 [ � � �ð Þ \ Xn Ymð Þ 2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since Q1 � Q2 � Q3 � � � � ; we have

Q1 \ Xn Ymð Þð Þ � Q2 \ Xn Ymð Þð Þ � Q3 \ Xn Ymð Þð Þ � � � � :

Now, since Q1 \ Xn Ymð Þð Þ, Q2 \ Xn Ymð Þð Þ; � � � 2 X; by (b),

Q1 [Q2 [Q3 [ � � �ð Þ \ Xn Ymð Þ ¼ Q1 \ Xn Ymð Þð Þ [ Q2 \ Xn Ymð Þð Þ [ � � �ð Þ 2 X;

and hence

Q1 [Q2 [Q3 [ � � �ð Þ \ Xn Ymð Þ 2 X:

II. Let Q1;Q2;Q3; � � � 2 M, and Q1 Q2 Q3 � � � : We have to show thatQ1 \Q2 \Q3 \ � � �ð Þ 2 M; that is 1. Q1 \Q2 \Q3 \ � � �ð Þ 2 S Tð Þ;2. for all positive integers n;m; Q1 \Q2 \Q3 \ � � �ð Þ \ Xn Ymð Þ 2 X.For 1: Since each Qk 2 M; each Qk 2 S Tð Þ, and each Qk \ Xn Ymð Þ 2 X:

Since each Qk 2 S Tð Þ; and S Tð Þ is a r-algebra, Q1 \Q2 \Q3 \ � � �ð Þ2 S Tð Þ:

For 2: Let us fix any positive integers m, n. We have to show that

Q1 \ Xn Ymð Þð Þ \ Q2 \ Xn Ymð Þð Þ \ � � � ¼ Q1 \Q2 \Q3 \ � � �ð Þ \ Xn Ymð Þ 2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since Q1 Q2 Q3 � � � ; we have

Xn Ymð Þ Q1 \ Xn Ymð Þð Þ Q2 \ Xn Ymð Þð Þ Q3 \ Xn Ymð Þð Þ � � � :

Now, since Q1 \ Xn Ymð Þð Þ, Q2 \ Xn Ymð Þð Þ; � � � 2 X; l Xnð Þ\1; andk Ymð Þ\1; by (d),

Q1 \Q2 \Q3 \ � � �ð Þ \ Xn Ymð Þ ¼ Q1 \ Xn Ymð Þð Þ \ Q2 \ Xn Ymð Þð Þ \ � � �ð Þ 2 X;


and hence

Q1 \Q2 \Q3 \ � � �ð Þ \ Xn Ymð Þ 2 X:

■)

f: Problem 3.232 M contains E; where E denotes the collection of all elementarysets.

(Solution Let A1 B1ð Þ [ � � � [ Ak Bkð Þð Þ 2 E; where A1; . . .;Ak 2 S,B1; . . .;Bk 2 T ; and A1 B1ð Þ; . . .; Ak Bkð Þ are pairwise disjoint. We have toshow that A1 B1ð Þ [ � � � [ Ak Bkð Þð Þ 2 M; that is1. A1 B1ð Þ [ � � � [ Ak Bkð Þð Þ 2 S Tð Þ;2. For all positice integers n;m;

A1 B1ð Þ \ Xn Ymð Þð Þ [ � � � [ Ak Bkð Þ \ Xn Ymð Þð Þ¼ A1 B1ð Þ [ � � � [ Ak Bkð Þð Þ \ Xn Ymð Þ 2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

For 1: Since S T is a r-algebra containing all measurable rectangles,A1 B1ð Þ [ � � � [ Ak Bkð Þð Þ 2 S Tð Þ:For 2: Let us fix any positive integers m; n: We have to show that

A1 \Xnð Þ B1 \ Ymð Þð Þ [ � � � [ Ak \Xnð Þ Bk \ Ymð Þð Þ¼ A1 B1ð Þ \ Xn Ymð Þð Þ [ � � � [ Ak Bkð Þ \ Xn Ymð Þð Þ 2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

By (a), A1 \Xnð Þ B1 \ Ymð Þð Þ; � � �, Ak \Xnð Þ Bk \ Ymð Þð Þ 2 X: SinceA1 B1ð Þ; . . .; Ak Bkð Þ are pairwise disjoint, A1 \Xnð Þ B1 \Ymð Þð Þ; . . .;Ak \Xnð Þ Bk \ Ymð Þð Þ are pairwise disjoint. Now, by (c1),

A1 \Xnð Þ B1 \ Ymð Þð Þ [ � � � [ Ak \Xnð Þ Bk \Ymð Þð Þð Þ 2 X:

■)

g: Problem 3.233 Q 2 X:

(Solution By (e) and (f),M is a monotone class containing E: Now, by Conclusion3.223, Q 2ð Þ S Tð Þ � M; and hence Q 2 M: It follows that eachQ\ Xn Ymð Þ 2 X: Let k; lð Þ 6¼ k1; l1ð Þ; where k; k1 2 1; . . .; nf g; and l; l1 21; . . .;mf g: It follows that either k 6¼ k1 or l 6¼ l1: For definiteness, let k 6¼ k1: It

follows that Xk \Xk1 ¼ ;; and hence

Q\ Xk Ylð Þð Þ \ Q\ Xk1 Yl1ð Þð Þ ¼ Q\ Xk \Xk1ð Þ Yl \ Yl1ð Þð Þ¼ Q\ ; Yl \ Yl1ð Þð Þ ¼ Q\; ¼ ;:


Thus, Q\ Xk Ylð Þ : k; lð Þ 2 1; . . .; nf g 1; . . .;mf gf g is a finite, disjointcollection of members in X: It follows, by (c1), that

Q ¼ Q\ X Yð Þ ¼ Q\ [ nk¼1Xk

� � Y

� �¼ Q\ [ n

k¼1 Xk Yð Þ� �

¼ Q\ [ nk¼1 Xk [ m

l¼1Yl� ��

¼ Q\ [ nk¼1 [ m

l¼1 Xk Ylð Þ� ��

¼ [ nk¼1 Q\ [ m

l¼1 Xk Ylð Þ� ��

¼ [ nk¼1 [ m

l¼1 Q\ Xk Ylð Þð Þ� �

2 X|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};and hence Q 2 X: ■)

Now, by the definition of X; uQ isS-measurable, wQ is T -measurable, andZX

uQdl ¼ZY

wQdk that is;ZX

k Qxð Þdl xð Þ ¼ZY

l Qyð Þdk yð Þ

0@ 1A:

Conclusion 3.234 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let Q 2S Tð Þ: Then

1. x 7! k Qxð Þ is an S-measurable mapping from X to 0;1½ �;2. y 7! l Qyð Þ is a T -measurable mapping from Y to 0;1½ �;3.RX k Qxð Þdl xð Þ ¼

RY l Qyð Þdk yð Þ:

Note 3.235 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let

l kð Þ : Q 7!ZX


l Qyð Þdk yð Þ

0@ 1Abe the mapping from the r-algebra S T to 0;1½ �:

Problem 3.236 l kð Þ is a measure.

(Solution Let Q1;Q2;Q3; � � � 2 S Tð Þ; that is for each positive integer n,x 7! k Qnð Þx

� �is an S-measurable mapping from X to 0;1½ �; y 7! l Qnð Þyð Þ is a T -

measurable mapping from Y to 0;1½ �; andZX

k Qnð Þx� �

dl xð Þ ¼ZY

l Qnð Þyð Þdk yð Þ:

Let Q1;Q2;Q3; . . .f g be a disjoint collection. We have to show that

X1n¼1

ZX

k Qnð Þx� �

dl xð Þ

0@ 1A ¼ZX

k Q1 [Q2 [Q3 [ � � �ð Þx� �

dl xð Þ:


RHS ¼ZX

k Q1 [Q2 [Q3 [ � � �ð Þx� �

dl xð Þ ¼ZX

k Q1ð Þx [ Q2ð Þx [ � � ��

dl xð Þ

¼ZX

k Q1ð Þx� �

þ k Q2ð Þx� �

þ � � ��

dl xð Þ ¼X1n¼1

ZX

k Qnð Þx� �

dl xð Þ

0@ 1A ¼ LHS:Þ

Here, l kð Þ is called the product of measures l and k:

Problem 3.237 l kð Þ : X Y ! 0;1½ � is a r-finite measure.

(Solution Since X;S; lð Þ is a r-finite measure space, X has r-finite measure, andhence there exists a countable collection X1;X2; . . .f g of members in S such thatX ¼ X1 [X2 [ � � � ; each l Xið Þ\1; and X1;X2; . . . are pairwise disjoint. Similarly,there exists a countable collection Y1; Y2; . . .f g of members in T such that Y ¼Y1 [ Y2 [ � � � ; each k Yið Þ\1; and Y1; Y2; . . . are pairwise disjoint. It suffices toshow that

1. [ Xn Ym : n;mð Þ 2 1; 2; . . .f g 1; 2; . . .f gf g ¼ X Y ;2. each Xn Ymð Þ 2 S Tð Þ;3. for all positive integers n, m, l kð Þ Xn Ymð Þ\1:

For 1: Here,

[ Xn Ym : n;mð Þ 2 1; 2; . . .f g 1; 2; . . .f gf g ¼ [1n¼1 [1

m¼1 Xn Ymð Þ� �

¼ [1n¼1 Xn [1

m¼1Ym� ��

¼ [1n¼1 Xn Yð Þ ¼ [1

n¼1Xn� �

Y ¼ X Y ;

so

[ Xn Ym : n;mð Þ 2 1; 2; . . .f g 1; 2; . . .f gf g ¼ X Y :

For 2: Since each Xn 2 S; and each Ym 2 T ; each Xn Ymð Þ is a measurablerectangle, and hence each Xn Ymð Þ 2 S Tð Þ:

For 3: For all positive integers n, m,

l kð Þ Xn Ymð Þ ¼ZX

k Xn Ymð Þx� �

dl xð Þ ¼ZX

k Ymð Þð Þ vXn

� �dl

¼ k Ymð Þð ÞZX

vXn

� �dl ¼ k Ymð Þð Þ l Xnð Þð Þ\1;

so, for all positive integers n;m;

l kð Þ Xn Ymð Þ\1:

■)


Conclusion 3.238 Let X;S; lð Þ, Y ; T ; kð Þ be r-finite measure spaces. Thenl kð Þ : X Y ! 0;1½ � is a r-finite measure.

Note 3.239 Let X;S; lð Þ, Y ; T ; kð Þ be r-finite measure spaces. Let f : X Yð Þ !0;1½ � be an S Tð Þ-measurable function.By Note 3.235, l kð Þ : X Y ! 0;1½ � is a r-finite measure. It follows thatRX Y f d l kð Þ

� �2 0;1½ �: Since f : X Yð Þ ! 0;1½ �; by Lemma 3.224, for

every x 2 X; fx : Y ! 0;1½ � is a T -measurable function, and henceRY fxdk

� �2

0;1½ �: Similarly,RX f

ydl� �

2 0;1½ �:

Definition By uf ; we shall mean the mapping uf : x 7!RY fxdk from X to 0;1½ �:

Similarly, by wf ; we shall mean the mapping wf : y 7!RX f

ydl from Y to 0;1½ �:

Problem 3.240

1. uf : X ! 0;1½ � is S-measurable,2. wf : Y ! 0;1½ � is T -measurable,3.RX uf dl ¼

RX Y f d l kð Þ ¼

RY wf dk:

(Solution Case I: when f ¼ vQ; where Q 2 S Tð Þ: By Conclusion 3.234,x 7! k Qxð Þ is an S-measurable mapping from X to 0;1½ �; y 7! l Qyð Þ is a T -mea-surable mapping from Y to 0;1½ �, andZ

X


l Qyð Þdk yð Þ:

Since

uf : x 7!ZY

fxdk|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZY

vQ� �

xdk ¼ZY

vQxdk ¼ k Qxð Þ;

and x 7! k Qxð Þ is an S-measurable mapping from X to 0;1½ �; uf : X ! 0;1½ � is S-measurable. Similarly, wf : Y ! 0;1½ � is T -measurable.

Next, ZX

uf dl ¼ZX


l Qyð Þdk yð Þ ¼ZY

wf dk;


so ZX

uf dl ¼ZY

wf dk:

Also,ZX Y

f d l kð Þ ¼Z

X Y

vQd l kð Þ ¼ l kð Þ Qð Þ ¼ZX

k Qxð Þdl xð Þ ¼ZX

uf dl:

Thus, ZX

uf dl ¼Z

X Y

f d l kð Þ ¼ZY

wf dk:

Case II: when f : X Yð Þ ! 0;1½ Þ is a simple function. It follows that there existdistinct real numbers a1; . . .; an such that f�1 a1ð Þ; . . .; f�1 anð Þ are members ofS Tð Þ; and


Here,

ua1v f�1 a1ð Þð Þ þ �� þ anv f�1 anð Þð Þ¼ uf : x 7!

ZY

fxdk|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ZY


� �xdk

¼ZY

a1v f�1 a1ð Þð Þx þ � � � þ anv f�1 anð Þð Þx

� �dk ¼ a1

ZY

v f�1 a1ð Þð Þxdkþ � � � þ an

ZY

v f�1 anð Þð Þxdk

¼ a1

ZY

vf�1 a1ð Þ

� �xdkþ � � � þ an

ZY

vf�1 anð Þ

� �xdk ¼ a1u

vf�1 a1ð Þ

� � xð Þþ � � � þ anu vf�1 anð Þ

� � xð Þ

¼ a1uvf�1 a1ð Þ

� �þ � � � þ anu vf�1 anð Þ

� �0@ 1A xð Þ;

so

ua1v f�1 a1ð Þð Þ þ �� þ anv f�1 anð Þð Þ¼ a1u

vf�1 a1ð Þ


� �:


By Case I, uvf�1 a1ð Þ

� �; . . .;uvf�1 anð Þ

� � are S-measurable functions, so

uf ¼ ua1v f�1 a1ð Þð Þ þ �� þ anv f�1 anð Þð Þ¼ a1u

vf�1 a1ð Þ


� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

is an S-measurable function, and hence uf is an S-measurable function. Similarly,wf is a T -measurable function.

Next, for each i 2 1; . . .; nf g; by Case I, uvf�1 aið Þ: X ! 0;1½ � is S-measurable,

wvf�1 aið Þ: Y ! 0;1½ � is T -measurable, andZ

X

uvf�1 aið Þdl ¼

ZX Y

vf�1 aið Þd l kð Þ ¼ZY

wvf�1 aið Þdk:

We have to show thatZX

ua1v f�1 a1ð Þð Þ þ �� þ anv f�1 anð Þð Þ

� �dl ¼Z

X Y


� �d l kð Þ

¼ZY

wa1v f�1 a1ð Þð Þ þ �� þ anv f�1 anð Þð Þ

� �dk;that is

ZX

a1uv f�1 a1ð Þð Þþ � � � þ anuv f�1 anð Þð Þ

� �dl ¼

ZX Y


� �d l kð Þ

¼ZY

a1wv f�1 a1ð Þð Þþ � � � þ anwv f�1 anð Þð Þ

� �dk;

that is

a1

ZX

uv f�1 a1ð Þð Þdlþ � � � þ an

ZX

uv f�1 anð Þð Þdl

¼ a1

ZX Y

v f�1 a1ð Þð Þd l kð Þþ � � � þ an

ZX Y

v f�1 anð Þð Þd l kð Þ

¼ a1

ZY

wv f�1 a1ð Þð Þdkþ � � � þ an

ZY

wv f�1 anð Þð Þdk:


This is clearly true, because, for each i 2 1; . . .; nf g;ZX

uvf�1 aið Þdl ¼

ZX Y

vf�1 aið Þd l kð Þ ¼ZY

wvf�1 aið Þdk:

Case III: when f : X Yð Þ ! 0;1½ � is a S Tð Þ-measurable function. ByLemma 1.98, there exists a sequence snf g of simple measurable functions sn :X Yð Þ ! 0;1½ Þ such that for every x; yð Þ 2 X Yð Þ; 0� s1 x; yð Þ� s2 x; yð Þ� � � � ; and limn!1 sn x; yð Þ ¼ f x; yð Þ: By Case II, for every positive integer n,usn : X ! 0;1½ � is S-measurable, wsn : Y ! 0;1½ � is T -measurable, and

ZX

usndl ¼Z

X Y

snd l kð Þ ¼ZY

wsndk:


ZX Y

f d l kð Þ ¼ limn!1

ZX Y

snd l kð Þ

0@ 1A:

Since for every x; yð Þ 2 X Yð Þ; 0� s1 x; yð Þ� s2 x; yð Þ� � � � ; for every x 2 X;0� s1ð Þx � s2ð Þx � � � � : Since for every x; yð Þ 2 X Yð Þ;

limn!1

snð Þx yð Þ ¼ limn!1

sn x; yð Þ ¼ f x; yð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fx yð Þ;

we have, for every x 2 X, limn!1 snð Þx¼ fx: Now, by Theorem 1.125,RY fxdk ¼ limn!1

RY snð Þxdk

� �:

Since

uf : x 7!ZY

fxdk|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ lim

n!1

ZY

snð Þxdk

0@ 1A ¼ limn!1

usn xð Þ;

we have uf ¼ limn!1

usn : Since each usn : X ! 0;1½ � is S-measurable,

uf ¼� �

limn!1 usn

� �is S-measurable, and hence uf is S-measurable. Similarly, wf

is T -measurable.


Since for every x 2 X, 0� s1ð Þx � s2ð Þx � � � � ; we have, for every x 2 X;

0�ZY

s1ð Þxdk�ZY

s2ð Þxdk� � � � ;

and hence for every x 2 X;

0�us1 xð Þ�us2 xð Þ� � � � :


limn!1

ZX

usndl

0@ 1A ¼ZX

limn!1

usn

� �dl ¼

ZX

uf dl;

so,RX uf dl ¼ limn!1

RX usndl

� �: Similarly,

RY wf dk ¼ limn!1

RY wsndk

� �: SinceR

X uf dl ¼ limn!1RX usndl

� �,

RX Y f d l kð Þ ¼ limn!1

RX Y snd l kð Þ

� �,R

Y wf dk ¼ limn!1RY wsndk

� �; and, for every positive integer n,

ZX

usndl ¼Z

X Y

snd l kð Þ ¼ZY

wsndk;

we have ZX

uf dl ¼Z

X Y

f d l kð Þ ¼ZY

wf dk:

■)

Conclusion 3.241 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let f :X Yð Þ ! 0;1½ � be a S Tð Þ-measurable function. Then

1. x 7!RY fxdk from X to 0;1½ � is S-measurable,

2. y 7!RX f

ydl from Y to 0;1½ � is T -measurable,3.RX

RY fxdk

� �dl xð Þ ¼


RY

RX f

ydl� �

dk yð Þ (in short we writeRX dl xð Þ

RY f x; yð Þdk yð Þ

� �¼RY dk yð Þ

RX f x; yð Þdl xð Þ

� �:)

This result, known as the Fubini Theorem, is due to G. Fubini (19.01.1879–06.06.1943, Italian). He was steered towards mathematics at an early age by hisfather, who was himself a teacher of mathematics. He studied with Dini andBianchi.


3.9 Convolution

Note 3.242 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let f : X Yð Þ !C be a S Tð Þ-measurable function.

It follows that fj j : X Yð Þ ! 0;1½ Þ is a S Tð Þ-measurable function. Now,by Conclusion 3.241,

1. u fj j : x 7!RY fj jxdk from X to 0;1½ � is S-measurable,

2. w fj j : y 7!RX fj jydl from Y to 0;1½ � is T -measurable,

3.RX u fj jdl ¼

RX Y fj jd l kð Þ ¼

RY w fj jdk:

LetRX u fj jdl 6¼ 1:

From Note 3.242(1),RX u fj jdl 2 0;1½ �; so


� � RX u fj jdl 2

0;1½ Þ, and henceRX Y fj jd l kð Þ 2 0;1½ Þ: This shows that f 2 L1 l kð Þ:

Conclusion 3.243 Let X;S; lð Þ, Y ; T ; kð Þ be r-finite measure spaces. Let f :X Yð Þ ! C be a S Tð Þ-measurable function. Let

u� : x 7!ZY

fj jxdk

be the mapping from X to 0;1½ �: LetRX u

�dl 6¼ 1: Then f 2 L1 l kð Þ:

Note 3.244 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let f : X Yð Þ !R be a S Tð Þ-measurable function. Let f 2 L1 l kð Þ; that isRX Y fj jd l kð Þ\1:

For every x 2 X; by Lemma 3.224, fx : Y ! R is a T -measurable function.

Problem 3.245 fx 2 L1 kð Þ for almost all x 2 X:

(Solution Since f : X Yð Þ ! R is a S Tð Þ-measurable function, we find thatf þ : X Yð Þ ! 0;1½ Þ is a S Tð Þ-measurable function, f� : X Yð Þ ! 0;1½ Þis a S Tð Þ-measurable function, f ¼ f þ � f�; and fj j ¼ f þ þ f�: Now, byConclusion 3.71,

1. u fj j : x 7!RY fj jxdk from X to 0;1½ � is S-measurable,

2. w fj j : y 7!RX fj jydl from Y to 0;1½ � is T -measurable,

3.RX u fj jdl ¼


RY w fj jdk:

4. uf þ : x 7!RY f þð Þxdk �

RY f þ þ f�ð Þxdk ¼

RY fj jxdk ¼ u fj j xð Þ

� �from X to

0;1½ � is S-measurable,5. wf þ : y 7!

RX f þð Þydl from Y to 0;1½ � is T -measurable,

6.RX uf þ dl ¼

RX Y f

þ d l kð Þ ¼RY wf þ dk;


7. uf� : x 7!RY f�ð Þxdk �

RY f þ þ f �ð Þxdk ¼

RY fj jxdk ¼ u fj j xð Þ

� �from X to

0;1½ � is S-measurable,8. wf� : y 7!

RX f�ð Þydl from Y to 0;1½ � is T -measurable,

9.RX uf�dl ¼

RX Y f

�d l kð Þ ¼RY wf�dk:

From 4, 0�uf þ �u fj j; soRX uf þ�� dl ¼

� � RX uf þ dl�

RX u fj jdl ¼

RX Y

�fj jd l kð Þ\1Þ; and hence uf þ 2 L1 lð Þ: Similarly, uf� 2 L1 lð Þ: Sinceuf þ 2 L1 lð Þ, l x : uf þ

� �xð Þ ¼ 1

� � �¼ 0: Similarly, l x : uf�

� �xð Þ ¼ 1

� � �¼

0: It follows that l x : uf þ� �

xð Þ ¼ 1 or uf��

xð Þ ¼ 1� � �

¼� �

l x : uf þ� ��

xð Þ ¼ 1g[ x : uf��

xð Þ ¼ 1�

Þ ¼ 0, that is

l x : uf þ� �


xð Þ ¼ 1� � �

¼ 0:

Let us take any

a 2 x : uf þ� �


xð Þ ¼ 1� c

; that isZY

f þð Þa�� dk ¼

ZY

f þð Þadk ¼ uf þ að Þ\1|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}; andZY

f�ð Þa�� dk ¼

ZY

f�ð Þadk ¼ uf� að Þ\1|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} :It suffices to show that fa 2 L1 kð Þ:Since

RY f þð Þa�� dk\1, f þð Þa2 L1 kð Þ: Similarly, f�ð Þa2 L1 kð Þ: It follows that

fa ¼ f þ � f�ð Þa¼� �

f þð Þa� f�ð Þa� �

2 L1 kð Þ: Thus, fa 2 L1 kð Þ: Similarly, f y 2L1 lð Þ for almost all y 2 Y : Since fx 2 L1 kð Þ for almost all x 2 X; we find thatuf : x 7!

RY fxdk is defined a.e. on X. Similarly, wf : y 7!

RX f

ydl is defined a.e. on

Y. Since uf þ 2 L1 lð Þ; and uf� 2 L1 lð Þ; uf ¼ u f þ�f�ð Þ ¼� �

uf þ � uf��

2 L1 lð Þ;and hence uf 2 L1 lð Þ: Similarly, wf 2 L1 kð Þ: Next,

ZX

uf dl ¼ZX

uf þ dl�ZX

uf�dl ¼Z

X Y

f þ d l kð Þ �ZX

uf�dl

¼Z

X Y

f þ d l kð Þ �Z

X Y

f�d l kð Þ ¼Z

X Y

f þ � f�ð Þd l kð Þ ¼Z

X Y

f d l kð Þ;

soRX uf dl ¼

RX Y f d l kð Þ: Similarly,

RY wf dk ¼

RX Y f d l kð Þ: Thus,

3.9 Convolution 567

ZX

uf dl ¼Z

X Y

f d l kð Þ ¼ZY

wf dk:

■)

Conclusion 3.246 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let f :X Yð Þ ! R be a S Tð Þ-measurable function. Let f 2 L1 l kð Þ; that isRX Y fj jd l kð Þ\1: Then

1. fx 2 L1 kð Þ for almost all x 2 X;2. f y 2 L1 lð Þ for almost all y 2 Y ;3. uf : x 7!

RY fxdk is defined a.e. on X,

4. wf : y 7!RX f

ydl is defined a.e. on Y,5. uf 2 L1 lð Þ;6. wf 2 L1 kð Þ;7.RX uf dl ¼


RY wf dk:

Note 3.247 Let X;S; lð Þ, Y ; T ; kð Þ ber-finite measure spaces. Let f : X Yð Þ ! C

be a S Tð Þ-measurable function. Let f 2 L1 l kð Þ; that isRX Y fj jd l kð Þ\1:

Since Re fð Þð Þþ i Im fð Þð Þ ¼ð Þf : X Yð Þ ! C is a S Tð Þ-measurable func-tion, Re fð Þð Þ : X Yð Þ ! R is a S Tð Þ-measurable function. SinceRe fð Þð Þþ i Im fð Þð Þ ¼ð Þf 2 L1 l kð Þ, Re fð Þð Þ 2 L1 l kð Þ: Now, by Conclusion

3.246,

1. Re fð Þð Þx2 L1 kð Þ for almost all x 2 X;2. Re fð Þð Þy2 L1 lð Þ for almost all y 2 Y ;3. u Re fð Þð Þ : x 7!

RY Re fð Þð Þxdk is defined a.e. on X,

4. w Re fð Þð Þ : y 7!RX Re fð Þð Þydl is defined a.e. on Y,

5. u Re fð Þð Þ 2 L1 lð Þ;6. w Re fð Þð Þ 2 L1 kð Þ;7.RX u Re fð Þð Þdl ¼

RX Y Re fð Þð Þd l kð Þ ¼

RY w Re fð Þð Þdk:

Similarly,

1′. Im fð Þð Þx2 L1 kð Þ for almost all x 2 X;2′. Im fð Þð Þy2 L1 lð Þ for almost all y 2 Y ;3′. u Im fð Þð Þ : x 7!

RY Im fð Þð Þxdk is defined a.e. on X,

4′. w Im fð Þð Þ : y 7!RX Im fð Þð Þydl is defined a.e. on Y,

5′. u Im fð Þð Þ 2 L1 lð Þ;6′. w Im fð Þð Þ 2 L1 kð Þ;7′.RX u Im fð Þð Þdl ¼

RX Y Im fð Þð Þd l kð Þ ¼

RY w Im fð Þð Þdk:


a. From (1), and (1′), fx ¼ Re fð Þð Þþ i Im fð Þð Þð Þx¼ Re fð Þð Þx þ i Im fð Þð Þx� ��

2 L1 kð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}for almost all x 2 X; and hence fx 2 L1 kð Þ for almost all x 2 X:

b. Similar to (a), f y 2 L1 lð Þ for almost all y 2 Y :c. From (3), and (3′),

uf ¼ u Re fð Þð Þþ i Im fð Þð Þð Þ

¼ u Re fð Þð Þ þ i u Im fð Þð Þ

� �� : x 7!

ZY

Re fð Þð Þxdk

0@ 1Aþ iZY

Im fð Þð Þxdk

0@ 1A0@ 1A|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ZY

Re fð Þð Þx þ i Im fð Þð Þx� ��

dk ¼ZY

Re fð Þð Þþ i Im fð Þð Þð Þxdk ¼ZY

fxdk

is defined a.e. on X, so uf : x 7!RY fxdk is defined a.e. on X.

d. Similar to (c), wf : y 7!RX f

ydl is defined a.e. on Y.e. From (5), and (5′),

uf ¼ u Re fð Þð Þþ i Im fð Þð Þð Þ ¼ u Re fð Þð Þ þ i u Im fð Þð Þ

� �2 L1 lð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence uf 2 L1 lð Þ:f. Similar to (e), wf 2 L1 kð Þ:g. From (7), and (7′),

ZX

uf dl ¼ZX

u Re fð Þð Þþ i Im fð Þð Þð Þ

� �dl ¼

ZX

u Re fð Þð Þ þ i u Im fð Þð Þ

� �� dl

¼ZX

u Re fð Þð Þdl

0@ 1Aþ iZX

u Im fð Þð Þdl

0@ 1A ¼Z

X Y

Re fð Þð Þd l kð Þ

0@ 1Aþ iZ

X Y

Im fð Þð Þd l kð Þ

0@ 1A|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼Z

X Y

Re fð Þð Þþ i Im fð Þð Þð Þd l kð Þ ¼Z

X Y

f d l kð Þ;

so, ZX

uf dl ¼Z

X Y

f d l kð Þ:

3.9 Convolution 569

Similarly,RY wf dk ¼

RX Y f d l kð Þ: Thus,

ZX

uf dl ¼Z

X Y

f d l kð Þ ¼ZYwf dk:

Conclusion 3.248 Let X;S; lð Þ; Y ; T ; kð Þ be r-finite measure spaces. Let f :X Yð Þ ! C be a S Tð Þ-measurable function. Let f 2 L1 l kð Þ: Then

a. fx 2 L1 kð Þ for almost all x 2 X;b. f y 2 L1 lð Þ for almost all y 2 Y ;c. uf : x 7!

RY fxdk is defined a.e. on X,

d. wf : y 7!RX f

ydl is defined a.e. on Y,e. uf 2 L1 lð Þ;f. wf 2 L1 kð Þ;g.RX uf dl ¼


RY wf dk:

Note 3.249 For every positive integer k, the Lebesgue measure on Rk will bedenoted by mk; the r-algebra of all Borel sets in Rk will be denoted by Bk; and ther-algebra of all Lebesgue measurable sets in Rk will be denoted by Mk:

a: Problem 3.350 For all positive integers r and s, Brþ s � Mr Ms:

(Solution Let G be an open subset of Rrþ s ffi Rr Rsð Þ: It follows that G can beexpressed as a countable union of sets of the form Gm Hn; where Gm is an openset in Rr; and Hn is an open set in Rs: It suffices to show that for all positive integersm and n, Gm Hn 2 Mr Ms:

Since each Gm is an open set in Rr; each Gm 2 Br � Mrð Þ; and hence eachGm 2 Mr: Similarly, each Hn 2 Ms: It follows that Gm Hn is a measurablerectangle in Rr Rs; and hence Gm Hnð Þ 2 Mr Ms: ■)

b: Problem 3.251 For all positive integers r and s, Mr Ms � Mrþ s:

(Solution Let E Fð Þ 2 Mr Ms; where E 2 Mr and F 2 Ms: It suffices toshow that

E Rsð Þ \ Rr Fð Þ ¼ E Fð Þ 2 Mrþ s|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}; that is, E Rsð Þ \ Rr Fð Þð Þ 2 Mrþ s:

Now, since Mrþ s is a r-algebra, it suffices that show thatE Rsð Þ; Rr Fð Þ 2 Mrþ s:Since E 2 Mr; by Conclusion 1.258(10), there exist sets A and B such that A is

an Fr in Rr; B is a Gd in Rr, A � E � B; and mr B� Að Þ ¼ 0: Since A is an Fr inRr, A Rs is an Fr in Rrþ s: Since B is a Gd in Rr, B Rs is a Gd in Rrþ s: SinceA � E � B; we have A Rs � E Rs � B Rs: Since mr B� Að Þ ¼ 0; we have


mrþ s B Rsð Þ � A Rsð Þð Þ ¼ mrþ s B� Að Þ Rsð Þ ¼ 0:

Since A Rs is an Fr in Rrþ s, B Rs is a Gd in Rrþ s, A Rs � E Rs �B Rs; and mrþ s B Rsð Þ � A Rsð Þð Þ ¼ 0; by Conclusion 1.258(10),E Rsð Þ 2 Mrþ s: Similarly, Rr Fð Þ 2 Mrþ s: ■)Since Rr;Mr;mrð Þ and Rs;Ms;msð Þ are r-finite measure spaces, by Note

3.238, the mapping

mr msð Þ : Q 7!ZRr

ms Qxð Þdmr xð Þ ¼ZRs

mr Qyð Þdms yð Þ

0@ 1Afrom the r-algebra Mr Ms to 0;1½ � is a r-finite measure. Also, mrþ s :Mrþ s ! 0;1½ � is a r-finite measure.

c: Problem 3.352 The r-finite measure mr msð Þ : Mr Msð Þ ! 0;1½ � istranslational invariant.

(Solution Let Q 2 Mr Msð Þ; and a; bð Þ 2 Rr Rs; where a 2 Rr; and b 2 Rs:We have to show that

mr msð Þ Qþ a; bð Þð Þ ¼ mr msð Þ Qð Þ:

Since Q 2 Mr Msð Þ; we have, for every x 2 Rr, Qx�a 2 Ms; and hencems Qx�a þ bð Þ ¼ ms Qx�að Þ:

LHS ¼ mr msð Þ Qþ a; bð Þð Þ ¼ZRr

ms Qþ a; bð Þð Þx� �

dmr xð Þ

¼ZRr

ms Qx�a þ bð Þdmr xð Þ ¼ZRr

ms Qx�að Þdmr xð Þ

¼ZRr

ms Qxð Þdmr xþ að Þ ¼ZRr

ms Qxð Þdmr xð Þ ¼ mr msð Þ Qð Þ ¼ RHS:

■)

d: Problem 3.253 There exists a nonnegative real number c such that, for everyQ 2 Brþ s;

mr msð Þ Qð Þ ¼ c mrþ s Qð Þð Þ:

(Solution Since Brþ s � Mr Ms; and mr msð Þ : Mr Msð Þ ! 0;1½ � is ameasure, mr msð Þ is a Borel measure. Also, by (c), mr msð Þ : Mr Msð Þ !0;1½ � is translational invariant. Now, by Conclusion 1.258(13), there exists a

3.9 Convolution 571

nonnegative real number c such that for every Q 2 Brþ s,mr msð Þ Qð Þ ¼ c mrþ s Qð Þð Þ: ■)

e. Let Q 2 Mr Ms:

Problem 3.254 mr msð ÞðQÞ ¼ mrþ sðQÞ:

(Solution Since Q 2 Mr Msð Þ; and Mr Msð Þ � Mrþ s; we have Q 2Mrþ s; and hence, by Conclusion 1.258(10), there exist sets A and B such that A isan Fr in Rrþ s; B is a Gd in Rrþ s; A � Q � B; and mrþ s B� Að Þ ¼ 0: Since A is anFr in Rrþ s; A 2 Brþ s: Similarly, B 2 Brþ s: Since A;B 2 Brþ s; and Brþ s is a r-algebra, we have B� Að Þ 2 Brþ s; and hence, by (d), 0� mr msð Þ Q� Að Þð� Þ mr msð Þ B� Að Þ ¼ c mrþ s B� Að Þð Þ ¼ c � 0 ¼ 0ð Þ: Thus, mr msð ÞðQ�AÞ ¼ 0: Since

0�mrþ s Q� Að Þ� mrþ s B� Að Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl};we have mrþ s Q� Að Þ ¼ 0:

Since

c ¼ c � 1 ¼ c mrþ s 0; 1½ �rþ s� �� ¼ mr msð Þ 0; 1½ �r 0; 1½ �sð Þ

¼ZRr

ms 0; 1½ �r 0; 1½ �sð Þx� �

dmrðxÞ ¼Z0;1½ �r

ms 0; 1½ �sð ÞdmrðxÞ

¼Z0;1½ �r

1 dmrðxÞ ¼ mr 0; 1½ �rð Þ ¼ 1;

we have c ¼ 1:

LHS ¼ mr msð Þ Qð Þ ¼ mr msð Þ A[ Q� Að Þð Þ¼ mr msð Þ Að Þþ mr msð Þ Q� Að Þ ¼ mr msð Þ Að Þþ 0

¼ mr msð Þ Að Þ ¼ c mrþ s Að Þð Þ ¼ 1 � mrþ s Að Þð Þ¼ mrþ s Að Þþ 0 ¼ mrþ s Að Þþmrþ s Q� Að Þ¼ mrþ s A[ Q� Að Þð Þ ¼ mrþ s Qð Þ ¼ RHS:

■)

f: Problem 3.255 The measure space Rrþ s;Mrþ s;mrþ sð Þ is a completion of themeasure space Rr Rs;Mr Ms;mr msð Þ: mrþ s is a completion of the productmeasure mr ms:


(Solution By Conclusion 1.258(5), mrþ s is complete. By (b),Mr Ms � Mrþ s:By (e), mrþ s is an extension of mr ms: Now, by the definition of completion,Rrþ s;Mrþ s;mrþ sð Þ is a completion of the measure spaceRr Rs;Mr Ms;mr msð Þ: ■)

Conclusion 3.256 mrþ s is a completion of the product measure mr ms:

Note 3.257 Let X;M; mð Þ be a measure space, where m : M ! 0;1½ �: Supposethat X;M�; m�ð Þ is the completion of X;M; mð Þ

(that is, M � M�; m� : M� ! 0;1½ � is an extension of M; and

F 2 M�; m� Fð Þ ¼ 0; andE � Fð Þ ) E 2 M�ð ÞÞ:Let f : X ! 0;1½ � be an M�-measurable function.By Lemma 1.98, there exists a sequence snf g of simple M�-measurable func-

tions sn : X ! 0;1½ Þ such that for every x in X, s0 xð Þ ¼ 0� s1 xð Þ� s2 xð Þ� � � � ;and limn!1 sn xð Þ ¼ f xð Þ: Hence,

s1 � s0ð Þþ s2 � s1ð Þþ s3 � s2ð Þþ � � � ¼ limn!1

sn ¼ f :

Thus,

f ¼ s1 � s0ð Þþ s2 � s1ð Þþ s3 � s2ð Þþ � � � :

Since each sn is a finite linear combination of characteristic functions, ands0 � s1 � s2 � � � � ; we find that each snþ 1 � snð Þ is a finite linear combination ofcharacteristic functions with positive coefficients, and hence we can write:

f ¼ð Þ s1 � s0ð Þþ s2 � s1ð Þþ s3 � s2ð Þþ � � � ¼ c1vE1þ c2vE2

þ � � � ;

where each ci [ 0; each Ei 2 M�; and E1;E2;E3; . . .f g is a pairwise disjointcollection of sets.

Since E1 2 M�; and X;M�; m�ð Þ is the completion of X;M; mð Þ; by Lemma1.141, there exist A1;B1 2 M satisfying A1 � E1 � B1, and m B1 � A1ð Þ ¼ 0:Similarly, there exist A2;B2 2 M satisfying A2 � E2 � B2, and m B2 � A2ð Þ ¼ 0;etc.

Since A1 � E1 � B1; we have, for every x 62 E1 � A1ð Þ; x 2 A1 if andðonlyif x 2 E1Þ: Thus, for every x 62 E1 � A1ð Þ, vA1

xð Þ ¼ vE1xð Þ: Similarly, for

every x 62 E2 � A2ð Þ; vA2xð Þ ¼ vE2

xð Þ; etc. Since E1;E2;E3; . . .f g is a pairwisedisjoint collection of sets, E1 � A1;E2 � A2;E3 � A3; . . .f g is a pairwise disjointcollection of sets. Since E1 � A1;E2 � A2;E3 � A3; . . .f g is a pairwise disjointcollection of sets, and for every positive integer n,

x 62 En � Anð Þ ) vAnxð Þ ¼ vEn

xð Þ;

3.9 Convolution 573

we have

x 62 E1 � A1ð Þ [ E2 � A2ð Þ [ � � �ð Þ ) c1vA1þ c2vA2

þ � � ��

xð Þ¼ c1vE1

þ c2vE2þ � � �

� �xð Þ ¼ f xð Þð Þ:

Since

0� m B1 � A1ð Þ [ B2 � A2ð Þ [ � � �ð Þ� m B1 � A1ð Þþ m B2 � A2ð Þþ � � �¼ 0þ 0þ � � � ¼ 0;

we have

m B1 � A1ð Þ [ B2 � A2ð Þ [ � � �ð Þ ¼ 0:

Now, since

E1 � A1ð Þ [ E2 � A2ð Þ [ � � � � B1 � A1ð Þ [ B2 � A2ð Þ [ � � � ;

and

x 62 E1 � A1ð Þ [ E2 � A2ð Þ [ � � �ð Þ ) c1vA1þ c2vA2

þ � � ��

xð Þ ¼ f xð Þ� �

;

we have f ¼ c1vA1þ c2vA2

þ � � ��

a.e. relative to m. Since each Ai 2 M; each vAiis

an M-measurable function, and hence c1vA1þ c2vA2

þ � � ��

is an M-measurablefunction.

Conclusion 3.258 Let X;M; mð Þ be a measure space, where m : M ! 0;1½ �:Suppose that X;M�; m�ð Þ is the completion of X;M; mð Þ: Let f : X ! 0;1½ � be anM�-measurable function. Then there exists an M-measurable function g : X !0;1½ � such that f ¼ g a.e. relative to m:

Theorem 3.259 Let X;M; mð Þ be a measure space, where m : M ! 0;1½ �:Suppose that X;M�; m�ð Þ is the completion of X;M; mð Þ: Let f : X ! C be anM�-measurable function. Then there exists an M-measurable function g : X ! C suchthat f ¼ g a.e. relative to m:

Proof Case I: when f : X ! R is an M�-measurable function. It follows that f þ :X ! 0;1½ Þ is an M�-measurable function, f� : X ! 0;1½ Þ is an M�-measurablefunction, and f ¼ f þ � f�: Now, by Conclusion 3.258, there exists an M-measur-able function g1 : X ! 0;1½ Þ such that f þ ¼ g1 a.e. relative to m: Also, there existsan M-measurable function g2 : X ! 0;1½ Þ such that f� ¼ g2 a.e. relative to m:

Since g1 : X ! 0;1½ Þ is M-measurable function, and g2 : X ! 0;1½ Þ is M-measurable function, g1 � g2ð Þ : X ! R is M-measurable function.

Since f þ ¼ g1 a.e. relative to m; and f� ¼ g2 a.e. relative to m;f ¼ð Þ f þ � f�ð Þ ¼ g1 � g2ð Þ a.e. relative to m; and hence f ¼ g1 � g2ð Þ a.e. relativeto m:


Case II: when f : X ! C is an M�-measurable function. It follows that Re fð Þ :X ! R is an M�-measurable function, and Im fð Þ : X ! R is an M�-measurablefunction. By Case I, there exists an M-measurable function g1 : X ! R such thatRe fð Þ ¼ g1 a.e. relative to m: Also, there exists an M-measurable function g2 :X ! R such that Im fð Þ ¼ g2 a.e. relative to m:

Since g1 : X ! R is an M-measurable function, and g2 : X ! R is an M-measurable function, g1 þ ig2ð Þ : X ! C is an M-measurable function. SinceRe fð Þ ¼ g1 a.e. relative to m; and Im fð Þ ¼ g2 a.e. relative to m,f ¼ð Þ Re fð Þþ i Im fð Þð Þð Þ ¼ g1 þ ig2ð Þ a.e. relative to m; and hence f ¼ g1 þ ig2ð Þ a.e. relative to m: ■

Note 3.260 Let f : R ! C; and g : R ! C be Borel functions. Let f ; g 2 L1 Rð Þ:

Problem 3.261 For almost all x 2 R,R1�1 f x� yð Þg yð Þj jdy\1:

(Solution Since f ; g 2 L1 Rð Þ; we have f1k k\1; and g1k k\1; and thereforef1k kð Þ g1k kð Þ\1:

Problem 3:262 The function

F : x; yð Þ 7! f x� yð Þg yð Þ

from R R ffi R2� �

to C is a Borel function on R2:

(Solution Since x; yð Þ 7! x� yð Þ is a continuous function from R2 to R; and f :R ! C is a Borel function, their composite x; yð Þ 7! f x� yð Þ is a Borel functionfrom R2 to C: Since x; yð Þ 7! y is a continuous function from R2 to R; and g :R ! C is a Borel function, their composite x; yð Þ 7! g yð Þ is a Borel function fromR2 to C: Since x; yð Þ 7! f x� yð Þ is a Borel function from R2 to C; and x; yð Þ 7! g yð Þis a Borel function from R2 to C; their product

F : x; yð Þ 7! f x� yð Þg yð Þ

from R2 to C is a Borel function on R2: ■)It follows that

Fj j : x; yð Þ 7! f x� yð Þg yð Þj j

from R R ffi R2� �

to C is a Borel function on R2: Now, by Conclusion 3.248,

Z1�1

Z1�1

Fj j x; yð Þdy

0@ 1Adx ¼Z1�1

Z1�1

Fj j x; yð Þdx

0@ 1Ady;

3.9 Convolution 575

and hence

Z1�1

Z1�1

f x� yð Þg yð Þj jdy

0@ 1Adx¼Z1�1

Z1�1

F x; yð Þj jdy

0@ 1Adx ¼Z1�1

Z1�1

F x; yð Þj jdx

0@ 1Ady

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Z1�1

Z1�1

f x� yð Þg yð Þj jdx

0@ 1Ady

¼Z1�1

Z1�1

f x� yð Þj j g yð Þj jdx

0@ 1Ady

¼Z1�1

Z1�1

f x� yð Þj jdx

0@ 1A g yð Þj jdy

¼Z1�1

Z1�1

f tð Þj jdt

0@ 1A g yð Þj jdy

¼Z1�1

fk k1� �

g yð Þj jdy ¼ f1k kð ÞZ1�1

g yð Þj jdy

¼ f1k kð Þ g1k kð Þ\1

Thus,

Z1�1

Z1�1


0@ 1Adx\1:

It follows that for almost all x 2 R,R1�1 f x� yð Þj j g yð Þj jdy\1: For those x, let

us put

h xð Þ �Z1�1

f x� yð Þg yð Þdy:


Now, since

Z1�1

h xð Þj jdx¼Z1�1

Z1�1

f x� yð Þg yð Þdy

��dx�

Z1�1

Z1�1


0@ 1Adx

¼Z1�1

Z1�1


0@ 1Adx ¼ fk k1� �

gk k1� �

\1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

we haveR1�1 h xð Þj jdx\1; and hence h 2 L1 Rð Þ: Now, since

hk k1¼� � Z1

�1

h xð Þj jdx� fk k1 gk k1;

we have

hk k1 � fk k1 gk k1:

■)

Conclusion 3.263 Let f : R ! C; and g : R ! C be Borel functions. Let f ; g 2L1 Rð Þ: Then,

1. for almost all x 2 R,R1�1 f x� yð Þg yð Þj jdy\1;

2. for those x, suppose that f � gð Þ : x 7!R1�1 f x� yð Þg yð Þdy: Then f � gð Þ 2

L1 Rð Þ; and

f � gk k1 � fk k1 gk k1:

Note 3.264 Let f : R ! C; and g : R ! C be Lebesgue measurable functions. Letf ; g 2 L1 Rð Þ:

There exist Borel functions f0 : R ! C; and g0 : R ! C such that f ¼ f0 a.e.,and g ¼ g0 a.e.

Since f ¼ f0 a.e., f0 2 L1 Rð Þ: Similarly, g0 2 L1 Rð Þ: Now, by Conclusion 3.263,

1. for almost all x 2 R,R1�1 f x� yð Þg yð Þj jdy ¼

� � R1�1 f0 x� yð Þg0 yð Þj jdy\1;

2. for those x, suppose that f0 � g0ð Þ : x 7!R1�1 f0 x� yð Þg0 yð Þdy ¼

R1�1 f x� yð Þg yð Þdy

� �:

Then f0 � g0ð Þ 2 L1 Rð Þ; and

f0 � g0k k1 � f0k k1 g0k k1 ¼ fk k1 gk k1� �

:

3.9 Convolution 577

Conclusion 3.265 Let f : R ! C; and g : R ! C be Lebesgue measurable func-tions. Let f ; g 2 L1 Rð Þ: Then

1. for almost all x 2 R;R1�1 f x� yð Þg yð Þj jdy\1;

2. for those x, suppose that f � gð Þ : x 7!R1�1 f x� yð Þg yð Þdy: Then f � gð Þ 2

L1 Rð Þ; and f � gk k1 � fk k1 gk k1:

Here f � g is called the convolution of f and g.

3.10 Distribution Function

Note 3.266 Let X;M; lð Þ be a r-finite, positive, measure space. Let f : X !0;1½ � be measurable.a. For every t 2 0;1½ Þ; t;1ð � is an open set in 0;1½ �: Now, since f : X ! 0;1½ �is measurable, for every t 2 0;1½ Þ; f�1 t;1ð �ð Þ 2 M; and hence for every t 20;1½ Þ; l f�1 t;1ð �ð Þð Þ 2 0;1½ �: Thus, u : t 7! l f�1 t;1ð �ð Þð Þ is a function from0;1½ Þ to 0;1½ �: It is clear that the function u : t 7! l f�1 t;1ð �ð Þð Þ from 0;1½ Þ to0;1½ � is decreasing. It follows that u is continuous at all points of 0;1½ Þ; except atcountable-many points. (cf. WR[1] Theorems 4.29 and 4.30). Now, since u :

0;1½ Þ ! 0;1½ � is decreasing, the function u : t 7! l f�1 t;1ð �ð Þð Þ from 0;1½ Þ to0;1½ � is Borel measurable.Here, the Borel-measurable function u : t 7! l f�1 t;1ð �ð Þð Þ from 0;1½ Þ to

0;1½ � is called the distribution function of f. For every t 2 0;1½ Þ;

u tð Þ ¼ l f�1 t;1ð �ð Þ� �

¼ l x : f xð Þ[ tf gð Þ� �

is denoted by l f [ tf g:

b: Problem 3.267 x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g is M mð Þ-measurable.

(Solution Case I: when f : X ! 0;1½ � is a simple measurable function. It followsthat there exist distinct positive real numbers a1; . . .; an such thatf�1 a1ð Þ; . . .; f�1 anð Þ are M-measurable sets, and


Hence,

x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g

¼ x; tð Þ : x 2 X; t 2 0;1½ �; t\a1v f�1 a1ð Þð Þ xð Þþ � � � þ anv f�1 anð Þð Þ xð Þn o

¼ f�1 a1ð Þ� �

0; a1½ Þ� �

[ � � � [ f�1 anð Þ� �

0; an½ Þ� �

:


Thus,

x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g

is a union of finitely-many measurable rectangles. It follows that

x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g

is M mð Þ-measurable.Case II: when f : X ! 0;1½ � is any measurable function. By Lemma 1.98, thereexists a sequence snf g of simple measurable functions sn : X ! 0;1½ Þ such that forevery x in X, 0� s1 xð Þ� s2 xð Þ� � � � ; and limn!1 sn xð Þ ¼ f xð Þ: By Case I, eachx; tð Þ : x 2 X; t 2 0;1½ �; t\sn xð Þf g is M mð Þ-measurable, so

[1n¼1 x; tð Þ : x 2 X; t 2 0;1½ �; t\sn xð Þf g

is M mð Þ-measurable. Since for every positive integer n, sn � snþ 1; we have

x; tð Þ : x 2 X; t 2 0;1½ �; t\s xð Þf g

¼ x; tð Þ : x 2 X; t 2 0;1½ �; t\ limn!1

sn xð Þn o

¼ [1n¼1 x; tð Þ : x 2 X; t 2 0;1½ �; t\sn xð Þf g|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

x; tð Þ : x 2 X; t 2 0;1½ �; t\s xð Þf g ¼ [1n¼1 x; tð Þ : x 2 X; t 2 0;1½ �; t\sn xð Þf g:

Now, since

[1n¼1 x; tð Þ : x 2 X; t 2 0;1½ �; t\sn xð Þf g

is M mð Þ-measurable,

x; tð Þ : x 2 X; t 2 0;1½ �; t\s xð Þf g

is M mð Þ-measurable.Thus, in all cases, x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g is M mð Þ-measurable.■)

Conclusion 3.268 Let X;M; lð Þ be a r-finite, positive measure space. Let f :X ! 0;1½ � be measurable. Then x; tð Þ : x 2 X; t 2 0;1½ �; t\f xð Þf g is M mð Þ-measurable.

3.10 Distribution Function 579

Note 3.269 Let X;M; lð Þ be a r-finite, positive measure space. Let f : X ! 0;1½ �be measurable. Let u : 0;1½ � ! 0;1½ � be monotonically increasing. Suppose thatfor every T 2 0;1ð Þ; u is absolutely continuous on 0; T½ �: Let u 0ð Þ ¼ 0; andlimt!1 u tð Þ ¼ u 1ð Þ:

Since for every T 2 0;1ð Þ;u is absolutely continuous on 0; T½ �; it follows thatu is continuous at all points of 0;1½ Þ: Since limt!1 u tð Þ ¼ u 1ð Þ; u is continuousat 1: Since u : 0;1½ � ! 0;1½ � is continuous at 1; and u is continuous at allpoints of 0;1½ Þ; u : 0;1½ � ! 0;1½ � is continuous, and hence u : 0;1½ � ! 0;1½ �is measurable. Since u : 0;1½ � ! 0;1½ � is continuous, and f : X ! 0;1½ � ismeasurable, their composite u � fð Þ : X ! 0;1½ � is measurable, and henceR

X u � fð Þdl� �

2 0;1½ �: Since for every T 2 0;1ð Þ; u � fð Þ : X ! 0;1½ � isabsolutely continuous on 0; T½ �; by the conclusion of Note 3.47, u : 0;1½ � !0;1½ � is differentiable a.e. on 0;1½ �:By Conclusion 3.268, the mapping t 7! l f [ tf g from 0;1½ Þ to 0;1½ � is Borel

measurable. Now, since u : 0;1½ � ! 0;1½ � is differentiable a.e. on 0;1½ �;

Z10

l f [ tf gð Þ u0 tð Þð Þdt

0@ 1A 2 0;1½ �:

Problem 3.270RX u � fð Þdl ¼

R10 l f [ tf gð Þ u0 tð Þð Þdt:

(Solution Here, the distribution function of f is t 7! l f�1 t;1ð �ð Þð Þ: By Conclusion3.268,

x; sð Þ : x 2 X; s 2 0;1½ �; s\f xð Þf g

is measurable, and hence for every t 2 0;1½ Þ;

x : t\f xð Þf g ¼ð Þ x; sð Þ : x 2 X; s 2 0;1½ �; s\f xð Þf gt

is measurable. It follows that

l f [ tf g ¼ð Þl x : t\f xð Þf gð Þ ¼ l x; sð Þ : x 2 X; s 2 0;1½ �; s\f xð Þf gt� �

;

and hence

l f [ tf g ¼ l x; sð Þ : x 2 X; s 2 0;1½ �; s\f xð Þf gt� �

:


Now,

RHS ¼Z10

l f [ tf gð Þ u0 tð Þð Þdt ¼Z10

l x; sð Þ : x 2 X; s 2 0;1½ �; s\f xð Þf gt� ��

u0 tð Þð Þdt

¼Z10

l x : t\f xð Þf gð Þð Þ u0 tð Þð Þdt ¼Z10

ZX

v x:t\f xð Þf gdl

0@ 1A u0 tð Þð Þdt

¼Z10

ZX

vf�1 t;1ð �ð Þ xð Þdl xð Þ

0@ 1A u0 tð Þð Þdt ¼Z10

ZX

vf�1 t;1ð �ð Þ xð Þ� �

u0 tð Þð Þdl xð Þ

0@ 1Adt

¼ZX

Z10


u0 tð Þð Þdt

0@ 1Adl xð Þ

¼ZX

Zf xð Þ

0


u0 tð Þð ÞdtþZ1f xð Þ


u0 tð Þð Þdt

0B@1CAdl xð Þ

¼ZX

Zf xð Þ

0

1ð Þ u0 tð Þð ÞdtþZ1f xð Þ

0ð Þ u0 tð Þð Þdt

0B@1CAdl xð Þ ¼

ZX

Zf xð Þ

0

u0 tð Þdt

0B@1CAdl xð Þ

¼ZX

u f xð Þð Þ � u 0ð Þð Þdl xð Þ ¼ZX

u � fð Þ xð Þ � u 0ð Þð Þdl xð Þ ¼ZX

u � fð Þ xð Þ � 0ð Þdl xð Þ

¼ZX

u � fð Þ xð Þdl xð Þ ¼ZX

u � fð Þdl ¼ LHS:

■)Conclusion 3.271 Let X;M; lð Þ be a r-finite, positive measure space. Let f :X ! 0;1½ � be measurable. Let u : 0;1½ � ! 0;1½ � be monotonically increasing.Suppose that for every T 2 0;1ð Þ; u is absolutely continuous on 0; T½ �: Letu 0ð Þ ¼ 0; and limt!1 u tð Þ ¼ u 1ð Þ: ThenZ

X

u � fð Þdl ¼Z10

l f [ tf gð Þ u0 tð Þð Þdt:

Note 3.272 Let p 2 1;1ð Þ: Let f : Rk ! 0;1½ Þ be a Lebesgue measurablefunction. Let f 2 Lp Rk

� �; and f 2 L1 Rk

� �: Let c 2 0; 1ð Þ:

For every t 2 0;1ð Þ; let us define a function gt : Rk ! 0;1½ Þ as follows: Forevery x 2 Rk;

gt xð Þ � f xð Þ if x 2 f�1 ct;1ð Þð Þ0 otherwise:

�


Clearly, for every t 2 0;1ð Þ; gt ¼ f � vf�1 ct;1ð Þð Þ: Since f : Rk ! 0;1½ Þ is a

Lebesgue measurable function, each f�1 ct;1ð Þð Þ is a measurable set, and henceeach vf�1 ct;1ð Þð Þ is Lebesgue measurable. Now, since f is Lebesgue measurable, eachproduct gt ¼ð Þf � vf�1 ct;1ð Þð Þ is Lebesgue measurable, and hence each gt is Lebesguemeasurable.

From the definition of gt; for every t 2 0;1ð Þ; f � gtð Þ : Rk ! 0; ct½ �: Now, byConclusion 2.20, for every t 2 0;1ð Þ; f � gtð Þ 2 L1 Rk

� �; and f � gtk k1 � ct:

Since for every t 2 0;1ð Þ; 0� gt � f ; and f 2 L1 Rk� �

; we have for every t 20;1ð Þ; gt 2 L1 Rk

� �: It follows that for every t 2 0;1ð Þ;

M gtð Þ : x 7! sup

RB x;rð Þ gtj jdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )¼ sup

RB x;rð Þ gtdm

m B x; rð Þð Þ : r 2 0;1ð Þ( ) !

is a measurable function from Rk to 0;1½ �: Since for every t 2 0;1ð Þ; gt 2L1 Rk� �

; and f 2 L1 Rk� �

; we have, for every t 2 0;1ð Þ, f � gtð Þ 2 L1 Rk� �

: Itfollows that for every t 2 0;1ð Þ;

M f � gtð Þ : x 7! sup

RB x;rð Þ f � gtj jdmm B x; rð Þð Þ : r 2 0;1ð Þ

( ):

Observe that

sup

RB x;rð Þ f � gtj jdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )¼ sup

RB x;rð Þ f � gtð Þdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )

� sup

RB x;rð Þ ctð Þdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )¼ sup

ctð Þ � m B x; rð Þð Þm B x; rð Þð Þ : r 2 0;1ð Þ

� �¼ ct:

Thus, M f � gtð Þ is a measurable function from Rk to 0;1½ �: Since

Mfð Þ : x 7! sup


( )

and


sup


( )¼ sup

RB x;rð Þ f dm

m B x; rð Þð Þ : r 2 0;1ð Þ( )

¼ sup

RB x;rð Þ f � gtð Þdmþ

RB x;rð Þ gtdm

m B x; rð Þð Þ : r 2 0;1ð Þ( )

¼ sup

RB x;rð Þ f � gtð Þdmm B x; rð Þð Þ þ

RB x;rð Þ gtdm

m B x; rð Þð Þ : r 2 0;1ð Þ( )

� sup

RB x;rð Þ f � gtð Þdmm B x; rð Þð Þ : r 2 0;1ð Þ

( )þ sup

RB x;rð Þ gtdm

m B x; rð Þð Þ : r 2 0;1ð Þ( )

¼ M f � gtð Þð Þ xð Þþ M gtð Þð Þ xð Þ� ctþ M gtð Þð Þ xð Þ

for every t 2 0;1ð Þ; Mf � ctþM gtð Þ: Hence, for every t 2 0;1ð Þ;

m Mf [ tf g ¼ m x : t\ Mfð Þ xð Þf gð Þ�m x : t\ctþ M gtð Þð Þ xð Þf gð Þ

¼ m x : t 1� cð Þ\ M gtð Þð Þ xð Þf gð Þ ¼ m M gtð Þð Þ�1 t 1� cð Þ;1ð �ð Þ� �

:

Since for every t 2 0;1ð Þ; gt 2 L1 Rk� �

; by Conclusion 3.115,

m Mf [ tf g� m M gtð Þð Þ�1 t 1� cð Þ;1ð �ð Þ� �

� 3kgtk k1

t 1� cð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 3k

RRk gtj jdmt 1� cð Þ ¼ 3k

RRk gtdmt 1� cð Þ ¼ 3k

Rf�1 ct;1ð Þð Þ gtdmþ

Rf�1 0;ct½ Þð Þ gtdm

t 1� cð Þ

¼ 3kRf�1 ct;1ð Þð Þ f dmþ

Rf�1 0;ct½ Þð Þ 0 dm

t 1� cð Þ ¼ 3kRf�1 ct;1ð Þð Þ f dm

t 1� cð Þ ;

and hence

m Mf [ tf g� 3kRf�1 ct;1ð Þð Þ f dm

t 1� cð Þ :

By cð Þ ) að Þ of Conclusion 3.156, for every T 2 0;1ð Þ; the mapping u :t 7! tp is absolutely continuous on 0; T½ �: Put u 1ð Þ ¼ 1: Now, by Conclusion3.271,

ZRk

Mfj jpdm ¼ZRk

Mfð Þpdm ¼ZRk

u � Mfð Þð Þdm ¼Z10

m Mf [ tf gð Þ u0 tð Þð Þdt


and

Z10

m Mf [ tf gð Þ u0 tð Þð Þdt ¼Z10

m Mf [ tf gð Þ ptp�1� �dt�

Z10

3kRf�1 ct;1ð Þð Þ f dm

t 1� cð Þ

!ptp�1� �

dt

¼ 3kp1� cð Þ

Z10

Zf�1 ct;1ð Þð Þ

f dm

0B@1CA tp�2� �

dt ¼ 3kp1� cð Þ

Z10


f xð Þdx

0B@1CA tp�2� �

dt

¼ 3kp1� cð Þ

Z10


f xð Þð Þ � tp�2� �dx

0B@1CAdt ¼ 3kp

1� cð Þ


Z10

f xð Þð Þ � tp�2� �dt

0@ 1Adx

¼ 3kp1� cð Þ


Zf xð Þc

0

f xð Þð Þ � tp�2� �dt

0B@1CAdx ¼ 3kp

1� cð Þ


f xð Þð ÞZf xð Þ

c

0

tp�2dt

0B@1CAdx

¼ 3kp1� cð Þ


f xð Þð Þ tp�1

p� 1

��t¼f xð Þc

t¼0

0@ 1Adx ¼ 3kp1� cð Þ p� 1ð Þ


f xð Þð Þ tp�1��t¼f xð Þ

c

t¼0

� dx

¼ 3kp1� cð Þ p� 1ð Þ


f xð Þð Þ f xð Þc

� p�1 !

dx ¼ 3kp1� cð Þ p� 1ð Þcp�1


fj jpdm

� 3kp1� cð Þ p� 1ð Þcp�1

ZRk

fj jpdm ¼ 3kp1� cð Þ p� 1ð Þcp�1 fk kp

� �p\1

so,RRk Mfj jpdm\1: It follows that Mfð Þ 2 Lp Rk

� �: Also,

Mfð Þk kp �3kp

1� cð Þ p� 1ð Þcp�1

� 1p

fk kp:

Conclusion 3.273 Let p 2 1;1ð Þ: Let f : Rk ! 0;1½ Þ be a Lebesgue measurablefunction. Let f 2 Lp Rk

� �; and f 2 L1 Rk

� �: Then Mfð Þ 2 Lp Rk

� �: Also, for every

c 2 0; 1ð Þ;

Mfð Þk kp �3kp


� 1p

fk kp:

Theorem 3.274 Let p 2 1;1ð Þ: Let f : Rk ! C be a Lebesgue measurablefunction. Let f 2 Lp Rk

� �; and f 2 L1 Rk

� �: Then Mfð Þ 2 Lp Rk

� �: Also, for every

c 2 0; 1ð Þ;


Mfð Þk kp �3kp


� 1p

fp:

Proof Since f : Rk ! C is a Lebesgue measurable function, fj j : Rk ! 0;1½ Þ is aLebesgue measurable function. Since f 2 Lp Rk

� �; we have fj j 2 Lp Rk

� �:

Similarly, fj j 2 L1 Rk� �

: Now, by Conclusion 3.273, M fj jð Þ 2 Lp Rk� �

: Also, forevery c 2 0; 1ð Þ;

M fj jð Þk kp �3kp


� 1p

fj jk kp:

Since

Mfð Þ : x 7! sup


( );

we have M fj j ¼ Mf : Also, fj jk kp¼ fk kp: It follows that Mfð Þ 2 Lp Rk� �

: Also, forevery c 2 0; 1ð Þ;

Mfk kð Þp �3kp


� 1p

fk kp:

■)Theorem 3.274 is due to G. H. Hardy (07.02.1877–01.12.1947, British) and

J. E. Littlewood (09.06.1885–06.09.1977, British).Hardy is known for his achievements in number theory and mathematical

analysis. In biology, he is known for his principle of population genetics. He is alsocredited for the discovery of the Indian mathematician Ramanujan.

Littlewood worked on topics relating to number theory, mathematical analysisand differential equation. He collaborated with Hardy for a long time.

Theorem 3.275 Let p 2 1;1ð Þ: Let q be the exponent conjugate to p: Let f :Rk ! C be a Lebesgue measurable function. Let f 2 Lp Rk

� �; and f 2 L1 Rk

� �:

Then

Mfð Þk kp � 3kpqe� �1

p fk kp:

Proof Since p 2 1;1ð Þ; and q is the exponent conjugate to p, 1� 1q ¼

� �1p 2

0; 1ð Þ; and hence 1q 2 0; 1ð Þ: In Theorem 3.275, let us take 1

q for c. We have


Mfð Þk kp �3kp

1� 1q

� �p� 1ð Þ 1

q

� �p�1

0B@1CA

1p

fk kp¼3kp

1� 1p

� �1� 1

p

� �p�1

0B@1CA

1p

fk kp

¼ 3kpqp

p� 1

� p�1 !1

p

fk kp

¼ 3kpq 1þ pp� 1

� 1� � p�1

!1p

fk kp¼ 3kpq 1þ 1p� 1

� p�1 !1

p

fk kp;

and hence

Mfð Þk kp � 3kpq 1þ 1p� 1

� p�1 !1

p

fk kp:

Problem 3:276 1þ 1p�1

� �p�1\e:

(Solution Since for x[ 0;

ddx

ln 1þ 1x

� � 11þ x

� ¼ 1

1þ 1x

�1x2

� þ 1

1þ xð Þ2¼ 1

1þ x1

1þ x� 1

x

� \0;

and

limx!1

ln 1þ 1x

� � 11þ x

� ¼ 0� 0 ¼ 0;

we have, for x[ 0;

ln 1þ 1x

� � 11þ x

� 0:

Next, for x[ 0;

ddx

1þ 1x

� x� ¼ d

dxexp ln 1þ 1

x

� x� � � ¼ d

dxexp x � ln 1þ 1

x

� � � ¼ exp x � ln 1þ 1

x

� � � 1 � ln 1þ 1

x

� þ x � 1

1þ 1x

� �1x2

!

¼ exp x � ln 1þ 1x

� � � ln 1þ 1

x

� � 11þ x

� � 0;

so, for x[ 0, x 7! 1þ 1x

� �x is an increasing function.


Let us fix an x[ 0: Let us take a positive integer n such that x\n: Now,

1þ 1x

� x

� 1þ 1n

� n

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 1þ 1þ 12!

1� 1n

� þ 1

3!1� 1

n

� 1� 2

n

� þ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

nþ 1ð Þterms

� 1þ 1þ 12!

þ 13!

þ � � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}nþ 1ð Þterms

\e;

so, 1þ 1x

� �x\e: Now, since 1\p; we have 1þ 1p�1

� �p�1\e: ■)

Thus,

Mfð Þk kp � 3kpqe� �1

p fk kp:

■)

3.11 Fourier Transforms

Note 3.277

Definition In the context of Fourier transforms, it is customary to denote1ffiffiffiffi2p

p Lebesgue measure onRð Þ by m.

Thus, the symbolR1�1 f xð Þdm xð Þ will stand for 1ffiffiffiffi

2ppR1�1 f xð Þdx:

Also, for every p 2 1;1½ Þ; fk kp will stand forR1�1 f xð Þj jpdm xð Þ

� �1p; and Lp will

stand for Lp Rð Þ: For every x 2 R; f � gð Þ xð Þ will stand forR1�1 f x� yð Þg yð Þdm yð Þ;

providedR1�1 f x� yð Þg yð Þdm yð Þ exists.

By Conclusion 3.265, if f ; g 2 L1; then f � gð Þ 2 L1 and f � gk k1 �ffiffiffiffiffiffi2p

pfk k1 gk k1:

Let f 2 L1: Hence f : R ! C is a Lebesgue measurable function satisfyingR1�1 f xð Þj jdx 2 0;1½ Þ: Since for every t 2 R; the mapping x 7! e�ixt from R to C iscontinuous, for every t 2 R; the mapping x 7! e�ixt from R to C is Lebesguemeasurable. It follows that for every t 2 R; the mapping x 7! f xð Þe�ixt from R to C

is Lebesgue measurable. Next,

Z1�1

f xð Þe�ixt�� dx ¼ Z1

�1

f xð Þj j e�ixt�� dx ¼ Z1

�1

f xð Þj j � 1 dx ¼Z1�1

f xð Þj jdx 2 0;1½ Þ;

so Z1�1

f xð Þe�ixt�� dx 2 0;1½ Þ;


and hence for every t 2 R; the mapping x 7! f xð Þe�ixt from R to C is a member ofL1: Thus,

f : t 7!Z1�1

f xð Þe�ixtdm xð Þ

is a function from R to C:

The mapping f 7! f defined on L1 is known as the Fourier transform. For everyf 2 L1; the function

f : t 7!Z1�1

f xð Þe�ixtdm xð Þ

from R to C is called the Fourier transform of f.

Conclusion 3.278 Let f 2 L1: Then f : t 7!R1�1 f xð Þe�ixtdm xð Þ is a function from

R to C:

Note 3.279 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1: Leta 2 R: Let g : x 7! f xð Þeiax be a function from R to C:

Since for every real x;

g xð Þj j ¼ f xð Þeiax�� ¼ f xð Þj j eiax

�� ¼ f xð Þj j1 ¼ f xð Þj j;

and f 2 L1; we have g 2 L1; and hence g : R ! C exists.

Problem 3.280 For every t 2 R; g tð Þ ¼ f t � að Þ:

(Solution Let us fix any t.

LHS ¼ g tð Þ ¼Z1�1

g xð Þe�ixtdm xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

g xð Þe�ixtdx

¼ 1ffiffiffiffiffiffi2p

pZ1�1

f xð Þeiax� �

e�ixtdx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f xð Þe�ix t�að Þdx ¼ f t � að Þ ¼ RHS:

■)

Conclusion 3.281 Let f 2 L1: Let a 2 R: Let g : x 7! f xð Þeiax be a function from R

to C: Then g 2 L1; and for every t 2 R; g tð Þ ¼ f t � að Þ:Let f : R ! C be a Lebesgue measurable function. Let f 2 L1: Let a 2 R: Let

h : x 7! f x� að Þ be a function from R to C: Since for every real x; h xð Þj j ¼


f x� að Þj j; and f 2 L1; we have h 2 L1; and hence h : R ! C exists.

Problem 3.282 For every t 2 R; h tð Þ ¼ f tð Þe�iat:

(Solution Let us fix any t:

LHS ¼ h tð Þ ¼Z1�1

h xð Þe�ixtdm xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

h xð Þe�ixtdx


pZ1�1

f x� að Þe�ixtdx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f yð Þe�i yþ að Þtdy


pZ1�1

f yð Þe�iyte�iatdy ¼ e�iat 1ffiffiffiffiffiffi2p

pZ1�1

f yð Þe�iytdy ¼ e�iat f tð Þ ¼ RHS:

■)

Conclusion 3.283 Let f 2 L1: Let a 2 R: Let h : x 7! f x� að Þ be a function fromR to C: Then h 2 L1; and for every t 2 R;

h tð Þ ¼ f tð Þe�iat:

Let f : R ! C be a Lebesgue measurable function. Let f 2 L1: Let g :

x 7! f �xð Þ be a function from R to C: Since for every real x; g xð Þj j ¼ f �xð Þ�� ¼

f �xð Þj j; and f 2 L1; we have g 2 L1; and hence g : R ! C exists.

Problem 3.284 For every t 2 R; g tð Þ ¼ f ðtÞ:

(Solution Let us fix any



pZ1�1

g xð Þe�ixtdx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f �xð Þeixte�ixtdx


pZ1�1

f �xð Þeixtdx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f �xð Þeixtdx

0@ 1A�


pZ�1

1

f yð Þei �yð Þt �1ð Þdy

0@ 1A�


pZ1�1

f yð Þe�iytdy

0@ 1A�


pZ1�1

f yð Þe�iytdy

0@ 1A�

¼ f tð Þ� ��¼ RHS:

■)

3.11 Fourier Transforms 589

Conclusion 3.285 Let f 2 L1: Let g : x 7! f �xð Þ be a function from R to C: Theng 2 L1; and for every t 2 R;

g tð Þ ¼ f tð Þ� ��

:

Let f : R ! C be a Lebesgue measurable function. Let f 2 L1: Let k be apositive real. Let g : x 7! f 1

k x� �

be a function from R to C: Since k is a positive real,f 2 L1; and, for every real x, g xð Þj j ¼ f 1

k x� �� ; we have g 2 L1; and hence g :

R ! C exists.

Problem 3.286 For every t 2 R, g tð Þ ¼ kf ktð Þ:

(Solution Let us fix any t:



pZ1�1

g xð Þe�ixtdx


pZ1�1

f1kx

� e�ixtdx ¼ 1ffiffiffiffiffiffi

2pp

Z1�1

f yð Þe�i kyð Þtkdy

¼ k1ffiffiffiffiffiffi2p

pZ1�1

f yð Þe�iy ktð Þdy

0@ 1A ¼ kf ktð Þ ¼ RHS:

■)

Conclusion 3.287 Let f 2 L1: Let k be a positive real. Let g : x 7! f 1k x� �

be afunction from R to C: Then g 2 L1; and for every t 2 R;

g tð Þ ¼ kf ktð Þ:

Let f : R ! C; and g : R ! C be Lebesgue measurable functions. Let f ; g 2 L1:

By Conclusion 3.265, f � gð Þ 2 L1; and hence df � gð Þ : R ! C exists.

Problem 3.288 For every t 2 R; df � gð Þ tð Þ ¼ f tð Þ � g tð Þ:

(Solution Let us fix any t.

LHS ¼ df � gð Þ tð Þ ¼Z1�1

f � gð Þ xð Þð Þe�ixtdm xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f � gð Þ xð Þð Þe�ixtdx


pZ1�1

Z1�1

f x� yð Þg yð Þdm yð Þ

0@ 1Ae�ixtdx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þg yð Þdy

0@ 1Ae�ixtdx


¼ 12p

Z1�1

Z1�1

f x� yð Þg yð Þe�ixtdy

0@ 1Adx ¼ 12p

Z1�1

Z1�1

f x� yð Þg yð Þe�ixtdx

0@ 1Ady


pZ1�1


pZ1�1

f x� yð Þe�ixtdx

0@ 1Ag yð Þdy ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f tð Þe�iyt� �

g yð Þdy

¼ f tð Þ 1ffiffiffiffiffiffi2p

pZ1�1

g yð Þe�iytdy

0@ 1A ¼ f tð Þg tð Þ ¼ RHS:

■)

Conclusion 3.289 Let f ; g 2 L1: Then f � gð Þ 2 L1; and, for every t 2 R;df � gð Þ tð Þ ¼ f tð Þ � g tð Þ:Let f : R ! C be a Lebesgue measurable function. Let f 2 L1: Let g : x 7! �

ixf xð Þ be a function from R to C: Let g 2 L1: Let a 2 R:

Problem 3.290 f� �0

að Þ ¼ g að Þ:

(Solution Let us take any sequence t1; t2; t3; . . .f g of real numbers such that each tnis different from a; and limn!1 tn ¼ a: It suffices to show that

�1ffiffiffiffiffiffi2p

p limn!1

Z1�1

f yð Þ 1� e�iy tn�að Þ

tn � a

� e�iyady

0@ 1A¼ 1ffiffiffiffiffiffi

2pp lim

n!1

1tn � a

Z1�1

f yð Þ e�iy tn�að Þ � 1� �

e�iyady


2pp lim

n!1

1tn � a

Z1�1

f yð Þe�iytn � f yð Þe�iya� �

dy

0@ 1A¼ lim

n!1

1tn � a


pZ1�1

f yð Þe�iytndy� 1ffiffiffiffiffiffi2p

pZ1�1

f yð Þe�iyady

0@ 1A¼ lim

n!1

1tn � a

f tnð Þ � f að Þ� �

¼ g að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1ffiffiffiffiffiffi

2pp

Z1�1

g yð Þe�iyady ¼ 1ffiffiffiffiffiffi2p

pZ1�1

�iyf yð Þð Þe�iyady

¼ �iffiffiffiffiffiffi2p

pZ1�1

yf yð Þe�iyady;


that is

2i limn!1

Z1�1

f yð Þsin y tn�að Þ

2

tn � ae�i y tn�að Þ

2ð Þ !

e�iyady

0@ 1A¼ lim

n!1

Z1�1

f yð Þ2 sin2 y tn�að Þ

2 þ i2 sin y tn�að Þ2 cos y tn�að Þ

2

tn � a

!e�iyady

0@ 1A¼ lim

n!1

Z1�1

f yð Þ 1� e�iy tn�að Þ

tn � a

� e�iyady

0@ 1A ¼ iZ1�1

yf yð Þe�iyady

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

limn!1

Z1�1


2


2ð Þ !

e�iyady

0@ 1A ¼ 12

Z1�1

yf yð Þe�iyady:

Since f 2 L1; and for sufficiently large positive integer n;


2


2ð Þ !

e�iya

�� ¼ f yð Þ

sin y tn�að Þ2

tn � a

!�� f yð Þj j yj j

2¼ 1

2yf yð Þj j;

by Theorem 1.136,

LHS ¼ limn!1

Z1�1


2


2ð Þ !

e�iyady

0@ 1A¼Z1�1

limn!1


2


2ð Þ !

e�iya

!dy

¼Z1�1

f yð Þ y21

� �e�iya

� �dy ¼ 1

2

Z1�1

yf yð Þe�iyady ¼ RHS:

■)

Conclusion 3.291 Let f ; g 2 L1: Let g : x 7! � ixf xð Þ be a function from R to C:Then,

f� �0

: x 7! g xð Þ



Note 3.292 Let p 2 1;1½ Þ: Let f : R ! C be a Lebesgue measurable function. Letf 2 Lp: For every y 2 R; let

fy : x 7! f x� yð Þ

be a function from R to C:Since for every y 2 R; the function x 7! x� yð Þ from R to R is continuous, and

f : R ! C is a Lebesgue measurable function, their composite fy : x 7! f x� yð Þfrom R to C is a Lebesgue measurable function. Since f 2 Lp; we haveR1�1 f xð Þj jpdx 2 0;1½ Þ; and hence for every y 2 R;

Z1�1

fy xð Þ�� pdx ¼ Z1

�1

f x� yð Þj jpdx ¼Z1�1

f tð Þj jpdt 2 0;1½ Þ:

Thus, for every y 2 R; and for every f 2 Lp; fy 2 Lp: It follows that, for fixedf 2 Lp; uf : y 7! fy is a mapping from metric space R to metric space Lp:

Let us fix any f 2 Lp:

Problem 3.293 uf : R ! Lp is uniformly continuous.

(Solution For this purpose, let us take any e[ 0: By Conclusion 2.50, Cc Rð Þ isdense in Lp Rð Þ ¼ Lpð Þ: Now, since f 2 Lp; there exists g 2 Cc Rð Þ � Lpð Þ such thatf � gk kp\ e

3 : Since g 2 Cc Rð Þ; g : R ! C is a continuous function such that

g�1 C� 0f gð Þð Þ�¼ð Þsupp gð Þ is a compact subset of R: Since supp gð Þ is a compactsubset of R, supp gð Þ is a bounded subset of R; and hence there exists a positive realnumber A such that supp gð Þ � �A;A½ �: Since g : R ! C is a continuous function,and �A;A½ � is a compact subset of R; the restriction of g to �A;A½ � is uniformlycontinuous. It follows that there exists d 2 0;Að Þ such that

s� tj j\d; and s; t 2 �A;A½ �ð Þ ) g sð Þ � g tð Þj j\ e

3 3Að Þ1p

:

Since supp gð Þ � �A;A½ �;

s; t 2 �A;A½ �c) g sð Þ ¼ 0 ¼ g tð Þð Þ:

It follows that

s� tj j\d; and s; t 2 Rð Þ ) g sð Þ � g tð Þj j\ e

3 3Að Þ1p

:

Let us fix any s; t 2 R satisfying 0\ s� tj j\d \Að Þ; and s\t: It suffices toshow that fs � ftk kp\e: Since s 2 R; and g 2 Lp; gs 2 Lp: Similarly, gt; fs; ft 2 Lp:Now, since fs � ftk kp � fs � gsk kp þ gs � gtk kp þ gt � ftk kp; it suffices to show that


fs � gsk kp þ gs � gtk kp þ ft � gtk kp\e:

Since

fs � gsð Þ : x 7! fs xð Þ � gs xð Þð Þ

and

fs xð Þ � gs xð Þð Þ ¼ f x� sð Þ � g x� sð Þð Þ ¼ f � gð Þ x� sð Þ ¼ f � gð Þs xð Þ;

we have

fs � gsð Þ ¼ f � gð Þs:

It follows that

fs � gsk kp ¼ f � gð Þs��

p¼Z1�1

f � gð Þs xð Þ�� pdx

0@ 1A1p

¼Z1�1

f � gð Þ x� sð Þj jpdx

0@ 1A1p

¼Z1�1

f � gð Þ yð Þj jpdy

0@ 1A1p

¼ f � gk kp\e3;

and hence fs � gsk kp\ e3 : Similarly, ft � gtk kp\ e

3 : Now, it suffices to show thatgs � gtk kp\ e

3 : Since

gs � gtð Þ : x 7! gs xð Þ � gt xð Þð Þ ¼ g x� sð Þ � g x� tð Þð Þð Þ;

we have

�ZA�s

�A�t

e

3 3Að Þ1p

!p

dx

0@ 1A1p

¼ e

3 3Að Þ1p

A� sð Þ � �A� tð Þð Þ1p

¼ e

3 3Að Þ1p

2Aþ s� tj jð Þ1p\

e

3 3Að Þ1p

2AþAð Þ1p¼ e

3;

and hence gs � gtk kp\ e3 : ■)

Conclusion 3.294 Let p 2 1;1½ Þ: Let f 2 Lp: Then the mapping uf : y 7! fy fromthe metric space R to metric space Lp is uniformly continuous.


Note 3.295

Definition Let f : R ! C be a function. If for every e[ 0; there exists a compactsubset K of R such that x 62 K ) f xð Þj j\eð Þ; then we say that f vanishes atinfinity. The collection of all continuous functions f : R ! C such that f vanishes atinfinity is denoted by C0 Rð Þ:

Problem 3.296 Cc Rð Þ � C0 Rð Þ:

(Solution Let f 2 Cc Rð Þ: We have to show that f 2 C0 Rð Þ: Since f 2 Cc Rð Þ;f : R ! C is a continuous function such that f�1 C� 0f gð Þð Þ�¼ð Þsupp fð Þ is acompact subset of R: It remains to show that f vanishes at infinity. For this purpose,let us take any e[ 0: Here, supp fð Þ is a compact subset of R: Next, let

x 62 supp fð Þð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ f�1 C� 0f gð Þ� �� f�1 C� 0f gð Þ:

It suffices to show that f xð Þj j\e: Since x 62 f �1 C� 0f gð Þ; f xð Þ ¼ 0; and hencef xð Þj j\e: ■)Let f : R ! C be a Lebesgue measurable function. Let f 2 L1:

Problem 3.297 The function f : t 7!R1�1 f xð Þe�ixtdm xð Þ from R to C is

continuous.

(Solution Let us fix any a 2 R: We have to show that f : R ! C is continuous ata: For this purpose, let us take any sequence t1; t2; t3; . . .f g of real numbers suchthat limn!1 tn ¼ a:


limn!1


pZ1�1

f yð Þe�iytndy

0@ 1A ¼ limn!1

f tnð Þ ¼ f að Þ|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f yð Þe�iyady;

that is

limn!1

Z1�1

f yð Þe�iytndy

0@ 1A ¼Z1�1

f yð Þe�iyady:

Since for every positive integer n, and for every y 2 R; f yð Þe�iytnj j ¼ f yð Þj j; andf 2 L1; by Theorem 1.80

LHS ¼ limn!1

Z1�1

f yð Þe�iytndy

0@ 1A ¼Z1�1

limn!1

f yð Þe�iytn� �

dy ¼Z1�1

f yð Þe�iyady ¼ RHS:

■)


Problem 3.298 f vanishes at infinity.

(Solution For this purpose, let us take any e[ 0: Since for every t 2 R� 0f gð Þ;

f tð Þ�� ¼ 1

2f tð Þþ f tð Þ�� ¼ 1


pZ1�1

f xð Þe�ixtdxþ 1ffiffiffiffiffiffi2p

pZ1�1

f xð Þe�ixtdx

��

¼ 12


pZ1�1

f xð Þe�ixtdx� 1ffiffiffiffiffiffi2p

pZ1�1

f xð Þe�ixt �1ð Þdx

��

¼ 12


pZ1�1


pZ1�1

f xð Þe�ixte�ipdx

��

¼ 12


pZ1�1


pZ1�1

f xð Þe�i xþ ptð Þtdx

��

¼ 12


pZ1�1


pZ1�1

f y� pt

� �e�iytdy

��

¼ 12


pZ1�1


pZ1�1

f x� pt

� �e�ixtdx

��

¼ 12


pZ1�1

f xð Þe�ixt � f x� pt

� �e�ixt

� �dx

��

¼ 1


pZ1�1

f xð Þ � f x� pt

� �� e�ixtdx

��

� 1


pZ1�1


� �� e�ixt

�� dx ¼ 1


pZ1�1


� �� dx¼ 1


pZ1�1

f xð Þ � fptxð Þ

�� dx ¼ 12


pZ1�1

f � fpt

� �xð Þ

�� dx0@ 1A

¼ 12

f � fpt

�� 1� f � fp

t

�� 1;

we have, for every t 2 R� 0f gð Þ; f tð Þ�� f � fp

t

�� 1: Since f 2 L1; by Conclusion

3.294, y 7! fy is a function continuous at 0. It follows that limt!1 fpt¼ f0 ¼ fð Þ in the

Banach space L1: It follows that there exists a real T1 [ 0 such that


t 2 T1;1ð Þ ) f � fpt

�� 1\e

� �: Since y 7! fy is a function continuous at 0,

limt!�1 fpt¼ f0 ¼ fð Þ in the Banach space L1; and hence there exists a real T2\0

such that

t 2 �1; T2ð Þ ) f � fpt

�� 1\e

� �:

Thus,

t 62 T2; T1½ � ) f tð Þ��

f � fpt

�� 1\e

� �:

Now, since T2; T1½ � is compact, and

t 62 T2; T1½ � ) f tð Þ�� \e

� �;

f vanishes at infinity. ■)Since the function f : t 7!

R1�1 f xð Þe�ixtdm xð Þ from R to C is continuous, and f

vanishes at infinity, f 2 C0 Rð Þ: Since f : R ! C is continuous, f : R ! C isLebesgue measurable function. Since for every t 2 R;

f tð Þ�� ¼ 1ffiffiffiffiffiffi

2pp

Z1�1

f xð Þe�ixtdx

�� ¼ 1ffiffiffiffiffiffi

2pp

Z1�1

f xð Þe�ixtdx

��

� 1ffiffiffiffiffiffi2p

pZ1�1

f xð Þe�ixt�� dx ¼ 1ffiffiffiffiffiffi

2pp

Z1�1

f xð Þj jdx ¼ fk k1;

we have, for every t 2 R; f tð Þ�� fk k1; and hence by Conclusion 2.20, f 2 L1 Rð Þ;

and f��

1 � fk k1:

Conclusion 3.299 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1:Then f 2 C0 Rð Þ: Also, f 2 L1 Rð Þ; and f

�� 1 � fk k1:

Note 3.300

Definition We define H : t 7! e� tj j to be a function from R to 0; 1ð � � 0;1ð Þð Þ:Let k 2 0;1ð Þ:Since for every x 2 R;


Z1�1

H ktð Þeitx�� dm tð Þ¼

Z1�1

H ktð Þj jdm tð Þ ¼Z1�1

H ktð Þdm tð Þ


pZ1�1

H ktð Þdt ¼ 1ffiffiffiffiffiffi2p

pZ1�1

e� ktj jdt ¼ 1ffiffiffiffiffiffi2p

pZ1�1

e�k tj jdt ¼ 1ffiffiffiffiffiffi2p

p 2Z10

e�ktdt

¼ffiffiffi2p

r Z10

e�ktdt ¼ffiffiffi2p

r1�k

e�kt��10 ¼

ffiffiffi2p

r1�k

0� 1ð Þ ¼ffiffiffi2p

r1k\1;

we have, for every x 2 R,R1�1 H ktð Þeitxj jdm tð Þ\1: This shows that for every

x 2 R; the function

t 7! H ktð Þeitx

from R to C is a member of L1: It follows that for every t 2 R;R1�1 H ktð Þeitxdm tð Þ

� �2 C:

Definition For every k 2 0;1ð Þ; we define hk : x 7!R1�1 H ktð Þeitxdm tð Þ to be a

function from R to C:For every k 2 0;1ð Þ; and for every x 2 R;

hk xð Þ ¼Z1�1

H ktð Þeitxdm tð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

H ktð Þeitxdt ¼ 1ffiffiffiffiffiffi2p

pZ1�1

e� ktj jeitxdt


pZ1�1

e�k tj jeitxdt ¼ 1ffiffiffiffiffiffi2p

pZ0�1

e�k tj jeitxdtþZ10

e�k tj jeitxdt


2pp

Z0�1

e�k �tð ÞeitxdtþZ10

e�kteitxdt

0@ 1A ¼ 1ffiffiffiffiffiffi2p

pZ0�1

ekteitxdtþZ10

e�kteitxdt


2pp

Z01

ek �sð Þei �sð Þx �1ð ÞdsþZ10

e�kteitxdt


pZ10

e�kse�isxdsþZ10

e�kteitxdt


2pp

Z10

e�ks�isxdsþZ10

e�ktþ itxdt


pZ10

e� kþ ixð ÞsdsþZ10

e� k�ixð Þtdt


2pp e� kþ ixð Þs

� kþ ixð Þ

��10þ e� k�ixð Þt

� k� ixð Þ

��10

!¼ 1ffiffiffiffiffiffi

2pp 1

� kþ ixð Þ 0� 1ð Þþ 1� k� ixð Þ 0� 1ð Þ

�


p 1kþ ixð Þ þ

1k� ixð Þ

� ¼ 1ffiffiffiffiffiffi

2pp 2k

k2þ x2� � ¼ ffiffiffi

2p

rk

1

k2 þ x2;

so, for every k 2 0;1ð Þ;

hk : x 7!ffiffiffi2p

rk

1

k2 þ x2


is a function from R to 0;1ð Þ: Since for every k 2 0;1ð Þ;Z1�1

hk xð Þj jdm xð Þ¼Z1�1

hk xð Þdm xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

hk xð Þdx


pZ1�1

ffiffiffi2p

rk

1

k2 þ x2dx

¼ kp

Z1�1

1

k2 þ x2dx ¼ k

p1ktan�1 x

k

��1�1

¼ 1p

p2� � p

2

� �� ¼ 1\1;

we have, for every k 2 0;1ð Þ;Z1�1

hk xð Þj jdm xð Þ\1;

and hence for every k 2 0;1ð Þ, hk 2 L1: Also, we have seen that for everyk 2 0;1ð Þ;

hkk k1¼� � Z1

�1

hk xð Þdm xð Þ ¼ 1:

Let f 2 L1:Since f 2 L1; and for every k 2 0;1ð Þ; hk 2 L1; by Conclusion 3.265 we have,

for every k 2 0;1ð Þ; f � hkð Þ 2 L1; and

f � hkk k1 � fk k1 hkk k1¼ fk k11 ¼ fk k1:

Thus, for every k 2 0;1ð Þ; f � hkk k1 � fk k1:

Problem 3.301 For every k 2 0;1ð Þ;

f � hkð Þ : x 7!Z1�1

H ktð Þf tð Þeixtdm tð Þ


(Solution Let us fix any k 2 0;1ð Þ; and x 2 R: We have to show that

f � hkð Þ xð Þ ¼Z1�1

H ktð Þf tð Þeixtdm tð Þ:


LHS ¼ f � hkð Þ xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þhk yð Þdy ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð ÞZ1�1

H ktð Þeitydm tð Þ

0@ 1Ady


pZ1�1

f x� yð Þ 1ffiffiffiffiffiffi2p

pZ1�1

H ktð Þeitydt

0@ 1Ady ¼ 12p

Z1�1

f x� yð ÞZ1�1

H ktð Þeitydt

0@ 1Ady

¼ 12p

Z1�1

Z1�1

f x� yð ÞH ktð Þeitydt

0@ 1Ady ¼ 12p

Z1�1

Z1�1

f x� yð ÞH ktð Þeitydy

0@ 1Adt

¼ 12p

Z1�1

Z1�1

f x� yð Þeitydy

0@ 1AH ktð Þdt ¼ 12p

Z1�1

Z�1

1

f sð Þeit x�sð Þ �1ð Þds

0@ 1AH ktð Þdt

¼ 12p

Z1�1

Z1�1

f sð Þeitxe�itsds

0@ 1AH ktð Þdt ¼ 1ffiffiffiffiffiffi2p

pZ1�1


pZ1�1

f sð Þe�istds

0@ 1AeitxH ktð Þdt


pZ1�1

f tð ÞeitxH ktð Þdt ¼Z1�1

H ktð Þf tð Þeixtdm tð Þ ¼ RHS:

■)

Conclusion 3.302 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1:Then for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þ ¼Z1�1

H ktð Þf tð Þeixtdm tð Þ:

Note 3.303 Let g : R ! C be a Lebesgue measurable function. Let g 2 L1 Rð Þ:Let a 2 R: Let g : R ! C be continuous at a:

Since g 2 L1 Rð Þ; we have gk k12 0;1½ Þ; and, by Conclusion 2.18,g sð Þj j � gk k1 a.e. on R: Since g : R ! C is a Lebesgue measurable function, andfor every x 2 R; y 7! x� yð Þ is a continuous function from R to R; for every x 2 R;their composite ~gx : y ! g x� yð Þ is a Lebesgue measurable function from R to C:

Now, let us fix any x 2 R:Since g 2 L1 Rð Þ; we have

1[ gk k1 ¼ inf a : a 2 0;1½ Þ; andm gj j�1 a;1ð �ð Þ� �

¼ 0n o

¼ inf a : a 2 0;1½ Þ; andm y : a\ g yð Þj jf gð Þ ¼ 0f g¼ inf a : a 2 0;1½ Þ; andm y : a\ ~gx x� yð Þj jf gð Þ ¼ 0f g¼ inf a : a 2 0;1½ Þ; andm x� s : a\ ~gx sð Þj jf gð Þ ¼ 0f g¼ inf a : a 2 0;1½ Þ; andm x� s : a\ ~gx sð Þj jf gð Þ ¼ 0f g¼ inf a : a 2 0;1½ Þ; andm s : a\ ~gx sð Þj jf gð Þ ¼ 0f g;

and hence ~gx 2 L1 Rð Þ:


Also, ~gxk k1¼ gk k1: Since ~gx 2 L1 Rð Þ; and for every k 2 0;1ð Þ; hk 2 L1; byLemma 2.23 we have for every k 2 0;1ð Þ;

g � hkð Þ xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

g x� yð Þhk yð Þdy ¼ 1ffiffiffiffiffiffi2p

pZ1�1

~gx yð Þhk yð Þdy

¼ ~gxhkk k1 � hkk k1 ~gxk k1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ hkk k1 gk k1¼ 1 gk k1¼ gk k1\1:

Thus, for every k 2 0;1ð Þ;

g � hkð Þ : R ! 0;1½ Þ

is a function.

Problem 3.304 limk!0 g � hkð Þ að Þð Þ ¼ g að Þ:

(Solution We have to show that


p limk!0

Z1�1

g a� yð Þhk yð Þdy

0@ 1A ¼ limk!0

Z1�1

g a� yð Þhk yð Þdm yð Þ

0@ 1A¼ lim

k!0g � hkð Þ að Þð Þ ¼ g að Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ g að Þ � 1&

¼ g að Þ �Z1�1

hk yð Þdm yð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

g að Þhk yð Þdy;

that is

limk!0

Z1�1

g a� ksð Þ � g að Þð Þh1 sð Þds

0@ 1A¼ lim

k!0

Z1�1

g a� yð Þ � g að Þð Þ h1yk

� �� 1kdy

0@ 1A¼ lim

k!0

Z1�1

g a� yð Þ � g að Þð Þffiffiffi2p

r1

1

12 þ yk

� �2 !

1kdy

0@ 1A


¼ limk!0

Z1�1

g a� yð Þ � g að Þð Þffiffiffi2p

rk

1

k2 þ y2dy

0@ 1A¼ lim

k!0

Z1�1

g a� yð Þ � g að Þð Þhk yð Þdy

0@ 1A¼ lim

k!0

Z1�1

g a� yð Þhk yð Þ � g að Þhk yð Þð Þdy

0@ 1A¼ lim

k!0

Z1�1

g a� yð Þhk yð Þdy�Z1�1

g að Þhk yð Þdy

0@ 1A ¼ 0

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

limk!0

Z1�1

g a� ksð Þ � g að Þð Þh1 sð Þds

0@ 1A ¼ 0:

For this purpose, let us take any sequence k1; k2; k3; . . .f g of positive realnumbers such that limn!1 kn ¼ 0: It suffices to show that

limn!1

Z1�1

g a� knsð Þ � g að Þð Þh1 sð Þds

0@ 1A ¼ 0:

Since for every positive integer n; and for every s 2 R;

g a� knsð Þ � g að Þð Þh1 sð Þj j ¼ g a� knsð Þ � g að Þj j h1 sð Þj j� g a� knsð Þj j þ g að Þj jð Þ h1 sð Þj j � gk k1 þ gk k1

� �h1 sð Þj j a:e:;

and h1 2 L1; by Theorem 1.136

LHS ¼ limn!1

Z1�1

g a� knsð Þ � g að Þð Þh1 sð Þds

0@ 1A¼Z1�1

limn!1

g a� knsð Þ � g að Þð Þh1 sð Þ� �

dy

¼Z1�1

g a� 0sð Þ � g að Þð Þh1 sð Þdy ¼Z1�1

0 dy ¼ 0 ¼ RHS:

■)


Conclusion 3.305 Let g : R ! C be a Lebesgue measurable function. Let g 2L1 Rð Þ: Let a 2 R: Let g : R ! C be continuous at a: Then

limk!0

g � hkð Þ að Þð Þ ¼ g að Þ:

Note 3.306 Let p 2 1;1ð Þ: Let f : R ! C be a Lebesgue measurable function. Letf 2 Lp Rð Þ:

Let q be the exponent conjugate to p. Now, since p 2 1;1ð Þ; we have q 21;1ð Þ: Since for every k 2 0;1ð Þ;


rk

1

k2 þ x2¼

ffiffiffi2p

r1k

1

1þ xk

� �2 �ffiffiffi2p

r1k

!

is a function from R to 0;ffiffi2p

q1k

� i; for every k 2 0;1ð Þ; we have

Z1�1

hk xð Þj jqdx¼Z1�1

hk xð Þð Þqdx ¼Z1�1

ffiffiffi2p

rk

1

k2 þ x2

!q

dx

¼Z1�1

ffiffiffi2p

r1k

1

1þ xk

� �2 !q

dx ¼ffiffiffi2p

r1k

!q Z1�1

1

1þ xk

� �2 !q

dx

¼ffiffiffi2p

r1k

!q Z1�1

11þ y2

� q

kdy ¼ffiffiffi2p

r1k

!q

kZ1�1

11þ y2ð Þq dy

�ffiffiffi2p

r1k

!q

kZ1�1

1

1þ y2ð Þ1dy ¼

ffiffiffi2p

r1k

!q

2kð ÞZ10

11þ y2

dy

¼ffiffiffi2p

r1k

!q

2kð Þtan�1 y��10 ¼

ffiffiffi2p

r1k

!q

2kð Þ p2� 0

� �¼

ffiffiffi2p

r1k

!q

kpð Þ\1;

and hence for every k 2 0;1ð Þ,R1�1 hk xð Þj jqdx\1: This shows that for every

k 2 0;1ð Þ; hk 2 Lq:Since f 2 Lp Rð Þ; we have fk kp2 0;1½ Þ: Since f : R ! C is a Lebesgue mea-

surable function, and for every x 2 R, y 7! ðx� yÞ is a continuous function from R

to R; for every x 2 R; their composite

~fx : y ! f x� yð Þ

is a Lebesgue measurable function from R to C: Next, for every x 2 R;


Z1�1

~fx yð Þ�� pdy ¼ Z1

�1

f x� yð Þj jpdy ¼Z�1

1

f sð Þj jp �1ð Þds ¼Z1�1

f sð Þj jpds

¼ fk kp� �p

\1;

so, for every x 2 R;

~fx 2 Lp Rð Þ; and ~fx��

p¼ fk kp:

Since for every x 2 R, ~fx 2 Lp Rð Þ; and for every k 2 0;1ð Þ; hk 2 Lq; byLemma 2.21 we have for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þj j ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þhk yð Þdy

��


pZ1�1

~fx yð Þhk yð Þdy

�� 1ffiffiffiffiffiffi

2pp

Z1�1

~fx yð Þhk yð Þ�� dy

¼ ~fxhk��

1 � ~fx��

p hkk kq|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fk kp hkk kq\1:

Thus, for every k 2 0;1ð Þ; f � hkð Þ : R ! C is a function.

Problem 3.307 For every k 2 0;1ð Þ; and for every x 2 R;

Z1�1

f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ

��p


pZ1�1

f x� yð Þhk yð Þdy� f xð Þ � 1ffiffiffiffiffiffi2p

pZ1�1

hk yð Þdy

��p


pZ1�1

f x� yð Þhk yð Þdy� f xð Þ � 1

��p

¼ f � hkð Þ xð Þ � f xð Þj jp�Z1�1

f x� yð Þ � f xð Þj jphk yð Þdm yð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

Z1�1


��p

�Z1�1

f x� yð Þ � f xð Þj jphk yð Þdm yð Þ;


that is

uZ1�1


0@ 1A�Z1�1

u f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ;

where u : t 7! tj jp is the function from R to 0;1½ Þ: Since

Z1�1

f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þ � f xð Þð Þhk yð Þdy


pZ1�1

f x� yð Þ � f xð Þð Þffiffiffi2p

rk

1

k2þ y2dy ¼ k

p

Z1�1

f x� yð Þ � f xð Þð Þ ddy

1ktan�1 y

k

� � dy

¼ kp

Zp2k

� p2k

f x� k tan ksð Þð Þ � f xð Þð Þds ¼ kp

Z 12

�12

f x� k tan ptð Þð Þ � f xð Þð Þpkdt

¼Z 1

2

�12

f x� k tan ptð Þð Þ � f xð Þð Þdt;

and

Z1�1

u f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

u f x� yð Þ � f xð Þð Þhk yð Þdy


pZ1�1

u f x� yð Þ � f xð Þð Þffiffiffi2p

rk

1

k2þ y2dy ¼ k

p

Z1�1

u f x� yð Þ � f xð Þð Þ 1

k2 þ y2dy

¼ kp

Z1�1

u f x� yð Þ � f xð Þð Þ ddy

1ktan�1 y

k

� � dy ¼ k

p

Zp2k

� p2k

u f x� k tan ksð Þð Þ � f xð Þð Þds

¼ kp

Z 12

�12

u f x� k tan ptð Þð Þ � f xð Þð Þ pkdt ¼

Z 12

�12

u f x� k tan ptð Þð Þ � f xð Þð Þdt;


uZ 1

2

�12

f x� k tan ptð Þð Þ � f xð Þð Þdt

0B@1CA�

Z 12

�12

u f x� k tan ptð Þð Þ � f xð Þð Þdt:


Since p[ 1; u : t 7! tj jp is a convex function from R to 0;1½ Þ; by Conclusion2.10, we have

uZ 1

2

�12


0B@1CA�

Z 12

�12


■)Since for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þ � f xð Þj jp �Z1�1

f x� yð Þ � f xð Þj jphk yð Þdm yð Þ;

we have, for every k 2 0;1ð Þ;

Z1�1

f � hkð Þ xð Þ � f xð Þj jpdx�Z1�1

Z1�1

f x� yð Þ � f xð Þj jphk yð Þdm yð Þ

0@ 1Adx

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1ffiffiffiffiffiffi

2pp

Z1�1

Z1�1

f x� yð Þ � f xð Þj jphk yð Þdy

0@ 1Adx


pZ1�1

Z1�1

f x� yð Þ � f xð Þj jphk yð Þdx

0@ 1Ady


pZ1�1

Z1�1

f x� yð Þ � f xð Þj jpdx

0@ 1Ahk yð Þdy


pZ1�1

Z1�1

fy xð Þ � f xð Þ�� pdx

0@ 1Ahk yð Þdy


pZ1�1

Z1�1

fy � f� �

xð Þ�� pdx

0@ 1Ahk yð Þdy ¼Z1�1

fy � f��

p

� �phk yð Þdy

and hence for every k 2 0;1ð Þ;

Z1�1


g yð Þhk yð Þdy;


where g : y 7! fy � fp�� p

is a function from R to 0;1½ Þ: By Conclusion 3.294,the mapping y 7! fy from metric space R to normed linear space Lp Rð Þ is uniformly

continuous, and f 2 Lp Rð Þ; g : y 7! fy � f��

p

� �pis a continuous function from R

to 0;1½ Þ: Since g : R ! 0;1½ Þ is a continuous function, g : R ! 0;1½ Þ is aLebesgue measurable.

Problem 3.308 For every y 2 R, g yð Þj j ¼ð Þg yð Þ� 2 fp�� p

:

(Solution For every y 2 R;

g yð Þð Þ1p ¼ fy � f

�� p � fy��

p þ fk kp¼1ffiffiffiffiffiffi2p

pZ1�1

fy sð Þ�� pds

0@ 1A1p

þ fk kp


pZ1�1

f s� yð Þj jpds

0@ 1A1p

þ fk kp¼1ffiffiffiffiffiffi2p

pZ1�1

f tð Þj jpdt

0@ 1A1p

þ fk kp

¼ fk kp þ fk kp¼ 2 fk kp;

so, for every y 2 R, g yð Þ� 2 fk kp� �p

: ■)

Since g : R ! 0;1½ Þ is Lebesgue measurable, and, for every y 2 R,

g yð Þj j � 2 fk kp� �p

\1ð Þ; g 2 L1 Rð Þ: Since g : R ! 0;1½ Þ is continuous and g 2L1 Rð Þ; by Conclusion 3.305,

limk!0

Z1�1

g sð Þ hk sð Þð Þdm sð Þ

0@ 1A ¼ limk!0


pZ1�1

g sð Þ hk �sð Þð Þds

0@ 1A¼ lim

k!0


pZ�1

1

g sð Þ hk �sð Þð Þ �1ð Þds

0@ 1A¼ lim

k!0


pZ1�1

g �yð Þhk yð Þdy

0@ 1A¼ lim

k!0

Z1�1

g 0� yð Þhk yð Þdm yð Þ

0@ 1A¼ lim

k!0g � hkð Þ 0ð Þð Þ ¼ g 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f0 � fk kp¼ f � fk kp¼ 0;


and hence

limk!0

Z1�1


0@ 1A ¼ 0:

It follows that for every k 2 0;1ð Þ;

Z1�1


�2 fk kp

�phk yð Þdy

¼ffiffiffiffiffiffi2p

p �2 fk kp

�p Z1�1

hk yð Þdm yð Þ ¼ffiffiffiffiffiffi2p

p �2 fk kp

�p � 1 ¼ffiffiffiffiffiffi2p

p �2 fk kp

�p\1;


Z1�1

f � hkð Þ � fð Þ xð Þj jpdx�ffiffiffiffiffiffi2p

p �2 fk kp

�p\1:

This shows that for every k 2 0;1ð Þ;

f � hkð Þ � fð Þ 2 Lp Rð Þ:

Now, since f 2 Lp Rð Þ; and Lp Rð Þ is a linear space, for every k 2 0;1ð Þ;f � hkð Þ 2 Lp Rð Þ:It follows that for every k 2 0;1ð Þ;

f � hkð Þ � fk kp� �p

�Z1�1

g yð Þhk yð Þdm yð Þ;

and hence

0�ð Þ limk!0

�f � hkð Þ � fk kp

�p � limk!0

Z1�1

g yð Þhk yð Þdm yð Þ

0@ 1A ¼ 0ð Þ:

Thus,

limk!0

�f � hkð Þ � fk kp

�p ¼ 0;

and hence limk!0 f � hkð Þ � fk kp¼ 0:


Conclusion 3.309 Let p 2 1;1ð Þ: Let f : R ! C be a Lebesgue measurablefunction. Let f 2 Lp Rð Þ: Then for every k 2 0;1ð Þ; f � hkð Þ 2 Lp Rð Þ; andlimk!0 f � hkð Þ � fk kp¼ 0:

Note 3.310 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1 Rð Þ:Since for every k 2 0;1ð Þ;


rk

1

k2 þ x2¼

ffiffiffi2p

r1k

1

1þ xk

� �2 �ffiffiffi2p

r1k

!

is a function from R to 0;ffiffi2p

q1k

� i; we have, for every k 2 0;1ð Þ; hkk k 2 L1; and

hkk k1 �ffiffi2p

q1k :

Since f 2 L1 Rð Þ; we have fk k12 0;1½ Þ: Since f : R ! C is a Lebesgue mea-surable function, and, for every x 2 R, y 7! x� yð Þ is a continuous function from R

to R, for every x 2 R, their composite

~fx : y ! f x� yð Þ

is a Lebesgue measurable function from R to C: Next, for every x 2 R,Z1�1

~fx yð Þ�� dy¼ Z1

�1

f x� yð Þj jdy ¼Z�1

1

f sð Þj j �1ð Þds

¼Z1�1

f sð Þj j ds ¼ffiffiffiffiffiffi2p

pfk k1\1;

so, for every x 2 R;

~fx 2 L1 Rð Þ; and ~fx��

1¼ fk k1:

Since for every x 2 R; ~fx 2 L1 Rð Þ; and for every k 2 0;1ð Þ; hk 2 L1; byLemma 2.23 we have, for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þj j ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þhk yð Þdy

��


pZ1�1

~fx yð Þhk yð Þdy

�� 1ffiffiffiffiffiffi

2pp

Z1�1

~fx yð Þhk yð Þ�� dy

¼ ~fxhk��

1 � ~fx��

1 hkk k1|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fk k1 hkk k1\1:

Thus, for every k 2 0;1ð Þ, f � hkð Þ : R ! C is a function.


Problem 3.311 For every k 2 0;1ð Þ; and for every x 2 R; f � hkð Þ xð Þ � f xð Þj j�R1�1 f x� yð Þ � f xð Þj jhk yð Þdm yð Þ.


Z1�1


�� ¼ 1ffiffiffiffiffiffi

2pp

Z1�1

f x� yð Þhk yð Þdy� f xð Þ � 1ffiffiffiffiffiffi2p

pZ1�1

hk yð Þdy

��


pZ1�1

f x� yð Þhk yð Þdy� f xð Þ � 1

��

¼ f � hkð Þ xð Þ � f xð Þj j �Z1�1

f x� yð Þ � f xð Þj jhk yð Þdm yð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

Z1�1


��

Z1�1

f x� yð Þ � f xð Þj jhk yð Þdm yð Þ;

that is

uZ1�1


0@ 1A�Z1�1

u f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ;

where u : t 7! tj j is the function from R to 0;1½ Þ: Since

Z1�1

f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þ � f xð Þð Þhk yð Þdy


pZ1�1

f x� yð Þ � f xð Þð Þffiffiffi2p

rk

1

k2 þ y2dy

¼ kp

Z1�1

f x� yð Þ � f xð Þð Þ ddy

1ktan�1 y

k

� � dy

¼ kp

Zp2k

� p2k

f x� k tan ksð Þð Þ � f xð Þð Þds


¼ kp

Z 12

�12

f x� k tan ptð Þð Þ � f xð Þð Þ pkdt

¼Z1

2

�12

f x� k tan ptð Þð Þ � f xð Þð Þdt;

and

Z1�1

u f x� yð Þ � f xð Þð Þhk yð Þdm yð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

u f x� yð Þ � f xð Þð Þhk yð Þdy


pZ1�1

u f x� yð Þ � f xð Þð Þffiffiffi2p

rk

1

k2 þ y2dy

¼ kp

Z1�1

u f x� yð Þ � f xð Þð Þ 1

k2 þ y2dy

¼ kp

Z1�1

u f x� yð Þ � f xð Þð Þ ddy

1ktan�1 y

k

� � dy

¼ kp

Zp2k

� p2k

u f x� k tan ksð Þð Þ � f xð Þð Þds

¼ kp

Z 12

�12

u f x� k tan ptð Þð Þ � f xð Þð Þ pkdt

¼Z 1

2

�12

u f x� k tan ptð Þð Þ � f xð Þð Þdt;


uZ 1

2

�12


0B@1CA�

Z 12

�12


Since u : t 7! tj j is a convex function from R to 0;1½ Þ; by the conclusion ofNote 2.1, we have


uZ 1

2

�12


0B@1CA�

Z 12

�12


■)Since for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þ � f xð Þj j �Z1�1

f x� yð Þ � f xð Þj jhk yð Þdm yð Þ;

we have, for every k 2 0;1ð Þ;

Z1�1

f � hkð Þ xð Þ � f xð Þj jdx�Z1�1

Z1�1

f x� yð Þ � f xð Þj jhk yð Þdm yð Þ

0@ 1Adx

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼ 1ffiffiffiffiffiffi

2pp

Z1�1

Z1�1

f x� yð Þ � f xð Þj jhk yð Þdy

0@ 1Adx ¼ 1ffiffiffiffiffiffi2p

pZ1�1

Z1�1

f x� yð Þ � f xð Þj jhk yð Þdx

0@ 1Ady


pZ1�1

Z1�1

f x� yð Þ � f xð Þj jdx

0@ 1Ahk yð Þdy ¼ 1ffiffiffiffiffiffi2p

pZ1�1

Z1�1

fy xð Þ � f xð Þ�� dx

0@ 1Ahk yð Þdy


pZ1�1

Z1�1

fy � f� �

xð Þ�� dx

0@ 1Ahk yð Þdy ¼Z1�1

fy � f��

1hk yð Þdy;


Z1�1


g yð Þhk yð Þdy;

where g : y 7! fy � f��

1 is a function from R to 0;1½ Þ: By Conclusion 3.294, themapping y 7! fy from metric space R to normed linear space L1 Rð Þ is uniformlycontinuous, and f 2 L1 Rð Þ; g : y 7! fy � f

�� 1 is a continuous function from R to

0;1½ Þ: Since g : R ! 0;1½ Þ is a continuous function, g : R ! 0;1½ Þ is aLebesgue measurable.

Problem 3.312 For every y 2 R, g yð Þj j ¼ð Þg yð Þ� 2 fk k1:

(Solution For every y 2 R,


g yð Þ ¼ fy � f��

1 � fy��

1 þ fk k1¼1ffiffiffiffiffiffi2p

pZ1�1

fy sð Þ�� dsþ fk k1


pZ1�1

f s� yð Þj jdsþ fk k1¼1ffiffiffiffiffiffi2p

pZ1�1

f tð Þj jdtþ fk k1¼ fk k1 þ fk k1¼ 2 fk k1;

so, for every y 2 R, g yð Þ� 2f1: ■)Since g : R ! 0;1½ Þ is Lebesgue measurable, and, for every y 2 R,

g yð Þj j � 2 fk k1 \1ð Þ; g 2 L1 Rð Þ: Since g : R ! 0;1½ Þ is continuous, and g 2L1 Rð Þ; by Conclusion 3.305

limk!0

Z1�1


0@ 1A ¼ limk!0


pZ1�1

g sð Þ hk �sð Þð Þds

0@ 1A¼ lim

k!0


pZ�1

1

g sð Þ hk �sð Þð Þ �1ð Þds

0@ 1A ¼ limk!0


pZ1�1

g �yð Þhk yð Þdy

0@ 1A¼ lim

k!0

Z1�1

g 0� yð Þhk yð Þdm yð Þ

0@ 1A ¼ limk!0

g � hkð Þ 0ð Þð Þ ¼ g 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ f0 � fk k1¼ f � fk k1¼ 0;

and hence

limk!0

Z1�1


0@ 1A ¼ 0:

It follows that, for every k 2 0;1ð Þ;

Z1�1


2 fk k1� �

hk yð Þdy

¼ffiffiffiffiffiffi2p

p2 fk k1� � Z1

�1

hk yð Þdm yð Þ ¼ffiffiffiffiffiffi2p

p2 fk k1� �

� 1 ¼ffiffiffiffiffiffi2p

p2 fk k1� �

\1;



Z1�1

f � hkð Þ � fð Þ xð Þj jdx�ffiffiffiffiffiffi2p

p2 fk k1� �

\1:

This shows that for every k 2 0;1ð Þ, f � hkð Þ � fð Þ 2 L1 Rð Þ: Now, sincef 2 L1 Rð Þ, and L1 Rð Þ is a linear space, for every k 2 0;1ð Þ, f � hkð Þ 2 L1 Rð Þ:

It follows that for every k 2 0;1ð Þ, f � hkð Þ � fk k1 �R1�1 g yð Þhk yð Þdm yð Þ; and

hence

0�ð Þ limk!0

f � hkð Þ � fk k1 � limk!0

Z1�1

g yð Þhk yð Þdm yð Þ

0@ 1A ¼ 0ð Þ:

Thus,

limk!0

f � hkð Þ � fk k1¼ 0:

Conclusion 3.313 Let f : R ! C be a Lebesgue measurable function. Letf 2 L1 Rð Þ. Then, for every k 2 0;1ð Þ, f � hkð Þ 2 L1 Rð Þ; and limk!0 f � hkð Þk�f k1 ¼ 0:

If we combine Conclusion 3.309 with the preceding result, we get following

Conclusion 3.314 Let p 2 1;1½ Þ: Let f : R ! C be a Lebesgue measurablefunction. Let f 2 Lp Rð Þ: Then, for every k 2 0;1ð Þ, f � hkð Þ 2 Lp Rð Þ; and

limk!0

f � hkð Þ � fk kp¼ 0:

3.12 Inversion Theorem

Note 3.315 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1 Rð Þ: Letf 2 L1 Rð Þ:

Since f 2 L1 Rð Þ; by Conclusion 3.299, the function f : t 7!R1�1 f xð Þe�ixtdm xð Þ

from R to C is a member of C0 Rð Þ: Also, f 2 L1 Rð Þ; and f��

1 � fk k1.Since f 2 L1 Rð Þ; we have f

�� 2 L1 Rð Þ: Since f 2 L1 Rð Þ; by Conclusion 3.89,for every k 2 0;1ð Þ; and for every x 2 R;

f � hkð Þ xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

e� ktj j f tð Þeixtdt:

Problem 3.316 For every x 2 R;


ffiffiffiffiffiffi2p

plimk!0

f � hkð Þ xð Þð Þ ¼�

limk!0

Z1�1

e� ktj j f tð Þeixtdt

0@ 1A ¼Z1�1

f tð Þeixtdt:

(Solution For this purpose, let us take any sequence k1; k2; k3; . . .f g of positivereal numbers such that limn!1 kn ¼ 0: Next, let us fix any x 2 R. It suffices toshow that

limn!1

Z1�1

e� kntj j f tð Þeixtdt

0@ 1A ¼Z1�1

f tð Þeixtdt:

Since for every positive integer n, and for every real t,

e� kntj j f tð Þeixt�� ¼ e� kntj j f tð Þ

�� 1 � f tð Þ�� ¼ f

�� tð Þ;and f

�� 2 L1 Rð Þ; by Theorem 1.136

LHS ¼ limn!1

Z1�1

e� kntj j f tð Þeixtdt

0@ 1A ¼Z1�1

limn!1

e� kntj j f tð Þeixt� ��

dt

¼Z1�1

e� 0tj j f tð Þeixt� �

dt ¼Z1�1

f tð Þeixtdt ¼ RHS:

■)It follows that for every x 2 R;

limk!0

f � hkð Þ xð Þð Þ ¼Z1�1

f tð Þeixtdm tð Þ:

Since f 2 L1 Rð Þ; by Conclusion 3.313 for every positive integer n, f � h1n

� �2

L1 Rð Þ; and limn!1 f � h1n

� �� f

�� 1¼ 0: Thus, the sequence f � h1

n

� �n oconverges

to f 2 L1 Rð Þð Þ: By Conclusion 2.33, there exists a subsequence f � h 1nk

� �n oof

f � h1n

� �n osuch that

Z1�1

f tð Þeixtdm tð Þ ¼ limk!0

f � hkð Þ xð Þð Þ ¼ limk!1

f � h 1nk

� �xð Þ

� �¼ f xð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} a:e: onR:

Thus, f xð Þ ¼ g xð Þ a.e., where

3.12 Inversion Theorem 615

g : x 7!Z1�1

f tð Þeixtdm tð Þ ¼Z1�1

f tð Þe�it �xð Þdm tð Þ ¼ ^f �xð Þ

0@ 1Ais a function from R to C:

Problem 3.317 g 2 C0 Rð Þ:

(Solution Since f 2 L1 Rð Þ; by Conclusion 3.299 ^f 2 C0 Rð Þ; and hence ^f : R ! C

is a continuous function, and ^f vanishes at infinity. Since ^f : R ! C is a continuous

function, the function x 7! ^f �xð Þ is continuous, and hence g : R ! C is a contin-

uous function. It suffices to show that x 7! ^f �xð Þ vanishes at infinity.For this purpose, let us take any e[ 0: Since ^f vanishes at infinity, there exists a

compact subset K of R such that x 62 K ) ^f xð Þ�� \e

� �: Since K is compact, �Kð Þ

is a compact set and hence K [ �Kð Þ is compact. Let x 62 K [ �Kð Þð Þ: It suffices toshow that ^f �xð Þ

�� \e:

Since x 62 K [ �Kð Þð Þ; we have x 62 �Kð Þ: Since x 62 �Kð Þ; we have �xð Þ 62 K;

and hence ^f �xð Þ�� \e: ■)

Conclusion 3.318 Let f : R ! C be a Lebesgue measurable function. Let f 2L1 Rð Þ: Let f 2 L1 Rð Þ: Let

g : x 7!Z1�1

f tð Þeixtdm tð Þ

be a function from R to C: Then f xð Þ ¼ g xð Þ a.e., and g 2 C0 Rð Þ:This result is known as the inversion theorem.

Theorem 3.319 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1 Rð Þ:Let f ¼ 0: Then f xð Þ ¼ 0 a.e. on R:

Proof Since 0 2 L1 Rð Þ; and f ¼ 0; we have f 2 L1 Rð Þ: Now, by Conclusion 3.318

f xð Þ ¼Z1�1

f tð Þeixtdm tð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Z1�1

0 tð Þð Þeixtdm tð Þ ¼Z1�1

0ð Þeixtdm tð Þ ¼ 0 a:e:;

and hence f xð Þ ¼ 0 a.e. on R: ■This result is known as the uniqueness theorem.


3.13 Plancherel Theorem

Note 3.320 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1 \ L2:Let ~f : x 7! f �xð Þ be a function from R to C:

Problem 3.321 ~f : R ! C is a Lebesgue measurable function.

(Solution Let G be an open subset of C: We have to show that

� f�1 �z : z 2 Gf gð Þ� �

¼ x : �x 2 f�1 �z : z 2 Gf gð Þ�

¼ x : f �xð Þ 2 �z : z 2 Gf gf g

¼ x : f �xð Þ 2 Gn o

¼ x : ~f xð Þ 2 G�

¼ ~f� ��1

Gð Þ|fflfflfflfflffl{zfflfflfflfflffl}is Lebesgue measurable set inR; that is� f�1 �z : z 2 Gf gð Þð Þ is Lebesgue measurableset in R. Since G is an open subset of C; �z : z 2 Gf g is open in C: Now, sincef : R ! C is Lebesgue measurable function, f�1 �z : z 2 Gf gð Þ is Lebesgue mea-surable set inR, and hence� f�1 �z : z 2 Gf gð Þð Þ is Lebesgue measurable set inR.■)

Problem 3.322 ~f 2 L1 \ L2 � L1ð Þ:

(Solution Since ~f : R ! C is a Lebesgue measurable function, it remains to show

thatR1�1

~f xð Þ�� dx 2 0;1½ Þ; and

R1�1

~f xð Þ�� 2dx 2 0;1½ Þ: Since f 2 L1 \ L2; we

have f 2 L1; and f 2 L2; and henceR1�1 f sð Þj jds 2 0;1½ Þ; and

R1�1 f sð Þj j2ds 2

0;1½ Þ: Since

Z1�1

~f xð Þ�� dx

¼Z1�1

f �xð Þ�� dx ¼ Z1

�1

f �xð Þj jdx ¼Z�1

1

f sð Þj j �1ð Þds

¼Z1�1

f sð Þj jds 2 0;1½ Þ;

we haveR1�1

~f xð Þ�� dx 2 0;1½ Þ: Similarly,

R1�1

~f xð Þ�� 2dx 2 0;1½ Þ: ■)

Since f 2 L1; and ~f 2 L1; by Conclusion 3.278 f � ~f� �

2 L1. Since f 2 L2; we have,for every x 2 R,

Z1�1

f�x sð Þj j2ds ¼Z1�1

f xþ sð Þj j2ds ¼Z1�1

f tð Þj j2dt 2 0;1½ Þ;

3.13 Plancherel Theorem 617

and hence for every x 2 R; f�x 2 L2: Since f 2 L2; for every x 2 R; f�x 2 L2; andL2 is a Hilbert space, for every x 2 R;

f � ~f� �

xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f x� yð Þ~f yð Þdy ¼ 1ffiffiffiffiffiffi2p

pZ�1

1

f x� yð Þf �yð Þ �1ð Þdy


pZ1�1

f xþ sð Þf sð Þds ¼ 1ffiffiffiffiffiffi2p

pZ1�1

f�x sð Þf sð Þds

¼Z1�1

f�x sð Þf sð Þdm sð Þ ¼ f�x; fð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

and hence

f � ~f� �

: x 7! f�x; fð Þ

is a function from R to C. Since f 2 L2; by Conclusion 3.294 y 7! fy from R to L2 iscontinuous, and hence the function y 7! f�y from R to L2 is continuous. Now, sinceL2 is a Hilbert space, by Lemma 2.67 f � ~f

� �: x 7! f�x; fð Þ from R to C is

continuous.

Problem 3.323 f � ~f� �

: x 7! f�x; fð Þ from R to C is a bounded function.

(Solution Let us take any x 2 R: Since L2 is a Hilbert space, by Schwarz inequality

f � ~f� �

xð Þ�� ¼ f�x; fð Þj j � fk k2 f�xk k2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ fk k2

Z1�1

f�x yð Þj j2dm yð Þ

0@ 1A12

¼ fk k2Z1�1

f xþ yð Þj j2dm yð Þ

0@ 1A12

¼ fk k2Z1�1

f sð Þj j2dm sð Þ

0@ 1A12

¼ fk k2 fk k2¼ fk k2� �2

;

and we have

f � ~f� �

xð Þ�� fk k2

� �2:

Thus, f � ~f� �

: x 7! f�x; fð Þ from R to C is a bounded function. ■)

Since f � ~f� �

2 L1; by Conclusion 3.302 we have, for every k 2 0;1ð Þ;


f � ~f� �

� hk� �

0ð Þ ¼Z1�1

H ktð Þ df � ~f� �tð Þei0tdm tð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Z1�1

H ktð Þ df � ~f� �tð Þdm tð Þ;


f � ~f� �

� hk� �

0ð Þ ¼Z1�1

H ktð Þ df � ~f� �tð Þdm tð Þ:

Since f 2 L1; and ~f 2 L1; by Conclusions 3.289 and 3.285, we have, for everyt 2 R;

df � ~f� �tð Þ ¼ f tð Þ � b~f tð Þ ¼ f tð Þ � f tð Þ ¼ f tð Þ

�� 2� �;

and hence for every t 2 R;

df � ~f� �tð Þ ¼ f tð Þ

�� 2 2 0;1½ Þð Þ:

Since f � ~f� �

: R ! C is a bounded function, f � ~f� �

2 L1: Now, since f � ~f� �

:

R ! C is continuous at 0, by Conclusion 3.305,

limk!0

f � ~f� �

� hk� �

0ð Þ� �

¼ f � ~f� �

0ð Þ ¼ f�0; fð Þ ¼ f ; fð Þ ¼ fk k2� �2� �

;

and hence

limk!0

Z1�1

f tð Þ�� 2e� tj jkdm tð Þ

0@ 1A ¼ limk!0

Z1�1

e� ktj j f tð Þ�� 2dm tð Þ

0@ 1A¼ lim

k!0

Z1�1

H ktð Þ f tð Þ�� 2dm tð Þ

0@ 1A ¼ limk!0

Z1�1

H ktð Þ df � ~f� �tð Þdm tð Þ

0@ 1A¼ lim

k!0f � ~f� �

� hk� �

0ð Þ� �

¼ fk k2� �2|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Thus,

limk!0

Z1�1


0@ 1A ¼ fk k2� �2 \1ð Þ:


Since for fixed real t, k 7! e� tj jk is a monotonically decreasing function from

0;1½ Þ to 0; 1ð �; we have, for fixed real t, k 7! f tð Þ�� 2e� tj jk is a monotonically

decreasing function from 0;1½ Þ to 0; f tð Þ�� 2� i

; and hence k 7!R1�1 f tð Þ�� 2

e� tj jkdm tð Þ is a monotonically decreasing function from 0;1½ Þ to 0;1½ �:

Problem 3.324 fk k2� �2¼� �

limk!0R1�1 f tð Þ�� 2e� tj jkdm tð Þ

� �¼ f

�� 2

� �2:

(Solution For this purpose, let us take any decreasing sequence k1; k2; k3; . . .f g ofpositive real numbers such that limn!1 kn ¼ 0: Since

k 7!Z1�1


is a monotonically decreasing function from 0;1½ Þ to 0;1½ �; it suffices to show that

limn!1

Z1�1

f tð Þ�� 2e� tj jkndm tð Þ

0@ 1A ¼ f��

2

� �2:

Since k1; k2; k3; . . .f g is decreasing sequence of positive real numbers, and

k 7! f tð Þ�� 2e� tj jk is a monotonically decreasing function from 0;1½ Þ to 0; f tð Þ

�� 2� i;

we have

f tð Þ�� 2e� tj jk1 � f tð Þ

�� 2e� tj jk2 � f tð Þ�� 2e� tj jk3 � � � � ;

and hence by Theorem 1.125,

1[ fk k2� �2 ¼ lim

n!1

Z1�1


0@ 1A ¼Z1�1

limn!1

f tð Þ�� 2e� tj jkn� ��

dm tð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Z1�1

f tð Þ�� 2e� tj j0� �

dm tð Þ ¼Z1�1

f tð Þ�� 2dm tð Þ:

Thus,R1�1 f tð Þ�� 2dm tð Þ\1; and hence f 2 L2: Also,

limn!1

Z1�1


0@ 1A ¼ f��

2

� �2:

■)

Since fk k2� �2¼ f

�� 2

� �2; we have f

�� 2¼ fk k2:


Conclusion 3.325 Let f : R ! C be a Lebesgue measurable function. Let f 2L1 \ L2: Then f 2 L2; and

f��

2¼ fk k2:

Note 3.326 Since L1 and L2 are linear spaces, L1 \ L2 is a linear subspace of L2: ByConclusion 3.325, U : f 7! f is a mapping from L1 \ L2 to L2:

Problem 3.327 The map U : L1 \ L2ð Þ ! L2 is linear.

(Solution Let us take any f ; g 2 L1 \ L2ð Þ: Let a; b 2 R: We have to show thataf þ bgð Þ^¼ af þ bg: For this purpose, let us take any t 2 R: We have to show that

af þ bgð Þ^ tð Þ ¼ af þ bg� �

tð Þ:

LHS ¼ af þ bgð Þ^ tð Þ ¼Z1�1

af þ bgð Þ xð Þð Þe�ixtdm xð Þ

¼Z1�1

af xð Þþ bg xð Þð Þe�ixtdm xð Þ ¼Z1�1

af xð Þe�ixt þ bg xð Þe�ixt� �

dm xð Þ

¼ aZ1�1

f xð Þe�ixt� �

dm xð Þþ bZ1�1

g xð Þe�ixt� �

dm xð Þ ¼ a f tð Þ� �

þ b g tð Þð Þ

¼ af þ bg� �

tð Þ ¼ RHS:

■)It follows that U L1 \ L2ð Þ is a linear subspace of Hilbert space L2: Now, by

Lemma 2.72,

f : f 2 L2; and for every g 2 U L1 \ L2ð Þ; f ; gð Þ ¼ 0�

¼� �

U L1 \ L2ð Þð Þ? is aclosed linear subspace of L2; and hence f : f 2 L2; and for

�every g 2

U L1 \ L2ð Þ; f ; gð Þ ¼ 0g is a closed linear subspace of L2:Let us fix any f 2 L2 such that for every g 2 U L1 \ L2ð Þ; g; fð Þ ¼ 0.

Problem 3.328 f ¼ 0:(Solution

Problem 3:329 For every a 2 R; and for every k 2 0;1ð Þ; the functionx 7! eiaxH kxð Þ from R to C is a member of L1 \ L2:

(Solution Since the function

x 7! eiaxH kxð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} ¼ eiaxe� kxj j ¼ eiaxe�k xj j ¼ eiax�k xj j


from R to C is continuous, the function

x 7! eiaxH kxð Þ

from R to C is Lebesgue measurable. So, it suffices the show that

Z1�1

eiaxH kxð Þ�� dm xð Þ 2 0;1½ Þ; and

Z1�1

eiaxH kxð Þ�� 2dm xð Þ 2 0;1½ Þ:

Here,

Z1�1

eiaxH kxð Þ�� dm xð Þ ¼

Z1�1

H kxð Þj jdm xð Þ ¼Z1�1

H kxð Þdm xð Þ ¼Z1�1

e� kxj jdm xð Þ

¼Z1�1

e�k xj jdm xð Þ ¼ 1ffiffiffiffiffiffi2p

pZ1�1

e�k xj jdx ¼ 1ffiffiffiffiffiffi2p

p 2Z10

e�kxdx ¼ffiffiffi2p

re�kx

�k

��10

¼ffiffiffi2p

r1�k

0� 1ð Þ ¼ffiffiffi2p

r1k2 0;1½ Þ;

so,R1�1 eiaxH kxð Þj jdm xð Þ 2 0;1½ Þ: Similarly,

R1�1 eiaxH kxð Þj j2dm xð Þ 2 0;1½ Þ: ■)

It follows, by Conclusion 3.325, that for every a 2 R; and for every k 2 0;1ð Þ;the function

t 7!Z1�1

eiaxH kxð Þ� �

e�ixtdm xð Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}¼Z1�1

H kxð Þ eix a�tð Þ� �

dm xð Þ ¼ hk a� tð Þ

from R to C is a member of U L1 \ L2ð Þ: Thus, for every a 2 R; and for everyk 2 0;1ð Þ; the function

kk;a : t 7! hk a� tð Þ

from R to C is a member of U L1 \ L2ð Þ: Now, by the given assumption, for everya 2 R; and for every k 2 0;1ð Þ;

hk � �fð Þ að Þ ¼Z1�1

hk a� tð Þ�f tð Þdm tð Þ ¼Z1�1

kk;a tð Þf tð Þdm tð Þ ¼ kk;a; f� �

¼ 0|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl};


and hence for every a 2 R, and for every k 2 0;1ð Þ; hk � �fð Þ að Þ ¼ 0: It followsthat for every k 2 0;1ð Þ; �f � hkð Þ ¼ð Þ hk � �fð Þ ¼ 0; and hence by Conclusion3.309,

fk k2¼ �f��

2¼ limk!0

0ð Þ � �f��

2¼ limk!0

�f � hkð Þ � �f��

2¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :Since f2 ¼ 0; we have f ¼ 0: ■)

Conclusion 3.330 Let U : f 7! f be a mapping from L1 \ L2 to L2: Then U :

L1 \ L2ð Þ ! L2 is linear, and

U L1 \ L2� �� ?¼ 0f g:

Lemma 3.331 Let H be a Hilbert space. Let Y be a linear subspace of H. LetY? ¼ 0f g: Then Y is dense in H.

Proof

Problem 3:332 Y is a linear subspace of H.

(Solution Let x; y 2 �Y : Let a; b 2 C: We have to show that axþ byð Þ 2 �Y : Sincex 2 �Y ; there exists a convergent sequence xnf g in Y such that limn!1 xn ¼ x:Similarly, there exists a convergent sequence ynf g in Y such that limn!1 yn ¼ y: Itfollows that limn!1 axn þ bynð Þ ¼ axþ byð Þ: Since for every positive integer n,xn; yn 2 Y , and Y is a linear subspace of H, we have, for every positive integer n,axn þ bynð Þ 2 Y . Thus, axn þ bynf g is a convergent sequence in Y such thatlimn!1 axn þ bynð Þ ¼ axþ byð Þ; and hence axþ byð Þ 2 �Y : ■)

Thus, �Y is a closed linear subspace of H. Since Y � �Y ; we have 0f g �ð Þ �Yð Þ?�Y? ¼ 0f gð Þ; and hence, �Yð Þ?¼ 0f g:

We have to show that �Y ¼ H:

If not, otherwise, let H 6� �Y : We have to arrive at a contradiction.Since H 6� �Y ; there exists a 2 H such that a 62 �Y : Since a 2 H; and �Y is a closed

linear subspace of H, by Conclusion 2.75 there exists x; y such that x 2 �Y ; y 2�Yð Þ?; and a ¼ xþ y: Since y 2 �Yð Þ? ¼ 0f gð Þ; we have a� x ¼ð Þy ¼ 0; and hencea ¼ x 2 �Yð Þ: Thus, a 2 �Y : This is a contradiction. ■

Lemma 3.333 Let U : f 7! f be a mapping from L1 \ L2 to L2: Then U L1 \L2ð Þ isdense in L2:

Proof By Conclusion 3.330, U L1 \L2ð Þð Þ?¼ 0f g:Now, sinceU L1 \ L2ð Þ is a linearsubspace of the Hilbert space L2; by Lemma 3.331 U L1 \L2ð Þ is dense in L2: ■

Note 3.334 Let X be a metric space with metric d. Let p1; p2; . . .f g, andq1; q2; . . .f g be Cauchy sequences in X.


a: Problem 3.335 d p1; q1ð Þ; d p2; q2ð Þ; . . .f g is a Cauchy sequence in R:

(Solution For this purpose, let us take any e[ 0: Since p1; p2; . . .f g is Cauchy,there exists positive integer N1 such that

m; n�N1 ) d pm; pnð Þ\ e2

� �:

Similarly, there exists positive integer N2 such that

m; n�N2 ) d qm; qnð Þ\ e2

� �:

It follows that for every m; n�max N1;N2f g;

d pm; qmð Þ � d pn; qnð Þ� d pm; pnð Þþ d pn; qnð Þþ d qn; qmð Þð Þ � d pn; qnð Þ

¼ d pm; pnð Þþ d qn; qmð Þ\ e2þ e

2¼ e;

and hence for every m; n�max N1;N2f g;

d pm; qmð Þ � d pn; qnð Þ\e:

Similarly, for every m; n�max N1;N2f g;

d pn; qnð Þ � d pm; qmð Þ\e:

It follows that for every m; n�max N1;N2f g;

d pn; qnð Þ � d pm; qmð Þj j\e:

Thus, d p1; q1ð Þ; d p2; q2ð Þ; . . .f g is a Cauchy sequence in R: ■)Since d p1; q1ð Þ; d p2; q2ð Þ; . . .f g is a Cauchy sequence in R, and R is complete,

there exists a nonnegative real number a such that limn!1 d pn; qnð Þ ¼ a: Iflimn!1 d pn; qnð Þ ¼ 0; then we shall write

pnf g� qnf g:

Let C be the collection of all Cauchy sequences in X.

b: Problem 3.336 � is an equivalence relation over C:(Solution

1. Reflexive: Let us take any p1; p2; . . .f g 2 C: We have to show that pnf g� pnf g;that is, limn!1 d pn; pnð Þ ¼ 0:

LHS ¼ limn!1

d pn; pnð Þ ¼ limn!1

0 ¼ 0 ¼ RHS:


2. Symmetry: Let us take any p1; p2; . . .f g; q1; q2; . . .f g 2 C: Let pnf g� qnf g: Wehave to show that qnf g� pnf g; that is limn!1 d qn; pnð Þ ¼ 0: Sincepnf g� qnf g; we have limn!1 d pn; qnð Þ ¼ 0:

LHS ¼ limn!1

d qn; pnð Þ ¼ limn!1

d pn; qnð Þ ¼ 0 ¼ RHS:

3. Transitive: Let us take any p1; p2; . . .f g; q1; q2; . . .f g; r1; r2; . . .f g 2 C: Letpnf g� qnf g; and qnf g� rnf g: We have to show that pnf g� rnf g; that is

limn!1 d pn; rnð Þ ¼ 0: Since pnf g� qnf g; we have limn!1 d pn; qnð Þ ¼ 0:Similarly, limn!1 d qn; rnð Þ ¼ 0: It follows that

limn!1

d pn; qnð Þþ d qn; rnð Þð Þ ¼ 0:


0� d pn; rnð Þ� d pn; qnð Þþ d qn; rnð Þð Þ;

and limn!1 d pn; qnð Þþ d qn; rnð Þð Þ ¼ 0; we have limn!1

d pn; rnð Þ ¼ 0: ■)

Since � is an equivalence relation over C; C is partitioned into equivalenceclasses.

Let us denote the collection of all equivalence classes by X�:

Now, let pnf g½ �; qnf g½ � 2 X�; where pnf g; qnf g 2 C: Let p0n� � �

; q0n� � �

2 X�;

where p0n�

; q0n�

2 C: Let pnf g½ � ¼ p0n� � �

; and qnf g½ � ¼ q0n� � �

:

c: Problem 3.337 limn!1 d pn; qnð Þ ¼ limn!1 d p0n; q0n

� �; that is limn!1

d pn; qnð Þ � d p0n; q0n

� �� ¼ 0:

(Solution Since pnf g½ �; p0n� � �

2 X�; and pnf g½ � ¼ p0n� � �

; we have pnf g� p0n�

;

and hence limn!1 d pn; p0n� �

¼ 0: Similarly, limn!1 d qn; q0n� �

¼ 0: It follows thatlimn!1 d pn; p0n

� �þ d qn; q0n� ��

¼ 0: Since for every positive integer n;


� �� d pn; p

0n

� �þ d p0n; q

0n

� �þ d q0n; qn� ��

� d p0n; q0n

� �¼ d pn; p

0n

� �þ d qn; q

0n

� �� ;

we have, for every positive integer n,


� �� d pn; p

0n

� �þ d qn; q

0n

� �� :

Similarly, for every positive integer n,

d p0n; q0n

� �� d pn; qnð Þ� d pn; p

0n

� �þ d qn; q

0n

� �� :

It follows that for every positive integer n,


0�ð Þ d pn; qnð Þ � d p0n; q0n

� �� d pn; p0n

� �þ d qn; q

0n

� �� :

Now, since

limn!1

d pn; p0n

� �þ d qn; q

0n

� �� ¼ 0;

we have

limn!1


� �� ¼ 0:

■)This shows that

D : pnf g½ �; qnf g½ �ð Þ 7! limn!1

d pn; qnð Þ

is a well-defined function from X� X� to 0;1½ Þ:

d: Problem 3.338 D is a metric in X�:

(Solution

1. Let us take any pnf g½ � 2 X�; where pnf g 2 C: We have to show thatD pnf g½ �; pnf g½ �ð Þ ¼ 0; that is limn!1 d pn; pnð Þ ¼ 0:

LHS ¼ limn!1

d pn; pnð Þ ¼ limn!1

0 ¼ 0 ¼ RHS:

2. Let us take any pnf g½ �; qnf g½ � 2 X�; where pnf g; qnf g 2 C: We have to showthat D pnf g½ �; qnf g½ �ð Þ ¼ D qnf g½ �; pnf g½ �ð Þ; that is

limn!1

d pn; qnð Þ ¼ limn!1

d qn; pnð Þ:

This is clear.3. Let us take any pnf g½ �; qnf g½ �; rnf g½ � 2 X�; where pnf g; qnf g; rnf g 2 C:We have

to show that D pnf g½ �; rnf g½ �ð Þ�D pnf g½ �; qnf g½ �ð ÞþD qnf g½ �; rnf g½ �ð Þ; that is

limn!1

d pn; rnð Þ� limn!1

d pn; qnð Þþ limn!1

d qn; rnð Þ;

that is

limn!1


d pn; qnð Þþ d qn; rnð Þð Þ:



d pn; rnð Þ� d pn; qnð Þþ d qn; rnð Þð Þ;

we have

limn!1


d pn; qnð Þþ d qn; rnð Þð Þ:

■)

e: Problem 3.339 X�;Dð Þ is a complete metric space.

(Solution For this purpose, let us take any Cauchy sequence

p 1ð Þ1 ; p 1ð Þ

2 ; . . .n oh i

; p 2ð Þ1 ; p 2ð Þ

2 ; . . .n oh i

; . . .n o

in X�; where p 1ð Þ1 ; p 1ð Þ

2 ; . . .n o

; p 2ð Þ1 ; p 2ð Þ

2 ; . . .n o

; � � � 2 C: Since p 1ð Þ1 ; p 1ð Þ

2 ; . . .n o

2 C;

p 1ð Þ1 ; p 1ð Þ

2 ; . . .n o

is a Cauchy sequence in X, and hence there exists a positive integer

N1 such that

m; n�N1 ) d p 1ð Þm ; p 1ð Þ

n

� �\

11

� :

Similarly, there exists a positive integer N2 such that

m; n�N2 ) d p 2ð Þm ; p 2ð Þ

n

� �\

12

� :

Also, there exists a positive integer N3 such that

m; n�N3 ) d p 3ð Þm ; p 3ð Þ

n

� �\

13

� ; etc:

It follows that

n�N1 ) d p 1ð ÞN1; p 1ð Þ

n

� �\

11

� ; n�N2 ) d p 2ð Þ

N2; p 2ð Þ

n

� �\

12

� ; n�N3 ) d p 3ð Þ

N3; p 3ð Þ

n

� �\

13

� ; etc:

Since the constant sequence p 1ð ÞN1; p 1ð Þ

N1; p 1ð Þ

N1; . . .

n ois a Cauchy sequence in X,

p 1ð ÞN1; p 1ð Þ

N1; p 1ð Þ

N1; . . .

n o2 C: Now, since p 1ð Þ

1 ; p 1ð Þ2 ; p 1ð Þ

3 ; . . .n o

2 C; limn!1 d p 1ð ÞN1; p 1ð Þ

n

� �exists. Since limn!1 d p 1ð Þ

N1; p 1ð Þ

n

� �; and


n�N1 ) d p 1ð ÞN1; p 1ð Þ

n

� �\

11

� ;

we have limn!1 d p 1ð ÞN1; p 1ð Þ

n

� �� 1

1 : Similarly, limn!1 d p 2ð ÞN2; p 2ð Þ

n

� �� 1

2,

limn!1 d p 3ð ÞN3; p 3ð Þ

n

� �� 1

3 ; etc. In short, for every positive integer n,

limk!1

d p nð ÞNn; p nð Þ

k

� �� 1

n:

Problem 3:340 p 1ð ÞN1; p 2ð Þ

N2; p 3ð Þ

N3; . . .

n ois a Cauchy sequence in X.

(Solution For this purpose, let us take any e[ 0: Since

p 1ð Þ1 ; p 1ð Þ

2 ; . . .n oh i

; p 2ð Þ1 ; p 2ð Þ

2 ; . . .n oh i

; . . .n o

is a Cauchy sequence in X�, there exists a positive integer N such that

m; n�N ) D p mð Þ1 ; p mð Þ

2 ; . . .n oh i

; p nð Þ1 ; p nð Þ

2 ; . . .n oh i� �

\e3

� �;

that is

m; n�N ) limk!1

d p mð Þk ; p nð Þ

k

� �\

e3

� :

Since for every positive integers m; n; k;

d p mð ÞNm

; p nð ÞNn

� �� d p mð Þ

Nm; p mð Þ

k

� �þ d p mð Þ

k ; p nð Þk

� �þ d p nð Þ

k ; p nð ÞNn

� �;

we have, for every positive integer m; n;

d p mð ÞNm

; p nð ÞNn

� �� lim

k!1d p mð Þ

Nm; p mð Þ

k

� �þ d p mð Þ

k ; p nð Þk

� �þ d p nð Þ

k ; p nð ÞNn

� �� ¼ lim

k!1d p mð Þ

Nm; p mð Þ

k

� �þ lim

k!1d p mð Þ

k ; p nð Þk

� �þ lim

k!1d p nð Þ

k ; p nð ÞNn

� �� 1

mþ lim

k!1d p mð Þ

k ; p nð Þk

� �þ lim

k!1d p nð Þ

Nn; p nð Þ

k

� �� 1

mþ lim

k!1d p mð Þ

k ; p nð Þk

� �þ 1

n;

and hence for every positive integer m; n;

d p mð ÞNm

; p nð ÞNn

� �� 1

mþ lim

k!1d p mð Þ

k ; p nð Þk

� �þ 1

n:

It follows that,


m; n�N ) d p mð ÞNm

; p nð ÞNn

� �\

1m

þ e3þ 1

n

� :

Since limn!11n ¼ 0; there exists a positive integer K such that k�K ) 1

k\e3 :

Now,

m; n�max N;Kf g ) d p mð ÞNm

; p nð ÞNn

� �\

e3þ e

3þ e

3¼ eð Þ;

so p 1ð ÞN1; p 2ð Þ

N2; p 3ð Þ

N3; . . .

n ois a Cauchy sequence in X. ■)

It follows that p 1ð ÞN1; p 2ð Þ

N2; p 3ð Þ

N3; . . .

n oh i2 X�:

Problem 3:341 limn!1

D p nð Þ1 ; p nð Þ

2 ; . . .n oh i

; p 1ð ÞN1; p 2ð Þ

N2; p 3ð Þ

N3; . . .

n oh i� �¼ 0:


limn!1

limk!1

d p nð Þk ; p kð Þ

Nk

� �� ¼

� limn!1

D p nð Þ1 ; p nð Þ

2 ; . . .n oh i

; p 1ð ÞN1; p 2ð Þ

N2; p 3ð Þ

N3; . . .

n oh i� �¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};

that is

limn!1

limk!1


Nk

� �� ¼ 0:

Since for every positive integer n, 0�ð Þ limk!1 d p nð ÞNn; p nð Þ

k

� �� 1

n ; and

limn!11n ¼ 0; we have

limn!1

limk!1


Nk

� �� ¼ 0:

■)Thus, X�;Dð Þ is a complete metric space. ■)Let

u : p 7! p; p; p; . . .f g½ �

be a mapping from X to X�:

f: Problem 3.342 u is an isometry, and hence u : X ! X� is 1-1.

(Solution Let us take any p; q 2 X: We have to show that


D u pð Þ;u qð Þð Þ ¼ d x; yð Þ:

LHS ¼ D u pð Þ;u qð Þð Þ ¼ D p; p; p; . . .f g½ �; q; q; q; . . .f g½ �ð Þ¼ lim

n!1d p; qð Þ ¼ d p; qð Þ ¼ RHS:

■)

g: Problem 3.343 u Xð Þ is a dense subset of X�:

(Solution Let p1; p2; p3; . . .f g½ � 2 X�; where p1; p2; p3; . . .f g 2 C: We have toshow that

p1; p2; p3; . . .f g½ � 2 u Xð Þ:

Since p1 2 X; and u : X ! X�; u p1ð Þ 2 X�: Similarly, u p2ð Þ 2 X�;u p3ð Þ 2 X�;etc. It suffices to show that

limn!1

pn; pn; pn; . . .f g½ � ¼ limn!1

u pnð Þ ¼ p1; p2; p3; . . .f g½ �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is

limn!1

limm

d pn; pmð Þ� �

¼ limn!1

D pn; pn; pn; . . .f g½ �; p1; p2; p3; . . .f g½ �ð Þð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl};that is

limn!1

limm


¼ 0:

For this purpose, let us take any e[ 0: Since p1; p2; p3; . . .f g 2 C;p1; p2; p3; . . .f g is a Cauchy sequence in X, there hence exists a positive integer

N such that

m; n�N ) d pn; pmð Þ\eð Þ:

It follows that, for every n�N;

m�N ) d pn; pmð Þ\eð Þ;

and hence for every n�N; limm d pn; pmð Þ� e: It follows that

0�ð Þ limn!1 limm


� e:


Thus, for every e[ 0; limn!1

limm d pn; pmð Þð Þ� �

2 0; e½ �; and hence

limn!1 limm d pn; pmð Þð Þ ¼ 0: ■)

h: Problem 3.344 If X is complete, then u Xð Þ ¼ X�:

(Solution Let X be complete. Let us take any p1; p2; p3; . . .f g½ � 2 X�; wherep1; p2; p3; . . .f g 2 C: It suffices to show that p1; p2; p3; . . .f g½ � 2 u Xð Þ:Since p1; p2; p3; . . .f g 2 C; p1; p2; p3; . . .f g is a Cauchy sequence in X. Now,

since X is complete, there exists p 2 X such that limn!1 pn ¼ p; and hence

D p1; p2; p3; . . .f g½ �;u pð Þð Þ ¼ D p1; p2; p3; . . .f g½ �; p; p; p; . . .f g½ �ð Þ¼ limn!1 d pn; pð Þ ¼ 0|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} :

Since

D p1; p2; p3; . . .f g½ �;u pð Þð Þ ¼ 0; p1; p2; p3; . . .f g½ � ¼ u pð Þ 2 u Xð Þð Þ;

we have p1; p2; p3; . . .f g½ � 2 u Xð Þ: ■)By (f), u : p 7! p; p; p; . . .f g½ � is an isometry from X to X�; so we can identify

X with u Xð Þ in X�: In view of (g), we say that X is a dense subset of X�: Also, by(h), if X is complete, then X ¼ X�:

Here X� is called the completion of X.

Conclusion 3.345 Let X be a metric space with metric d. Then there exists acompletion X� of X. If X is complete, then X� ¼ X:

Note 3.346 Let H be a Hilbert space. Let X and Y be linear subspaces of H. Let

f : X �!½onto

�Y be linear isometry. Let X; Y be dense subsets of H.

Let us take any p 2 H ¼ �Xð Þ:It follows that there exists a sequence pnf g in X such that limn!1 pn ¼ p: Let

p0n�

be another sequence in X such that limn!1 p0n ¼ p:

Problem 3.347 limn!1 f pnð Þ ¼ limn!1 f p0n� �

:

(Solution Since limn!1 pn ¼ p; pnf g is a Cauchy sequence in X. Now, sincef : X ! Y is an isometry, f pnð Þf g is a Cauchy sequence in Y � Hð Þ: Since H is aHilbert space, there exists b 2 H such that limn!1 f pnð Þ ¼ b: Similarly, there existsb0 2 H such that limn!1 f p0n

� �¼ b0: We have to show that b ¼ b0:

Since limn!1 f pnð Þ ¼ b; and limn!1 f p0n� �

¼ b0, limn!1 f pn � p0n� ��

¼Þ limn!1 f pnð Þ � f p0n

� �� ¼ b� b0: Since limn!1 pn ¼ p; and limn!1 p0n ¼ p,

limn!1 pn � p0n� �

¼ 0 2 Xð Þ: Since each pn 2 X; each p0n 2 X; and X is a linearsubspace of H, each pn � p0n

� �2 X: Since f : X ! Y is an isometry, f : X ! Y is

continuous. Since f : X ! Y is continuous, each pn � p0n� �

2 X; andlimn!1 pn � p0n

� �¼ 0 2 Xð Þ; we have


b� b0 ¼ limn!1

f pn � p0n� ��

¼ f 0ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0;

and hence b ¼ b0: ■)This shows that ~f : p 7! limn!1 f pnð Þ is a well-defined mapping from H to H,

where pnf g is a sequence in X such that limn!1 pn ¼ p:

Problem 3.348 ~f is an extension of f.

(Solution Let us take any p 2 X: We have to show that ~f pð Þ ¼ f pð Þ: Since p 2 X;the constant sequence p; p; p; . . .f g is a sequence in X such that limn!1 p ¼ p; andhence ~f pð Þ ¼ limn!1 f pð Þ ¼ f pð Þð Þ: Thus, ~f pð Þ ¼ f pð Þ: ■)

Problem 3.349 ~f : H ! H is linear.

(Solution Let p; q 2 H: Let a; b 2 C: We have to show that ~f apþ bqð Þ ¼a ~f pð Þ� �

þ b ~f qð Þ� �

: Since p 2 H; there exists a sequence pnf g in X such thatlimn!1 pn ¼ p: Similarly, there exists a sequence qnf g in X such that limn!1 qn ¼q: Now, since X is a linear subspace of H, apn þ bqnf g is a sequence in X. Also,limn!1 apn þ bqnð Þ ¼ apþ bq: It follows that

~f apþ bqð Þ ¼ limn!1

f apn þ bqnð Þð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ limn!1

a f pnð Þð Þþ b f qnð Þð Þð Þ

¼ a limn!1

f pnð Þð Þ� �

þ b limn!1

f qnð Þð Þ� �

¼ a ~f pð Þ� �


:

Thus,

~f apþ bqð Þ ¼ a ~f pð Þ� �


:

■)

Problem 3.350 The linear map ~f : H ! H is isometric.

(Solution Let p 2 H: We shall try to show that ~f pð Þ�� ¼ pk k: Since p 2 H; and

X is a dense subset of H, there exists a sequence pnf g in X such that limn!1 pn ¼ p:Since pnf g is a sequence in X such that limn!1 pn ¼ p; we have ~f pð Þ ¼limn!1 f pnð Þ: Since limn!1 pn ¼ p; we have

~f pð Þ�� ¼ lim

n!1f pnð Þ

�� ¼ limn!1

f pnð Þk k ¼ limn!1

pnk k ¼ pk k|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl};so


~f pð Þ�� ¼ pk k:

■)Since ~f : H ! H is isometric, ~f : H ! H is continuous. Since ~f : H ! H is an

extension of

f : X !onto

Y ;

X; Y are dense subsets of complete space H, by Lemma 2.105 ~f is an isometry fromH onto H. Now, since ~f : H ! H is linear, ~f is a Hilbert space isomorphism fromH onto H.

Conclusion 3.351 Let H be a Hilbert space. Let X and Y be linear subspaces of

H. Let f : X �!½onto

�Y be linear isometry. Let X; Y be dense subsets of H. Then there

exists an extension ~f : H ! H such that ~f is Hilbert space isomorphism fromH onto H.

Note 3.352 By Conclusion 3.325, the mapping U : f 7! f from L1 \L2ð Þ to L2 is anL2-isometry, so we can identify L1 \ L2ð Þ with U L1 \ L2ð Þ: By Lemma 3.333,U L1 \ L2ð Þ is dense in L2; so L1 \L2ð Þ is dense in L2: Thus, U is an L2-isometricmapping from one dense subset L1 \ L2ð Þ of L2 to another dense subset U L1 \ L2ð Þof L2:

Now, since L2 is a Hilbert space, by Conclusion 3.351, there exists an extensioneU : L2 ! L2 such that eU is Hilbert space isomorphism from L2 onto L2:

Conclusion 3.353 There exists a Hilbert space isomorphism eU from L2 onto L2

such that eU is an extension of U : f 7! f from L1 \ L2ð Þ to L2:

For every f 2 L2, eUðf Þ 2 L2ð Þ is denoted by f : Thus, f 7! f is a Hilbert spaceisomorphism from L2 onto L2: It follows that, for every f 2 L2, f

�� 2¼ fk k2:

The above conclusion, known as the Plancherel theorem, is due to M.Plancherel (16.01.1885–04.03.1967, Swiss). He gave courses not only in mathe-matics, but also in electric engineering. He is best known for his theorem in har-monic analysis.

Exercises

3:1 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! C be acomplex measure on M: Show that the total variation lj j : M ! 0;1½ � is apositive measure on M:


3:2 Let X be any nonempty set. Let M be a r-algebra in X. Let k1 : M ! C; andk2 : M ! C be complex measures on M: Let k1 ? k2: Show that k1j j ? k2j j:

3:3 Let X be any nonempty set. Let M be a r-algebra in X. Let k : M ! C be acomplex measure on M: Let l : M ! 0;1½ Þ be a positive measure. Letk l: Show that kj j l:

3:4 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ �be a positive measure on M: Show that

a. for every g 2 L2 lð Þ; Ug : f 7!RX f � gð Þdl is a bounded linear functional

on the normed linear space L2 lð Þ;b. Ug

�� gk kq:

3:5 Let M be the r-algebra of Borel sets in R: Let l : M ! C be a complexmeasure on M: Let f : x 7! l �1; xð Þð Þ be the function from R to C: Leta 2 R; and A 2 C: Suppose that f 0 að Þ 6¼ A: Let n be a positive integer. Showthat there exists an open interval I which contains a satisfying m Ið Þ\n; and


�� 1:

3:6 Suppose that f : R2 ! C is continuous at a, and f 2 L1 R2� �

: Show that a is aLebesgue point of f.

3:7 Let f : �1; 1½ � ! R be any function. Let f be monotonically increasing. Letf be absolutely continuous on a; b½ �: Let M be the set of all Lebesgue mea-surable subsets of R: Let E be a subset of �1; 1½ � such that E 2 M; andm Eð Þ ¼ 0: Show that

a. f Eð Þ 2 M;b. m f Eð Þð Þ ¼ 0:

3:8 Let T : �1; 1ð Þ �1; 1ð Þ ! R2 be a continuous map. Let T be differentiableat 0; 0ð Þ: Show that

limr!0

m T B 0; 0ð Þ; rð Þð Þð Þm B 0; 0ð Þ; rð Þð Þ ¼ det T 0 0; 0ð Þð Þj j:

3:9 Let f : R ! C be a Lebesgue measurable function. Let f 2 L1 \ L2: Let ~f :x 7! f �xð Þ be a function from R to C: Show that

a. ~f : R ! C is a Lebesgue measurable function,b. ~f 2 L1 \ L2:


3:10 Let X be any nonempty set. Let M be a r-algebra in X. Let l : M ! 0;1½ Þbe a positive measure on M: Let U : L2 lð Þ ! C be a bounded linear func-tional. Show that there exists a function g : X ! C such that

a. g 2 L2 lð Þ;b. for every f 2 L2 lð Þ;U fð Þ ¼

RX f � gð Þdl;

c. Uk k ¼ gk k2:


Bibliography

1. Ahlfors, L.V.: Complex Analysis. McGraw-Hill, India (2006)2. Apostol, T.M.: Introduction to Analytic Number Theory. Springer (1998)3. Conway, J.B.: Functions of One Complex Variable. Springer (1978)4. Jacobson, N.: Lectures on Abstract Algebra I: Basic Concepts. Springer (2013)5. Rudin, W.: Principles of Mathematical Analysis, 3rd edn. McGraw-Hill Book Company

(2006)6. Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Education (1986)7. Ullrich, D.C.: Complex Made Simple. AMS (2008)

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Rajnikant Sinha Real and Complex Analysis