The Kurzweil{Henstock integral for undergraduatesfonda/p2017_book_KH.pdf · The integral in question is indeed more general than Lebesgue’s in RN, but its construction is rather

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The Kurzweil–Henstock integralfor undergraduates

A promenade alongthe marvelous theory of integration

Alessandro Fonda

May 6, 2018


To Sofia, Marcello, and Elisa


Contents

Introduction vii

1 Functions of one real variable 1

1.1 P-partitions and Riemann sums . . . . . . . . . . . . . . . . . . . . . 1

1.2 The notion of δ-fineness . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Integrable functions on a compact interval . . . . . . . . . . . . . . . 5

1.4 Elementary properties of the integral . . . . . . . . . . . . . . . . . . 7

1.5 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . 10

1.6 Primitivable functions . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.7 Primitivation by parts and by substitution . . . . . . . . . . . . . . . 16

1.8 The Cauchy criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.9 Integrability on sub-intervals . . . . . . . . . . . . . . . . . . . . . . 22

1.10 R-integrable functions and continuous functions . . . . . . . . . . . . 24

1.11 The Saks–Henstock theorem . . . . . . . . . . . . . . . . . . . . . . . 28

1.12 L-integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.13 The monotone convergence theorem . . . . . . . . . . . . . . . . . . 35

1.14 The dominated convergence theorem . . . . . . . . . . . . . . . . . . 39

1.15 Integration on non-compact intervals . . . . . . . . . . . . . . . . . . 42

1.16 The Hake Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.17 Integrals and series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2 Functions of several real variables 53

2.1 Integrability on rectangles . . . . . . . . . . . . . . . . . . . . . . . . 53

2.2 Integrability on a bounded set . . . . . . . . . . . . . . . . . . . . . . 57

2.3 The measure of a bounded set . . . . . . . . . . . . . . . . . . . . . . 60

2.4 The Cebicev inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 62

2.5 Negligible sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.6 A characterization of bounded measurable sets . . . . . . . . . . . . 65

2.7 Continuous functions and L-integrable functions . . . . . . . . . . . 69

2.8 Limits and derivatives under the integration sign . . . . . . . . . . . 72

v

vi CONTENTS

2.9 The reduction formula . . . . . . . . . . . . . . . . . . . . . . . . . . 762.10 Change of variables in the integral . . . . . . . . . . . . . . . . . . . 842.11 Change of measure by diffeomorphisms . . . . . . . . . . . . . . . . . 902.12 The general theorem on the change of variables . . . . . . . . . . . . 922.13 Some useful transformations in R2 . . . . . . . . . . . . . . . . . . . 952.14 Cylindrical and spherical coordinates in R3 . . . . . . . . . . . . . . 972.15 The integral on unbounded sets . . . . . . . . . . . . . . . . . . . . . 100

3 Differential forms 1113.1 The vector spaces ΩM (RN ) . . . . . . . . . . . . . . . . . . . . . . . 1113.2 Differential forms in RN . . . . . . . . . . . . . . . . . . . . . . . . . 1133.3 External product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1143.4 External differential . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.5 Differential forms in R3 . . . . . . . . . . . . . . . . . . . . . . . . . 1173.6 M-surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1203.7 The integral of a differential form . . . . . . . . . . . . . . . . . . . . 1263.8 Scalar functions and M -superficial measure . . . . . . . . . . . . . . 1303.9 The oriented boundary of a rectangle . . . . . . . . . . . . . . . . . . 1353.10 The Gauss formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1373.11 Oriented boundary of a M -surface . . . . . . . . . . . . . . . . . . . 1393.12 The Stokes–Cartan formula . . . . . . . . . . . . . . . . . . . . . . . 1423.13 Analogous results in R2 . . . . . . . . . . . . . . . . . . . . . . . . . 1463.14 Exact differential forms . . . . . . . . . . . . . . . . . . . . . . . . . 149

A Differential calculus in RN 155A.1 The differential of a scalar-valued function . . . . . . . . . . . . . . . 155A.2 Twice differentiable scalar-valued functions . . . . . . . . . . . . . . 158A.3 The differential of a vector-valued function . . . . . . . . . . . . . . 160A.4 Some computational rules . . . . . . . . . . . . . . . . . . . . . . . . 161A.5 The implicit function theorem . . . . . . . . . . . . . . . . . . . . . . 163A.6 Local diffeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . 169

B Stokes–Cartan and Poincare theorems 171

C On differentiable manifolds 179

D The Banach–Tarski paradox 185

E A brief historical note 191

References 198

Introduction

This book is the outcome of the beginners’ courses held over the past few yearsto my undergraduate students. The aim was to provide them with a general andsufficiently easy to grasp theory of the integral. The integral in question is indeedmore general than Lebesgue’s in RN , but its construction is rather simple, since itmakes use of Riemann sums which, being geometrically viewable, are easily under-standable.

This approach to the theory of the integral was developed independently byJaroslav Kurzweil and Ralph Henstock since 1957 (cf. [8, 5]). A number of booksare now available [1, 4, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21]. However, Ifeel that most of these monographs address to an expert reader, rather than to abeginner student. This is why I wanted to maintain here the exposition at a verydidactical level, trying to avoid as much as possible unnecessary technicalities.

The book is divided into three main chapters and five appendices.

The first chapter outlines the theory for functions of one real variable. I havedone my best to keep the explanation as simple as can be, following as far as possiblethe lines of the theory of the Riemann integral. However, there are some interestingpeculiarities.

• The Fundamental Theorem of differential and integral calculus is very general andnatural: one only has to assume the given function to be primitivable, i.e., to bethe derivative of a differentiable function. The proof is simple and clearly shows thelink between differentiability and integrability.

• The generalized integral, on a bounded but not compact interval, is indeed a stan-dard integral: In fact, Hake’s Theorem shows that a function having a generalizedintegral on such an interval can be extended to a function which is integrable in thestandard sense on the closure of its domain.

• Integrable functions according to Lebesgue are those functions which are inte-grable, and whose absolute value is integrable, too.

In the second chapter, the theory is extended to real functions of several realvariables. No difficulties are encountered while considering functions defined onrectangles. When the functions are defined on more general domains, however, anobstacle arises concerning the property of additivity on subdomains. It is thennecessary to limit one’s attention to functions which are integrable according toLebesgue, after having introduced the concept of measurable set. On the other hand,the Fubini Reduction Theorem has a rather technical but conceptually simple proof,which only makes use of the definition. In the Theorem on the Change of Variables

vii

viii INTRODUCTION

in the integral, once again complications may arise (see, e.g., [2]), so that I againdecided to limit the discussion only to functions which are integrable according toLebesgue. The same goes for functions which are defined on unbounded sets. Thesedifficulties are intrinsic, not only at an expository level, and research on some ofthese issues is still being carried out.

The third chapter illustrates the theory of differential forms. The aim is to provethe classical theorems carrying the name of Stokes, and Poincare’s theorem on exactdifferential forms. Dimension 3 has been considered closely: indeed, the theoremsby Stokes–Cartan and Poincare are proved in this chapter only in this case, and thereader is referred to Appendix B for the general proof. Also, I opted to discuss onlythe theory for M -surfaces, without generalizing and extending it to more complexgeometrical objects (see Appendix C). In some parts of this chapter the regularityassumptions could be weakened, but I did not want to enter into a topic touching astill ongoing research.

In Appendix A the basic facts about differential calculus in RN are reviewed.

In Appendix B the theorems by Stokes–Cartan and Poincare are proved. Theproofs are rather technical but do not present great conceptual difficulties.

In Appendix C one can find a brief introduction to the theory of differentiablemanifolds, with particular emphasis on the corresponding version of the Stokes–Cartan theorem. I did not want to deal with this argument extensively, and theproofs are only sketched. For a more complete treatment, we refer to [20].

In Appendix D one of the most surprising results of modern mathematics isreported, the so-called Banach–Tarski paradox. It states that a three-dimensionalball can be divided into a certain number of subsets which, after some well-chosenrotations and translations, finally give two identical copies of the starting ball. Whyreporting on this in a book about integration? Well, the Banach–Tarski paradoxshows the existence of sets which are not measurable (a rotation and a translationmaintain the measure of a set, provided this set is measurable!), and it does this ina very spectacular way.

Appendix E entails a short historical note on the evolution of the concept ofintegral. This note is by no means complete. The aim is to give an idea of the roleplayed by the Riemann sums in the different stages of the history of the integral.

Note. A preliminary version of this book was published in Italian under the titleLezioni sulla teoria dell’integrale. It has been revised here, extending and improvingmost of the arguments.

Chapter 1

Functions of one real variable

Along this chapter, we denote by I a compact interval of the real line R, i.e., aninterval of the type [a, b].

1.1 P-partitions and Riemann sums

Let us start by introducing the notion of P-partition of the interval I.

Definition 1.1 A P-partition of the interval I = [a, b] is a set

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am]) ,

whose elements appear as couples (xj , [aj−1, aj ]), where [aj−1, aj ] is a subset of I andxj is a point in it. Precisely, we have

a = a0 < a1 < . . . < am−1 < am = b ,

and, for every j = 1, . . . ,m,xj ∈ [aj−1, aj ] .

Example. Consider the interval [0, 1]. As examples of P-partitions of I we have the following sets:

Π =

(1

6, [0, 1]

)Π =

(0,

[0,

1

3

]),

(1

2,

[1

3, 1

])Π =

(1

3,

[0,

1

3

]),

(1

3,

[1

3,

2

3

]),

(2

3,

[2

3, 1

])Π =

(1

8,

[0,

1

4

]),

(3

8,

[1

4,

1

2

]),

(5

8,

[1

2,

3

4

]),

(7

8,

[3

4, 1

]).

1

2 CHAPTER 1. FUNCTIONS OF ONE REAL VARIABLE

We consider now a function f , defined on the interval I, having real values. Toeach P-partition of the interval I we can associate a real number, in the followingway.

Definition 1.2 Let f : I → R be a function and

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am])

a P-partition of I. We call Riemann sum associated to I, f and Π the real numberS(I, f,Π) defined by

S(I, f,Π) =

m∑j=1

f(xj)(aj − aj−1) .

In order to better understand this definition, assume for simplicity the functionf to be positive on I. Then, to each P-partition of I we associate the sum of theareas of the rectangles having base [aj−1, aj ] and height [0, f(xj)].

f (x)

a bx1 xa1 x2 x3 x4a2 a3

If f is not positive on I, the areas will be considered with positive or negativesign depending on whether f(xj) be positive or negative, respectively. If f(xj) = 0,the j−th term of the sum will obviously be zero.

Example. Let I = [0, 1], f(x) = 4x2 − 1, and

Π =

(1

8,

[0,

1

4

]),

(1

2,

[1

4,

3

4

]),

(7

8,

[3

4, 1

]).

Then,

S(I, f,Π) = −15

16· 1

4+ 0 · 1

2+

33

16· 1

4=

9

32.

Now we ask whether, taking the P-partitions finer and finer, the Riemann sumsassociated to them will converge to some value. When this happens for a positive

1.2. THE NOTION OF δ-FINENESS 3

function f, such a value can be visualized as the area of the region in the cartesianplane which is confined between the graph of f and the horizontal axis. To be ableto analyze this question, we need to specify what we mean for a P-partition to be“fine”.

1.2 The notion of δ-fineness

Let us introduce the notion of “fineness” for the P-partition Π previously defined.For brevity, we call gauge on I every function δ : I → R such that δ(x) > 0 forevery x ∈ I. Such a function will be useful for having a control on the amplitude ofthe various intervals determined by the points of the P-partition.

Definition 1.3 Given a gauge δ on I, we say that the P-partition Π introducedabove is δ-fine if, for every j = 1, . . . ,m,

xj − aj−1 ≤ δ(xj) , and aj − xj ≤ δ(xj) .

Equivalently, we may write

[aj−1, aj ] ⊆ [xj − δ(xj), xj + δ(xj)] ,

or elsexj − δ(xj) ≤ aj−1 ≤ xj ≤ aj ≤ xj + δ(xj) .

We will show now that it is always possible to find a δ-fine P-partition of theinterval I, whatever the gauge δ. In the following theorem, due to P. Cousin, thecompactness of the interval I plays an essential role.

Theorem 1.4 Given a compact interval I, for every gauge δ on I there is a δ-fineP-partition of I.

Proof We proceed by contradiction. Assume there exists a gauge δ on I for whichit is impossible to find any δ-fine P-partition of I. Let us divide the interval I intwo equal sub-intervals, having the mid point of I as common extremum. Then,at least one of the two sub-intervals does not have any δ-fine P-partition. Let uschoose it, and divide it again in two equal sub-intervals. Continuing this way, weconstruct a sequence (In)n of bottled sub-intervals, whose lengths tend to zero, eachof which does not have any δ-fine P-partitions. By the Cantor Theorem there is apoint c ∈ I belonging to all of these intervals. Moreover, it is clear that, from acertain n thereof, every In will be contained in [c − δ(c), c + δ(c)]. Choose one ofthese, e.g. In. Then the set Π = (c, In), whose only element is the couple (c, In),is a δ-fine P-partition of In, in contradiction with the above.


Examples. Let us see, as examples, some δ-fine P-partitions of the interval [0, 1]. We start with aconstant gauge: δ(x) = 1

5. Since the previous theorem does not give any information on how to find

a δ-fine P-partition, we will proceed by guessing. As a first guess, we choose the aj equally distantand the xj as the middle points of the intervals [aj−1, aj ]. Hence:

aj =j

m, xj =

2j − 1

2m(j = 1, . . . ,m) .

For the corresponding P-partition to be δ-fine, it has to be

xj − aj−1 =1

2m≤ 1

5, aj − xj =

1

2m≤ 1

5.

These inequalities are satisfied choosing m ≥ 3. If m = 3, we have the δ-fine P-partition(1

6,

[0,

1

3

]),

(1

2,

[1

3,

2

3

]),

(5

6,

[2

3, 1

]).

If, instead of taking the points xj in the middle of the respective intervals, we would like to choosethem, for example, at the left extremum, i.e. xj = j−1

m, in order to have a δ-fine P-partition we

should ask that

xj − aj−1 = 0 ≤ 1

5, aj − xj =

1

m≤ 1

5.

These inequalities are verified if m ≥ 5. For instance, if m = 5, we have the δ-fine P-partition(0,

[0,

1

5

]),

(1

5,

[1

5,

2

5

]),

(2

5,

[2

5,

3

5

]),

(3

5,

[3

5,

4

5

]),

(4

5,

[4

5, 1

]).

Notice that, with such a choice of the aj , if m ≥ 5, the points xj can actually be taken arbitrarilyin the respective intervals [aj−1, aj ], still obtaining δ-fine P-partitions.

The previous example shows how it is possible to construct δ-fine P-partitions in the case ofa gauge δ which is constant with value 1

5. It is clear that a similar procedure can be used for a

constant gauge with arbitrary value. Consider now the case when δ is a continuous function. Then,the Weierstrass Theorem says that there is for δ(x) a minimum positive value: let it be δ. Considerthen the constant gauge with value δ, and construct a δ-fine P-partition with the procedure we haveseen above. Clearly, such a P-partition has to be δ-fine, as well. It is thus seen how the case of acontinuous gauge can be reduced to that of a constant gauge.

Consider now the following non-continuous gauge:

δ(x) =

1

2if x = 0 ,

x

2if x ∈ ]0, 1] .

As before, we proceed by guessing. Let us try, as above, taking the aj equally distant and the xjas the middle points of the intervals [aj−1, aj ]. This time, however, we are going to fail; indeed, weshould have

x1 = x1 − a0 ≤ δ(x1) =x12,

which is clearly impossible if x1 > 0. The only way to solve this problem is to choose x1 = 0. Wedecide then, for instance, to take the xj to coincide with aj−1, as was also done above. We thusfind a δ-fine P-partition: (

0,

[0,

1

2

]),

(1

2,

[1

2,

3

4

]),

(3

4,

[3

4, 1

]).

1.3. INTEGRABLE FUNCTIONS ON A COMPACT INTERVAL 5

Notice that a more economic choice could have been(0,

[0,

1

2

]),

(1,

[1

2, 1

]).

The choice x1 = 0 is however unavoidable.Consider now the following gauge: once fixed a point c ∈ ]0, 1[ ,

δ(x) =

c− x2

if x ∈ [0, c[ ,

1

5if x = c ,

x− c2

if x ∈ ]c, 1] .

Similar considerations to those made in the previous case lead to the conclusion that, in order tohave a δ-fine P-partition, it is necessary that one of the xj be chosen as to be the point c. Forexample, if c = 1

2, a possible choice is the following:(

0,

[0,

1

4

]),

(1

4,

[1

4,

3

8

]),

(1

2,

[3

8,

5

8

]),

(3

4,

[5

8,

3

4

]),

(1,

[3

4, 1

]).

1.3 Integrable functions on a compact interval

Consider a function f , defined on the interval I = [a, b]. We are now in the positionto define what we mean by convergence of the Riemann sum when the P-partitionsbecome “finer and finer”.1

Definition 1.5 A function f : I → R is said to be integrable if there is a realnumber A with the following property: given ε > 0, it is possible to find a gauge δon I such that, for every δ-fine P-partition Π of I, it is

|S(I, f,Π)−A| ≤ ε .

We will also say that f is integrable on I.

Let us prove that there is at most one A ∈ R which verifies the conditions of thedefinition. If there were a second one, say A′, we would have that, for every ε > 0there would be two gauges δ and δ′ on I associated respectively to A and A′ by thedefinition. Define the gauge

δ′′(x) = minδ(x), δ′(x) .

Once a δ′′-fine P-partition Π of I is chosen, we have that Π is both δ-fine and δ′-fine,hence

|A−A′| ≤ |A− S(I, f,Π)|+ |S(I, f,Π)−A′| ≤ 2ε .

Since this holds for every ε > 0, it necessarily has to be A = A′.

1The following definition is due to J. Kurzweil and R. Henstock: see Appendix E for an historicaloverview.


If f : I → R is an integrable function, the only element A ∈ R verifying theconditions of the definition is called the integral of f on I and is denoted by oneof the following symbols:∫

If ,

∫ b

af ,

∫If(x) dx ,

∫ b

af(x) dx .

The presence of the letter x in the notation introduced here has no independentimportance. It could be replaced by any other letter t, u, α, . . . , or by any othersymbol, unless already used with another meaning. For reasons to be explained lateron, we set, moreover, ∫ a

bf = −

∫ b

af , and

∫ a

af = 0 .

Examples. Consider the function f : [a, b]→ R, with 0 ≤ a < b, defined by f(x) = xn, where n isa natural number. In case n = 0, we have a constant function of value 1. In that case, the Riemannsums are all equal to b− a, and one easily verifies that the function is integrable and∫ b

a

1 = b− a .

If n = 1, given a P-partition Π of [a, b], the Riemann sum associated is

S(I, f,Π) =

m∑j=1

xj(aj − aj−1) .

In order to find a candidate for the integral, let us consider a particular P-partition where the xjare the middle points of the intervals [aj−1, aj ]. In this particular case, we have

m∑j=1

xj(aj − aj−1) =

m∑j=1

aj−1 + aj2

(aj − aj−1) =1

2

m∑j=1

(a2j − a2j−1) =1

2(b2 − a2).

We want to prove now that the function f(x) = x is integrable on [a, b] and that its integral is really12(b2 − a2). Fix ε > 0. Taken any P-partition Π, we have:∣∣∣∣S(I, f,Π)− 1

2(b2 − a2)

∣∣∣∣=

∣∣∣∣∣m∑j=1

xj(aj − aj−1)−m∑j=1

aj−1 + aj2

(aj − aj−1)

∣∣∣∣∣≤

m∑j=1

∣∣∣xj − aj−1 + aj2

∣∣∣ (aj − aj−1)

≤m∑j=1

aj − aj−1

2(aj − aj−1) .

If we choose the gauge δ to be constant with value εb−a , then, for every δ-fine P-partition Π we

have: ∣∣∣∣S(I, f,Π)− 1

2(b2 − a2)

∣∣∣∣ ≤ m∑j=1

aj − aj−1

22δ =

ε

b− a

m∑j=1

(aj − aj−1) = ε .

1.4. ELEMENTARY PROPERTIES OF THE INTEGRAL 7

The condition of the definition is thus verified with this choice of the gauge, and we have provedthat ∫ b

a

x dx =1

2(b2 − a2) .

If n = 2, it is more difficult to find a candidate for the integral. It can be found by choosing aparticular P-partition where the xj are [ 1

3(a2j−1 +aj−1aj+a2j )]

1/2; indeed, in this case, the Riemannsum is given by

m∑j=1

a2j−1 + aj−1aj + a2j3

(aj − aj−1) =1

3

m∑j=1

(a3j − a3j−1) =1

3(b3 − a3) .

At this point, it is possible to proceed like in the case n = 1 to prove that the function f(x) = x2 isintegrable on [a, b] and that its integral is 1

3(b3 − a3) : once fixed ε > 0, choose the constant gauge

δ = ε2(b2−a2) so that, for any δ-fine P-partition,∣∣∣∣S(I, f,Π)− 1

3(b3 − a3)

∣∣∣∣≤ m∑j=1

∣∣∣∣x2j − a2j−1 + aj−1aj + a2j3

∣∣∣∣ (aj − aj−1)

≤m∑j=1

(a2j − a2j−1)2δ

=ε

b2 − a2m∑j=1

(a2j − a2j−1) = ε .

We have thus proved that ∫ b

a

x2 dx =1

3(b3 − a3) .

In general, it is possible to prove in an analogous way that every function f(x) = xn is integrableon [a, b], and ∫ b

a

xn dx =1

n+ 1(bn+1 − an+1) .

1.4 Elementary properties of the integral

Let f, g be two real functions defined on I = [a, b], and α ∈ R be a constant. It iseasy to verify that, for every P-partition Π of I,

S(I, f + g,Π) = S(I, f,Π) + S(I, g,Π) ,

andS(I, αf,Π) = αS(I, f,Π) .

These linearity properties are inherited by the integral, as will be proved in thefollowing two propositions.

Proposition 1.6 If f and g are integrable on I, then f + g is integrable on I andone has: ∫

I(f + g) =

∫If +

∫Ig .


Proof Set A1 =∫I f and A2 =

∫I g. Being ε > 0 fixed, there are two gauges δ1 and

δ2 on I such that, for every P-partition Π of I, if Π is δ1-fine, then

|S(I, f,Π)−A1| ≤ε

2,

while if Π is δ2-fine, then

|S(I, g,Π)−A2| ≤ε

2.

Let us define the gauge δ on I as δ(x) = minδ1(x), δ2(x). Let Π be a δ-fine P-partition of I. It is thus both δ1-fine and δ2-fine, and we have:

|S(I, f + g,Π)− (A1 +A2)|= |S(I, f,Π)−A1 + S(I, g,Π)−A2|≤ |S(I, f,Π)−A1|+ |S(I, g,Π)−A2|

≤ ε

2+ε

2= ε .

This completes the proof.

Proposition 1.7 If f is integrable on I and α ∈ R, then αf is integrable on I andone has: ∫

I(αf) = α

∫If .

Proof If α = 0, the result is obvious. If α 6= 0, set A =∫I f and fix ε > 0. There is

a gauge δ on I such that

|S(I, f,Π)−A| ≤ ε

|α|,

for every δ-fine P-partition Π of I. Then, for every δ-fine P-partition Π of I, we have

|S(I, αf,Π)− αA| = |αS(I, f,Π)− αA| ≤ |α| ε|α|

= ε ,

and the proof is thus completed.

We have just proved that the set of integrable functions on [a, b] is a vector spaceand that the integral is a linear function on it.

Example. Every polynomial function is integrable on an interval [a, b]. If for instance f(x) =2x2 − 3x+ 7, we have∫ b

a

f = 2

∫ b

a

x2 dx− 3

∫ b

a

x dx+ 7

∫ b

a

1 dx =2

3(b3 − a3)− 3

2(b2 − a2) + 7(b− a) .

We now study the behavior of the integral with respect to the order relationin R.

1.4. ELEMENTARY PROPERTIES OF THE INTEGRAL 9

Proposition 1.8 If f is integrable on I and f(x) ≥ 0 for every x ∈ I, then∫If ≥ 0 .

Proof Fix ε > 0. There is a gauge δ on I such that∣∣∣∣S(I, f,Π)−∫If

∣∣∣∣ ≤ ε ,for every δ-fine P-partition Π of I. Hence,∫

If ≥ S(I, f,Π)− ε ≥ −ε ,

being clearly S(I, f,Π) ≥ 0. Since this is true for every ε > 0, it has to be∫I f ≥ 0,

thus proving the result.

Corollary 1.9 If f and g are integrable on I and f(x) ≤ g(x) for every x ∈ I, then∫If ≤

∫Ig .

Proof It is sufficient to apply the preceding proposition to the function g − f.

Corollary 1.10 If f and |f | are integrable on I, then∣∣∣∣∫If

∣∣∣∣ ≤ ∫I|f | .

Proof Applying the preceding corollary to the inequalities

−|f | ≤ f ≤ |f | ,

we have

−∫I|f | ≤

∫If ≤

∫I|f | ,

whence the conclusion.


1.5 The Fundamental Theorem

The following theorem constitutes a link between the differential and the integralcalculus. It is called the Fundamental Theorem of differential an integralcalculus. More briefly, we will call it the Fundamental Theorem.

Theorem 1.11 Let F : [a, b] → R, be a differentiable function, and let f be itsderivative: F ′(x) = f(x) for every x ∈ [a, b]. Then, f is integrable on [a, b], and∫ b

af = F (b)− F (a) .

Proof Fix ε > 0. For every x ∈ I, since F ′(x) = f(x), there is a δ(x) > 0 such that,for every u ∈ I ∩ [x− δ(x), x+ δ(x)], one has

|F (u)− F (x)− f(x)(u− x)| ≤ ε

b− a|u− x| .

We thus have a gauge δ on I. Consider now a δ-fine P-partition of I,

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am]) .

Since, for every j = 1, . . . ,m,

xj − δ(xj) ≤ aj−1 ≤ xj ≤ aj ≤ xj + δ(xj) ,

one has

|F (aj)− F (aj−1)− f(xj)(aj − aj−1)| == |F (aj)− F (xj)− f(xj)(aj − xj) + [F (xj)− F (aj−1) + f(xj)(aj−1 − xj)]|≤ |F (aj)− F (xj)− f(xj)(aj − xj)|+ |F (aj−1)− F (xj)− f(xj)(aj−1 − xj)|

≤ ε

b− a(|aj − xj |+ |aj−1 − xj |)

=ε

b− a(aj − xj + xj − aj−1)

=ε

b− a(aj − aj−1) .

We deduce that∣∣∣∣∣F (b)− F (a)−m∑j=1

f(xj)(aj − aj−1)

∣∣∣∣∣ =

=

∣∣∣∣∣m∑j=1

[F (aj)− F (aj−1)]−m∑j=1


∣∣∣∣∣

1.6. PRIMITIVABLE FUNCTIONS 11

=

∣∣∣∣∣m∑j=1

[F (aj)− F (aj−1)− f(xj)(aj − aj−1)]

∣∣∣∣∣≤

m∑j=1

∣∣∣F (aj)− F (aj−1)− f(xj)(aj − aj−1)∣∣∣

≤m∑j=1

ε

b− a(aj − aj−1) = ε ,

and the theorem is proved.

1.6 Primitivable functions

We introduce the concept of primitive of a given function.

Definition 1.12 A function f : I → R is said to be primitivable if there is adifferentiable function F : I → R such that F ′(x) = f(x) for every x ∈ I. Such afunction F is called a primitive of f .

The Fundamental Theorem establishes that all primitivable functions definedon a compact interval I = [a, b] are integrable, and that their integral is easilycomputable, once a primitive is known. It can be reformulated as follows.

Theorem 1.13 Let f : [a, b]→ R be primitivable and let F be a primitive. Then fis integrable on [a, b] and ∫ b

af = F (b)− F (a) .

It is sometimes useful to denote the difference F (b)− F (a) with the symbols

[F ]ba , [F (x)]x=bx=a ,

or variants of these as, for instance, [F (x)]ba, when no ambiguities arise.

Example. Consider the function f(x) = xn. It is easy to see that F (x) = 1n+1

xn+1 is a primitive.The Fundamental Theorem tells us that∫ b

a

xn dx =

[1

n+ 1xn+1

]ba

=1

n+ 1(bn+1 − an+1) ,

a result we already obtained in a direct way in the case 0 ≤ a < b.


The fact that the difference F (b)−F (a) does not depend from the chosen prim-itive is explained by the following proposition.

Proposition 1.14 Let f : I → R be a primitivable function, and let F be one of itsprimitives. Then, a function G : I → R is a primitive of f if and only if F − G isa constant function on I.

Proof If F −G is constant, then

G′(x) = (F + (G− F ))′(x) = F ′(x) + (G− F )′(x) = F ′(x) = f(x) ,

for every x ∈ I, and hence G is a primitive of f. On the other hand, if G is a primitiveof f , we have

(F −G)′(x) = F ′(x)−G′(x) = f(x)− f(x) = 0 ,

for every x ∈ I. Consequently, F −G is constant on I.

Notice that, if f : I → R is a primitivable function, it is also primitivable onevery sub-interval of I. In particular, it is integrable on every interval [a, x] ⊆ I, andtherefore it is possible to define a function

x 7→∫ x

af ,

which we call the indefinite integral of f . We denote this function by one of thefollowing symbols: ∫ ·

af ,

∫ ·

af(t) dt

(notice that in this last notation it is convenient to use a different letter from x forthe variable of f ; for instance, we have chosen here the letter t). The FundamentalTheorem tells us that, if F is a primitive of f, then, for every x ∈ [a, b],∫ x

af = F (x)− F (a) .

This fact yields, taking into account Proposition 1.14, that the function∫ ·a f is itself

a primitive of f. We thus have the following

Corollary 1.15 Let f : [a, b] → R be a primitivable function. Then, the indefiniteintegral

∫ ·a f is one of its primitives: it is a function defined on [a, b], differentiable

and, for every x ∈ [a, b], we have(∫ ·

af

)′(x) = f(x) .


Notice that the choice of the point a in the definition of∫ ·a f is by no way

necessary. One could take any point ω ∈ I and consider the function∫ ·ω f. The

conventions made on the integral with exchanged extrema are such that the abovestated theorem still holds. Indeed, if F is a primitive of f, even if x < ω we have∫ x

ωf = −

∫ ω

xf = −(F (ω)− F (x)) = F (x)− F (ω) ,

so that∫ ·ω f is still a primitive of f.

We will denote the set of all primitives of f with one of the following symbols:∫f ,

∫f(x) dx .

Concerning the use of x, an observation analogous to the one made for the integralcan be made here, as well: it can be replaced by any other letter or symbol, withthe due precautions. When applying the theory to practical problems, however, ifF denotes a primitive of f, instead of correctly writing∫

f = F + c : c ∈ R ,

it is common to find improper expressions of the type∫f(x) dx = F (x) + c ,

where c ∈ R stands for an arbitrary constant; we will adapt to this habit, too. Letus make a list of primitives of some elementary functions:∫

ex dx= ex + c ,∫sinx dx=− cosx+ c ,∫cosx dx= sinx+ c ,∫xα dx=

xα+1

α+ 1+ c (α 6= −1) ,∫

1

xdx= ln |x|+ c ,∫

1

1 + x2dx= arctanx+ c ,∫

1√1− x2

dx= arcsinx+ c .


Notice that the definition of primitivable function makes sense even in some caseswhere f is not necessarily defined on a compact interval, and indeed the formulasabove are valid on the natural domains of the considered functions.

Example. Using the Fundamental Theorem, we find:∫ π

0

sinx dx = [− cosx]π0 = − cosπ + cos 0 = 2 .

Notice that the presence of the arbitrary constant c can sometimes lead to ap-parently different results. For example, it is readily verified that one also has∫

1√1− x2

dx = − arccosx+ c .

This is explained by the fact that arcsinx = π2 − arccosx for every x ∈ [−1, 1], and

one should not think that here c refers to the same constant as one appearing in thelast formula of the above list.

One should be careful with the notation introduced for the primitives, whichlooks similar to that for the integral, even if the two concepts are completely different.Their relation comes from the Fundamental Theorem: we have∫ ·

ωf ∈

∫f ,

with any ω ∈ I, and ∫ b

af =

[∫ ·

ωf

]ba

.

One could be tempted to write∫ b

af =

[∫f(x) dx

]ba

;

actually the left term is a real number, while the right term is something we have noteven defined (it could be the set whose only element is

∫ ba f ). In the applications,

however, one often abuses of these notations.

From the known properties of derivatives, one can easily prove the followingproposition.

Proposition 1.16 Let f and g be two functions, primitivable on the interval I, andα ∈ R be arbitrary. Let F and G be two primitives of f and g, respectively. Then


1. f+g is primitivable on I and F +G is one of its primitives; we will briefly write2∫(f + g) =

∫f +

∫g ;

2. αf is primitivable on I and αF is one of its primitives; we will briefly write∫(αf) = α

∫f .

As a consequence of this proposition we have that the set of primitivable functionson I is a vector space.

We conclude this section exhibiting an interesting class of integrable functionswhich are not primitivable. Let the function f : [a, b]→ R be such that the set

E = x ∈ [a, b] : f(x) 6= 0

is finite or countable (for instance, a function which is zero everywhere except at apoint, or the Dirichlet function, defined by f(x) = 1 if x is rational, and f(x) = 0 ifx is irrational).

Let us prove that such a function is integrable, with∫ ba f = 0. Assume for defi-

niteness that E be infinite (the case when E is finite can be treated in an analogousway). Being countable, we can write E = en : n ∈ N. Once ε > 0 has been fixed,we construct a gauge δ on [a, b] in this way: if x 6∈ E, we set δ(x) = 1; if instead fora certain n it is x = en, we set

δ(en) =ε

2n+3|f(en)|.

Let now Π = (x1, [a0, a1]), . . . , (xm, [am−1, am]) be a δ-fine P-partition of [a, b]. Bythe way f is defined, the associated Riemann sum becomes

S([a, b], f,Π) =∑

1≤j≤m :xj∈E

f(xj)(aj − aj−1) .

Now, [aj−1, aj ] ⊆ [xj − δ(xj), xj + δ(xj)], so that aj − aj−1 ≤ 2δ(xj), and if xj is inE it is xj = en, for some n ∈ N. To any such en can however correspond one or twopoints xj , so that we we will have∣∣∣∣∣ ∑

1≤j≤m :xj∈E


∣∣∣∣∣ ≤ 2∞∑n=0

|f(en)|2δ(en) = 4∞∑n=0

ε

2n+3= ε .

This shows that f is integrable on [a, b] and that∫ ba f = 0.

2Here and in the following we use in an intuitive way the algebraic operations involving sets. Tobe precise, the sum of two sets A and B is defined as

A+B = a+ b : a ∈ A, b ∈ B .


Let us see now that, if E is non-empty, then f is not primitivable on [a, b]. Indeed,if it were, its indefinite integral

∫ ·a f should be one of its primitives. Proceeding as

above, one sees that, for every x ∈ [a, b], one has∫ xa f = 0. Then, being the derivative

of a constant function, f should be identically zero, which is false.

1.7 Primitivation by parts and by substitution

We introduce here two methods frequently used for determining the primitives ofcertain functions. The first is known as the method of primitivation by parts.

Proposition 1.17 Let F,G : I → R be two differentiable functions, and let f, g bethe respective derivatives. One has that fG is primitivable on I if and only if suchis Fg, in which case a primitive of fG is obtained subtracting from FG a primitiveof Fg; we will briefly write: ∫

fG = FG−∫Fg .

Proof Being F and G differentiable, such is FG, as well, and we have

(FG)′ = fG+ Fg .

Being (FG)′ primitivable on I with primitive FG, the conclusion follows from Propo-sition 1.16.

Example. We would like to find a primitive of the function h(x) = xex. Define the followingfunctions: f(x) = ex, G(x) = x, and consequently F (x) = ex, g(x) = 1. Applying the formula givenby the above proposition, we have:∫

exx dx = exx−∫ex dx = xex − ex + c ,

where c stands, as usual, for an arbitrary constant.

As an immediate consequence of Proposition 1.17, we have the rule of integra-tion by parts: ∫ b

afG = F (b)G(b)− F (a)G(a)−

∫ b

aFg .

Examples. Applying the formula directly to the function h(x) = xex of the previous example, weobtain ∫ 1

0

exx dx = e1 · 1− e0 · 0−∫ 1

0

ex dx = e− [ex]10 = e− (e1 − e0) = 1 .

Notice that we could attain the same result using the Fundamental Theorem, having already foundthat a primitive of h is given by H(x) = xex − ex :∫ 1

0

exx dx = H(1)−H(0) = (e− e)− (0− 1) = 1 .

1.7. PRIMITIVATION BY PARTS AND BY SUBSTITUTION 17

Let us see some more examples. Let h(x) = sin2 x. With the obvious choice of the functions fand G, we find ∫

sin2 x dx=− cosx sinx+

∫cos2 x dx

=− cosx sinx+

∫(1− sin2 x) dx

= x− cosx sinx−∫

sin2 x dx ,

from which we get ∫sin2 x dx =

1

2(x− cosx sinx) + c .

Consider now the case of the function h(x) = lnx, with x > 0. In order to apply the formula ofprimitivation by parts, we choose the functions f(x) = 1, G(x) = lnx. In this way, we find∫

lnx dx = x lnx−∫x

1

xdx = x lnx−

∫1 dx = x lnx− x+ c .

The second method we want to study is known as the method of primitivationby substitution.

Proposition 1.18 Let ϕ : I → R be a differentiable function and f : ϕ(I) → R bea primitivable function on the interval ϕ(I), with primitive F. Then, the function(f ϕ)ϕ′ is primitivable on I, and one of its primitives is given by F ϕ. We willbriefly write: ∫

(f ϕ)ϕ′ =

(∫f

) ϕ .

Proof The composite function F ϕ is differentiable on I and

(F ϕ)′ = (F ′ ϕ)ϕ′ = (f ϕ)ϕ′ .

It follows that (f ϕ)ϕ′ is primitivable on I, with primitive F ϕ.

As an example, we look for a primitive of the function h(x) = xex2

. Defining ϕ(x) = x2,f(t) = 1

2et (it is advisable to use different letters to indicate the variables of ϕ and f), we have that

h = (f ϕ)ϕ′. Since a primitive of f is seen to be F (t) = 12et, a primitive of h is F ϕ, i.e.∫

xex2

dx = F (ϕ(x)) + c =1

2ex

2

+ c .

The formula of primitivation by substitution is often written in the form∫f(ϕ(x))ϕ′(x) dx =

∫f(t) dt

∣∣∣∣t=ϕ(x)

,


where, if F is a primitive of f, the right term should be read as∫f(t) dt

∣∣∣∣t=ϕ(x)

= F (ϕ(x)) + c ,

where c ∈ R is arbitrary. Formally, there is a “change of variable” t = ϕ(x), and thesymbol dt joins the game to replace ϕ′(x) dx (the Leibniz notation dt

dx = ϕ′(x) maybe used as a mnemonic rule).

Example. In order to find a primitive of the function h(x) = ln xx, we can choose ϕ(x) = lnx, apply

the formula ∫lnx

xdx =

∫t dt

∣∣∣∣t=ln x

,

and thus find 12(lnx)2 + c (in this case, writing t = lnx, one has that the symbol dt replaces 1

xdx).

As a consequence of the above formulas, we have the rule of integration bysubstitution: ∫ b

af(ϕ(x))ϕ′(x) dx =

∫ ϕ(b)

ϕ(a)f(t) dt .

Indeed, if F is a primitive of f on ϕ(I), by the Fundamental Theorem, we have∫ b

a(f ϕ)ϕ′ = (F ϕ)(b)− (F ϕ)(a) = F (ϕ(b))− F (ϕ(a)) =

∫ ϕ(b)

ϕ(a)f.

Example. Taking the function h(x) = xex2

defined above, we have∫ 2

0

xex2

dx =

∫ 4

0

1

2et dt =

1

2[et]40 =

e4 − 1

2.

Clearly, the same result is obtainable directly by the Fundamental Theorem, once we know that a

primitive of h is given by H(x) = 12ex

2

. Indeed, we have∫ 2

0

xex2

dx = H(2)−H(0) =1

2e4 − 1

2e0 =

e4 − 1

2.

In case the function ϕ : I → ϕ(I) be invertible, one can also write∫f(t) dt =

∫f(ϕ(x))ϕ′(x) dx

∣∣∣∣x=ϕ−1(t)

,

with the corresponding formula for the integral:∫ β

αf(t) dt =

∫ ϕ−1(β)

ϕ−1(α)f(ϕ(x))ϕ′(x) dx .

1.7. PRIMITIVATION BY PARTS AND BY SUBSTITUTION 19

Example. Looking for a primitive of f(t) =√

1− t2, with t ∈ ]− 1, 1[ , we may try to consider thefunction ϕ : ]0, π[→]− 1, 1[ defined as ϕ(x) = cosx, so that

f(ϕ(x))ϕ′(x) =√

1− cos2 x (− sinx) = − sin2 x .

As we have already proved, this last function is primitivable, so we can write∫ √1− t2 dt=−

∫sin2 x dx

∣∣x=arccos t

=− 1

2(x− sinx cosx)

∣∣∣∣x=arccos t

+ c

=−1

2

(arccos t− t

√1− t2

)+ c .

We are now in the position to compute primitives and integrals for a large classof functions. Some of these are proposed in the exercises below.

Exercises

1. Making use of the known rules for the computation of the primitives, recover thefollowing formulas: ∫

1

(2 + 3x)7dx = − 1

18(2 + 3x)6+ c ,

∫ √x+ 7 dx =

2

3

√(x+ 7)3 + c ,

∫x2 + 3x− 2√

xdx =

2

5

√x5 + 2

√x3 − 4

√x+ c ,

∫1

√x−√x+ 1

dx = −2

3

(√(x+ 1)3 +

√x3)

+ c ,

∫1

x2 − 5x+ 6dx = ln |x− 3| − ln |x− 2|+ c ,

∫1

x2 + 4x+ 5dx = arctan(x+ 2) + c ,

∫1

sin2 x cos2 xdx = tanx− 1

tanx+ c ,

∫1

coshxdx = 2 arctan(ex) + c ,

∫lnx

xdx =

1

2(lnx)2 + c .


2. Primitivation by parts gives the following:∫x sinx dx = sinx− x cosx+ c ,

∫ √1− x2

x2dx = −

√1− x2

x+ arcsinx+ c ,∫

(lnx)2 dx = x[(lnx)2 − 2 lnx+ 2

]+ c ,∫

arcsinx dx = x arcsinx+√

1− x2 + c .

3. Let f : R → R be a primitivable T -periodic function. Provide a criterion toensure that its primitives are T -periodic, as well.

4. Prove that, if f : R→ R is any primitivable function, then∫ 2π

0f(sinx) cosx dx = 0 ,

∫ 2π

0f(cosx) sinx dx = 0 .

5. Given a primitivable function f : R→ R, prove that:

a) if f is an odd function, then all its primitives are even functions;b) if f is an even function, then one of its primitives is an odd function;

c) if∫ ba f = 0 for every a, b ∈ R, then f is identically equal to zero.

1.8 The Cauchy criterion

We have the following characterization for a function to be integrable.

Theorem 1.19 A function f : I → R is integrable if and only if for every ε > 0there is a gauge δ on I such that, taking two δ-fine P-partitions Π, Π of I, one has

|S(I, f,Π)− S(I, f, Π)| ≤ ε .

Proof Let us see first the necessary condition. Let f be integrable on I with integralA, and fix ε > 0. Then, there is a gauge δ on I such that, for every δ-fine P-partitionΠ of I, it is

|S(I, f,Π)−A| ≤ ε

2.

If Π and Π are two δ-fine P-partitions, we have:

|S(I, f,Π)− S(I, f, Π)| ≤ |S(I, f,Π)−A|+ |A− S(I, f, Π)| ≤ ε

2+ε

2= ε .

1.8. THE CAUCHY CRITERION 21

Let us see now the sufficient condition. Once assumed the stated condition, let uschoose ε = 1 so that we find a gauge δ1 on I such that

|S(I, f,Π)− S(I, f, Π)| ≤ 1 ,

whenever Π and Π are δ1-fine P-partitions of I. Taking ε = 1/2, we can find a gaugeδ2 on I, that we can choose so that δ2(x) ≤ δ1(x) for every x ∈ I, such that

|S(I, f,Π)− S(I, f, Π)| ≤ 1

2,

whenever Π and Π are δ2-fine P-partitions of I. We can continue this way, choosingε = 1/k, with k a positive integer, and find a sequence (δk)k of gauges on I suchthat, for every x ∈ I,

δ1(x) ≥ δ2(x) ≥ . . . ≥ δk(x) ≥ δk+1(x) ≥ . . . ,

and such that

|S(I, f,Π)− S(I, f, Π)| ≤ 1

k,

whenever Π and Π are δk-fine P-partitions of I.

Let us fix, for every k, a δk-fine P-partition Πk of I. We want to show that(S(I, f,Πk))k is a Cauchy sequence of real numbers. Let ε > 0 be given. Let uschoose a positive integer m such that mε ≥ 1. If k1 ≥ m and k2 ≥ m, assuming forinstance k2 ≥ k1, we have

|S(I, f,Πk1)− S(I, f,Πk2)| ≤ 1

k1≤ 1

m≤ ε .

This proves that (S(I, f,Πk))k is a Cauchy sequence. Hence, it has a finite limit,which we denote by A.

Now we show that A is just the integral of f on I. Fix ε > 0; let n be a positiveinteger such that nε ≥ 1, and consider the gauge δ = δn. For every δ-fine P-partitionΠ of I and for every k ≥ n, it is

|S(I, f,Π)− S(I, f,Πk)| ≤1

n≤ ε .

Letting k tend to +∞, we have that S(I, f,Πk) tends to A, and consequently

|S(I, f,Π)−A| ≤ ε .

The proof is thus completed.


1.9 Integrability on sub-intervals

In this section we will see that if a function is integrable on an interval I = [a, b], itis also integrable on any of its sub-intervals. In particular, it is possible to considerits indefinite integral. Moreover, we will see that, if a function is integrable on twocontiguous intervals, it is also integrable on their union. More precisely, we have thefollowing property of additivity on sub-intervals.

Theorem 1.20 Given three points a < c < b, let f : [a, b]→ R be a function. Then,f is integrable on [a, b] if and only if it is integrable both on [a, c] and on [c, b]. Inthis case, ∫ b

af =

∫ c

af +

∫ b

cf .

f(x)

a c b x

Proof First assume f to be integrable on [a, b]. Let us choose for example the firstsub-interval, [a, c], and prove that f is integrable on it, using the Cauchy criterion.Fix ε > 0; being f integrable on [a, b], it verifies the Cauchy condition on [a, b], andhence there is a gauge δ on [a, b] such that

|S([a, b], f,Π)− S([a, b], f, Π)| ≤ ε ,

for every two δ-fine P-partitions Π, Π of [a, b]. The restrictions of δ to [a, c] and [c, b]are two gauges δ1 and δ2 on the respective sub-intervals. Let now Π1 and Π1 be twoδ1-fine P-partitions of [a, c]. Let us fix a δ2-fine P-partition Π2 of [c, b] and considerthe P-partition Π of [a, b] made by Π1 ∪Π2, and the P-partition Π of [a, b] made byΠ1 ∪Π2. It is clear that both Π and Π are δ-fine. Moreover, we have

|S([a, c], f,Π1)− S([a, c], f, Π1)| = |S([a, b], f,Π)− S([a, b], f, Π)| ≤ ε ;

the Cauchy condition thus holds, so that f is integrable on [a, c]. Analogously it canbe proved that f is integrable on [c, b].

Suppose now that f be integrable on [a, c] and on [c, b], and let us prove then

that f is integrable on [a, b] with integral∫ ca f +

∫ bc f. Once ε > 0 is fixed, there are

1.9. INTEGRABILITY ON SUB-INTERVALS 23

a gauge δ1 on [a, c] and a gauge δ2 on [c, b] such that, for every δ1-fine P-partitionΠ1 of [a, c] it is ∣∣∣∣S([a, c], f,Π1)−

∫ c

af

∣∣∣∣ ≤ ε

2,

and for every δ2-fine P-partition Π2 of [c, b] it is∣∣∣∣S([c, b], f,Π2)−∫ b

cf

∣∣∣∣ ≤ ε

2.

We define now a gauge δ on [a, b] in this way:

δ(x) =

min

δ1(x), c−x2

if a ≤ x < c

minδ1(c), δ2(c) if x = c

minδ2(x), x−c2

if c < x ≤ b .

Let nowΠ = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am])

be a δ-fine P-partition of [a, b]. Notice that, because of the particular choice of thegauge δ, there must be a certain for which x = c. Hence, we have

S([a, b], f,Π) = f(x1)(a1 − a0) + . . .+ f(x−1)(a−1 − a−2) +

+f(c)(c− a−1) + f(c)(a − c) +

+f(x+1)(a+1 − a) + . . .+ f(xm)(am − am−1) .

Let us set

Π1 = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (x−1, [a−2, a−1]), (c, [a−1, c])

andΠ2 = (c, [c, a]), (x+1, [a, a+1]), . . . , (xm, [am−1, am])

(but in case a−1 or a coincide with c we will have to take away an element). ThenΠ1 is a δ1-fine P-partition of [a, c] and Π2 is a δ2-fine P-partition of [c, b], and wehave

S([a, b], f,Π) = S([a, c], f,Π1) + S([c, b], f,Π2) .

Consequently,∣∣∣∣S([a, b], f,Π)−(∫ c

af +

∫ b

cf

)∣∣∣∣ ≤≤∣∣∣∣S([a, c], f,Π1)−

∫ c

af

∣∣∣∣+

∣∣∣∣S([c, b], f,Π2)−∫ b

cf

∣∣∣∣≤ ε

2+ε

2= ε ,

which completes the proof.


Example. Consider the function f : [0, 2]→ R defined by

f(x) =

2 if x ∈ [0, 1] ,

3 if x ∈ ]1, 2] .

Being f constant on [0, 1], it is integrable there, and∫ 1

0f = 2. Moreover, on the interval [1, 2] the

function f differs from a constant only in one point: we have that f(x)− 3 is zero except for x = 1.For what have been proved somewhat before, the function f − 3 is integrable on [1, 2] with zerointegral and so, being f = (f − 3) + 3, even f is integrable and

∫ 2

1f = 3. In conclusion,∫ 2

0

f(x) dx =

∫ 1

0

f(x) dx+

∫ 2

1

f(x) dx = 2 + 3 = 5 .

It is easy to see from the theorem just proved above that if a function is integrableon an interval I, it still is on any sub-interval of I. Moreover, we have the following

Corollary 1.21 If f : I → R is integrable, for any three arbitrarily chosen pointsu, v, w in I one has ∫ w

uf =

∫ v

uf +

∫ w

vf .

Proof The case u < v < w follows immediately from the previous theorem. Theother possible cases are easily obtained using the conventions on the integrals withexchanged or equal extrema.

1.10 R-integrable functions and continuous functions

Let us introduce an important class of integrable functions.

Definition 1.22 We say that an integrable function f : I → R is R-integrable (orintegrable according to Riemann), if among all possible gauges δ : I → R which verifythe definition of integrability it is always possible to choose one which is constanton I.

It is immediate to see, repeating the proofs, that the set of R-integrable functionsis a vector subspace of the space of integrable functions.3 Moreover, the followingCauchy criterion holds for R-integrable functions, whenever one considers only con-stant gauges.

Theorem 1.23 A function f : I → R is R-integrable if and only if for every ε > 0there is a constant δ > 0 such that, taken two δ-fine P-partitions Π, Π of I, one has

|S(I, f,Π)− S(I, f, Π)| ≤ ε .3It is also possible to show that any R-integrable function is L-integrable, but we prefer not

entering into details.

1.10. R-INTEGRABLE FUNCTIONS AND CONTINUOUS FUNCTIONS 25

The Cauchy criterion permits to prove the integrability of continuous functions.To simplify the expressions to come, we will denote by µ(K) the length of a boundedinterval K. In particular,

µ([a, b]) = b− a .It will be useful, moreover, to make the convention that the length of the empty setis equal to zero.

Theorem 1.24 Every continuous function f : I → R is R-integrable.

Proof Fix ε > 0. Being f continuous on a compact interval, it is uniformly contin-uous there, so that there is a δ > 0 such that, for x and x′ in I,

|x− x′| ≤ 2δ ⇒ |f(x)− f(x′)| ≤ ε

b− a.

We will verify the Cauchy condition for the R-integrability by considering the con-stant gauge δ. Let

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am])

andΠ = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am])

be two δ-fine P-partitions of I. Let us define the intervals (perhaps empty or reducedto a single point)

Ij,k = [aj−1, aj ] ∩ [ak−1, ak] .

Then, we have

aj − aj−1 =m∑k=1

µ(Ij,k) , ak − ak−1 =m∑j=1

µ(Ij,k) ,

and, if Ij,k is non-empty, |xj − xk| ≤ 2δ. Hence,

|S(I, f,Π)− S(I, f, Π)| =

∣∣∣∣∣∣m∑j=1

m∑k=1

f(xj)µ(Ij,k)−m∑k=1

m∑j=1

f(xk)µ(Ij,k)

∣∣∣∣∣∣=

∣∣∣∣∣∣m∑j=1

m∑k=1

[f(xj)− f(xk)]µ(Ij,k)

∣∣∣∣∣∣≤

m∑j=1

m∑k=1

|f(xj)− f(xk)|µ(Ij,k)

≤m∑j=1

m∑k=1

ε

b− aµ(Ij,k) = ε .

Therefore, the Cauchy condition holds true, and the proof is completed.


Concerning the continuous functions, even the following holds.

Theorem 1.25 Every continuous function f : [a, b]→ R is primitivable.

Proof Being continuous, f is integrable on every sub-interval of [a, b], so that wecan consider the function

∫ ·a f, indefinite integral of f. Let us prove that

∫ ·a f is a

primitive of f, i.e., that taken a point x0 in [a, b], the derivative of∫ ·a f in x0 coincides

with f(x0). We first consider the case when x0 ∈ ]a, b[ . We want to prove that

limh→0

1

h

(∫ x0+h

af −

∫ x0

af

)= f(x0) .

f (x)

a x0 x0+ h b x

Equivalently, since

1

h

(∫ x0+h

af −

∫ x0

af

)− f(x0) =

1

h

∫ x0+h

x0

(f(x)− f(x0)) dx ,

we will show that

limh→0

1

h

∫ x0+h

x0

(f(x)− f(x0)) dx = 0 .

Fix ε > 0. Being f continuous in x0, there is a δ > 0 such that, for every x ∈ [a, b]satisfying |x− x0| ≤ δ, one has |f(x)− f(x0)| ≤ ε. Taking h such that 0 < |h| ≤ δ,we distinguish two cases. If 0 < h ≤ δ, then∣∣∣∣1h

∫ x0+h

x0

(f(x)− f(x0)) dx

∣∣∣∣ ≤ 1

h

∫ x0+h

x0

|f(x)− f(x0)| dx ≤ 1

h

∫ x0+h

x0

ε dx = ε ;

on the other hand, if −δ ≤ h < 0, we have∣∣∣∣1h∫ x0+h

x0

(f(x)− f(x0)) dx

∣∣∣∣ ≤ 1

−h

∫ x0

x0+h|f(x)− f(x0)| dx ≤ 1

−h

∫ x0

x0+hε dx = ε ,

and the proof is competed when x0 ∈ ]a, b[ . In case x0 = a or x0 = b, we proceedanalogously, considering the right or the left derivative, respectively.

1.10. R-INTEGRABLE FUNCTIONS AND CONTINUOUS FUNCTIONS 27

Notice that it is not always possible to find an elementary expression for theprimitive of a continuous function. As an example, the function f(x) = ex

2is

primitivable, being continuous, but it can be proved that there is no elementaryformula defining any of its primitives.4

Let us now prove that the Dirichlet function f is not R-integrable on any interval[a, b] (remember that f is 1 on the rationals and 0 on the irrationals). We will showthat the Cauchy condition is not verified. Take δ > 0 and let a = a0 < a1 < . . . <am = b be such that, for every j = 1, . . . ,m, one has aj − aj−1 ≤ δ. In every interval[aj−1, aj ] we can choose a rational number xj and an irrational number xj . The twoP-partitions

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am]) ,

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am]) ,

are δ-fine, and by the very definition of f we have

S([a, b], f, Π)− S([a, b], f,Π) =

m∑j=1

[f(xj)− f(xj)](aj − aj−1)

=m∑j=1

(aj − aj−1) = b− a .

Since δ > 0 has been taken arbitrarily, the Cauchy condition for R-integrability doesnot hold, so that f cannot be R-integrable on [a, b].

Exercises

1. Recalling that

sin(2θ) =2 tan θ

1 + tan2 θ, cos(2θ) =

1− tan2 θ

1 + tan2 θ,

recover the following formulas:∫1

sinxdx = ln

∣∣∣tanx

2

∣∣∣+ c ,

∫1

cosxdx = ln

∣∣∣∣1 + tan x2

1− tan x2

∣∣∣∣+ c .

Alternatively, ∫1

cosxdx =

1

2ln

∣∣∣∣1 + sinx

1− sinx

∣∣∣∣+ c .

4By “elementary formula” I mean here an analytic formula where only polynomials, exponentials,logarithms and trigonometric functions appear.


2. Show that, if f : [0, 1]→ R is defined as

f(x) =

1 if x /∈ Q ,3 if x ∈ Q ,

then∫ 1

0 f(x) dx = 1.

3. Let f : R→ R be continuous and odd. Then,∫ c−c f = 0, for every c ∈ R.

4. Prove that, if f : [a, b]→ R is monotone, then it is R-integrable.

5. Show that every R-integrable function f : [a, b]→ R is necessarily bounded.

1.11 The Saks–Henstock theorem

Let us go back to analyze the definition of integrability for a function f : I → R.The function f is integrable on I with integral

∫I f if, for every ε > 0, there is a

gauge δ such that, for every δ-fine P-partition

Π = (x1, [a0, a1]), (x2, [a1, a2]), . . . , (xm, [am−1, am])

of I, the following holds: ∣∣∣∣S(I, f,Π)−∫If

∣∣∣∣ ≤ ε .In this situation, since

S(I, f,Π) =

m∑j=1

f(xj)(aj − aj−1) ,

∫If =

m∑j=1

∫ aj

aj−1

f ,

we have ∣∣∣∣∣∣m∑j=1

(f(xj)(aj − aj−1)−

∫ aj

aj−1

f

)∣∣∣∣∣∣ ≤ ε .This fact tells us that the sum of all “errors” (f(xj)(aj−aj−1)−

∫ ajaj−1

f) is arbitrarily

small, provided that the P-partition be sufficiently fine. Notice that those “errors”may be either positive or negative, so that in the sum they could compensate withone another. The following Saks–Henstock’s theorem tells us that even the sumof all absolute values of those “errors” can be made arbitrarily small.

Theorem 1.26 Let f : I → R be an integrable function and let δ be a gauge on Isuch that, for every δ-fine P-partition Π of I, it happens that |S(I, f,Π)−

∫I f | ≤ ε.

Then, for such P-partitions Π = (x1, [a0, a1]), . . . , (xm, [am−1, am]) we also have

m∑j=1

∣∣∣∣∣f(xj)(aj − aj−1)−∫ aj

aj−1

f

∣∣∣∣∣ ≤ 4ε .

1.11. THE SAKS–HENSTOCK THEOREM 29

Proof We consider separately in the sum the positive and the negative terms. Let usprove that the sum of the positive terms is less than or equal to 2ε. In an analogousway one can proceed for the negative terms. Rearranging the terms in the sum, wecan assume that the positive ones be the first q terms (f(xj)(aj − aj−1)−

∫ ajaj−1

f),

with j = 1, . . . , q, i.e.,

f(x1)(a1 − a0)−∫ a1

a0

f , . . . , f(xq)(aq − aq−1)−∫ aq

aq−1

f .

Consider the remaining m− q intervals [ak−1, ak], with k = q + 1, . . . ,m :

[aq, aq+1] , . . . , [am−1, am] .

Being f integrable on these intervals, there exist some gauges δk on [ak−1, ak], re-spectively, which we can choose such that δk(x) ≤ δ(x) for every x ∈ [ak−1, ak], forwhich ∣∣∣∣∣S([ak−1, ak], f,Πk)−

∫ ak

ak−1

f

∣∣∣∣∣ ≤ ε

m− q,

for every δk-fine P-partition Πk of [ak−1, ak]. Consequently, the family Π made bythe couples (x1, [a0, a1]), . . . , (xq, [aq−1, aq]) and by the elements of the families Πk,with k varying from q + 1 to m, is a δ-fine P-partition of I such that

S(I, f, Π) =

q∑j=1

f(xj)(aj − aj−1) +m∑

k=q+1

S([ak−1, ak], f,Πk) .

Then, we have:

q∑j=1

(f(xj)(aj − aj−1)−

∫ aj

aj−1

f

)=

q∑j=1

f(xj)(aj − aj−1)−q∑j=1

∫ aj

aj−1

f

=

(S(I, f, Π)−

m∑k=q+1

S([ak−1, ak], f,Πk)

)−

(∫If −

m∑k=q+1

∫ ak

ak−1

f

)

≤∣∣∣∣S(I, f, Π)−

∫If

∣∣∣∣+

m∑k=q+1

∣∣∣∣∣S([ak−1, ak], f,Πk)−∫ ak

ak−1

f

∣∣∣∣∣≤ ε+ (m− q) ε

m− q= 2ε .

Proceeding similarly for the negative terms, the conclusion follows.

The following corollary will be useful in the next section to study the integrabilityof the absolute value of an integrable function.


Corollary 1.27 Let f : I → R be an integrable function and let δ be a gauge on Isuch that, for every δ-fine P-partition Π of I, it happens that |S(I, f,Π)−

∫I f | ≤ ε.

Then, for such P-partitions Π = (x1, [a0, a1]), . . . , (xm, [am−1, am]) we also have∣∣∣∣∣∣S(I, |f |,Π)−m∑j=1

∣∣∣∣∣∫ aj

aj−1

f

∣∣∣∣∣∣∣∣∣∣∣ ≤ 4ε .

Proof Using the well known inequalities for the absolute value, we have:∣∣∣∣∣∣S(I, |f |,Π)−m∑j=1

∣∣∣∣∣∫ aj

aj−1

f

∣∣∣∣∣∣∣∣∣∣∣=∣∣∣∣∣∣m∑j=1

[|f(xj)|(aj − aj−1)−

∣∣∣∣∣∫ aj

aj−1

f

∣∣∣∣∣]∣∣∣∣∣∣

≤m∑j=1

∣∣∣∣∣|f(xj)(aj − aj−1)| −

∣∣∣∣∣∫ aj

aj−1

f

∣∣∣∣∣∣∣∣∣∣

≤m∑j=1

∣∣∣∣∣f(xj)(aj − aj−1)−∫ aj

aj−1

f

∣∣∣∣∣ ≤ 4ε .


In the sequel it will be useful to consider even the so-called sub-P-partitions of theinterval I. A sub-P-partition is a set of the type Ξ = (ξj , [αj , βj ]) : j = 1, . . . ,m,where the intervals [αj , βj ] are non-overlapping, but not necessarily contiguous, andξj ∈ [αj , βj ] for every j = 1, . . . ,m. Using the Cousin’s theorem, it is easy to seethat every sub-P-partition can be extended to a P-partition of I.

For a sub-P-partition Ξ, it is still meaningful to consider the associated Riemannsum:

S(I, f,Ξ) =m∑j=1

f(ξj)(βj − αj) .

Moreover, given a gauge δ on I, the sub-P-partition Ξ is δ-fine if, for every j, onehas

ξj − αj ≤ δ(ξj) e βj − ξj ≤ δ(ξj) .

The Saks–Henstock’s theorem can then be generalized to the sub-P-partitions, sim-ply considering the fact that every sub-P-partition is a subset of a P-partition. Wecan thus obtain the following equivalent statement of the Saks–Henstock’stheorem.

Theorem 1.28 Let f : I → R be an integrable function and let δ be a gauge on Isuch that, for every Π of I, it happens that |S(I, f,Π) −

∫I f | ≤ ε. Then, for every

1.12. L-INTEGRABLE FUNCTIONS 31

δ-fine sub-P-partition Ξ = (ξj , [αj , βj ]) : j = 1, . . . ,m of I, we have

m∑j=1

∣∣∣∣∣f(ξj)(βj − αj)−∫ βj

αj

f

∣∣∣∣∣ ≤ 4ε .

Notice that, as a consequence of this last statement, for such sub-P-partitions Ξwe have, in particular, ∣∣∣∣∣∣S(I, f,Ξ)−

m∑j=1

∫ βj

αj

f

∣∣∣∣∣∣ ≤ 4ε .

1.12 L-integrable functions

In this section, we introduce another important class of integrable functions on theinterval I = [a, b].

Definition 1.29 We say that an integrable function f : I → R is L-integrable (orintegrable according to Lebesgue), if even |f | happens to be integrable on I.

It is clear that every positive integrable function is L-integrable. Moreover, everycontinuous function on [a, b] is L-integrable there, since |f | is still continuous. Wehave the following characterization of L-integrability.

Proposition 1.30 Let f : I → R be an integrable function, and consider the set Sof all real numbers

q∑i=1

∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ ,obtained choosing c0, c1, . . . , cq in I in such a way that a = c0 < c1 < . . . < cq = b.The function f is L-integrable on I if and only if S is bounded above. In that case,we have ∫

I|f | = supS .

Proof Assume first f to be L-integrable on I. If a = c0 < c1 < . . . < cq = b, then fand |f | are integrable on every sub-interval [ci−1, ci], and we have

q∑i=1

∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ ≤q∑i=1

∫ ci

ci−1

|f | =∫I|f | .

Consequently, the set S is bounded above: supS ≤∫I |f | < +∞.


On the other hand, assume now S to be bounded above and let us prove that inthat case |f | is integrable on I and

∫I |f | = supS. Fix ε > 0. Let δ1 be a gauge such

that, for every δ1-fine P-partition Π of I, one has

|S(I, f,Π)−∫If | ≤ ε

8.

Letting A = supS, by the properties of the sup there surely are a = c0 < c1 < . . . <cq = b such that

A− ε

2≤

q∑i=1

∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ ≤ A .We construct the gauge δ2 in such a way that, for every x ∈ I, it has to be that[x− δ2(x), x+ δ2(x)] meets only those intervals [ci−1, ci] to which x belongs. In thisway,• if x belongs to the interior of one of the intervals [ci−1, ci], we have that [x −

δ2(x), x+ δ2(x)] is contained in ]ci−1, ci[ ;• if x coincides with one of the ci in the interior of [a, b], then [x−δ2(x), x+δ2(x)]

is contained in ]ci−1, ci+1[ ;• if x = a, then [x, x+ δ2(x)] is contained in [a, c1[ ;• if x = b, then [x− δ2(x), x] is contained in ]cq−1, b].

Set, for every x ∈ I, δ(x) = minδ1(x), δ2(x). Once taken a δ-fine P-partitionΠ = (x1, [a0, a1]), . . . , (xm, [am−1, am]) of I, consider the intervals (possibly emptyor reduced to a point)

Ij,i = [aj−1, aj ] ∩ [ci−1, ci] .

The choice of the gauge δ2 yields that, if Ij,i has a positive length, then xj ∈ Ij,i.Indeed, if xj 6∈ [ci−1, ci], then

[aj−1, aj ] ∩ [ci−1, ci] ⊆ [xj − δ2(xj), xj + δ2(xj)] ∩ [ci−1, ci] = Ø .

Therefore, taking those Ij,i, the set

Π = (xj , Ij,i) : j = 1, . . . ,m , i = 1, . . . , q , µ(Ij,i) > 0

is a δ-fine P-partition of I, and we have:

S(I, |f |,Π) =m∑j=1

|f(xj)|(aj − aj−1) =m∑j=1

q∑i=1

|f(xj)|µ(Ij,i) = S(I, |f |, Π) .

Moreover,

A− ε

2≤

q∑i=1

∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ =

q∑i=1

∣∣∣∣∣∣m∑j=1

∫Ij,i

f

∣∣∣∣∣∣ ≤q∑i=1

m∑j=1

∣∣∣∣∣∫Ij,i

f

∣∣∣∣∣ ≤ A ,

1.12. L-INTEGRABLE FUNCTIONS 33

and by Corollary 1.27,∣∣∣∣∣∣S(I, |f |, Π)−q∑i=1

m∑j=1

∣∣∣∣∣∫Ij,i

f

∣∣∣∣∣∣∣∣∣∣∣ ≤ 4

ε

8=ε

2.

Consequently, we have

|S(I, |f |,Π)−A| = |S(I, |f |, Π)−A|

≤

∣∣∣∣∣∣S(I, |f |, Π)−q∑i=1

m∑j=1

∣∣∣∣∣∫Ij,i

f

∣∣∣∣∣∣∣∣∣∣∣+

∣∣∣∣∣∣q∑i=1

m∑j=1

∣∣∣∣∣∫Ij,i

f

∣∣∣∣∣−A∣∣∣∣∣∣

≤ ε

2+ε

2= ε ,

which is what was to be proved.

We have a series of corollaries.

Corollary 1.31 Let f, g : I → R be two integrable functions such that, for everyx ∈ I,

|f(x)| ≤ g(x) ;

then f is L-integrable on I.

Proof Take c0, c1, . . . , cq in I so that a = c0 < c1 < . . . < cq = b. Being −g(x) ≤f(x) ≤ g(x) for every x ∈ I, we have that

−∫ ci

ci−1

g ≤∫ ci

ci−1

f ≤∫ ci

ci−1

g ,

i.e. ∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ ≤∫ ci

ci−1

g ,

for every 1 ≤ i ≤ q. Hence,

q∑i=1

∣∣∣∣∣∫ ci

ci−1

f

∣∣∣∣∣ ≤q∑i=1

∫ ci

ci−1

g =

∫Ig .

Then, the set S is bounded above by∫I g, so that f is L-integrable on I.

Corollary 1.32 Let f, g : I → R be two L-integrable functions and α ∈ R be aconstant. Then f + g and αf are L-integrable on I.


Proof By the assumption, f, |f | and g, |g| are integrable on I. Then, such are alsof + g, |f |+ |g|, αf, and |α||f |. On the other hand, for every x ∈ I, it is

|(f + g)(x)| ≤ |f(x)|+ |g(x)| ,

|αf(x)| ≤ |α| |f(x)| .

Corollary 1.31 then guarantees that f + g and αf are L-integrable on I.

We have thus proved that the L-integrable functions make up a vector subspaceof the space of integrable functions.

Corollary 1.33 Let f1, f2 : I → R be two L-integrable functions. Then minf1, f2and maxf1, f2 are L-integrable on I.

Proof It follows immediately from the formulas

minf1, f2 =1

2(f1 + f2 − |f1 − f2|) ,

maxf1, f2 =1

2(f1 + f2 + |f1 − f2|)

and from Corollary 1.32.

Corollary 1.34 A function f : I → R is L-integrable if and only if both its positivepart f+ = maxf, 0 and its negative part f− = max−f, 0 are integrable on I. Inthat case,

∫I f =

∫I f

+ −∫I f−.

Proof It follows immediately from Corollary 1.33 and the formulas f = f+ − f−,|f | = f+ + f−.

We want to see now an example of an integrable function which is not L-integrable. Let f : [0, 1]→ R be defined by

f(x) =1

xsin

1

x2,

if x 6= 0, and f(0) = 0. Let us define the two auxiliary functions g : [0, 1] → R andh : [0, 1]→ R such that, if x 6= 0,

g(x) =1

xsin

1

x2+ x cos

1

x2, h(x) = −x cos

1

x2,

and g(0) = h(0) = 0. It is easily seen that g is primitivable on [0, 1] and that one ofits primitives G : [0, 1]→ R is given by

G(x) =x2

2cos

1

x2,

1.13. THE MONOTONE CONVERGENCE THEOREM 35

if x 6= 0, and G(0) = 0. Moreover, h is continuous on [0, 1], hence it is primitivablethere, too. Hence, even the function f = g + h is primitivable on [0, 1]. By theFundamental Theorem, f is then integrable on [0, 1]. We will show now that |f | isnot integrable on [0, 1]. Consider the intervals [((k+ 1)π)−1/2, (kπ)−1/2], with k ≥ 1.The function |f | is continuous on these intervals, hence it is primitivable there. Bythe substitution y = 1/x2, we obtain∫ (kπ)−1/2

((k+1)π)−1/2

1

x

∣∣∣∣sin 1

x2

∣∣∣∣ dx =

∫ (k+1)π

kπ

1

2y| sin y| dy .

On the other hand,∫ (k+1)π

kπ

1

2y| sin y| dy ≥ 1

2(k + 1)π

∫ (k+1)π

kπ| sin y| dy =

1

(k + 1)π.

If |f | were integrable on [0, 1], we would have that, for every n ≥ 1,∫ 1

0|f |=

∫ ((n+1)π)−1/2

0|f |+

n∑k=1

∫ (kπ)−1/2

((k+1)π)−1/2

|f |+∫ 1

π−1/2

|f |

≥n∑k=1

∫ (kπ)−1/2

((k+1)π)−1/2

|f |

≥n∑k=1

1

(k + 1)π,

which is impossible, since the series∑∞

k=11

(k+1)π diverges. Hence, f is not L-

integrable on [0, 1].

1.13 The monotone convergence theorem

In this section and in the next one, as well, we will consider the situation where asequence of integrable functions (fk)k converges pointwise to a function f : for everyx ∈ I,

limk→∞

fk(x) = f(x) .

The question is whether ∫If = lim

k→∞

∫Ifk ,

i.e., whether the following formula holds:∫I

(limk→∞

fk(x))dx = lim

k→∞

∫Ifk(x) dx .


Example. Let us first show that in some cases the answer could be in the negative. Consider thefunctions fk : [0, π]→ R, with k = 1, 2, . . ., defined by

fk(x) =

k sin(kx) if x ∈ [0, π

k] ,

0 otherwise.

For every x ∈ [0, π], we have limk→∞ fk(x) = 0, while∫ π

0

fk(x) dx =

∫ π/k

0

k sin(kx) dx =

∫ π

0

sin(t) dt = 2 .

We will see now that the formula holds true if the sequence of functions ismonotone, or bounded in some way. Let us start with the following result, knownas the monotone convergence theorem, due to B. Levi.

Theorem 1.35 We are given a function f : I → R and a sequence of functionsfk : I → R, with k ∈ N, verifying the following conditions:

1. the sequence (fk)k converges pointwise to f ;

2. the sequence (fk)k is monotone;

3. each function fk is integrable on I;

4. the real sequence (∫I fk)k has a finite limit.

Then, f is integrable on I, and ∫If = lim

k→∞

∫Ifk .

Proof We assume for definiteness that the sequence (fk)k be increasing on I; there-fore, we have

fk(x) ≤ fk+1(x) ≤ f(x) ,

for every k ∈ N and every x ∈ I. Let us set

A = limk→∞

∫Ifk .

We will prove that f is integrable on I and that A is its integral. Fix ε > 0. Beingevery fk integrable on I, there are some gauges δ∗k on I such that, if Πk is a δ∗k-fineP-partition of I, ∣∣∣∣S(I, fk,Πk)−

∫Ifk

∣∣∣∣ ≤ ε

3 · 2k+3.

Moreover, there is a k ∈ N such that, for every k ≥ k, it is

0 ≤ A−∫Ifk ≤

ε

3,

1.13. THE MONOTONE CONVERGENCE THEOREM 37

and since the sequence (fk)k converges pointwise on I to f, for every x ∈ I there isa natural number n(x) ≥ k, such that, for every k ≥ n(x), one has

|fk(x)− f(x)| ≤ ε

3(b− a).

Let us define the gauge δ in the following way: for every x ∈ I,

δ(x) = δ∗n(x)(x) .

Let now Π = (x1, [a0, a1]), . . . , (xm, [am−1, am]) be a δ-fine P-partition of I. Wehave:

|S(I, f,Π) −A| =

∣∣∣∣∣∣m∑j=1

f(xj)(aj − aj−1)−A

∣∣∣∣∣∣≤

∣∣∣∣∣∣m∑j=1

[f(xj)− fn(xj)(xj)](aj − aj−1)

∣∣∣∣∣∣+

+

∣∣∣∣∣∣m∑j=1

[fn(xj)(xj)(aj − aj−1)−

∫ aj

aj−1

fn(xj)

]∣∣∣∣∣∣+

∣∣∣∣∣∣m∑j=1

∫ aj

aj−1

fn(xj) −A

∣∣∣∣∣∣ .Estimation of the first term gives∣∣∣∣∣∣

m∑j=1

[f(xj)− fn(xj)(xj)](aj − aj−1)

∣∣∣∣∣∣ ≤m∑j=1

|f(xj)− fn(xj)(xj)|(aj − aj−1)

≤m∑j=1

ε

3(b− a)(aj − aj−1) =

ε

3.

In order to estimate the second term, set

r = min1≤j≤m

n(xj) , s = max1≤j≤m

n(xj) ,

and notice that, putting together the terms whose indices n(xj) coincide with a samevalue k, by the second statement of Saks–Henstock’s theorem (Theorem 1.28), weobtain ∣∣∣∣∣∣

m∑j=1

[fn(xj)(xj)(aj − aj−1)−

∫ aj

aj−1

fn(xj)

]∣∣∣∣∣∣ =


=

∣∣∣∣∣∣s∑

k=r

∑1≤j≤m : n(xj)=k

[fk(xj)(aj − aj−1)−

∫ aj

aj−1

fk

]∣∣∣∣∣∣

≤s∑

k=r

∑1≤j≤m : n(xj)=k

∣∣∣∣∣fk(xj)(aj − aj−1)−∫ aj

aj−1

fk

∣∣∣∣∣≤

s∑k=r

4ε

3 · 2k+3≤ ε

3.

Concerning the third term, since r ≥ k, using the monotonicity of the sequence (fk)kwe have

0 ≤ A−∫Ifs = A−

m∑j=1

∫ aj

aj−1

fs ≤

≤ A−m∑j=1

∫ aj

aj−1

fn(xj) ≤

≤ A−m∑j=1

∫ aj

aj−1

fr = A−∫Ifr ≤

ε

3,

from which ∣∣∣∣∣∣m∑j=1

∫ aj

aj−1

fn(xj) −A

∣∣∣∣∣∣ ≤ ε

3.

Hence,

|S(I, f,Π)−A| ≤ ε

3+ε

3+ε

3= ε ,


As an immediate consequence of the monotone convergence theorem, we havethe analogous statement for a series of functions.

Corollary 1.36 We are given a function f : I → R and a sequence of functionsfk : I → R, with k ∈ N, verifying the following conditions:

1. the series∑

k fk converges pointwise to f ;

2. for every k ∈ N and every x ∈ I, it is fk(x) ≥ 0;


4. the series∑

k(∫I fk) converges.

Then, f is integrable on I, and ∫If =

∞∑k=0

∫Ifk .

1.14. THE DOMINATED CONVERGENCE THEOREM 39

Example. Consider the Taylor series associated to the function f(x) = ex2

,

ex2

=

∞∑k=0

x2k

k!.

The functions fk(x) = x2k

k!satisfy the assumptions 1 and 2 of the corollary. Moreover, they are

integrable on I = [a, b] and∫ b

a

fk(x) dx =

[x2k+1

(2k + 1)k!

]ba

=b2k+1 − a2k+1

(2k + 1)k!,

so that it can be seen that the series∑k(∫Ifk) converges. It is then possible to apply the corollary,

thus obtaining ∫ b

a

ex2

dx =

∞∑k=0

b2k+1 − a2k+1

(2k + 1)k!.

In particular, considering the indefinite integral∫ ·0f, we find an expression for the primitives of ex

2

,i.e., ∫

ex2

dx =

∞∑k=0

x2k+1

(2k + 1)k!+ c .

1.14 The dominated convergence theorem

We start by proving the following preliminary result.

Lemma 1.37 Let f1, f2, . . . , fn : I → R be integrable functions. If there exists anintegrable function g : I → R such that, for every x ∈ I and 1 ≤ k ≤ n it happensthat

g(x) ≤ fk(x) ,

then minf1, f2, . . . , fn and maxf1, f2, . . . , fn are integrable on I.

Proof Consider the case n = 2. The functions f1−g and f2−g, being integrable andnon-negative, are L-integrable. Hence, minf1 − g, f2 − g and maxf1 − g, f2 − gare L-integrable, by Corollary 1.33. The conclusion then follows from the fact that

minf1, f2 = minf1 − g, f2 − g+ g ,

maxf1, f2 = maxf1 − g, f2 − g+ g .

The general case can be easily obtained by induction.

We are now ready to state and prove the following important result due to H.Lebesgue, known as the dominated convergence theorem.

Theorem 1.38 We are given a function f : I → R and a sequence of functionsfk : I → R, with k ∈ N, verifying the following conditions:


1. the sequence (fk)k converges pointwise to f ;


3. there are two integrable functions g, h : I → R for which

g(x) ≤ fk(x) ≤ h(x) ,

for every k ∈ N and x ∈ I.Then, the sequence (

∫I fk)k has a finite limit, f is integrable on I, and∫

If = lim

k→∞

∫Ifk .

Proof For any couple of natural numbers k, `, define the functions

φk,` = minfk, fk+1, . . . , fk+` , Φk,` = maxfk, fk+1, . . . , fk+` .

By the above proved lemma, all φk,` and Φk,` are integrable on I. Moreover, for anyfixed k, the sequence (φk,`)` is decreasing and bounded below by g, and the sequence(Φk,`)` is increasing and bounded above by h. hence, these sequences converge totwo functions φk and Φk, respectively:

lim`→∞

φk,` = φk = inffk, fk+1, . . . ,

lim`→∞

Φk,` = Φk = supfk, fk+1, . . . .

Furthermore, the sequence (∫I φk,`)` is decreasing and bounded below by

∫I g, while

the sequence (∫I Φk,`)` is increasing and bounded above by

∫I h. The monotone

convergence theorem then guarantees that the functions φk and Φk are integrableon I.

Now, the sequence (φk)k is increasing, and the sequence (Φk)k is decreasing; aslimk→∞ fk = f, we must have

limk→∞

φk = lim infk→∞

fk = f ,

limk→∞

Φk = lim supk→∞

fk = f .

Moreover, the sequence (∫I φk)k is increasing and bounded above by

∫I h, while the

sequence (∫I Φk)k is decreasing and bounded below by

∫I g. We can then apply again

the monotone convergence theorem, from which we deduce that f is integrable on Iand ∫

If = lim

k→∞

∫Iφk = lim

k→∞

∫I

Φk .

Being φk ≤ fk ≤ Φk, we have∫I φk ≤

∫I fk ≤

∫I Φk, and the conclusion follows.

1.14. THE DOMINATED CONVERGENCE THEOREM 41

Example. Consider, for k ≥ 1, the functions fk : [0, 3]→ R defined by fk(x) = arctan(kx− k2

k+1

).

We have the following situation:

limk→∞

fk(x) =

−π

2if x ∈ [0, 1[ ,

0 if x = 1 ,

π

2if x ∈ ]1, 3] .

Moreover,

|fk(x)| ≤ π

2,

for every k ∈ N and x ∈ [0, 3]. The assumptions of the theorem are then satisfied, taking the twoconstant functions g(x) = −π

2, h(x) = π

2. We can then conclude that

limk→∞

∫ 3

0

arctan

(kx− k2

k + 1

)dx = −π

2+ 2

π

2=π

2.

Exercises

1. By the use of the dominated convergence theorem, prove that

limk→∞

∫ 1

0

sin(ekx)√k

dx = 0 , limk→∞

∫ 1

−1

√k arctan

(xk

)dx = 0 .

2. Let fk : [0, 1]→ R be defined as

fk(x) =

0 if x ∈

[0 ,

1

k

[,

1

kxif x ∈

[1

k, 1

].

Prove that

limk→∞

∫ 1

0fk(x) dx = 0 ,

both by a direct computation and by the monotone or the dominated convergencetheorems. Let now

fk(x) =

0 if x ∈

[0 ,

1

k2

[,

1

kxif x ∈

[1

k2, 1

].

Explain why, in this case, the monotone or dominated convergence theorems cannotbe applied.

3. Compute the following limit:

limk→∞

∫ π

0k3(

sin(xk

)− x

k

)dx .


1.15 Integration on non-compact intervals

We begin by considering a function f : [a, b[→ R, where b ≤ +∞. Assume that f beintegrable on every compact interval of the type [a, c], with c ∈ ]a, b[ . This happens,for instance, when f is continuous on [a, b[ .

Definition 1.39 We say that a function f : [a, b[→ R is integrable if f is inte-grable on [a, c] for every c ∈ ]a, b[ , and the limit

limc→b−

∫ c

af

exists and is finite. In that case, the above limit is called the integral of f on [a, b[

and it is denoted by∫ ba f , or

∫ ba f(x) dx.

In particular, if b = +∞, we will write:∫ +∞a f , or

∫ +∞a f(x) dx.

Examples. Let a > 0; it is readily seen that the function f : [a,+∞[→ R, defined by f(x) = x−α,is integrable if and only if α > 1, in which case we have∫ +∞

a

dx

xα=a1−α

α− 1.

Consider now the case a < b < +∞. It can be verified that the function f : [a, b[→ R, defined byf(x) = (b− x)−β , is integrable if and only if β < 1, in which case we have∫ b

a

dx

(b− x)β=

(b− a)1−β

1− β .

a a b

One often speaks of improper integral when considering functions which aredefined on non-compact intervals. One also says that the integral converges if thefunction f is integrable on [a, b[ , i.e., when the limit limc→b−

∫ ca f exists and is finite.

If the limit does not exist, it is said that the integral is undetermined. If it exists andequals +∞ or −∞, it is said that the integral diverges to +∞ or to −∞, respectively.

It is clear that the convergence of the integral depends solely on the behaviorof the function near the point b. In other words, modifying the function outside aneighborhood of b, the convergence of the integral is by no means compromised.

1.15. INTEGRATION ON NON-COMPACT INTERVALS 43

Let us now state the Cauchy convergence criterion.

Theorem 1.40 Let f : [a, b[→ R be a function, which is integrable on [a, c], forevery c ∈ ]a, b[ . A necessary and sufficient condition for f to be integrable on [a, b[is that for every ε > 0 there is a c ∈ ]a, b[ such that, taking as c′ and c′′ any twonumbers in [c, b[ , it is ∣∣∣∣∣

∫ c′′

c′f

∣∣∣∣∣ ≤ ε .Proof It is a direct consequence of the Cauchy criterion for the limit, when appliedto the function F : [a, b[→ R defined as F (c) =

∫ ca f .

From the Cauchy criterion we deduce the following comparison criterion.

Theorem 1.41 Let f : [a, b[→ R be a function, which is integrable on [a, c], forevery c ∈ ]a, b[ . If there is an integrable function g : [a, b[→ R such that, for everyx ∈ [a, b[ ,

|f(x)| ≤ g(x) ,

then f is integrable on [a, b[ , too.

Proof Once fixed ε > 0, there is a c ∈ ]a, b[ such that, taking arbitrarily c′, c′′ in

[c, b[ , it is |∫ c′′c′ g| ≤ ε. If for example c′ ≤ c′′, being −g ≤ f ≤ g, one has

−∫ c′′

c′g ≤

∫ c′′

c′f ≤

∫ c′′

c′g ,

and therefore ∣∣∣∣∣∫ c′′

c′f

∣∣∣∣∣ ≤∫ c′′

c′g ≤ ε .

The Cauchy condition then holds, whence the conclusion.

As an immediate consequence, we have the following.

Corollary 1.42 Let f : [a, b[→ R be a function, which is integrable on [a, c], forevery c ∈ ]a, b[ . If |f | is integrable on [a, b[ , then also f is such, and∣∣∣∣∫ b

af

∣∣∣∣ ≤ ∫ b

a|f | .

In the case when both f and |f | are integrable on [a, b[ , we say that f is L-integrable, or absolutely integrable, on [a, b[ .


Example. Consider the function f : [π,+∞[→ R defined by f(x) = sin xx

. We will see that it isintegrable on [π,+∞[ , but not absolutely integrable there. To see that it is integrable, take c > πand integrate by parts: ∫ c

π

sinx

xdx =

[− cosx

x

]cπ−∫ c

π

cosx

x2dx .

We find that

limc→+∞

∫ c

π

f = − 1

π− limc→+∞

∫ c

π

cosx

x2dx ,

and this last limit is finite by the comparison theorem, since∣∣∣cosx

x2

∣∣∣ ≤ x−2 .

Hence, f is integrable on [π,+∞[ . Assume by contradiction that it was also absolutely integrable.In that case, for every integer n ≥ 2, we would have∫ nπ

π

∣∣∣∣ sinxx∣∣∣∣ dx=

n−1∑k=1

∫ (k+1)π

kπ

| sinx|x

dx

≥n−1∑k=1

1

(k + 1)π

∫ (k+1)π

kπ

| sinx| dx

=2

π

n−1∑k=1

1

k + 1.

but this is impossible, since the series∑∞k=1

1k+1

diverges.

Let us now state a corollary of the comparison criterion which is often used inpractice.

Corollary 1.43 Let f, g : [a, b[→ R be two functions with positive values, which areintegrable on [a, c] for every c ∈ ]a, b[ . Assume that the following limit exists:

L = limx→b−

f(x)

g(x).

Then, the following conclusions hold:

a) if L = 0 and g is integrable on [a, b[ , then such is f, as well;

b) if 0 < L < +∞, then f is integrable on [a, b[ if and only if such is g;

c) if L = +∞ and g is not integrable on [a, b[ , then neither f is such.

Example. Consider the function f : [0,+∞[→ R defined by

f(x) = e1/(x2+1) − 1 .

As a comparison function, I would like to take g(x) = x−2. A technical problem arises, however,since g is not defined on [0,+∞[ . We can proceed in two different ways: either we restrict f to aninterval of the type [a,+∞[ , with a > 0, and we observe that this operation does not modify the

1.15. INTEGRATION ON NON-COMPACT INTERVALS 45

convergence (or the non convergence) of the integral, since f is continuous on [0, a]; or we adapt tothis situation the function g : for instance, we can choose

g(x) =

1 if x ∈ [0, 1] ,

x−2 if x ≥ 1 .

Once this has been done, observe that

limx→+∞

f(x)

g(x)= limx→+∞

x2(e1/(x2+1) − 1) = lim

x→+∞

x2

x2 + 1

e1/(x2+1) − 1

1/(x2 + 1)= 1 .

Since g is integrable on [0,+∞[ , such is f, as well.

We consider now a function f : ]a, b]→ R, with a ≥ −∞. There is an analogousdefinition of its improper integral.

Definition 1.44 We say that a function f : ]a, b] → R is integrable if f is inte-grable on [c, b] for every c ∈ ]a, b[ , and the limit

limc→a+

∫ b

cf

exists and is finite. In that case, the above limit is called the integral of f on ]a, b]

and it is denoted by∫ ba f , or

∫ ba f(x) dx.

Given the function f : ]a, b] → R, it is possible to consider the function g :[a′, b′[→ R, with a′ = −b and b′ = −a, defined by g(x) = f(−x). It is easy to seethat f is integrable on ]a, b] if and only if g is integrable on [a′, b′[ . In this way weare reconducted to the previous theory.

We will also define the integral of a function f : ]a, b[→ R, with −∞ ≤ a < b ≤+∞, in this way:

Definition 1.45 We say that f : ]a, b[→ R is integrable if, once we fix a pointp ∈ ]a, b[ , the function f is integrable on [p, b[ and on ]a, p]. In that case, the integralof f on ]a, b[ is defined by ∫ b

af =

∫ p

af +

∫ b

pf .

It is easy to verify that the given definition does not depend on the choice ofp ∈ ]a, b[ .

Examples. If a, b ∈ R, one can verify that the function

f(x) = ((x− a)(b− x))−β


is integrable on ]a, b[ if and only if β < 1. In this case, it is possible to choose, for instance,p = (a+ b)/2. Another case arises when a = −∞ and b = +∞. For example, one easily verifies thatthe function f(x) = (x2 + 1)−1 is integrable on ]−∞,+∞[ . Taking for instance p = 0, we have:∫ +∞

−∞

1

x2 + 1dx =

∫ 0

−∞

1

x2 + 1dx+

∫ +∞

0

1

x2 + 1dx = π .

A further case one could face in the applications is when a function happens notto be defined in an interior point of an interval.

Definition 1.46 Given a < q < b, we say that a function f : [a, b] \ q → R isintegrable if f is both integrable on [a, q[ and on ]q, b]. In that case, we set∫ b

af =

∫ q

af +

∫ b

qf .

For example, if a < 0 < b, the function f(x) =√|x|/x is integrable on [a, b]\0,

and ∫ b

a

√|x|x

dx =

∫ 0

a

−1√−x

dx+

∫ b

0

1√xdx = 2

√b− 2

√−a .

On the other hand, the function f(x) = 1/x is not integrable on [a, b] \ 0, evenif the fact that f is odd could lead someone to define the integral on symmetricintervals with respect to the origin as being equal to zero. However, by doing so,some important properties of the integral would be lost, as for example the additivityon sub-intervals.

Different situations can be faced combining together those treated above. I prefernot to go deeper into these details; in each single case, the choice of the appropriatemethod will be made by the right guess.

1.16 The Hake Theorem

Recall that a function f : [a, b[→ R is said to be integrable if it is integrable on[a, c], for every c ∈ ]a, b[ , and the limit

limc→b−

∫ c

af

exists and is finite. We want to prove the following result due to H. Hake.

Theorem 1.47 Let b < +∞, and assume f : [a, b[→ R to be a function which isintegrable on [a, c], for every c ∈ ]a, b[ . Then, the function f is integrable on [a, b[ ifand only if f is the restriction of an integrable function f : [a, b]→ R. In that case,∫ b

af =

∫ b

af .

1.16. THE HAKE THEOREM 47

Proof Assume first that f be the restriction of an integrable function f : [a, b]→ R.Fix ε > 0; we want to find a γ > 0 such that, if c ∈ ]a, b[ and b− c ≤ γ, then∣∣∣∣∫ c

af −

∫ b

af

∣∣∣∣ ≤ ε .Let δ be a gauge such that, for every δ-fine P-partition of [a, b], it is |S(I, f ,Π) −∫ ba f | ≤

ε8 . We choose a positive constant γ ≤ δ(b) such that γ|f(b)| ≤ ε

2 . If c ∈ ]a, b[and b − c ≤ γ, by the Saks–Henstock theorem, taking the δ-fine sub-P-partitionΠ = (b, [c, b]), we have ∣∣∣∣f(b)(b− c)−

∫ b

cf

∣∣∣∣ ≤ 4ε

8=ε

2,

and hence∣∣∣∣∫ c

af −

∫ b

af

∣∣∣∣ =

∣∣∣∣∫ b

cf

∣∣∣∣ ≤ ε

2+ |f(b)(b− c)| ≤ ε

2+ |f(b)|γ ≤ ε

2+ε

2= ε .

Let us prove now the other implication. Assume f to be integrable on [a, b[ , andlet A be its integral. We extend f to a function f defined on the whole interval [a, b],by setting, for instance, f(b) = 0. In order to prove that f is integrable on [a, b] withintegral A, fix ε > 0. There is a γ > 0 such that, if c ∈ ]a, b[ and b− c ≤ γ, then∣∣∣∣∫ c

af −A

∣∣∣∣ ≤ ε

2.

Consider the sequence (ci)i of points in [a, b[ , given by

ci = b− b− ai+ 1

.

Notice that it is strictly increasing, it converges to b, and it is c0 = a. Since f isintegrable on each interval [ci−1, ci], we can consider, for each i ≥ 1, a gauge δi on[ci−1, ci] such that, for every δi-fine P-partition Πi of [ci−1, ci], one has∣∣∣∣∣S([ci−1, ci], f,Πi)−

∫ ci

ci−1

f

∣∣∣∣∣ ≤ ε

2i+4.

We define a gauge δ on [a, b] by setting

δ(x) =

min

δi(x), x−ci−1

2 , ci−x2

if x ∈ ]ci−1, ci[ ,

minδ1(a), c1−a2

if x = a ,

minδi(ci), δi+1(ci),

ci−ci−1

2 , ci+1−ci2

if x = ci and i ≥ 1 ,

γ if x = b .


Let Π = (xj , [aj−1, aj ]) : j = 1, . . . ,m, be a δ-fine P-partition of [a, b]. Denote byq be the smallest integer for which cq+1 ≥ am−1. The choice of the gauge permits tosplit the Riemann sum, similarly as in the proof of the theorem on the additivity ofthe integral on subintervals, and the sum S([a, b], f ,Π) will thus contain

• q Riemann sums on [ci−1, ci], with i = 1, . . . , q;

• a Riemann on [cq, am−1];

• a last term f(xm)(b− am−1).

To better clarify what we just said, assume for example that q = 2; then there mustbe a 1 for which x1 = c1, and a 2 for which x2 = c2. Then,

S([a, b], f ,Π) = [f(x1)(a1 − a) +. . .+ f(x1−1)(a1−1 − a1−2) + f(c1)(c1 − a1−1)]

+[f(c1)(a1 − c1) +. . .+ f(x2−1)(a2−1 − a2−2) + f(c2)(c2 − a2−1)]

+[f(c2)(a2 − c2) + . . .+ f(xm−1)(am−1 − am−2)]

+f(xm)(b− am−1) .

In general, we will have

S([a, b], f ,Π) =

q∑i=1

S([ci−1, ci], f,Πi) + S([cq, am−1], f,Πq+1) +

+f(xm)(b− am−1) ,

where, for i = 1, . . . , q, Πi is a δi-fine P-partition of [ci−1, ci], while Πq+1 is a δq+1-fineP-partition of [cq, am−1], and hence a δq+1-fine sub-P-partition of [cq, cq+1]. By thechoice of the gauge δ, it has to be xm = b and hence f(xm) = 0. Moreover, sinceδ(xm) = γ, it is b− am−1 ≤ γ. Using the fact that∫ am−1

af =

q∑i=1

∫ ci

ci−1

f +

∫ am−1

cq

f ,

by the Saks–Henstock theorem we have

|S([a, b], f ,Π)−A| ≤∣∣∣∣S([a, b], f ,Π)−

∫ am−1

af

∣∣∣∣+

∣∣∣∣∫ am−1

af −A

∣∣∣∣≤

q∑i=1

∣∣∣∣∣S([ci−1, ci], f,Πi)−∫ ci

ci−1

f

∣∣∣∣∣+

+

∣∣∣∣∣S([cq, am−1], f,Πq+1)−∫ am−1

cq

f

∣∣∣∣∣+

∣∣∣∣∫ am−1

af −A

∣∣∣∣≤

q∑i=1

ε

2i+4+ 4

ε

2q+4+ε

2

≤ ε

4+ε

4+ε

2= ε ,

1.17. INTEGRALS AND SERIES 49


The above theorem suggests that even when f : [a,+∞[→ R, the definition ofthe improper integral could be reduced to that of a usual integral. Indeed, fixingarbitrarily b > a, we could define a continuously differentiable strictly increasingfunction ϕ : [a, b[→ R such that ϕ(a) = a and limu→b− ϕ(u) = +∞; for example,take ϕ(u) = a+ ln b−a

b−u . A formal change of variables then gives∫ +∞

af(x) dx =

∫ b

af(ϕ(u))ϕ′(u) du ,

and to this last integral Hake’s theorem applies.

With this idea in mind, it is possible to prove that f : [a,+∞[→ R is integrableand its integral is a real number A if and only if for every ε > 0 there is a gauge δ,defined on [a,+∞[ , and a positive constant α such that, if

a = a0 < a1 < . . . < am−1 , with am−1 ≥ α ,

and, for every j = 1, . . . ,m− 1,

xj − δ(xj) ≤ aj−1 ≤ xj ≤ aj ≤ xj + δ(xj) ,

then ∣∣∣∣∣m−1∑j=1


∣∣∣∣∣ ≤ ε .We refer to the book of Bartle [1] for a complete treatment of this case.

Needless to say, similar considerations can be made in the case when the functionf is defined on an interval of the type ]a, b], with −∞ ≤ a.

1.17 Integrals and series

We now prove a theorem which shows the close connection between the theory ofnumerical series and that of the improper integral.

Theorem 1.48 Let f : [1,+∞[→ R be a function which is positive, decreasing andintegrable on [1, c], for every c > 1. Then f is integrable on [1,+∞[ if and only ifthe series

∑∞k=1 f(k) converges. Moreover, we have

∞∑k=2

f(k) ≤∫ +∞

1f ≤

∞∑k=1

f(k) .


Proof For x ∈ [k, k + 1], it has to be f(k + 1) ≤ f(x) ≤ f(k). Hence,

f(k + 1) ≤∫ k+1

kf ≤ f(k) .

Summing up, we obtain

n∑k=1

f(k + 1) ≤∫ n+1

1f ≤

n∑k=1

f(k) .

Being f positive, the sequence (∑n

k=1 f(k))n and the function c 7→∫ c

1 f are both in-creasing and therefore have a limit. The conclusion now follows from the comparisontheorem for limits.

1 2 3 4 5

Observations. It is clear that the choice a = 1 in the theorem just proved is byno way necessary. Notice moreover that this theorem is often used to determine theconvergence of a series, giving the estimate∫ +∞

1f ≤

∞∑k=1

f(k) ≤ f(1) +

∫ +∞

1f .

Example. Consider the series∑∞k=1 k

−3; in this case:∫ +∞

1

1

x3dx ≤

∞∑k=1

1

k3≤ 1 +

∫ +∞

1

1

x3dx ,

and then1

2≤∞∑k=1

1

k3≤ 3

2.

A greater accuracy is easily attained by computing the sum of a few first terms and then using theestimate given by the integral. For example, separating the first two terms, we have:

∞∑k=1

1

k3= 1 +

1

8+

∞∑k=3

1

k3,

with ∫ +∞

3

1

x3dx ≤

∞∑k=3

1

k3≤ 1

27+

∫ +∞

3

1

x3dx .

1.17. INTEGRALS AND SERIES 51

We thus have the following estimate:

255

216≤∞∑k=1

1

k3≤ 263

216.

We conclude this chapter with a series of exercises on this final part.

Exercises

1. Establish whether the following improper integrals converge:∫ +∞

0

√x

x3 + 1dx

∫ +∞

0

1√x (1 +

√x+ x)

dx

2. For which α ∈ R the integrals∫ +∞

π

1

x(lnx)αdx ,

∫ +∞

π

1

x(lnx)(ln(lnx))αdx

converge?

3. The following improper integrals converge?∫ 2

0

lnx√x cosx

dx

∫ 1

0

1

x lnxdx

4. The following series converge?

∞∑k=1

1

k ln k,

∞∑k=1

1

k ln k ln(ln k)


Chapter 2

Functions of several realvariables

In this chapter we extend the theory developed in the previous one to functions ofseveral variables defined on subsets of RN, with values in R. In order to simplify theexposition, we will often concentrate on the case N = 2. It will not be so difficultfor the reader to extend the various results to the case of a generic dimension N .

2.1 Integrability on rectangles

We begin by considering the case of functions defined on rectangles. A rectangleof RN is a set of the type [a1, b1] × . . . × [aN , bN ]. This word is surely familiar inthe case N = 2. If N = 1, a rectangle happens to be a compact interval while, ifN = 3, usually one prefers to call it “rectangle parallelepiped”. In the followingexposition, we concentrate for simplicity on the two-dimensional case. The generalcase is perfectly similar and does not involve greater difficulties, except for thenotations.

We consider the rectangle I = [a1, b1]× [a2, b2] ⊆ R2. Let us define the measureof I :

µ(I) = (b1 − a1)(b2 − a2) .

We say that two rectangles are non-overlapping if their interiors are disjoint.

A P-partition of the rectangle I is a set

Π = (x1, I1), (x2, I2), . . . , (xm, Im) ,

where the Ij are non-overlapping rectangles whose union is I and, for every j =1, . . . ,m, the point xj = (xj , yj) belongs to Ij .

53

54 CHAPTER 2. FUNCTIONS OF SEVERAL REAL VARIABLES

Example. If I = [0, 10]× [0, 6], a possible P-partition is the following:

Π = ((1, 1), [0, 7]× [0, 2]), ((0, 5), [0, 3]× [2, 6]),

((5, 4), [3, 10]× [4, 6]), ((10, 0), [7, 10]× [0, 4]), ((5, 3), [3, 7]× [2, 4]) .

I2

I1

I3

I5I4

x2

x1

x5

x3

x4

Let us now consider a function f defined on the rectangle I, with values in R,and let Π = (xj , Ij) : j = 1, . . . ,m be a P-partition of I. We call Riemann sumassociated to I, f and Π the real number S(I, f,Π) defined by

S(I, f,Π) =m∑j=1

f(xj)µ(Ij) .

Whenever f happens to be positive, this number is the sum of the volumes of theparallelepipeds having as base Ij and height [0, f(xj)].

x1

x2

x3

We now introduce the notion of fineness for the P-partition Π defined above. Wecall gauge on I every function δ : I → R such that δ(x) > 0 for every x ∈ I. Given

2.1. INTEGRABILITY ON RECTANGLES 55

a gauge δ on I, we say that the P-partition Π introduced above is δ-fine if, for everyj = 1, . . . ,m,

Ij ⊆ [xj − δ(xj , yj), xj + δ(xj , yj)]× [yj − δ(xj , yj), yj + δ(xj , yj)] .

In the following, given x = (x, y) ∈ I and r > 0, in order to shorten the notationswe will write

B[x, r] = [x− r, x+ r]× [y − r, y + r] ;

the P-partition Π will then be δ-fine if, for every j = 1, . . . ,m,

Ij ⊆ B[xj , δ(xj)] .

Example. Let I = [0, 1]× [0, 1] and δ be the gauge defined as follows:

δ(x, y) =

x+ y

3if (x, y) 6= (0, 0) ,

1

2if (x, y) = (0, 0) .

We want to find a δ-fine P-partition of I. Similarly as was seen in the case N = 1 we have inthis case that one of the points xj necessarily has to coincide with (0, 0). We can then choose, forexample,

Π =

((0, 0),

[0,

1

2

]×[0,

1

2

]),

((1

2, 1

),

[0,

1

2

]×[

1

2, 1

]),((

1,1

2

),

[1

2, 1

]×[0,

1

2

]),

((1, 1),

[1

2, 1

]×[

1

2, 1

]).

It is interesting to observe that it is not always possible to construct δ-fine P-partitions by onlyjoining points on the edges of I. The reader can be convinced by trying to do this with the followinggauge:

δ(x, y) =

x+ y

16if (x, y) 6= (0, 0) ,

1

2if (x, y) = (0, 0) .

As is the one-dimensional case, one can prove that for every gauge δ on I thereexists a δ-fine P-partition of I (Cousin’s Theorem). The following definition isidentical to the one seen in Chapter 1.

Definition 2.1 A function f : I → R is said to be integrable (on the rectangle I)if there is a real number A with the following property: given ε > 0, it is possible tofind a gauge δ on I such that, for every δ-fine P-partition Π of I, it is

|S(I, f,Π)−A| ≤ ε .

We briefly overview all the properties which can be obtained from the givendefinition in the same way as was done in the case of a function of a single variable.


There is at most one A ∈ R which verifies the conditions of the definition. Sucha number is called the integral of f on I and is denoted by one of the followingsymbols: ∫

If ,

∫If(x) dx ,

∫If(x, y) dx dy .

The set of integrable functions is a vector space, and the integral is a linear functionon it: ∫

I(f + g) =

∫If +

∫Ig ,

∫I(αf) = α

∫If

(with α ∈ R); it preserves the order:

f ≤ g ⇒∫If ≤

∫Ig .

The Cauchy criterion of integrability holds. Moreover, we have the followingversion of the theorem on additivity on subrectangles.

Theorem 2.2 Let f : I → R be a function and K1,K2, . . . ,Kl be non overlappingsub-rectangles of I whose union is I. Then, f is integrable on I if and only if it isintegrable on each of the Ki. In that case, we have∫

If =

l∑i=1

∫Ki

f .

In particular, if a function is integrable on a rectangle, it still is on every sub-rectangle. The proof of the theorem is similar to that given in the one-dimensionalcase, and is based on the possibility of constructing a gauge which permits to splitthe Riemann sums on I in the sum of Riemann sums on the single subrectangles.

We say that an integrable function on I is R-integrable there (or integrableaccording to Riemann) if, among all possible gauges δ which verify the definition ofintegrability, it is always possible to choose one which is constant on I. The set ofR-integrable functions is a vector subspace of the space of integrable functions andcontains the subspace of continuous functions.

We say that an integrable function f : I → R is L-integrable (or integrableaccording to Lebesgue) if |f | is integrable on I, as well. The L-integrable functionsmake up a vector subspace of the space of integrable functions. If f and g aretwo L-integrable functions on I, then the functions minf, g and maxf, g areL-integrable on I, too. A function f is L-integrable on I if and only if such are itspositive part f+ = maxf, 0 and its negative part f− = max−f, 0.

The Saks–Henstock’s theorem, the monotone convergence theorem ofB. Levi and the dominated convergence theorem of H. Lebesgue extend, withstatements and proofs perfectly analogous to those given in the first chapter, to theintegrable functions on a rectangle.

2.2. INTEGRABILITY ON A BOUNDED SET 57

Exercises

1. Let f : [a, b] × [c, d] → R be defined as f(x, y) = x. Prove that f is integrableand compute ∫

[a,b]×[c,d]f(x, y) dx dy .

2. Let f : [0, 1]× [0, 1]→ R be the function defined as

f(x, y) =

3 if x ∈

[0, 1

2

],

5 if x ∈[

12 , 1]

.

Prove that f is integrable with∫[0,1]×[0,1]

f(x, y) dx dy = 4 .

3. Prove that, if f : I → R is equal to 0 for every (x, y) /∈ Q×Q, then∫I f = 0.

4. By the use of the dominated convergence theorem, prove that

limk→∞

∫[0,1]×[0,1]

cos(ek(x2+y2))√k

dx dy = 0 ,

limk→∞

∫[−1,1]×[−1,1]

√k sinh

(x+ y

k

)dx dy = 0 .

2.2 Integrability on a bounded set

Given a bounded set E and a function f whose domain contains E, we define thefunction fE as follows:

fE(x) =

f(x) if x ∈ E ,

0 if x 6∈ E .

We can prove the following

Proposition 2.3 Let I1 and I2 be two rectangles containing the set E. Then, fE isintegrable on I1 if and only if it is integrable on I2. In that case, we have

∫I1fE =∫

I2fE .

Proof We consider for simplicity the case N = 2. Assume that fE be integrableon I1. Let K be a rectangle containing both I1 and I2. We can construct somenon-overlapping rectangles K1, . . . ,Kr, also non-overlapping with I1, such that I1 ∪


K1 ∪ . . .∪Kr = K. We now prove that fE is integrable on each of the subrectanglesK1, . . . ,Kr, and that the integrals

∫K1fE , . . . ,

∫KrfE are all equal to zero. Notice

that fE , restricted to each of these subrectangles, is zero everywhere except perhapson one of their edges. We are thus led to prove the following lemma, which willpermit us to conclude the proof.

Lemma 2.4 Let K be a rectangle and g : K → R be a function which is zeroeverywhere except perhaps on one edge of K. Then g is integrable on K and

∫K g = 0.

Proof We first assume that the function g be bounded on K, i.e., that there is aconstant C > 0 for which

|g(x, y)| ≤ C ,

for every (x, y) ∈ K. Fix ε > 0. Let L be the edge of the rectangle K on which g canbe nonzero, and denote by ` its length. Define the constant gauge δ = ε

C` . Then, forevery δ-fine P-partition Π = (x1, I1), (x2, I2), . . . , (xm, Im) of K, we have:

|S(K, g,Π)| ≤m∑j=1

|g(xj)|µ(Ij)

=∑

j : xj∈L

|g(xj)|µ(Ij)

≤C∑

j : xj∈L

µ(Ij)

≤Cδ` = ε .

This proves that g is integrable on K and∫K g = 0 in the case when g is bounded

on K. If g is not such, assume that it has non-negative values. Define the followingsequence (gk)k of functions:

gk(x) = ming(x), k .

Being the functions gk bounded, for what have been seen above we have∫K gk = 0,

for every k. It is easily seen that the sequence thus defined satisfies the conditionsof the monotone convergence theorem and converges pointwise to g. It then followsthat g is integrable on K and ∫

Kg = lim

k

∫Kgk = 0 .

If g does not have only non-negative values, it is always possible to consider g+ andg−. From the above,

∫K g

+ =∫K g− = 0, and then

∫K g = 0, which is what was to

be proved.

2.2. INTEGRABILITY ON A BOUNDED SET 59

End of the proof. Having proved that fE is integrable on each of the K1, . . . ,Kr andthat the integrals

∫K1fE , . . . ,

∫KrfE are equal to zero, by the theorem of additivity

on subrectangles we have that, being fE integrable on I1, it is such on K, and∫KfE =

∫I1

fE +

∫K1

fE + . . .+

∫Kr

fE =

∫I1

fE .

But then fE is integrable on every sub-interval of K, and in particular on I2. We cannow construct, analogously to what has just been done for I1, some non-overlappingrectangles J1, . . . , Js, also non-overlapping with I2, such that I2 ∪ J1 ∪ . . .∪ Js = K.Similarly, we will have∫

KfE =

∫I2

fE +

∫J1

fE + . . .+

∫Js

fE =

∫I2

fE ,

which proves that∫I1fE =

∫I2fE . To see that the condition is necessary and suffi-

cient, just exchange the roles of I1 and I2 in the above proof.

We are thus led to the following.

Definition 2.5 Given a bounded set E, we say that the function f : E → R isintegrable (on E) if there is a rectangle I containing the set E on which fE isintegrable. In that case, we set ∫

Ef =

∫IfE .

When f is integrable on E according to the previous definition, one has that fEis integrable on any rectangle containing the set E, and the integral of fE remainsthe same on each such rectangle.

With the given definition, all the properties of the integral seen beforeeasily extend. There is an exception concerning the additivity, since it is not truein general that a function which is integrable on a bounded set remains integrableon any of its subsets. Indeed, take a function f : E → R which is integrable but notL-integrable. We consider the subset

E′ = x ∈ E : f(x) ≥ 0 ,

and we show that f cannot be integrable on E′. If it was, then f+ would be integrableon E. But then also f− = f+ − f would be integrable on E, and therefore f shouldbe L-integrable on E, in contradiction with the assumption.

We will see that, with respect to additivity, the L-integrable functions have asomewhat better behavior.


2.3 The measure of a bounded set

Definition 2.6 A bounded set E is said to be measurable if the constant function 1is integrable on E. In that case, the number

∫E 1 is said to be the measure of E

and is denoted by µ(E).

The measure of a measurable set is thus a non-negative number. The empty setis assumed to be measurable, and its measure is equal to 0. In the case of a subsetof R2, its measure is also called the area of the set. If E = [a1, b1] × [a2, b2] is arectangle, it is easily seen that

µ(E) =

∫E

1 = (b1 − a1)(b2 − a2) ,

so that the notation is in accordance with the one already introduced for rectangles.For a subset of R3, the measure is also called the volume of the set.

Not every set is measurable. It is shown in Appendix C that, when dealing withnon-measurable sets, some paradoxical situations can arise. In the following, we willbe careful to always consider measurable sets.

Let us analyze some properties of the measure. It is useful to introduce thecharacteristic function of a set E, defined by

χE(x) =

1 if x ∈ E ,0 if x 6∈ E .

If I is a rectangle containing the set E, we thus have

µ(E) =

∫IχE .

Proposition 2.7 Let A and B be two bounded and measurable sets. The followingproperties hold:

(a) if A ⊆ B, then B \A is measurable, and

µ(B \A) = µ(B)− µ(A) ;

in particular, µ(A) ≤ µ(B).

(b) A ∪B and A ∩B are measurable, and

µ(A ∪B) + µ(A ∩B) = µ(A) + µ(B) ;

in particular, if A and B are disjoint, then µ(A ∪B) = µ(A) + µ(B).

Proof Let I be a rectangle containing A∪B. If A ⊆ B, then χB\A = χB −χA, andproperty (a) follows by integrating on I.

2.3. THE MEASURE OF A BOUNDED SET 61

Being χA∪B = maxχA, χB and χA∩B = minχA, χB, we have that χA∪B andχA∩B are integrable on I. Moreover,

χA∪B + χA∩B = χA + χB ,

and integrating on I we have (b).

The following proposition states the property of complete additivity of themeasure.

Proposition 2.8 If (Ak)k≥1 is a sequence of bounded and measurable sets, whoseunion A = ∪k≥1Ak is bounded, then A is measurable, and

µ(A) ≤∞∑k=1

µ(Ak) .

If the sets Ak are pairwise disjoint, then equality holds.

Proof Assume first that the sets Ak be pairwise disjoint. Let I be a rectanglecontaining their union A. Then, for every x ∈ I,

χA(x) =∞∑k=1

χAk(x) .

Moreover, since for every positive integer q it is

q∑k=1

µ(Ak) = µ

(q⋃

k=1

Ak

)≤ µ(I) ,

the series∑∞

k=1

∫I χAk =

∑∞k=1 µ(Ak) converges. By the corollary following the

monotone convergence theorem, we have that A is measurable and

µ(A) =

∫IχA =

∫I

∞∑k=1

χAk =∞∑k=1

∫IχAk =

∞∑k=1

µ(Ak) .

When the sets Ak are not pairwise disjoint, consider the sets B1 = A1, B2 = A2 \A1

and, in general, Bk = Ak \ (A1 ∪ . . . ∪Ak−1). The sets Bk are measurable, pairwisedisjoint, and ∪k≥1Bk = ∪k≥1Ak. The conclusion then follows from what has beenproved above.

We have a similar proposition concerning the intersection of a countable familyof sets.


Proposition 2.9 If (Ak)k≥1 is a sequence of bounded and measurable sets, theirintersection A = ∩k≥1Ak is a measurable set.

Proof Let I be a rectangle containing the set A. Then,

⋂k≥1

Ak = I \

⋃k≥1

(I \ (Ak ∩ I))

,

and the conclusion follows from the two previous propositions.

The following two propositions will provide us with a large class of measurablesets.

Proposition 2.10 Every open and bounded set is measurable.

Proof Consider for simplicity the case N = 2. Let A be an open set contained ina rectangle I. We divide the rectangle I in four rectangles of equal areas using theaxes of its edges. Then we proceed analogously with each of these four rectangles,thus obtaining sixteen smaller rectangles, and so on. Being A open, for every x ∈ Athere is a small rectangle among those just constructed which contains x and iscontained in A. In this way, it is seen that the set A is covered by a countable familyof rectangles; being the union of a countable family of measurable sets, it is thereforemeasurable.

Proposition 2.11 Every compact set is measurable.

Proof Let B be a compact set, and let I be a rectangle whose interior I

contains

B. Being I

and I\B open and hence measurable, we have that B = I

\ (I\B) is

measurable.

Example. The setE = (x, y) ∈ R2 : 1 < x2 + y2 ≤ 4

is measurable, being the difference of the closed disks with radius 2 and 1 centered at the origin:

E = (x, y) ∈ R2 : x2 + y2 ≤ 4 \ (x, y) ∈ R2 : x2 + y2 ≤ 1 .

2.4 The Cebicev inequality

Theorem 2.12 Let E be a bounded set, and f : E → R be an integrable function,with non-negative values. Then, for every r > 0, the set

Er = x ∈ E : f(x) > r

2.5. NEGLIGIBLE SETS 63

is measurable, and

µ(Er) ≤1

r

∫Ef .

Proof Let I be a rectangle containing E. Once fixed r > 0, we define the followingfunctions on I :

fk(x) = min1, kmaxfE(x)− r, 0 .

They make up an increasing sequence of L-integrable functions which pointwiseconverges to χEr . Clearly,

0 ≤ fk(x) ≤ 1 ,

for every k and every x ∈ I. The monotone (or dominated, as well) convergencetheorem guarantees that χEr is integrable on I, i.e., that Er is measurable. Since,for every x ∈ E, it is rχEr(x) ≤ f(x), integrating we obtain the inequality we arelooking for.

Corollary 2.13 Let E be a bounded and measurable set, and f : E → R an in-tegrable function with non-negative values. Then, taken two real numbers r, s suchthat 0 ≤ r < s, the set

Er,s = x ∈ E : r ≤ f(x) < s

is measurable.

Proof Let E′r = x ∈ E : f(x) ≥ r. With the notations of the previous theorem,if r > 0, we have

E′r =⋂k> 1

r

Er− 1k,

while, if r = 0, we have E′r = E. In any case, E′r is measurable. Being Er,s = E′r \E′s,the conclusion follows.

2.5 Negligible sets

Definition 2.14 We say that a bounded set is negligible if it is measurable andits measure is equal to zero.

Every set made of a single point is negligible. Consequently, all finite or countablebounded sets are negligible. The edge of a rectangle in R2 is a negligible set, as shownby Lemma 2.4. Another interesting example of a negligible and not countable set isgiven by the Cantor set.


By the complete additivity of the measure, the union of any sequence of negligiblesets, if it is bounded, is always a negligible set.

Theorem 2.15 If E is a bounded set and f : E → R is equal to zero except for anegligible set, then f is integrable on E and

∫E f = 0.

Proof Let T be the negligible set on which f in non-zero. Assume first that thefunction f be bounded, i.e., that there is a constant C > 0 such that

|f(x)| ≤ C ,

for every x ∈ E. We consider a rectangle I containing E and prove that∫I fE = 0.

Fix ε > 0. Since T has zero measure, there is a gauge δ such that, for every δ-fineP-partition Π = (xj , Ij), j = 1, . . . ,m of I,

S(I, χT ,Π) =∑

j : xj∈T

µ(Ij) ≤ε

C,

so that|S(I, fE ,Π)| ≤

∑j : xj∈T

|f(xj)|µ(Ij) ≤ C∑

j : xj∈T

µ(Ij) ≤ ε .

Hence, in case f is bounded, it is integrable on E and∫E f = 0. If f is not

bounded, assume first that it has non-negative values. Define on E a sequenceof functions (fk)k :

fk(x) = minf(x), k .Being the functions fk bounded and zero except on T, for what has just been seenthey are integrable on E with

∫E fk = 0, for every k. It is easily seen that the

defined sequence satisfies the conditions of the monotone convergence theorem andconverges pointwise to f. Hence, f is integrable on E, and∫

Ef = lim

k

∫Efk = 0 .

If f does not have non-negative values, it is sufficient to consider f+ and f−, andapply to them what has been said above.

Theorem 2.16 If f : E → R is an integrable function on a bounded set E, havingnon negative values, with

∫E f = 0, then f is equal to zero except on a negligible set.

Proof Using the Cebicev inequality, we have that, for every positive integer k,

µ(E 1k) ≤ k

∫Ef = 0 .

Hence, every E 1k

is negligible, and since their union is just the set where f is non-

zero, we have the conclusion.

2.6. A CHARACTERIZATION OF BOUNDED MEASURABLE SETS 65

Definition 2.17 Let E be a bounded set. We say that a proposition is true almosteverywhere on E (or for almost every point of E) if the set of points for which itis false is negligible.

The results proved above have the following simple consequence.

Corollary 2.18 If two functions f and g, defined on the bounded set E, are equalalmost everywhere, then f is integrable on E if and only if such is g. In that case,∫E f =

∫E g.

This last corollary permits us to consider some functions which are defined almosteverywhere, and to define their integral.

Definition 2.19 A function f, defined almost everywhere on E, with real values, issaid to be integrable on E if it can be extended to an integrable function g : E → R.In this case, we set

∫E f =

∫E g.

It can be seen that all the properties and theorems seen before remaintrue for such functions. The reader is invited to verify this.

2.6 A characterization of bounded measurable sets

The following covering lemma will be useful in what follows.

Lemma 2.20 Let E be a set contained in a rectangle I, and let δ be a gauge on E.Then, there is a finite or countable family of nonoverlapping rectangles Jk, containedin I, whose union covers the set E, with the following property: in each of the setsJk there is a point xk belonging to E such that Jk ⊆ B[xk, δ(xk)].

Proof We consider for simplicity the case N = 2. Let us divide the rectangle Iin four rectangles, having the same areas, by the axes of its edges. We proceedanalogously with each of these four rectangles, obtaining sixteen smaller rectangles,and so on. We thus obtain a countable family of smaller and smaller rectangles. Forevery point x of E we can choose one of these rectangles which contains x and isitself contained in B[x, δ(x)]. These rectangles would satisfy the properties of thestatement, if they were non overlapping.

In order that the sets Jk be non-overlapping, it is necessary to make a carefulchoice of them, and here is how to do it. We first choose those from the beginningfour, if there are any, which contain a point xk belonging to E such that Jk ⊆B[xk, δ(xk)]; once this choice has been made, we eliminate all the smaller rectanglescontained in them. We consider then the sixteen smaller ones and, among the ones


which remained after the first elimination procedure, we choose those, if there areany, which contain a point xk belonging to E such that Jk ⊆ B[xk, δ(xk)]; once thischoice has been made, we eliminate all the smaller rectangles contained in them;and so on.

Remark. Notice that if, in the assumptions of the covering lemma, it happens thatE is contained in an open set which is itself contained in I, then all the rectanglesJk can be chosen so to be all contained in that open set.

We can now prove the following characterization of the bounded measurablesets.1

Proposition 2.21 Let E be a bounded set, contained in a rectangle I. The threefollowing propositions are equivalent:

(i) the set E is measurable;

(ii) for every ε > 0 there are two finite or countable families (Jk) and (J ′k), eachmade of [non-overlapping] rectangles contained in I, such that

E ⊆⋃k

Jk , I \ E ⊆⋃k

J ′k and µ

((⋃k

Jk

)∩

(⋃k

J ′k

))≤ ε ;

(iii) there are two sequences (En)n≥1 and (E′n)n≥1 of bounded and measurablesubsets such that

E′n ⊆ E ⊆ En , limn→∞

(µ(En)− µ(E′n)) = 0 .

In that case, we have:

µ(E) = limn→∞

µ(En) = limn→∞

µ(E′n) .

Proof Let us first prove that (i) implies (ii). Assume that E be measurable, andfix ε > 0. By the Saks–Henstock’s theorem, there is a gauge δ on I such that, forevery δ-fine sub-P-partition Π = (xj ,Kj) : j = 1, . . . ,m of I,

m∑j=1

∣∣∣∣∣χE(xj)µ(Kj)−∫Kj

χE

∣∣∣∣∣ ≤ ε

2.

By the covering lemma, there is a family of non-overlapping rectangles Jk, containedin I, whose union covers E and in each Jk there is a point xk belonging to Esuch that Jk ⊆ B[xk, δ(xk)]. Let us fix a positive integer N and consider only

1In the following statements, the words in squared brakets may be omitted.

2.6. A CHARACTERIZATION OF BOUNDED MEASURABLE SETS 67

(x1, J1), . . . , (xN , JN ). They make up a δ-fine sub-P-partition of I. From the aboveinequality we then deduce that

N∑k=1

∣∣∣∣µ(Jk)−∫Jk

χE

∣∣∣∣ ≤ ε

2,

whenceN∑k=1

µ(Jk) ≤N∑k=1

∫Jk

χE +ε

2≤∫IχE +

ε

2= µ(E) +

ε

2.

Since this holds for every positive integer N, we have thus constructed a family (Jk)of non-overlapping rectangles, such that

E ⊆⋃k

Jk ,∑k

µ(Jk) ≤ µ(E) +ε

2.

Consider now I \ E, which is measurable, as well. We can repeat the same proce-dure as above replacing E by I \ E, thus finding a family (J ′k) of non-overlappingrectangles, contained in I, such that

I \ E ⊆⋃k

J ′k ,∑k

µ(J ′k) ≤ µ(I \ E) +ε

2.

Consequently,

I \

(⋃k

J ′k

)⊆ E ⊆

⋃k

Jk ,

and hence

µ

((⋃k

Jk

)∩

(⋃k

J ′k

))= µ

((⋃k

Jk

)\

(I \

(⋃k

J ′k

)))

= µ

(⋃k

Jk

)− µ

(I \

(⋃k

J ′k

))

= µ

(⋃k

Jk

)− µ(I) + µ

(⋃k

J ′k

)≤(µ(E) +

ε

2

)− µ(I) +

(µ(I \ E) +

ε

2

)= ε ,

and the implication is thus proved.


Taking ε = 1n , it is easy to see that (ii) implies (iii). Let us prove now that (iii)

implies (i). Consider the measurable sets

E′ =⋃n≥1

E′n , E =⋂n≥1

En ,

for which it has to be

E′ ⊆ E ⊆ E , µ(E′) = µ(E) .

Equivalently, we have

χE′ ≤ χE ≤ χE ,

∫I(χE− χ

E′) = 0 ,

so that χE′ = χE = χ

Ealmost everywhere. Then, E is measurable and µ(E) =

µ(E′) = µ(E). Moreover,

0 ≤ limn

[µ(E)− µ(E′n)] ≤ limn

[µ(En)− µ(E′n)] = 0 ,

hence µ(E) = limn µ(E′n). Analogously it is seen that µ(E) = limn µ(En), and theproof is thus completed.

Proposition 2.22 Let E be a bounded set. Then, E is negligible if and only if, forevery ε > 0 there is a finite or countable family (Jk) of [non-overlapping] rectanglessuch that

E ⊆⋃k

Jk ,∑k

µ(Jk) ≤ ε .

Proof The necessary condition is proved in the first part of the previous proposition.

Let us prove the sufficiency. Once fixed ε > 0, assume there exists a family (Jk)with the given properties. Let I be a rectangle containing the set E. On the otherhand, consider a family (J ′k) whose elements all coincide with I. The conditions ofthe previous proposition are then satisfied, so that E is indeed measurable. Then,

µ(E) ≤ µ

(⋃k

Jk

)≤∑k

µ(Jk) ≤ ε ;

being ε arbitrary, it has to be µ(E) = 0.

Remark. Observe that if E is contained in an open set which is itself contained ina rectangle I, all the rectangles Jk can be chosen so to be all contained in that openset.

2.7. CONTINUOUS FUNCTIONS AND L-INTEGRABLE FUNCTIONS 69

As a consequence of the previous proposition, it is not difficult to prove thefollowing.

Corollary 2.23 If IN−1 is a rectangle in RN−1 and T is a negligible subset of R,then IN−1 × T is negligible in RN .

Proof Fix ε > 0 and, according to Proposition 2.22, let (Jk) be a finite or countablefamily of intervals in R such that

T ⊆⋃k

Jk ,∑k

µ(Jk) ≤ε

µ(IN−1).

Defining the rectangles Jk = IN−1 × Jk, we have that

IN−1 × T ⊆⋃k

Jk ,∑k

µ(Jk) = µ(IN−1)∑k

µ(Jk) ≤ µ(IN−1)ε

µ(IN−1)= ε ,

and Proposition 2.22 applies again.

2.7 Continuous functions and L-integrable functions

We begin this section by showing that the continuous functions are L-integrable oncompact sets.

Theorem 2.24 Let E ⊆ RN be a compact set and f : E → R be a continuousfunction. Then, f is L-integrable on E.

Proof We consider for simplicity the case N = 2. Being f continuous on a compactset, there is a constant C > 0 such that

|f(x)| ≤ C ,

for every x ∈ E. Let I be a rectangle containing E. First we divide I into fourrectangles, by tracing the segments joining the mid points of its edges; we denote byU1,1, U1,2, U1,3, U1,4 these subrectangles. We now divide again each of these rectanglesin the same way, thus obtaining sixteen smaller subrectangles, which we denote byU2,1, U2,2, . . . , U2,16. Proceeding in this way, for every k we will have a subdivisionof the rectangle I in 22k small rectangles Uk,j , with j = 1, . . . , 22k. Whenever E has

non-empty intersection with U

k,j , we choose and fix a point xk,j ∈ E ∩ U

k,j . Definenow the function fk in the following way:

• if E ∩ U

k,j is non-empty, fk is constant on U

k,j with value f(xk,j);

• if E ∩ U

k,j is empty, fk is constant on U

k,j with value 0.


The functions fk are thus defined almost everywhere on I, not being defined onlyon the points of the grid made up by the above constructed segments, which form acountable family of negligible sets. The functions fk are integrable on each subrect-angle Uk,j , being constant in its interior. By the property of additivity on subrect-angles, these functions are therefore integrable on I. Moreover,

|fk(x)| ≤ C ,

for almost every x ∈ I and every k ≥ 1. Let us see now that fk converges pointwisealmost everywhere to fE . Indeed, taking a point x ∈ I not belonging to the grid,

for every k there is a j = j(k) for which x ∈ U

k,j(k); we have two possibilities:

a) x 6∈ E; in this case, being E closed, we have that, for k sufficiently large,

U

k,j(k) (whose dimensions tend to zero as k →∞) will have empty intersection with

E, and then fk(x) = 0 = fE(x).

b) x ∈ E; in this case, if k → +∞, we have that xk,j(k) → x (again using the

fact that U

k,j(k) has dimensions tending to zero). By the continuity of f, we havethat

fk(x) = f(xk,j(k))→ f(x) = fE(x) .

The dominated convergence theorem then yields the conclusion.

We now see that the L-integrability is conserved on measurable subsets.

Theorem 2.25 Let f : E → R be a L-integrable function on a bounded set E. Then,f is L-integrable on every measurable subset of E.

Proof Assume first f to have non-negative values. Let S be a measurable subsetof E, and define on E the functions fk = minf, kχS. They form an increasingsequence of L-integrable functions, since such are both f and kχS , which convergespointwise to fS . Moreover, it is ∫

Efk ≤

∫Ef ,

for every k. The monotone convergence theorem then guarantees that f is integrableon S in this case. In the general case, f being L-integrable, both f+ and f− areL-integrable on E. Hence, by the above, they are both L-integrable on S, and thensuch is f , too.

Let us prove now the property of complete additivity of the integral forL-integrable functions. We will say that two bounded measurable subsets arenon-overlapping if their intersection is a negligible set.

2.7. CONTINUOUS FUNCTIONS AND L-INTEGRABLE FUNCTIONS 71

Theorem 2.26 Let (Ek) be a finite or countable family of measurable non overlap-ping sets whose union is a bounded set E. Then f is L-integrable on E if and onlyif the two following conditions hold:

(a) f is L-integrable on each Ek ;

(b)∑

k

∫Ek|f(x)| dx < +∞.

In that case, we have: ∫Ef =

∑k

∫Ek

f .

Proof Observe that

f(x) =∑k

fEk(x) , |f(x)| =∑k

|fEk(x)| ,

for almost every x ∈ E. If f is L-integrable on E, from the preceding theorem (a)follows. Moreover, it is obvious that (b) holds whenever the sets Ek are in a finitenumber. If instead they are infinite, for any fixed n, we have

n∑k=1

∫Ek

|f(x)| dx =n∑k=1

∫E|fEk(x)| dx ≤

∫E|f(x)| dx ,

and (b) follows.Assume now that (a) and (b) hold. If the sets Ek are in a finite number, it is

sufficient to integrate on E both terms in the equation f =∑

k fEk . If instead theyare infinite, assume first that f has non-negative values. In this case, the corollaryfollowing the monotone convergence theorem, when applied to the series

∑k fEk ,

yields the conclusion. In the general case, it is sufficient to consider, as usual, thepositive and the negative parts of f .

Exercises

1. Compute the area of the set

(x, y) ∈ R2 : 0 ≤ x ≤ 1 , 0 ≤ y ≤ x .

2. Compute the following integral:∫[0,1]×[0,1]

x2 dx dy .

3. Prove that the function f : ]0, 1]× [0, 1]→ R, defined as

f(x, y) =√k if (x, y) ∈

]1

k + 1,

1

k

]× [0, 1] ,

is integrable, and give an estimate of its integral.


2.8 Limits and derivatives under the integration sign

Let X be a metric space, Y a bounded subset of RN, and consider a function f :X × Y → R. (For simplicity, we may think of X and Y as subsets of R.) The firstquestion we want to face is the following: when does the formula

limx→x0

(∫Yf(x, y) dy

)=

∫Y

(limx→x0

f(x, y))dy

hold? The following is a generalization of the dominated convergence theorem.

Theorem 2.27 Let x0 be an accumulation point of X, and assume that:

(i) for every x ∈ X \ x0, the function f(x, ·) is integrable on Y, so that we candefine the function

F (x) =

∫Yf(x, y) dy ;

(ii) for almost every y ∈ Y the limit limx→x0 f(x, y) exists and is finite, so thatwe can define almost everywhere the function

η(y) = limx→x0

f(x, y) ;

(iii) there are two integrable functions g, h : Y → R such that

g(y) ≤ f(x, y) ≤ h(y) ,

for every x ∈ X \ x0 and almost every y ∈ Y.Then, η is integrable on Y, and we have:

limx→x0

F (x) =

∫Yη(y) dy .

Proof Let us take a sequence (xk)k in X \ x0 which tends to x0. Define, for everyk, the functions fk : Y → R such that fk(y) = f(xk, y). By the assumptions (i), (ii)and (iii) we can apply the dominated convergence theorem, so that

limkF (xk) = lim

k

(∫Yfk(y) dy

)=

∫Y

(limkfk(y)

)dy =

∫Yη(y) dy .

The conclusion then follows from the characterization of the limit by the use ofsequences.

2.8. LIMITS AND DERIVATIVES UNDER THE INTEGRATION SIGN 73

We have the following consequence of the proposition just proved.

Corollary 2.28 If X is a subset of RM, Y ⊆ RN is compact, and f : X × Y → Ris continuous, then the function F : X → R, defined by

F (x) =

∫Yf(x, y) dy ,

is continuous.

Proof The function F (x) is well defined, being f(x, ·) continuous on the compactset Y. Let us fix x0 ∈ X and prove that F is continuous at x0. By the continuityof f,

η(y) = limx→x0

f(x, y) = f(x0, y) .

Moreover, once taken a compact neighborhood U of x0, there is a constant C > 0such that |f(x, y)| ≤ C, for every (x, y) ∈ U × Y. The previous theorem can then beapplied, and we have:

limx→x0

F (x) =

∫Yf(x0, y) dy = F (x0) ,

thus proving that F is continuous at x0.

Let now X be a subset of R. The second question we want to face is the following:when does the formula

d

dx

(∫Yf(x, y) dy

)=

∫Y

(∂f

∂x(x, y)

)dy

hold? The following result is often quoted as the Leibniz rule.

Theorem 2.29 Let X be an interval in R containing x0, and assume that:

(i) for every x ∈ X, the function f(x, ·) is integrable on Y, so that we can definethe function

F (x) =

∫Yf(x, y) dy ;

(ii) for every x ∈ X and almost every y ∈ Y, the partial derivative ∂f∂x (x, y)

exists;

(iii) there are two integrable functions g, h : Y → R such that

g(y) ≤ ∂f

∂x(x, y) ≤ h(y) ,

for every x ∈ X and almost every y ∈ Y.


Then, the function ∂f∂x (x, ·), defined almost everywhere on Y, is integrable there, the

derivative of F in x0 exists, and we have:

F ′(x0) =

∫Y

(∂f

∂x(x0, y)

)dy .

Proof We define, for x ∈ X different from x0, the function

ψ(x, y) =f(x, y)− f(x0, y)

x− x0.

For every x ∈ X \ x0, the function ψ(x, ·) is integrable on Y. Moreover, for almostevery y ∈ Y, it is

limx→x0

ψ(x, y) =∂f

∂x(x0, y) .

By the Lagrange Mean Value Theorem, for (x, y) as above there is a ξ ∈ X betweenx0 and x such that

ψ(x, y) =∂f

∂x(ξ, y) .

By assumption (iii), we then have

g(y) ≤ ψ(x, y) ≤ h(y) ,

for every x ∈ X \x0 and almost every y ∈ Y. We are then in the hypotheses of theprevious theorem, and we can conclude that the function ∂f

∂x (x0, ·), defined almosteverywhere on Y, is integrable there, and

limx→x0

(∫Yψ(x, y) dy

)=

∫Y

(∂f

∂x(x0, y)

)dy .

On the other hand,

limx→x0

(∫Yψ(x, y) dy

)= limx→x0

(∫Y

f(x, y)− f(x0, y)

x− x0dy

)= limx→x0

1

x− x0

(∫Yf(x, y) dy −

∫Yf(x0, y) dy

)= limx→x0

F (x)− F (x0)

x− x0,

so that F is differentiable at x0 and the conclusion holds true.

2.8. LIMITS AND DERIVATIVES UNDER THE INTEGRATION SIGN 75

Corollary 2.30 If X is an interval in R, Y is a compact subset of RN, and thefunction f : X × Y → R is continuous and has a continuous partial derivative∂f∂x : X × Y → R, then the function F : X → R, defined by

F (x) =

∫Yf(x, y) dy ,

is differentiable with a continuous derivative.

Proof The function F (x) is well defined, being f(x, ·) continuous on the compactset Y. Taking a point x0 ∈ X and a nontrivial compact interval U ⊆ X, containingit, there is a constant C > 0 such that |∂f∂x (x, y)| ≤ C, for every (x, y) ∈ U × Y. Bythe preceding theorem, F is differentiable at x0. The same argument holds replacingx0 by any x ∈ X, and

F ′(x) =

∫Y

(∂f

∂x(x, y)

)dy .

The continuity of F ′ : X → R now follows from Corollary 2.28.

Example. Consider, for x ≥ 0, the function

f(x, y) =e−x

2(y2+1)

y2 + 1.

We want to see if the corresponding function F (x) =∫ 1

0f(x, y) dy is differentiable and, in this case,

to find its derivative. We have that

∂f

∂x(x, y) = −2xe−x

2(y2+1) ,

which, for y ∈ [0, 1] and x ≥ 0, is such that

−√

2

e≤ −2xe−x

2

≤ −2xe−x2(y2+1) ≤ 0 .

We can then apply the Leibniz rule, so that

F ′(x) = −2x

∫ 1

0

e−x2(y2+1) dy .

Let us make a digression, so to present a nice formula. By the change of variable t = xy, onehas

−2x

∫ 1

0

e−x2(y2+1) dy = −2e−x

2∫ x

0

e−t2

dt = − d

dx

(∫ x

0

e−t2

dt

)2

.

Taking into account that F (0) = π/4, we have

F (x) =π

4−(∫ x

0

e−t2

dt

)2

.

We would like now to pass to the limit for x→ +∞. Since, for x ≥ 0, one has

0 ≤ e−x2(y2+1)

y2 + 1≤ 1 ,


we can pass to the limit under the sign of integration, thus obtaining

limx→+∞

∫ 1

0

e−x2(y2+1)

y2 + 1dy =

∫ 1

0

(lim

x→+∞

e−x2(y2+1)

y2 + 1

)dy = 0 .

Hence, (∫ +∞

0

e−t2

dt

)2

=π

4

and, by symmetry, ∫ +∞

−∞e−t

2

dt =√π ,

which is a very useful formula in various applications.

2.9 The reduction formula

Before stating the main theorem, it is useful to first prove a preliminary result.

Proposition 2.31 Let f : I → R be an integrable function on the rectangle I =[a1, b1]× [a2, b2]. Then, for almost every x ∈ [a1, b1], the function f(x, ·) is integrableon [a2, b2].

Proof Let T ⊆ [a1, b1] be the set of those x ∈ [a1, b1] for which f(x, ·) is notintegrable on [a2, b2]. Let us prove that T is a negligible set. For each x ∈ T , theCauchy condition does not hold. Hence, if we define the sets

Tn =

x :

for every gauge δ2 on [a2, b2] there are two

δ2-fine P-partitions Π2 and Π2 of [a2, b2] such that

S([a2, b2], f(x, ·),Π2)− S([a2, b2], f(x, ·), Π2) > 1n

,

we have that each x ∈ T belongs to Tn, if n is sufficiently large. So, T is the unionof all Tn, and if we prove that any Tn is negligible, by the properties of the measurewe will also have that T is such. In order to do so, let us consider a certain Tn andfix ε > 0. Being f integrable on I, there is a gauge δ on I such that, taken two δ-fineP-partitions Π and Π of I, it is

|S(I, f,Π)− S(I, f, Π)| ≤ ε

n.

The gauge δ on I determines, for every x ∈ [a1, b1], a gauge δ(x, ·) on [a2, b2]. Wenow associate to each x ∈ [a1, b1] two δ(x, ·)-fine P-partitions Πx

2 and Πx2 of [a2, b2]

in the following way:- if x ∈ Tn, we can choose Πx

2 and Πx2 such that

S([a2, b2], f(x, ·),Πx2)− S([a2, b2], f(x, ·), Πx

2) >1

n;

- if instead x 6∈ Tn, we take Πx2 and Πx

2 equal to each other.

2.9. THE REDUCTION FORMULA 77

Let us write the two P-partitions Πx2 and Πx

2 thus determined:

Πx2 = (yxj ,Kx

j ) : j = 1, . . . ,mx , Πx2 = (yxj , Kx

j ) : j = 1, . . . , mx .

We define a gauge δ1 on [a1, b1], setting

δ1(x) = minδ(x, yx1 ), . . . , δ(x, yxmx), δ(x, yx1 ), . . . , δ(x, yxmx) .

Let now Π1 = (xi, Ji) : i = 1, . . . , k be a δ1-fine P-partition of [a1, b1]. We want toprove that S([a1, b1], χTn ,Π1) ≤ ε, i.e.∑

i : xi∈Tn

µ(Ji) ≤ ε .

To this aim, define the following two P-partitions of I which make use of the elementsof Π1:

Π = ((xi, yxij ), Ji ×Kxij ) : i = 1, . . . , k , j = 1, . . . ,mxi ,

Π = ((xi, yxij ), Ji × Kxij ) : i = 1, . . . , k , j = 1, . . . , mxi .

They are δ-fine, and hence

|S(I, f,Π)− S(I, f, Π)| ≤ ε

n.

On the other hand, we have

|S(I, f,Π)−S(I, f, Π)| =

=

∣∣∣∣∣k∑i=1

mxi∑j=1

f(xi, yxij )µ(Ji ×Kxi

j )−k∑i=1

mxi∑j=1

f(xi, yxij )µ(Ji × Kxi

j )

∣∣∣∣∣=

∣∣∣∣∣∣k∑i=1

µ(Ji)

mxi∑j=1

f(xi, yxij )µ(Kxi

j )−mxi∑j=1

f(xi, yxij )µ(Kxi

j )

∣∣∣∣∣∣=

∣∣∣∣∣k∑i=1

µ(Ji)[S([a2, b2], f(xi, ·),Πxi2 )− S([a2, b2], f(xi, ·), Πxi

2 )]

∣∣∣∣∣=

∑i : xi∈Tn

µ(Ji)[S([a2, b2], f(xi, ·),Πxi2 )− S([a2, b2], f(xi, ·), Πxi

2 )] .

Recalling that

S([a2, b2], f(xi, ·),Πxi2 )− S([a2, b2], f(xi, ·), Πxi

2 ) >1

n,


we conclude that

ε

n≥ |S(I, f,Π)− S(I, f, Π)| > 1

n

∑i : xi∈Tn

µ(Ji) ,

and hence S([a1, b1], χTn ,Π1) ≤ ε, which is what we wanted to prove. All this showsthat the sets Tn are negligible, and therefore T is negligible, too.

The following Reduction Theorem, due to G. Fubini, permits to compute theintegral of an integrable function of two variables by performing two integrations offunctions of one variable.

Theorem 2.32 Let f : I → R be an integrable function on the rectangle I =[a1, b1]× [a2, b2]. Then:

(i) for almost every x ∈ [a1, b1], the function f(x, ·) is integrable on [a2, b2];

(ii) the function∫ b2a2f(·, y) dy, defined almost everywhere on [a1, b1], is integrable

there;

(iii) we have: ∫If =

∫ b1

a1

(∫ b2

a2

f(x, y) dy

)dx .

Proof We have already proved (i) in the preliminary proposition. Let us now prove(ii) and (iii). Let T be the negligible subset of [a1, b1] such that, for x ∈ T, thefunction f(x, ·) is not integrable on [a2, b2]. Being T × [a2, b2] negligible in I, we canmodify on that set the function f without changing the integrability properties. Wecan set, for example, f = 0 on that set. In this way, we can assume without loss ofgenerality that T be empty. Let us define

F (x) =

∫ b2

a2

f(x, y) dy .

We want to prove that F is integrable on [a1, b1] and that∫ b1

a1

F =

∫If .

Let ε > 0 be fixed. Because of the integrability of f on I, there is a gauge δ on Isuch that, for every δ-fine P-partition Π of I,∣∣∣∣S(I, f,Π)−

∫If

∣∣∣∣ ≤ ε

2.


We now associate to each x ∈ [a1, b1] a δ(x, ·)-fine P-partition Πx2 of [a2, b2] such

that

|S([a2, b2], f(x, ·),Πx2)− F (x)| ≤ ε

2(b1 − a1)

(this is possible since f(x, ·) is integrable on [a2, b2] with integral F (x)). Let us writethe P-partitions thus determined:

Πx2 = (yxj ,Kx

j ) : j = 1, . . . ,mx .

We define a gauge δ1 on [a1, b1], by setting

δ1(x) = minδ(x, yx1 ), . . . , δ(x, yxmx) .

Consider now a δ1-fine P-partition of [a1, b1] :

Π1 = (xi, Ji) : i = 1, . . . , n ,

and construct the following δ-fine P-partition of I, making use of the elements of Π1 :

Π = ((xi, yxij ), Ji ×Kxij ) : i = 1, . . . , n , j = 1, . . . ,mxi .

We have the following inequalities:∣∣∣∣S([a1, b1], F,Π1)−∫If

∣∣∣∣ ≤≤ |S([a1, b1], F,Π1)− S(I, f,Π)|+

∣∣∣∣S(I, f,Π)−∫If

∣∣∣∣≤

∣∣∣∣∣n∑i=1

F (xi)µ(Ji)−n∑i=1

mxi∑j=1

f(xi, yxij )µ(Ji ×Kxi

j )

∣∣∣∣∣+ε

2

≤n∑i=1

∣∣∣∣∣F (xi)−mxi∑j=1

f(xi, yxij )µ(Kxi

j )

∣∣∣∣∣µ(Ji) +ε

2

≤n∑i=1

ε

2(b1 − a1)µ(Ji) +

ε

2= ε .

This proves that F is integrable on [a1, b1] and∫ b1

a1

F =

∫If .



Example. Consider the function f(x, y) = x2 sin y on the rectangle I = [−1, 1] × [0, π]. Being fcontinuous on a compact set, it is integrable there, so that:∫

I

f =

∫ 1

−1

(∫ π

0

x2 sin y dy

)dx

=

∫ 1

−1

x2[− cos y]π0 dx = 2

∫ 1

−1

x2 dx = 2

[x3

3

]1−1

=4

3.

Clearly, the following version of Fubini Theorem holds, which is symmetric withrespect to the preceding one.

Theorem 2.33 Let f : I → R be an integrable function on the rectangle I =[a1, b1]× [a2, b2]. Then:

(i) for almost every y ∈ [a2, b2], the function f(·, y) is integrable on [a1, b1];

(ii) the function∫ b1a1f(x, ·) dx, defined almost everywhere on [a2, b2], is integrable

there;

(iii) we have: ∫If =

∫ b2

a2

(∫ b1

a1

f(x, y) dx

)dy .

As an immediate consequence, we have that, if f is integrable on I = [a1, b1] ×[a2, b2], then ∫ b1

a1

(∫ b2

a2

f(x, y) dy

)dx =

∫ b2

a2

(∫ b1

a1

f(x, y) dx

)dy .

Therefore, if the above equality does not hold, then the function f is not integrableon I.

Examples. Consider the function

f(x, y) =

x2 − y2

(x2 + y2)2if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0) ,

on the rectangle I = [0, 1]× [0, 1]. If x 6= 0, it is∫ 1

0

x2 − y2

(x2 + y2)2dy =

[y

x2 + y2

]y=1

y=0

=1

x2 + 1,

so that ∫ 1

0

(∫ 1

0

x2 − y2

(x2 + y2)2dy

)dx =

∫ 1

0

1

x2 + 1dx = [arctanx]10 =

π

4.

Analogously, we see that ∫ 1

0

(∫ 1

0

x2 − y2

(x2 + y2)2dx

)dy = −π

4,

and we thus conclude that f is not integrable on I.


As a further example, consider the function

f(x, y) =

xy

(x2 + y2)2if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0) ,

on the rectangle I = [−1, 1]× [−1, 1]. In this case, if x 6= 0, we have∫ 1

−1

xy

(x2 + y2)2dy =

[−x

2(x2 + y2)

]y=1

y=−1

= 0 ,

so that ∫ 1

−1

(∫ 1

−1

xy

(x2 + y2)2dy

)dx = 0 .

Analogously, we see that ∫ 1

−1

(∫ 1

−1

xy

(x2 + y2)2dx

)dy = 0 .

Nevertheless, we are not allowed to conclude that f is integrable on I. Truly, it is not at all. Indeed,if f were integrable, it should be such on every subrectangle, and in particular on [0, 1]× [0, 1]. But,if x 6= 0, we have ∫ 1

0

xy

(x2 + y2)2dy =

[−x

2(x2 + y2)

]y=1

y=0

=1

2x(x2 + 1),

which is not integrable with respect to x on [0, 1].

When the function f is defined on a bounded subset E of R2, it is possible tostate the Reduction Theorem for the function fE . Let I = [a1, b1] × [a2, b2] be arectangle containing E. Let us define the sections of E :

Ex = y ∈ [a2, b2] : (x, y) ∈ E , Ey = x ∈ [a1, b1] : (x, y) ∈ E ,

and the projections of E :

P1E = x ∈ [a1, b1] : Ex 6= Ø , P2E = y ∈ [a2, b2] : Ey 6= Ø .

Ex

P1E

E

x

We can then reformulate Fubini Theorem in the following way.


Theorem 2.34 Let f : E → R be an integrable function on the bounded set E.Then:(i) for almost every x ∈ P1E, the function fE(x, ·) is integrable on the set Ex;(ii) the function x 7→

∫Exf(x, y) dy, defined almost everywhere on P1E, is integrable

there;(iii) we have: ∫

Ef =

∫P1E

(∫Ex

f(x, y) dy

)dx .

Analogously, the function y 7→∫Eyf(x, y) dx, defined almost everywhere on P2E, is

integrable there, and ∫Ef =

∫P2E

(∫Ey

f(x, y) dx

)dy .

Example. Consider the function f(x, y) = |xy| on the set

E = (x, y) ∈ R2 : 0 ≤ x ≤ 1,−x2 ≤ y ≤ x2 .

Being f continuous and E compact, the theorem applies; we have P1E = [0, 1] and, for everyx ∈ P1E, Ex = [−x2, x2]. Hence:∫

E

f =

∫ 1

0

(∫ x2

−x2|xy| dy

)dx

=

∫ 1

0

|x|[y2

2

]x2−x2

dx =

∫ 1

0

x5 dx =

[x6

6

]10

=1

6.

As a corollary, we have a method to compute the measure of a bounded measur-able set.

Corollary 2.35 If E ⊆ R2 is a bounded and measurable set, then:

(i) for almost every x ∈ P1E, the set Ex is measurable;

(ii) the function x 7→ µ(Ex), defined almost everywhere on P1E, is integrablethere;

(iii) we have:

µ(E) =

∫P1E

µ(Ex) dx .

Analogously, the function y 7→ µ(Ey), defined almost everywhere on P2E, is inte-grable there, and

µ(E) =

∫P2E

µ(Ey) dy .


Example. Let us compute the area of the disk with radius R > 0 : let E = (x, y) ∈ R2 : x2 + y2 ≤R2. Being E a compact set, it is measurable. We have that P1E = [−R,R] and, for every x ∈ P1E,Ex = [−

√R2 − x2,

√R2 − x2]. Hence:

µ(E) =

∫ R

−R2√R2 − x2 dx =

∫ π/2

−π/22R2 cos2 t dt

= R2 [t+ cos t sin t]π/2

−π/2 = πR2 .

In the case of functions of more than two variables, analogous results to thepreceding ones hold true, with the same proofs. One simply needs to separate thevariables in two different groups, calling x the first group and y the second one, andthe same formulas hold true.

Example. We want to compute the volume of the three-dimensional ball with radius R > 0 : letE = (x, y, z) ∈ R3 : x2 + y2 + z2 ≤ R2. Let us group together the variables (y, z) and considerthe projection on the x-axis: P1E = [−R,R]. The sections Ex then are disks of radius

√R2 − x2,

and we have:

µ(E) =

∫ R

−Rπ(R2 − x2) dx = π(2R3)− π

[x3

3

]R−R

=4

3πR3 .

Another way to compute the same volume is to group the variables (x, y) and consider P1E =(x, y) : x2 + y2 = R2. For every (x, y) ∈ P1E, it is

E(x,y) =[−√R2 − x2 − y2,

√R2 − x2 − y2

],

so that

µ(E) =

∫P1E

2√R2 − x2 − y2 dx dy =

∫ R

−R

(∫ √R2−x2

−√R2−x2

2√R2 − x2 − y2 dy

)dx

=

∫ R

−R

(∫ π/2

−π/22(R2 − x2) cos2 t dt

)dx =

∫ R

−Rπ(R2 − x2) dx =

4

3πR3 ,

by the change of variable t = arcsin(y/√R2 − x2

).

Iterating the above reduction procedure, it is possible to prove, for a function ofN variables which is integrable on a rectangle

I = [a1, b1]× [a2, b2]× . . .× [aN , bN ] ,

formulas like∫If =

∫ b1

a1

(∫ b2

a2

(. . .

∫ bN

aN

f(x1, x2, . . . , xN ) dxN . . .

)dx2

)dx1 .

Exercises


1. Compute the following integrals:∫[0,1]×[2,3]

x2y dx dy ,

∫[−1,1]×[1,2]

y exy2dx dy ,

∫[−1,1]×[−2,2]

sin(x3) arctan(x2y) dx dy .

2. Compute, for any α ≥ 1, the area of the set

E = (x, y) ∈ R2 : 1 ≤ x ≤ α, x2 − y2 ≥ 1 .

3. Compute the integral ∫E

sin2(xy) dx dy ,

with the same set E defined above.

4. Compute the integral ∫E

(x2 + y2) dx dy ,

where E = (x, y) ∈ R2 : x2 + y2 ≤ R2, for some R > 0.

5. Compute the volume of the set(x, y, z) ∈ R3 :

x2

a2+y2

b2+z2

c2≤ 1

,

where a, b, c are positive constants (geometrically corresponding to the three semi-axes of an ellipsoid).

2.10 Change of variables in the integral

In this section we look for an analogue to the formula of integration by substitution,which was proved in Chapter 1 for functions of a single variable. The proof ofthat formula was based on the Fundamental Theorem. Since we do not have sucha powerful tool for functions of several variables, actually we will not be able tocompletely generalize that formula.

2.10. CHANGE OF VARIABLES IN THE INTEGRAL 85

For example, the function ϕ will be not only assumed to be differentiable, but wewill need it to be a diffeomorphism between two open sets A and B of RN . In otherwords, ϕ : A → B will be continuously differentiable, invertible, and ϕ−1 : B → Acontinuously differentiable, as well. It is useful to recall that a diffeomorphismtransforms open sets into open sets, closed sets into closed sets and, for every pointq ∈ A, the Jacobian matrix ϕ′(q) is invertible: it is detϕ′(q) 6= 0. Moreover, we willneed the following property.

Lemma 2.36 Let A ⊆ RN be an open set, and ϕ : A → RN be a C1-function; if Sis a subset of A of the type

S = [a1, b1]× . . .× [aN−1, bN−1]× c ,

then ϕ(S) is negligible.

Proof For simplicity, let us concentrate on the case of a subset of R2 of the type

S = [0, 1]× 0 .

Consider the rectangles (indeed squares)

Jk,n =

[k − 1

n,k

n

]×[− 1

2n,

1

2n

],

with k = 1, . . . , n. For n large enough, they are contained in a rectangle R which isitself contained in A. Being R a compact set, there is a constant C > 0 such that‖ϕ′(q)‖ ≤ C, for every q ∈ R. Then, ϕ is Lipschitz continuous on R with Lipschitzconstant C. Since the sets Jk,n have as diameter 1

n

√2, the sets ϕ(Jk,n) are surely

contained in some squares Jk,n whose sides’ lengths are equal to Cn

√2. We then have

that ϕ(S) is covered by the rectangles Jk,n, and

n∑k=1

µ(Jk,n) ≤ n(C

n

√2

)2

=2C2

n.

Since this quantity can be made arbitrarily small, the conclusion follows from Corol-lary 2.22.

As a consequence of the above lemma, it is easy to see that the image of theboundary of a rectangle through a diffeomorphism ϕ is a negligible set. In particular,given two non-overlapping rectangles, their images are non-overlapping sets.

We are now ready to prove a first version of the Theorem on the Change ofVariables in the integral, which will be generalized in a later section.


Theorem 2.37 Let ϕ be a diffeomorphism between two open and bounded sets Aand B = ϕ(A), and f : B → R be a continuous function. Then, for every closedsubset D of A, we have:∫

ϕ(D)f(x) dx =

∫Df(ϕ(u)) |detϕ′(u)| du .

Proof Notice first of all that the integrals appearing in the formula are both mean-ingful, being the sets D and ϕ(D) compact and the considered functions continuous.We will proceed by induction on the dimension N. Let us first consider the caseN = 1.

First of all, using the method of integration by substitution, one verifies thatthe formula is true when D is a compact interval [a, b] : it is sufficient to considerthe two possible cases in which ϕ is increasing or decreasing, and recall that ev-ery continuous function is primitivable. For instance, if ϕ is decreasing, we haveϕ([a, b]) = [ϕ(b), ϕ(a)], so that:∫

ϕ([a,b])f(x) dx =

∫ ϕ(a)

ϕ(b)f(x) dx

=

∫ a

bf(ϕ(u))ϕ′(u) du

=

∫ b

af(ϕ(u))|ϕ′(u)| du

=

∫[a,b]

f(ϕ(u))|ϕ′(u)| du .

Let now R be a closed subset of A whose interior R

contains D. Being both fand (f ϕ)|ϕ′| continuous, they are integrable on the compact sets ϕ(R) and R,

respectively. The open sets R

and R\D can each be split into a countable union of

non-overlapping compact intervals, whose images through ϕ also are non-overlappingclose intervals. By the complete additivity of the integral, the formula holds true

for R

and R\D : ∫

ϕ(R

)f(x) dx =

∫Rf(ϕ(u))|ϕ′(u)| du ,∫

ϕ(R\D)

f(x) dx =

∫R\Df(ϕ(u))|ϕ′(u)| du .

Hence, ∫ϕ(D)

f(x) dx=

∫ϕ(R\(R\D))

f(x) dx


=

∫ϕ(R

)f(x) dx−

∫ϕ(R\D)

f(x) dx

=

∫Rf(ϕ(u))|ϕ′(u)| du−

∫R\Df(ϕ(u))|ϕ′(u)| du

=

∫Df(ϕ(u))|ϕ′(u)| du ,

so that the formula is proved in the case N = 1.

Assume now that the formula holds for the dimension N, and let us prove thatit also holds for N + 1.2 Once we fix a point u ∈ A, at least one of the partialderivatives ∂ϕi

∂uj(u) is non-zero. We can assume without loss of generality that it is

∂ϕN+1

∂uN+1(u) 6= 0. Consider the function

α(u1, . . . , uN+1) = (u1, . . . , uN , ϕN+1(u1, . . . , uN+1)) .

Being detα′(u) =∂ϕN+1

∂uN+1(u) 6= 0, we have that α is a diffeomorphism between an

open neighborhood U of u and an open neighborhood V of α(u). Assume first thatD be contained in U, and set D = α(D).

We define on V the function β = ϕ α−1, which is of the form

β(v1, . . . , vN+1) = (β1(v1, . . . , vN+1), . . . , βN (v1, . . . , vN+1), vN+1) ,

where, for j = 1, . . . , N, it is

βj(v1, . . . , vN+1) = ϕj(v1, . . . , vN , [ϕN+1(v1, . . . , vN , ·)]−1(vN+1)) .

Such a function β is a diffeomorphism between the open sets V and W = ϕ(U).

Consider the sections

Vt = (v1, . . . , vN ) : (v1, . . . , vN , t) ∈ V ,

and the projectionPN+1V = t : Vt 6= Ø .

For t ∈ PN+1V , define the function

βt(v1, . . . , vN ) = (β1(v1, . . . , vN , t), . . . , βN (v1, . . . , vN , t)) ,

which happens to be a diffeomorphism, defined on the open set Vt, whose image isthe open set

Wt = (x1, . . . , xN ) : (x1, . . . , xN , t) ∈W .2At a first reading, it is advisable to consider the transition from N = 1 to N + 1 = 2.


Moreover, detβ′t(v1, . . . , vN ) = detβ′(v1, . . . , vN , t). Consider also the sections

Dt = (v1, . . . , vN ) : (v1, . . . , vN , t) ∈ D ,

and the projection

PN+1D = t : Dt 6= Ø .

Analogously, we consider β(D)t and PN+1β(D). By the definition of β, it is

β(D)t = βt(Dt) , PN+1β(D) = PN+1D .

Using the Reduction Theorem and the inductive assumption, we have:∫β(D)

f =

∫PN+1β(D)

(∫βt(Dt)

f(x1, . . . , xN , t) dx1 . . . dxN

)dt

=

∫PN+1D

(∫Dt

f(βt(v1, . . . , vN ), t) |detβ′t(v1, . . . , vN )| dv1 . . . dvN

)dt

=

∫PN+1D

(∫Dt

f(β(v1, . . . , vN , t)) | detβ′(v1, . . . , vN , t)| dv1 . . . dvN

)dt

=

∫Df(β(v)) | detβ′(v)| dv .

Consider now the function f : V → R defined as

f(v) = f(β(v)) |detβ′(v)| .

Define the sections

Du1,...,uN = uN+1 : (u1, . . . , uN , uN+1) ∈ D ,

and the projection

P1,...,ND = (u1, . . . , uN ) : Du1,...,uN 6= Ø .

In an analogous way we define α(D)u1,...,uN and P1,...,Nα(D). They are all closed setsand, by the definition of α, we have

α(D)u1,...,uN = ϕN+1(u1, . . . , uN , Du1,...,uN ) , P1,...,Nα(D) = P1,...,ND .

Moreover, for every (u1, . . . , uN ) ∈ P1,...,ND, the function defined by

t→ ϕN+1(u1, . . . , uN , t)


is a diffeomorphism of one variable between the open sets Uu1,...,uN and Vu1,...,uN ,sections of U and V, respectively. Using the Reduction Theorem and the one-dimensional formula of change of variables proved above, we have that∫

α(D)f =

∫P1,...,Nα(D)

(∫α(D)u1,...,uN

f(v1, . . . , vN+1) dvN+1

)dv1 . . . dvN

=

∫P1,...,ND

(∫ϕN+1(u1,...,uN ,Du1,...,uN )

f(v1, . . . , vN+1) dvN+1

)dv1 . . . dvN

=

∫P1,...,ND

(∫Du1,...,uN

f(u1, . . . , uN , ϕN+1(u1, . . . , uN+1))·

·

∣∣∣∣∣∂ϕN+1

∂uN+1(u1, . . . , uN+1)

∣∣∣∣∣ duN+1

)du1 . . . duN

=

∫Df(α(u)) | detα′(u)| du .

Hence, being ϕ = β α, we have:∫ϕ(D)

f(x) dx=

∫β(D)

f(x) dx

=

∫Df(β(v)) | detβ′(v)| dv

=

∫α(D)

f(v) dv

=

∫Df(α(u)) | detα′(u)| du

=

∫Df(β(α(u))) | detβ′(α(u))| |detα′(u)| du

=


We have then proved that, for every u ∈ A, there is a δ(u) > 0 such that the thesisholds true when D is contained in B[u, δ(u)]. A gauge δ is thus defined on A. ByLemma 2.20, we can now cover A with a countable family (Jk) of non overlappingrectangles, each contained in a rectangle of the type B[u, δ(u)], so that the formulaholds for the closed sets contained in any of these rectangles.

At this point let us consider an arbitrary closed subset D of A. Then, the formulaholds for each D ∩ Jk and, by the complete additivity of the integral and the factthat the sets ϕ(D ∩ Jk) are non-overlapping (as a consequence of Lemma 2.36), we


have: ∫ϕ(D)

f(x) dx=∑k

∫ϕ(D∩Jk)

f(x) dx

=∑k

∫D∩Jk

f(ϕ(u)) | detϕ′(u)| du

=


The theorem is thus completely proved.

Remark. The formula on the change of variables is often written, setting ϕ(D) = E,in the equivalent form∫

Ef(x) dx =

∫ϕ−1(E)

f(ϕ(u)) | detϕ′(u)| du .

Example. Consider the set

E = (x, y) ∈ R2 : −1 ≤ x ≤ 1, x2 ≤ y ≤ x2 + 1 ,

and let f(x, y) = x2y be a function on it. Defining ϕ(u, v) = (u, v + u2), we have a diffeomorphismwith detϕ′(u, v) = 1. Being ϕ−1(E) = [−1, 1]× [0, 1], by the Theorem on the Change of Variablesand the use of Fubini Theorem we have:∫

E

x2y dx dy =

∫ 1

−1

(∫ 1

0

u2(v + u2) dv

)du =

∫ 1

−1

(u2

2+ u4

)du =

11

15.

2.11 Change of measure by diffeomorphisms

In this section we study how the measure is changed by the action of a diffeomor-phism.

Theorem 2.38 Let ϕ be a diffeomorphism between two open and bounded sets Aand B. If D is a measurable subset of A, then ϕ(D) is measurable, |detϕ′| is inte-grable on D, and

µ(ϕ(D)) =

∫D|detϕ′(u)| du .

Proof By the preceding theorem, the formula holds true whenever D is closed. Sinceevery open set can be written as the union of a countable family of non overlapping(closed) rectangles, by the complete additivity and the fact that A is bounded, theformula holds true even if D is an open set.

2.11. CHANGE OF MEASURE BY DIFFEOMORPHISMS 91

Assume now that D is a measurable set whose closure D is contained in A. Let Rbe a closed subset of A whose interior R

contains D. Then, there is a constant C > 0

such that | detϕ′(u)| ≤ C for every u ∈ R. By Proposition 2.21, for every ε > 0there are two finite or countable families (Jk) and (J ′k), each made of non-overlapping

rectangles contained in R, such that

R\

(⋃k

J ′k

)⊆ D ⊆

⋃k

Jk , µ

((⋃k

Jk

)∩

(⋃k

J ′k

))≤ ε .

Since the formula to be proved holds both on the open sets and on the closed sets,it certainly holds on each rectangle Jk and J ′k; then, it holds on ∪kJk, on ∪kJ ′k, and

since it holds even on R, it has to be true on R

\ (∪kJ ′k), as well. We have thus that

ϕ(∪kJk) and ϕ(R\ (∪kJ ′k)) are measurable,

ϕ

(R\

(⋃k

J ′k

))⊆ ϕ(D) ⊆ ϕ

(⋃k

Jk

),

and

µ

(ϕ

(⋃k

Jk

))− µ

(ϕ

(R\

(⋃k

J ′k

)))=

=

∫∪kJk| detϕ′(u)| du−

∫R\(∪kJ ′k)

|detϕ′(u)| du

=

∫(∪kJk)∩(∪kJ ′k)

| detϕ′(u)| du

≤Cµ

((⋃k

Jk

)∩

(⋃k

J ′k

))≤Cε .

Taking ε = 1n , we find in this way two sequences Dn = ∪kJk,n and D′n = R

\(∪kJ ′k,n)

with the above properties. By Proposition 2.21, we have that ϕ(D) is measurableand µ(ϕ(D)) = limn µ(ϕ(Dn)) = limn µ(ϕ(D′n)). Moreover, since χDn convergesalmost everywhere to χD, by the dominated convergence theorem we have:

µ(ϕ(D)) = limnµ(ϕ(Dn))

= limn

∫Dn

| detϕ′(u)| du

= limn

∫R| detϕ′(u)|χDn(u) du


=

∫R|detϕ′(u)|χD(u) du

=

∫D|detϕ′(u)| du .

We can now consider the case of an arbitrary measurable set D in A. Being Aopen we can consider a sequence of non-overlapping rectangles (Kn) contained in Awhose union is A. The formula holds for each of the sets D∩Kn, by the above. Thecomplete additivity of the integral and the fact that A is bounded then permit usto conclude.

Example. Consider the set

E = (x, y) ∈ R2 : x < y < 2x, 3x2 < y < 4x2 .

One sees that E is measurable, being an open set. Taking

ϕ(u, v) =

(u

v,u2

v

),

we have a diffeomorphism between the set D = ]1, 2[× ]3, 4[ and E = ϕ(D). Moreover,

detϕ′(u, v) = det

(1/v −u/v2

2u/v −u2/v2

)=u2

v3.

Applying the formula on the change of measure and Fubini Theorem, we have:

µ(E) =

∫ 2

1

(∫ 4

3

u2

v3dv

)du =

∫ 2

1

7

288u2 du =

49

864.

2.12 The general theorem on the change of variables

We are now interested in generalizing the Theorem on the Change of Variablesassuming f not necessarily continuous, but only L-integrable on a measurable set.In order to do this, it will be useful to prove the following important relation betweenthe integral of a function having non-negative values and the measure of its epigraph.

Proposition 2.39 Let E be a bounded and measurable set and f : E → R be abounded function with non-negative values. Let Gf be the set thus defined:

Gf = (x, t) ∈ E × R : 0 ≤ t ≤ f(x) .

Then, f is integrable on E if and only if Gf is measurable, in which case

µ(Gf ) =

∫Ef .

2.12. THE GENERAL THEOREM ON THE CHANGE OF VARIABLES 93

Proof Assume first Gf to be measurable. By Fubini Theorem, since P1Gf = E, the

sections being (Gf )x = [0, f(x)], we have that the function x 7→∫ f(x)

0 1 = f(x) isintegrable on E, and

µ(Gf ) =

∫Gf

1 =

∫E

(∫ f(x)

01 dt

)dx =

∫Ef(x) dx .

Assume now f to be integrable on E. Let C > 0 be a constant such that 0 ≤ f(x) <C, for every x ∈ E. Taken a positive integer n, we divide the interval [0, C] in nequal parts and consider, for j = 1, . . . , n, the sets

Ejn =

x ∈ E :

j − 1

nC ≤ f(x) <

j

nC

;

by Corollary 2.13 they are measurable, non overlapping and their union is E. Wecan then define on E the function ψn in the following way:

ψn =

n∑j=1

j

nCχ

Ejn,

and so

Gψn =n⋃j=1

(Ejn ×

[0,j

nC

]).

By Proposition 2.21, it is easy to see that, being the sets Ejn measurable, such are

the sets Ejn ×[0, jnC

], too. Consequently, the sets Gψn are measurable. Moreover,

sinceGf =

⋂n≥1

Gψn ,

even Gf is measurable, and the proof is thus completed.

We are now in the position to prove the second version of the Theorem on theChange of Variables in the integral.

Theorem 2.40 Let ϕ be a diffeomorphism between two bounded and open sets Aand B = ϕ(A) of RN , D a measurable subset of A and f : ϕ(D) → R a function.Then, f is L-integrable on ϕ(D) if and only if (f ϕ) |detϕ′| is L-integrable on D,in which case we have:∫

ϕ(D)f(x) dx =

∫Df(ϕ(u)) | detϕ′(u)| du .

Proof Assume that f be L-integrable on E = ϕ(D). We first consider the case whenf is bounded with non-negative values.


Let C > 0 be such that 0 ≤ f(x) < C, for every x ∈ E. We define the open sets

A = A× ]− C,C[ , B = B× ]− C,C[ ,

and the function ϕ : A→ B in the following way:

ϕ(u1, . . . , un, t) = (ϕ1(u1, . . . , un), . . . , ϕn(u1, . . . , un), t) .

This function is a diffeomorphism and det ϕ′(u, t) = detϕ′(u), for every (u, t) ∈ A.Let Gf be the epigraph of f :

Gf = (x, t) ∈ E × R : 0 ≤ t ≤ f(x) .

Being f L-integrable and E measurable, by the preceding proposition we have thatGf is a measurable set. Moreover,

ϕ−1(Gf ) = (u, t) ∈ D × R : 0 ≤ t ≤ f(ϕ(u)) .

Using the formula on the change of measure and Fubini Theorem, we have

µ(Gf ) =

∫ϕ−1(Gf )

|det ϕ′(u, t)| du dt

=

∫ϕ−1(Gf )

|detϕ′(u)| du dt

=

∫D

(∫ f(ϕ(u))

0|detϕ′(u)| dt

)du

=


On the other hand, by Proposition 2.39, we have that µ(Gf ) =∫ϕ(D) f, and this

proves that the formula holds in case f is bounded with non-negative values.

In the case when f is not bounded but still has non-negative values, we considerthe functions

fk(x) = minf(x), k .For each of them, the formula holds true, and using the monotone convergencetheorem one proves that the formula holds for f even in this case.

When f does not have non-negative values, it is sufficient to consider its positiveand negative parts, apply for them the formula and then subtract.

In order to obtain the opposite implication, it is sufficient to consider (f ϕ) | detϕ′| instead of f and ϕ−1 instead of ϕ, and to apply what has been justproved.

We recall here the equivalent formula∫Ef(x) dx =

∫ϕ−1(E)

f(ϕ(u)) | detϕ′(u)| du .

2.13. SOME USEFUL TRANSFORMATIONS IN R2 95

2.13 Some useful transformations in R2

There are some transformations which do not change the measure of any measurableset. We consider here some of those which are most frequently used in applications.

Translations. We call translation by a given vector a = (a1, a2) ∈ R2, the trans-formation defined by

ϕ(u, v) = (u+ a1, v + a2) .

x

y

It is readily seen that ϕ is a diffeomorphism with detϕ′ = 1, so that, given abounded measurable set D and a L-integrable function f on ϕ(D), we have:∫

ϕ(D)f(x, y) dx dy =

∫Df(u+ a1, v + a2) du dv .

Reflections. A reflection with respect to one of the cartesian axes is defined by

ϕ(u, v) = (−u, v) , or ϕ(u, v) = (u,−v) .

Here detϕ′ = −1, so that, taking for example the first case, we have:∫ϕ(D)

f(x, y) dx dy =

∫Df(−u, v) du dv .

x

y

Rotations. A rotation around the origin by a fixed angle α is given by

ϕ(u, v) = (u cosα− v sinα , u sinα+ v cosα) .


It is a diffeomorphism, with

detϕ′(u, v) = det

(cosα − sinαsinα cosα

)= (cosα)2 + (sinα)2 = 1 .

x

y

Hence, given a measurable set D and a L-integrable function3 f on ϕ(D), we have:∫ϕ(D)

f(x, y) dx dy =

∫Df(u cosα− v sinα , u sinα+ v cosα) du dv .

Another useful transformation is the function ψ : [0,+∞[×[0, 2π[→ R2 given by

ψ(ρ, θ) = (ρ cos θ, ρ sin θ) ,

x

yρ

θ

which defines the so-called polar coordinates in R2. Taken a bounded measurablesubset of R2, let BR be an open ball centered at the origin with radius R whichcontains it. Consider the open sets

A = ]0, R[× ]0, 2π[ , B = BR \ ([0,+∞[×0) .

The function ϕ : A→ B defined by ϕ(ρ, θ) = ψ(ρ, θ) happens to be a diffeomorphismand it is easily seen that detϕ′(ρ, θ) = ρ. We can apply the Theorem on the Change

3Let us mention here that reference [2] contains an ingenious example of a integrable functionin R2 whose rotation by α = π/4 is not integrable. This is why we have restricted our attentiononly to L-integrable functions.

2.14. CYLINDRICAL AND SPHERICAL COORDINATES IN R3 97

of Variables to the set E = E ∩B. Since E and ϕ−1(E) differ from E and ψ−1(E),respectively, by negligible sets, we obtain the following formula on the change ofvariables in polar coordinates:∫

Ef(x, y) dx dy =

∫ψ−1(E)

f(ψ(ρ, θ))ρ dρ dθ .

Example. Let f(x, y) = xy be defined on

E = (x, y) ∈ R2 : x ≥ 0, y ≥ 0, x2 + y2 < 9 .

By the formula on the change of variables in polar coordinates, it is ψ−1(E) = [0, 3[×[0, π2

]; by theReduction Theorem, we can then compute∫

E

f =

∫ π/2

0

(∫ 3

0

ρ3 cos θ sin θ dρ

)dθ =

81

4

∫ π/2

0

cos θ sin θ dθ =81

8.

2.14 Cylindrical and spherical coordinates in R3

We consider the function ξ : [0,+∞[×[0, 2π[×R→ R3 defined by

ξ(ρ, θ, z) = (ρ cos θ, ρ sin θ, z) ,

xρθ

y

z

which gives us the so-called cylindrical coordinates in R3. Taken a bounded andmeasurable set E of R3, let CR× ]−H,H[ be a cylinder containing it, having as basisthe open disk CR centered at the origin with radius R. Consider the open sets

A = ]0, R[× ]0, 2π[× ]−H,H[ ,

B = (CR \ ([0,+∞[×0)× ]−H,H[ .

The function ϕ : A→ B defined by ϕ(ρ, θ, z) = ξ(ρ, θ, z) happens to be a diffeomor-phism and it is easily seen that detϕ′(ρ, θ, z) = ρ. We can then apply the Theoremon the Change of Variables to the set E = E ∩ B. Since E and ϕ−1(E) differ from


E and ξ−1(E), respectively, by negligible sets, we obtain the following formula onthe change of variables in cylindrical coordinates:∫

Ef(x, y, z) dx dy dz =

∫ξ−1(E)

f(ξ(ρ, θ, z))ρ dρ dθ dz .

Example. Let us compute the integral∫Ef, where f(x, y, z) = x2 + y2 and

E = (x, y, z) ∈ R3 : x2 + y2 ≤ 1, 0 ≤ z ≤ x+ y +√

2 .

Passing to cylindrical coordinates, we notice that

ρ cos θ + ρ sin θ +√

2 ≥ 0 ,

for every θ ∈ [0, 2π[ and every ρ ∈ [0, 1]. By the Theorem on the Change of Variables, using FubiniTheorem, we compute∫

E

(x2 + y2) dx dy dz =

∫ξ−1(E)

ρ3 dρ dθ dz

=

∫ 1

0

(∫ 2π

0

(∫ ρ cos θ+ρ sin θ+√2

0

ρ3 dz

)dθ

)dρ

=

∫ 1

0

(∫ 2π

0

ρ3(ρ cos θ + ρ sin θ +√

2) dθ

)dρ

= 2π

∫ 1

0

ρ3√

2 dρ

=π√

2

2.

Consider now the function σ : [0,+∞[×[0, 2π[×[0, π]→ R3 defined by

σ(ρ, θ, φ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ,

ρ

θ

φ

x

y

z

which defines the so-called spherical coordinates in R3. Taken a bounded andmeasurable subset E of R3, let BR be an open three-dimensional ball containing it,centered at the origin with radius R. Consider the open sets

A = ]0, R[× ]0, 2π[× ]0, π[ , B = BR \ ([0,+∞[×0 × R) .

2.14. CYLINDRICAL AND SPHERICAL COORDINATES IN R3 99

The function ϕ : A → B defined by ϕ(ρ, θ, φ) = σ(ρ, θ, φ) happens to be a diffeo-morphism and it can be easily checked that detϕ′(ρ, θ, φ) = ρ2 sinφ. We can thenapply the Theorem on the Change of Variables to E = E ∩B. Since E and ϕ−1(E)differ from E and σ−1(E), respectively, by negligible sets, we obtain the followingformula on the change of variables in spherical coordinates:∫

Ef(x, y, z) dx dy dz =

∫σ−1(E)

f(σ(ρ, θ, φ))ρ2 sinφdρ dθ dφ .

Example. Let us compute the volume of the set

E =

(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ 1, z ≥√x2 + y2

.

We have:

µ(E) =

∫E

1 dx dy dz

=

∫σ−1(E)

ρ2 sinφdρ dθ dφ

=

∫ 1

0

(∫ π/4

0

(∫ 2π

0

ρ2 sinφdθ

)dφ

)dρ

= 2π

∫ 1

0

(∫ π/4

0

ρ2 sinφdφ

)dρ

= 2π

(1−√

2

2

)∫ 1

0

ρ2 dρ

=

(1−√

2

2

)2π

3.

Exercises

1. Let E be a planar set contained in [0,+∞[×R, and define the set

E rot =

(x, y, z) ∈ R3 : (x,√y2 + z2 ) ∈ E

(i.e., the set obtained rotating E around the x-axis). Prove that

µ(E rot) = 2π

∫Ex dx dy .

2. Use the above formula to compute the volume of a sphere: V = 43πR

3. Moreover,prove that the volume of the torus with minor radius r and major radius R is equalto 2π2Rr2.


3. By some modified cylindrical coordinates, prove that the volume of the cone(x, y, z) ∈ R3 :

√(xa

)2+(yb

)2≤ z

h≤ 1

,

where a, b and h are positive constants, is equal to 13(πab)h.

4. By some modified spherical coordinates, compute again the volume of the ellipsoid(x, y, z) ∈ R3 :

x2

a2+y2

b2+z2

c2≤ 1

.

5. Compute the integral ∫B1

√x2 + y2 + z2 dx dy dz ,

where B1 is the three-dimensional open ball centered at the origin, with radius 1.

2.15 The integral on unbounded sets

While for L-integrable functions the extension of the theory to unbounded sets doesnot encounter great difficulties, it seems not to exist a satisfactory general definitionof integrability for functions of several variables. This is the reason why, in thefollowing, we will concentrate only on the theory for L-integrable functions. We willuse the notation

B[0, r] =

(x1, . . . , xN ) ∈ RN : max|x1|, . . . , |xN | ≤ r.

Definition 2.41 Given a set E ⊆ RN, not necessarily bounded, a function f : E →R is said to be L-integrable (on E) if it is L-integrable on each of the bounded setsE ∩B[0, r], with r > 0, and the two following limits exist and are finite:

limr→+∞

∫E∩B[0,r]

f , limr→+∞

∫E∩B[0,r]

|f | .

In this case, the first of these limits is said to be the integral of f on E and isdenoted by the symbol

∫E f.

Equivalently, f is L-integrable on E if the two following limits exist and arefinite: ∫

Ef+ = lim

r→+∞

∫E∩B[0,r]

f+ ,

∫Ef− = lim

r→+∞

∫E∩B[0,r]

f− ;

in that case, we have∫E f =

∫E f

+ −∫E f−.

2.15. THE INTEGRAL ON UNBOUNDED SETS 101

It is not difficult to prove that the set of L-integrable functions is a vector space,and the integral is a linear function on it which preserves the order. Moreover, oneeasily verifies that a function f is L-integrable on a set E if and only if the functionfE is L-integrable on RN .

Definition 2.42 A set E ⊆ RN is said to be measurable if E ∩B[0, r] is measur-able, for every r > 0. In that case, we set

µ(E) = limr→+∞

µ(E ∩B[0, r]) .

Notice that µ(E), in some cases, can be +∞. It is finite if and only if the constantfunction 1 is L-integrable on E, i.e., the characteristic function of E is L-integrableon RN . The properties of bounded measurable sets extend easily to unbounded sets.In particular, all open sets and all closed sets are measurable.

The monotone convergence theorem of B. Levi attains the following generalform.

Theorem 2.43 We are given a function f and a sequence of functions fk, withk ∈ N, defined almost everywhere on a subset E of RN, with real values, verifyingthe following conditions:

1. the sequence (fk)k converges pointwise to f , almost everywhere on E;

2. the sequence (fk)k is monotone;

3. each function fk is L-integrable on E;

4. the real sequence (∫E fk)k has a finite limit.

Then, f is L-integrable on E, and∫Ef = lim

k→∞

∫Efk .

Proof Assume, for definiteness, that the sequence (fk)k is increasing. By consideringthe sequence (fk−f0)k instead of (fk)k, we can assume without loss of generality thatall the functions have almost everywhere non-negative values. Let A = limk(

∫E fk);

for every r > 0, we can apply the monotone convergence theorem on the boundedset E ∩B[0, r], so that f is integrable on E ∩B[0, r] and∫

E∩B[0,r]f = lim

k→∞

∫E∩B[0,r]

fk ≤ limk→∞

∫Efk = A .

Let us prove that the limit of∫E∩B[0,r] f exists, as r → +∞, and that it is equal

to A. Fix ε > 0; there is a k ∈ N such that, for k ≥ k,

A− ε

2≤∫Efk ≤ A ;


being moreover ∫Efk = lim

r→∞

∫E∩B[0,r]

fk ,

there is a r > 0 such that, for r ≥ r,

A− ε ≤∫E∩B[0,r]

fk ≤ A .

Then, since the sequence (fk)k is increasing, we have that, for every k ≥ k and everyr ≥ r, it is

A− ε ≤∫E∩B[0,r]

fk ≤ A .

Passing to the limit as k → +∞, we obtain, for every r ≥ r,

A− ε ≤∫E∩B[0,r]

f ≤ A .


As an immediate consequence there is the analogous statement for the series offunctions.

Corollary 2.44 We are given a function f and a sequence of functions fk, withk ∈ N, defined almost everywhere on a subset E of RN, with real values, verifyingthe following conditions:

1. the series∑

k fk converges pointwise to f , almost everywhere on E;

2. for every k ∈ N and almost every x ∈ E, it is fk(x) ≥ 0;


4. the series∑

k(∫E fk) converges.

Then, f is L-integrable on E and∫Ef =

∞∑k=0

∫Efk .

From the monotone convergence theorem we deduce, in complete analogy to thatseen for bounded sets, the dominated convergence theorem of H. Lebesgue.

Theorem 2.45 We are given a function f and a sequence of functions fk, withk ∈ N, defined almost everywhere on a subset E of RN, with real values, verifyingthe following conditions:

1. the sequence (fk)k converges pointwise to f , almost everywhere on E;



3. there are two functions g, h, defined almost everywhere and L-integrable on E,

such that

g(x) ≤ fk(x) ≤ h(x) ,

for every k ∈ N and almost every x ∈ E.Then, the sequence (

∫E fk)k has a finite limit, f is L-integrable on E, and∫

Ef = lim

k→∞

∫Efk .

As a direct consequence we have the property of complete additivity of theintegral for L-integrable functions:

Theorem 2.46 Let (Ek) be a finite or countable family of pairwise non-overlappingmeasurable subsets of RN, whose union is a set E. Then, f is L-integrable on E ifand only if the two following conditions hold:

(a) f is L-integrable on each set Ek ;

(b)∑

k

∫Ek|f(x)| dx < +∞.

In that case, we have ∫Ef =

∑k

∫Ek

f .

As another consequence, we have the Leibniz rule for not necessarily boundedsubsets Y of RN , which is stated as follows.

Theorem 2.47 Let f : X ×Y → R be a function, where X is a non-trivial intervalof R containing x0, and Y is a subset of RN , such that:

(i) for every x ∈ X, the function f(x, ·) is L-integrable on Y, so that we candefine the function

F (x) =

∫Yf(x,y) dy ;

(ii) for every x ∈ X and almost every y ∈ Y, the partial derivative ∂f∂x (x,y)

exists;

(iii) there are two L-integrable functions g, h : Y → R such that

g(y) ≤ ∂f

∂x(x,y) ≤ h(y) ,

for every x ∈ X and almost every y ∈ Y.


Then, the function ∂f∂x (x, ·), defined almost everywhere on Y, is L-integrable there,

the derivative of F in x0 exists, and we have:

F ′(x0) =

∫Y

(∂f

∂x(x0,y)

)dy .

Also the Reduction Theorem of G. Fubini extends to functions defined ona not necessarily bounded subset E of RN . Let N = N1 + N2 and write RN =RN1 × RN2 . For every (x,y) ∈ RN1 × RN2 , consider the sections of E :

Ex = y ∈ RN2 : (x,y) ∈ E , Ey = x ∈ RN1 : (x,y) ∈ E ,

and the projections of E :

P1E = x ∈ RN1 : Ex 6= Ø , P2E = y ∈ RN2 : Ey 6= Ø .

We can then reformulate the theorem in the following form.

Theorem 2.48 Let f : E → R be a L-integrable function. Then:

(i) for almost every x ∈ P1E, the function f(x, ·) is L-integrable on the set Ex ;

(ii) the function x 7→∫Exf(x,y) dy, defined almost everywhere on P1E, is L-

integrable there;

(iii) we have: ∫Ef =

∫P1E

(∫Ex

f(x,y) dy

)dx .

Analogously, the function y 7→∫Eyf(x,y) dx, defined almost everywhere on P2E,

is L-integrable there, and we have:∫Ef =

∫P2E

(∫Ey

f(x,y) dx

)dy .

Proof Consider for simplicity the case N1 = N2 = 1, the general case being per-fectly analogous. Assume first that f has non-negative values. By Fubini The-orem for bounded sets, once fixed r > 0, we have that, for almost every x ∈P1E∩ [−r, r], the function f(x, ·) is L-integrable on Ex∩ [−r, r]; the function gr(x) =∫Ex∩[−r,r] f(x, y) dy, defined almost everywhere on P1E∩[−r, r], is L-integrable there,

and ∫E∩B[0,r]

f =

∫P1E∩[−r,r]

gr(x) dx .

In particular, ∫P1E∩[−r,r]

gr(x) dx ≤∫Ef ,


so that, if 0 < s ≤ r, one has that gr is L-integrable on P1E ∩ [−s, s], and∫P1E∩[−s,s]

gr(x) dx ≤∫Ef .

Keeping s fixed, we let r tend to +∞. Since f has non-negative values, gr(x) will beincreasing with respect to r. Consequently, for almost every x ∈ P1E ∩ [−s, s], thelimit limr→+∞ gr(x) exists (possibly infinite), and we set

g(x) = limr→+∞

gr(x) = limr→+∞

∫Ex∩[−r,r]

f(x, y) dy .

Let T = x ∈ P1E ∩ [−s, s] : g(x) = +∞; let us prove that T is negligible. Wedefine the sets

Ern = x ∈ P1E ∩ [−s, s] : gr(x) > n.

By Theorem 2.12 these are measurable sets and the Cebicev inequality yields

µ(Ern) ≤ 1

n

∫P1E∩[−s,s]

gr(x) dx ≤ 1

n

∫Ef .

Hence, since the sets Ern increase with r, also the sets Fn = ∪rErn are measurable,and we have that µ(Fn) ≤ 1

n

∫E f . Being T ⊆ ∩nFn, we deduce that T is measurable,

with µ(T ) = 0.

Hence, for almost every x ∈ P1E ∩ [−s, s], the function f(x, ·) is L-integrable onthe set Ex and, by definition, ∫

Ex

f(x, y) dy = g(x) .

Moreover, if we take r in the set of natural numbers and apply the monotone conver-gence theorem to the functions gr, it follows that g is L-integrable on P1E ∩ [−s, s],and ∫

P1E∩[−s,s]g = lim

r→∞

∫P1E∩[−s,s]

gr ,

so that ∫P1E∩[−s,s]

(∫Ex

f(x, y) dy

)dx ≤

∫Ef .

Letting now s tend to +∞, we see that the limit

lims→+∞

∫P1E∩[−s,s]

(∫Ex

f(x, y) dy

)dx


exists and in finite; therefore, the function x 7→∫Exf(x, y) dy, defined almost every-

where on P1E, is L-integrable there, and its integral is the preceding limit. Moreover,from the above proved inequality, passing to the limit, we have that∫

P1E

(∫Ex

f(x, y) dy

)dx ≤

∫Ef .

On the other hand,∫E∩B[0,r]

f =

∫P1E∩[−r,r]

(∫Ex∩[−r,r]

f(x, y) dy

)dx

≤∫P1E∩[−r,r]

(∫Ex

f(x, y) dy

)dx

≤∫P1E

(∫Ex

f(x, y) dy

)dx ,

so that, passing to the limit as r → +∞,∫Ef ≤

∫P1E

(∫Ex

f(x, y) dy

)dx .

In conclusion, equality must hold, and the proof is thus completed in the case when fhas non-negative values. In the general case, just consider f+ and f−, and subtractthe corresponding formulas.

The analogous corollary for the computation of the measure holds.

Corollary 2.49 Let E be a measurable set. Then, E has a finite measure if andonly if:

(i) for almost every x ∈ P1E, the set Ex is measurable and has a finite measure;

(ii) the function x 7→ µ(Ex), defined almost everywhere on P1E, is L-integrablethere;

(iii) we have:

µ(E) =

∫P1E

µ(Ex) dx .

With a symmetric statement, if E has a finite measure, we also have

µ(E) =

∫P2E

µ(Ey) dy .

The Theorem on the Change of Variables in the integral also extends tounbounded sets, with the same statement.


Theorem 2.50 Let ϕ be a diffeomorphism between two open sets A and B = ϕ(A)of RN, D be a measurable subset of A, and f : ϕ(D) → R be a function. Then, fis L-integrable on ϕ(D) if and only if (f ϕ) | detϕ′| is L-integrable on D, in whichcase we have: ∫

ϕ(D)f(x) dx =


Proof Assume first that f be L-integrable on E = ϕ(D) with non-negative values.Then, for every r > 0,∫

D∩B[0,r]f(ϕ(u)) |detϕ′(u)| du =

∫ϕ(D∩B[0,r])

f(x) dx

≤∫ϕ(D)

f(x) dx ,

so that the limit

limr→+∞

∫D∩B[0,r]

f(ϕ(u)) |detϕ′(u)| du

exists and is finite. Then, (f ϕ) |detϕ′| is L-integrable on D and we have∫Df(ϕ(u)) |detϕ′(u)| du ≤

∫ϕ(D)

f(x) dx .

On the other hand, for every r > 0,∫E∩B[0,r]

f =

∫ϕ−1(E∩B[0,r])

(f ϕ) |detϕ′| ≤∫ϕ−1(E)

(f ϕ) |detϕ′| ,

so that, passing to the limit,∫Ef(x) dx = lim

r→+∞

∫E∩B[0,r]

f(x) dx

≤∫ϕ−1(E)

f(ϕ(u)) |detϕ′(u)| du .

The formula is thus proved when f has non-negative values. In general, just proceedas usual, considering f+ and f−.

To obtain the opposite implication, it is sufficient to consider (f ϕ) |detϕ′|instead of f and ϕ−1 instead of ϕ, and to apply the above.

Concerning the change of variables in polar coordinates in R2 or in cylindricalo spherical coordinates in R3, the same type of considerations we have made forbounded sets extend to the general case, as well.


Example. Let E = (x, y) ∈ R2 : x2 + y2 ≥ 1 and f(x, y) = (x2 + y2)−α, with α > 0. We have∫E

1

(x2 + y2)αdx dy =

∫ 2π

0

(∫ +∞

1

1

ρ2αρ dρ

)dθ

= 2π

∫ +∞

1

ρ1−2α dρ .

It is thus seen that f is integrable on E if and only if α > 1, in which case the integral is πα−1

.

Example. Let us compute the three-dimensional measure of the set

E =

(x, y, z) ∈ R3 : x ≥ 1,

√y2 + z2 ≤ 1

x

.

Using Fubini Theorem, grouping together the variables (y, z) we have

µ(E) =

∫ +∞

1

π1

x2dx = π .

Example. Consider the function f(x, y) = e−(x2+y2), and let us make a change of variables in polarcoordinates: ∫

R2

e−(x2+y2) dx dy =

∫ 2π

0

(∫ +∞

0

e−ρ2

ρ dρ

)dθ = 2π

[−1

2e−ρ

2]+∞0

= π .

Notice that, using Fubini Theorem, we have:∫R2

e−(x2+y2) dx dy =

∫ +∞

−∞

(∫ +∞

−∞e−x

2

e−y2

dx

)dy

=

(∫ +∞

−∞e−x

2

dx

)(∫ +∞

−∞e−y

2

dy

)=

(∫ +∞

−∞e−x

2

dx

)2

,

and we thus find again that ∫ +∞

−∞e−x

2

dx =√π .

Exercises

1. Prove that the sets Q2, Q× R and R×Q are negligible in R2.

2. Compute the integral ∫R2

1

(1 + x2 + y2)√x2 + y2

dx dy .

3. For what values of α > 0 is the integral∫R3

1

(1 + x2 + y2 + z2)αdx dy dz

well defined as a real number? What is its value?


4. LetEγ = (x, y) ∈ R2 : x ≥ 1, 0 ≤ y ≤ xγ ,

for some γ ∈ R. When is the function f(x, y) = xy integrable on Eγ? For thosevalues of γ, compute ∫

Eγ

xy dx dy .

5. Compute the measure of the four-dimensional ball

BR = (x1, x2, x3, x4) ∈ R4 : x21 + x2

2 + x23 + x2

4 ≤ R2 .


Chapter 3

Differential forms

In this chapter we develop a theory leading to important extensions to functions ofseveral variables of the formula given by the Fundamental Theorem. Nevertheless,we will not be able to completely generalize that theorem, because we need toassume, for those functions, a somewhat greater regularity.

3.1 The vector spaces ΩM(RN)

Consider, for every positive integer M, the sets ΩM (RN ) made by the M -linearantisymmetric functions on RN, with real values. It is well known that these arevector spaces on R. We also adopt the convention that Ω0(RN ) = R.

If we choose the indices i1, . . . , iM in the set 1, . . . , N, we can define the M -linear antisymmetric function dxi1,...,iM : it is the function which associates to thevectors

v(1) =

v(1)1...

v(1)N

, . . . , v(M) =

v(M)1...

v(M)N

,

the real number

det

v

(1)i1. . . v

(M)i1

... · · ·...

v(1)iM. . . v

(M)iM

.

Notice that, whenever two indices coincide, we have the zero function. If two indicesare exchanged, the function changes sign. Let us recall the following result fromelementary algebra.

Proposition 3.1 If 1 ≤M ≤ N, the space ΩM (RN ) has dimension(NM

). A basis is

given by (dxi1,...,iM )1≤i1<...<iM≤N . If M > N, then ΩM (RN ) = 0.

111

112 CHAPTER 3. DIFFERENTIAL FORMS

We are mostly interested in the case N = 3. Let us give a closer look to thespaces Ω1(R3), Ω2(R3) and Ω3(R3).

Consider Ω1(R3), the space of linear functions defined on R3, with values in R.We denote by dx1, dx2, dx3 the following linear functions:

dx1 :

v1

v2

v3

7→ v1 , dx2 :

v1

v2

v3

7→ v2 , dx3 :

v1

v2

v3

7→ v3 .

the space Ω1(R3) has dimension 3 and (dx1, dx2, dx3) is one of its bases.

Consider Ω2(R3), the space of bilinear antisymmetric functions defined on R3×R3,with values in R. It has dimension 3, and a basis is given by (dx1,2, dx1,3, dx2,3), where

dx1,2 :

v1

v2

v3

,

v′1v′2v′3

7→ det

(v1 v

′1

v2 v′2

)= v1v

′2 − v2v

′1 ,

dx1,3 :

v1

v2

v3

,

v′1v′2v′3

7→ det

(v1 v

′1

v3 v′3

)= v1v

′3 − v3v

′1 ,

dx2,3 :

v1

v2

v3

,

v′1v′2v′3

7→ det

(v2 v

′2

v3 v′3

)= v2v

′3 − v3v

′2 .

It is useful to recall that

dx1,1 = dx2,2 = dx3,3 = 0 ,

dx2,1 = −dx1,2 , dx3,1 = −dx1,3 , dx3,2 = −dx2,3 .

Consider Ω3(R3), the space of trilinear antisymmetric functions defined on R3×R3 × R3, with values in R. We denote by dx1,2,3 the following trilinear function:

dx1,2,3 :

v1

v2

v3

,

v′1v′2v′3

,

v′′1v′′2v′′3

7→ det

v1 v′1 v′′1

v2 v′2 v′′2

v3 v′3 v′′3

.

every element of the vector space Ω3(R3) is a scalar multiple of dx1,2,3 : the spaceΩ3(R3) has dimension 1. Recall that

dx1,2,3 = dx2,3,1 = dx3,1,2 = −dx3,2,1 = −dx2,1,3 = −dx1,3,2

and, when two indices coincide, we have the zero function.

3.2. DIFFERENTIAL FORMS IN RN 113

3.2 Differential forms in RN

Definition 3.2 Given an open subset U of RN, we call differential form of de-gree M (or M -differential form) a function

ω : U → ΩM (RN ) .

If M ≥ 1, once we consider the basis (dxi1,...,iM )1≤i1<...<iM≤N , the componentsof the M -differential form ω will be denoted by fi1,...,iM : U → R. We will then write

ω(x) =∑

1≤i1<...<iM≤Nfi1,...,iM (x) dxi1,...,iM .

Hence, the M -linear antisymmetric function ω(x) is determined by the(NM

)−dim-

ensional vectorF (x) = (fi1,...,iM (x))1≤i1<...<iM≤N .

A 0−differential form is nothing else than a function defined on U with values in R.We will say that a M -differential form is of class Ck if all its components are such.

It is possible to define the sum of two M -differential forms: if ω is as above andω is also defined on U and

ω(x) =∑

1≤i1<...<iM≤Ngi1,...,iM (x) dxi1,...,iM ,

we define in a natural way ω + ω as follows:

(ω + ω)(x) =∑

1≤i1<...<iM≤N(fi1,...,iM (x) + gi1,...,iM (x)) dxi1,...,iM .

Moreover, if c ∈ R, we define c ω, the product of the scalar c by the M -differentialform ω, in the following way:

(c ω)(x) =∑

1≤i1<...<iM≤Ncfi1,...,iM (x) dxi1,...,iM .

With these definitions, it can be checked that the set of differential forms of degreeM becomes a vector space.

Let us give a closer look to the case N = 3. Denoting by ωM a M -differentialform, with M = 1, 2, 3, we can write

ω1(x) = f1(x) dx1 + f2(x) dx2 + f3(x) dx3 ,

ω2(x) = f1,2(x) dx1,2 + f1,3(x) dx1,3 + f2,3(x)dx2,3 ,

ω3(x) = f1,2,3(x) dx1,2,3 .


Notice that ω1(x) and ω2(x) are determined by the three-dimensional vectors

F (x) = (f1(x), f2(x), f3(x)) , and G(x) = (f12(x), f13(x), f23(x)) ,

respectively.

3.3 External product

Given two differential forms ω : U → ΩM (RN ), ω : U → ΩM (RN ), of degrees M and

M, respectively, we want to define the differential form ω ∧ ω, of degree M + M,which is called external product of ω and ω. If

ω(x) =∑

1≤i1<...<iM≤Nfi1,...,iM (x) dxi1,...,iM ,

and

ω(x) =∑

1≤j1<...<jM≤Ngj1,...,jM (x) dxj1,...,jM ,

we set

(ω ∧ ω)(x) =∑

1≤i1<...<iM≤N1≤j1<...<jM≤N

fi1,...,iM (x)gj1,...,jM (x) dxi1,...,iM ,j1,...,jM .

Usually the symbol ∧ is omitted when one of the two is a 0−differential form, sincethe external product is, in this case, similar to the product with a scalar. Noticethat, in the above sum, all elements with a repeating index will be zero. Let us seenow some properties.

Proposition 3.3 If ω, ω, ˜ω are three differential forms of degrees M, M, ˜M, respec-tively, then

ω ∧ ω = (−1)MMω ∧ ω ,

(ω ∧ ω) ∧ ˜ω = ω ∧ (ω ∧ ˜ω) ;

if c ∈ R, then

(c ω) ∧ ω = ω ∧ (c ω) = c(ω ∧ ω) ;

moreover, when M = M,

(ω + ω) ∧ ˜ω = (ω ∧ ˜ω) + (ω ∧ ˜ω) ,

˜ω ∧ (ω + ω) = (˜ω ∧ ω) + (˜ω ∧ ω) .

3.3. EXTERNAL PRODUCT 115

Proof Assume that ω and ω are written as above, and let

˜ω(x) =∑

1≤k1<...<k ˜M≤N

hk1,...,k ˜M(x) dxk1,...,k ˜M

.

The first identity is obtained observing that, in order to arrive from the sequence ofindices i1, . . . , iM , j1, . . . , jM to the one j1, . . . , jM , i1, . . . , iM , one has first to movej1 towards the left making M exchanges, then the same has to be done for j2, if thereis one, and so on, till jM is reached. In the total, it is then necessary to operate MMexchanges of indices. Taking into account the fact that the differential form changessign each time there is an exchange, we have the formula we wanted to prove.

The proof of the second identity (associative property) shows no great difficulties,as well as for the identities where the constant c appears.

Concerning the distributive property, when M = M we have

((ω + ω) ∧ ˜ω)(x) =

=∑

1≤i1<...<iM≤N1≤k1<...<k ˜M

≤N

(fi1,...,iM (x) + gi1,...,iM (x))hk1,...,k ˜M(x) dxi1,...,iM ,k1,...,k ˜M

=∑

1≤i1<...<iM≤N1≤k1<...<k ˜M

≤N

(fi1,...,iM (x)hk1,...,k ˜M(x) +

+gi1,...,iM (x)hk1,...,k ˜M(x)) dxi1,...,iM ,k1,...,k ˜M

= ((ω ∧ ˜ω) + (ω ∧ ˜ω))(x) .

The last identity is proved either in an analogous way, or using the first and thefourth identities.

If we consider the particular case of the two constant differential forms

ω(x) = dx1 , ω(x) = dx2 , for every x ∈ U,

we will have that (ω ∧ ω)(x) = dx1,2, for every x ∈ U. We can then write

dx1 ∧ dx2 = dx1,2 .

More generally, in view of the associative property of the external product, we canwrite

dxi1 ∧ . . . ∧ dxiM = dxi1,...,iM .

In the following, we will use indifferently the one or the other notation.


3.4 External differential

Given a M -differential form ω if class C1, we want to define the differential formdexω, of degree M + 1, which is said to be the external differential of ω.

If ω is a 0−differential form, ω = f : U → R, its external differential dexω(x) isjust the differential df(x), which is a linear function defined on RN, with values inR. Being, for every v = (v1, . . . , vN ),

df(x)v =∂f

∂x1(x) v1 + . . .+

∂f

∂xN(x) vN ,

we have

df(x) =∂f

∂x1(x) dx1 + . . .+

∂f

∂xN(x) dxN =

N∑m=1

∂f

∂xm(x) dxm .

In the general case, if

ω(x) =∑

1≤i1<...<iM≤Nfi1,...,iM (x) dxi1 ∧ . . . ∧ dxiM ,

we setdexω(x) =

∑1≤i1<...<iM≤N

dfi1,...,iM (x) ∧ dxi1 ∧ . . . ∧ dxiM ,

or, equivalently,

dexω(x) =∑

1≤i1<...<iM≤N

N∑m=1

∂fi1,...,iM∂xm

(x) dxm ∧ dxi1 ∧ . . . ∧ dxiM .

In the following, in order to simplify the notations, we will always write dω insteadof dexω. Let us see some properties of the external differential.

Proposition 3.4 If ω and ω are two differential form of class C1, of degrees M andM, respectively, then

d(ω ∧ ω) = dω ∧ ω + (−1)Mω ∧ dω ;

if M = M and c ∈ R, it isd(ω + ω) = dω + dω ,

d(c ω) = c dω ;

if ω is of class C2, thend(dω) = 0 .

3.5. DIFFERENTIAL FORMS IN R3 117

Proof Concerning the first identity, if ω and ω are as above, we have:

d(ω ∧ ω)(x) =∑

1≤i1<...<iM≤N1≤j1<...<jM≤N

N∑m=1

∂

∂xm(fi1,...,iM gj1,...,jM )(x) dxm,i1,...,iM ,j1,...,jM

=∑

1≤i1<...<iM≤N1≤j1<...<jM≤N

N∑m=1

(∂fi1,...,iM∂xm

gj1,...,jM+

+fi1,...,iM∂gj1,...,jM∂xm

)(x) dxm,i1,...,iM ,j1,...,jM

=∑

1≤i1<...<iM≤N1≤j1<...<jM≤N

N∑m=1

(∂fi1,...,iM∂xm

gj1,...,jM

)(x) dxm,i1,...,iM ,j1,...,jM +

+(−1)M∑

1≤i1<...<iM≤N1≤j1<...<jM≤N

N∑m=1

(fi1,...,iM

∂gj1,...,jM∂xm

)(x) dxi1,...,iM ,m,j1,...,jM

= (dω ∧ ω)(x) + (−1)M (ω ∧ dω)(x) .

The second and third identities follow easily from the linearity of the derivative.

Concerning the last identity, we can see that

d(dω)(x) =∑

1≤i1<...<iM≤N

N∑k=1

N∑m=1

∂

∂xk

∂fi1,...,iM∂xm

(x) dxk,m,i1,...,iM .

Since∂

∂xk

∂fi1,...,iM∂xm

=∂

∂xm

∂fi1,...,iM∂xk

,

taking into account the fact that dxk ∧ dxm = −dxm ∧ dxk, it is seen that all theterms in the sums pairwise eliminate one another, so that d(dω)(x) = 0.

3.5 Differential forms in R3

When N = 3, if ω1 and ω1 are two 1-differential forms, e.g.,

ω1(x) = f1(x) dx1 + f2(x) dx2 + f3(x) dx3 ,

ω1(x) = g1(x) dx1 + g2(x) dx2 + g3(x) dx3 ,

using the associative and the distributive properties, we have that

ω1 ∧ ω1 = (f1g2 − f2g1) dx1,2 + (f1g3 − f3g1) dx1,3 + (f2g3 − f3g2) dx2,3 .


On the other hand, if ω1 is a 1-differential form and ω2 is a 2-differential form, e.g.,

ω1(x) = f1(x)dx1 + f2(x)dx2 + f3(x)dx3 ,

ω2(x) = g1,2(x) dx1,2 + g1,3(x) dx1,3 + g2,3(x) dx2,3 ,

it is

ω1 ∧ ω2 = (f1g2,3 − f2g1,3 + f3g1,2) dx1,2,3 .

If we have a 0-differential form ω0 = f : U → R, then

dω0(x) =∂f

∂x1(x) dx1 +

∂f

∂x2(x) dx2 +

∂f

∂x3(x) dx3 .

Taking a 1-differential form

ω1(x) = f1(x) dx1 + f2(x) dx2 + f3(x) dx3 ,

we have

dω1(x) =

(∂f2

∂x1(x)− ∂f1

∂x2(x)

)dx1,2 +

+

(∂f3

∂x1(x)− ∂f1

∂x3(x)

)dx1,3 +

+

(∂f3

∂x2(x)− ∂f2

∂x3(x)

)dx2,3 .

If we consider a 2-differential form

ω2(x) = g1,2(x) dx1,2 + g1,3(x) dx1,3 + g2,3(x) dx2,3 ,

then

dω2(x) =

(∂g2,3

∂x1(x)− ∂g1,3

∂x2(x) +

∂g1,2

∂x3(x)

)dx1,2,3 .

At this point, it is time to observe that, in view of future applications, a wiserchoice for the basis of the vector space Ω2(R3) could be the following:

(dx2,3 , dx3,1 , dx1,2) .

Indeed, in this way, associating,

• to each scalar function f : U → R,

either a 0−differential form ω0 = f ,

or a 3-differential form ω3 = f dx1,2,3 ;

3.5. DIFFERENTIAL FORMS IN R3 119

• to each vector field F = (F1, F2, F3) : U → R3,either a 1-differential form ω1 = F1 dx1 + F2 dx2 + F3 dx3 ,or a 2-differential form ω2 = F1 dx2,3 + F2 dx3,1 + F3 dx1,2 ,

we have the following:

dω0 corresponds to the gradient of f :

grad f =

(∂f

∂x1,∂f

∂x2,∂f

∂x3

);

dω1 corresponds to the curl of F :

curlF =

(∂F3

∂x2− ∂F2

∂x3,∂F1

∂x3− ∂F3

∂x1,∂F2

∂x1− ∂F1

∂x2

);

dω2 corresponds to the divergence of F :

divF =∂F1

∂x1+∂F2

∂x2+∂F3

∂x3.

Then, given two vector fields F and F , once we consider the associated 1-diff-erential forms

ω1 = F1 dx1 + F2 dx2 + F3 dx3 , ω1 = F1 dx1 + F2 dx2 + F3 dx3 ,

we have that ω1 ∧ ω1 corresponds to the vector product of F and F :

F × F = (F2F3 − F3F2 , F3F1 − F1F3 , F1F2 − F2F1) ;

if instead of ω1 we take the associated 2-differential form

ω2 = F1 dx2,3 + F2 dx3,1 + F3 dx1,2 ,

we have that ω1 ∧ ω2 corresponds to the scalar product of F and F :

〈F , F 〉 = F1F1 + F2F2 + F3F3 .

The properties of the external product and those of the external differential leadto formulas involving the gradient, the curl and the divergence. Taking f : U → R,f : U → R, F : U → R3 and F : U → R3, we have, for example, the following:

curl(grad f) = 0 ,

div(curlF ) = 0 ,

grad(ff) = f(grad f) + f(grad f) ,

curl(fF ) = (grad f)× F + f(curlF ) ,

div(fF ) = 〈grad f , F 〉+ f(div F ) ,

div(F × F ) = 〈curlF , F 〉 − 〈F , curl F 〉 .

The proofs are left to the reader, who might also enjoy the following exercises.


Exercises

1. Let ω, ω : R2 → Ω1(R2) be defined as

ω(x, y) = y2 dx− x2 dy , ω(x, y) = xy dx+ (x2 + y2) dy .

Compute ω ∧ ω, dω, dω, d(ω ∧ ω) and dω ∧ dω.

2. Let ω : R3 → Ω2(R3) be defined as

ω(x, y, z) = x2yz dy ∧ dz + xy2z dz ∧ dx+ xyz2 dx ∧ dy .

Compute dω : R3 → Ω3(R3).

3. Let f : R3 → R be the function defined as f(x, y, z) = xy2z3. Compute

∆f := (div(gradf)) .

4. Let F : R3 → R3 be the vector field defined by

F (x, y, z) = (x− y + z2, x2 + yz , x+ y2 − 2z) .

Compute

divF , curlF , grad(divF ) , curl(curlF ) .

5. Compute the exterior differential of ω : R4 → Ω1(R4), defined as

ω(x1, x2, x3, x4) = x2x3x4 dx1 + x1x3x4 dx2 + x1x2x4 dx3 + x1x2x3 dx4 .

3.6 M-surfaces

We denote by I a rectangle in RM, where 1 ≤M ≤ N.

Definition 3.5 We call M-surface in RN a function1 σ : I → RN of class C1. IfM = 1, σ is also said to be a curve; if M = 2, we will simply say surface. The setσ(I) is called the support of the M -surface σ. We will say that the M -surface σ is

regular if, for every u ∈ I, the Jacobian matrix σ′(u) has rank M.

1The partial derivatives of σ must be continuous on the whole I, and in the points of theboundary they are interpreted, if necessary, as right or left derivatives. Equivalently, σ could beextended to a C1-function defined on an open set containing I. In this perspective, the domain ofσ could be a more general set than a rectangle, as e.g. the closure of any bounded and open set, sothat the differential be well defined at the boundary points, as well. Analogous considerations canbe made on the domains of the considered differential forms.

3.6. M-SURFACES 121

Consider for example the case N = 3. A curve in R3 is a function σ : [a, b]→ R3,σ = (σ1, σ2, σ3). The curve is regular if, for every t ∈ ]a, b[ , the vector σ′(t) =(σ′1(t), σ′2(t), σ′3(t)) is not zero. In that case, it is possible to define the followingtangent versor at the point σ(t) :

τσ(t) =σ′(t)

‖σ′(t)‖.

σ(b )

σ(a )

σ(t )

τσ(t )

σ( t ) + τσ(t )

Example. The curve σ : [0, 2π]→ R3, defined by

σ(t) = (R cos(2t), R sin(2t), 0) ,

has as support the circle(x, y, z) : x2 + y2 = R2, z = 0

(which is covered twice). Being σ′(t) = (−2R sin(2t), 2R cos(2t), 0), it is a regular curve, and

τσ(t) = (− sin(2t), cos(2t), 0) .

A surface in R3 is a function σ : [a1, b1] × [a2, b2] → R3. The surface is regu-lar if, for every (u, v) ∈ ]a1, b1[× ]a2, b2[ , the vectors ∂σ

∂u (u, v), ∂σ∂v (u, v) are linearlyindependent. In that case, they determine a plane, called the tangent plane tothe surface at the point σ(u, v), and it is possible to define the following normalversor: σ(u ,v) + νσ(u ,v)

νσ(u ,v)

σ(u ,v)

νσ(u, v) =∂σ∂u (u, v)× ∂σ

∂v (u, v)

‖∂σ∂u (u, v)× ∂σ∂v (u, v)‖

.


Examples. 1. The surface σ : [0, π]× [0, π]→ R3, defined by

σ(φ, θ) = (R sinφ cos θ,R sinφ sin θ,R cosφ) ,

has as support the semi-sphere

(x, y, z) : x2 + y2 + z2 = R2, y ≥ 0 .

Being

∂σ

∂φ(φ, θ) = (R cosφ cos θ,R cosφ sin θ,−R sinφ) ,

∂σ

∂θ(φ, θ) = (−R sinφ sin θ,R sinφ cos θ, 0) ,

we compute

∂σ

∂φ(φ, θ)× ∂σ

∂θ(φ, θ) = (R2 sin2 φ cos θ,R2 sin2 φ sin θ,R2 sinφ cosφ) .

We thus see that it is a regular surface, and

νσ(φ, θ) = (sinφ cos θ, sinφ sin θ, cosφ) .

2. The surface σ : [0, 2π]× [0, 2π]→ R3, defined by

σ(u, v) = ((R+ r cosu) cos v, (R+ r cosu) sin v, r sinu) ,

where 0 < r < R, has as support a torus

(x, y, z) : (√x2 + y2 −R)2 + z2 = r2 .

Even in this case, one can verify that it is a regular surface.

3.6. M-SURFACES 123

A 3-surface in R3 is also called a volume.

Example. The function σ : [0, R]× [0, π]× [0, 2π]→ R3, defined by

σ(ρ, φ, θ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ,

has as support the closed ball

(x, y, z) : x2 + y2 + z2 ≤ R2 .

In this case, detσ′(ρ, φ, θ) = ρ2 sinφ, so that it is a regular volume.

Definition 3.6 Two M -surfaces σ : I → RN and σ : J → RN are said to beequivalent if they have the same support and there are two open sets A ⊆ I, B ⊆ J,and a diffeomorphism ϕ : A → B with the following properties: the sets I \ A andJ \ B are negligible and σ(u) = σ(ϕ(u)), for every u ∈ A. We say that σ and σhave the same orientation if detϕ′(u) > 0, for every u ∈ A; they have oppositeorientation if detϕ′(u) < 0, for every u ∈ A.

Examples. Given a curve σ : [a, b] → RN, an equivalent curve with opposite orientation is, forexample, σ : [a, b]→ RN defined by

σ(t) = σ(a+ b− t) .

If σ is regular, an interesting example of an equivalent curve with the same orientation isobtained by considering the function

ϕ(t) =

∫ t

a

‖σ′(τ)‖ dτ .

Since ϕ′(t) = ‖σ′(t)‖ > 0, for every t ∈ ]a, b[ , setting ι1 = ϕ(b), we have that ϕ : [a, b] → [0, ι1] isbijective and the curve σ1 : [0, ι1] → RN , defined as σ1(s) = σ(ϕ−1(s)) is equivalent to σ. Noticethat, for every s ∈ ]0, ι1[ , it is

‖σ′1(s)‖ = ‖σ′(ϕ−1(s))(ϕ−1)′(s)‖

=

∥∥∥∥σ′(ϕ−1(s))1

ϕ′(ϕ−1(s))

∥∥∥∥=

∥∥∥∥σ′(ϕ−1(s))1

‖σ′(ϕ−1(s))‖

∥∥∥∥ = 1 .


Given a surface σ : [a1, b1] × [a2, b2] → R3, an equivalent surface with opposite orientation is,for example, σ : [a1, b1]× [a2, b2]→ R3 defined by

σ(u, v) = σ(u, a2 + b2 − v) ,

or by

σ(u, v) = σ(a1 + b1 − u, v) .

As will be seen in the sequel, two M -surfaces with the same support are notnecessarily equivalent. Let us introduce a particular class of M -surfaces for whichthis inconvenience does not happen.

Definition 3.7 A M -surface σ : I → RN is a M-parametrization of a set Mif it is regular, injective on I

, and σ(I) = M. We say that a subset of RN is

M-parametrizable if there is a M -parametrization of it.

Examples. The circle M = (x, y) ∈ R2 : x2 + y2 = 1 is 1-parametrizable and σ : [0, 2π] → R2,given by σ(t) = (cos t, sin t), is a 1-parametrization of it.

A 2-parametrization of the sphere M = (x, y, z) ∈ R3 : x2 + y2 + z2 = 1 is, for example,σ : [0, π]× [0, 2π]→ R3, defined by

σ(φ, θ) = (sinφ cos θ, sinφ sin θ, cosφ) .

The following theorem is crucial for the treatment of the measure of M -parame-trizable M -surfaces.

Theorem 3.8 Two M -parametrizations of the same set are always equivalent.

Proof Let M be the subset of RN taken in consideration, and let σ : I → RN andσ : J → RN be two of its M−parametrizations. We define the sets

A = I∩ σ−1(M\ (σ(∂I) ∪ σ(∂J))) , B = J

∩ σ−1(M\ (σ(∂I) ∪ σ(∂J))) .

Then, for every u ∈ A, since σ(u) ∈ M \ (σ(∂I) ∪ σ(∂J)) and σ(J) = M, there

exists a v ∈ J

such that σ(v) = σ(u). Clearly, σ(v) ∈M\ (σ(∂I)∪ σ(∂J)), so that

v ∈ B. Moreover, since σ is injective on J, there is a unique v in J

with such a

property. We can thus define ϕ : A → B by setting ϕ(u) = v. Hence, for u ∈ Aand v ∈ B,

ϕ(u) = v ⇔ σ(u) = σ(v) .

This function ϕ : A → B is invertible: a symmetrical argument may be used todefine its inverse ϕ−1 : B → A.

3.6. M-SURFACES 125

Let us verify that the set A is open. Since σ, σ are continuous functions and∂I, ∂J are compact sets, we have that σ(∂I) ∪ σ(∂J) is compact, hence closed.ThenM\ (σ(∂I)∪ σ(∂J)) is relatively open inM, and σ−1(M\ (σ(∂I)∪ σ(∂J)) is

relatively open in I, so that its intersection with I

is an open set. In an analogousway it can be seen that B is an open set, as well.

Let us take a v0 ∈ J, and set x0 = σ(v0). The Jacobian matrix σ′(v0) has rank

M , and we may assume without loss of generality that the first M lines be linearlyindependent. Being RN ' RM × RN−M , we will write every point x ∈ RN in theform x = (x1,x2), with x1 ∈ RM and x2 ∈ RN−M . However, not to have doubleindices below, we will write x0 = (x0

1,x02).

Let Φ : J × RN−M → RN be defined as

Φ(v,z) = σ(v) + (0,z) .

Then Φ′(v0,0) is invertible, so that Φ is a local diffeomorphism: there are an openneighborhood V0 of v0, an open neighborhood Ω0 of 0 in RN−M , and open neigh-borhood W0 of x0 such that Φ : V0 × Ω0 → W0 is a diffeomorphism. Moreover,

we can assume that V0 ⊆ J. Let Ψ = Φ−1 : W0 → V0 × Ω0. We will write

Ψ(x) = (Ψ1(x),Ψ2(x)), with Ψ1(x) ∈ V0 and Ψ2(x) ∈ Ω0.

We now prove that ϕ is of class C1. Take u0 ∈ A, and set x0 = σ(u0) andv0 = ϕ(u0). Assume v0 as above, with σ′(v0) having the first M lines linearlyindependent, so that the local diffeomorphism Ψ : W0 → V0 × Ω0 can be defined.Take an open neighborhood U0 of u0, contained in A, such that σ(U0) ⊆W0. Then,for u ∈ U0 and v ∈ B,

ϕ(u) = v ⇔ σ(u) = Φ(v,0) ⇔ (v,0) = Ψ(σ(u)) .

Hence, ϕ coincides with Ψ1 σ on the open set U0, yielding that ϕ is continuouslydifferentiable.

In a symmetric way it is proved that ϕ−1 : B → A is of class C1, so that ϕhappens to be a diffeomorphism.

We now prove that the sets I \A and J \B are negligible. Let us consider, e.g.,the second one:

J \B = ∂J ∪ (J\B) = ∂J ∪ v ∈ J

: σ(v) ∈ σ(∂I) ∪ v ∈ J

: σ(v) ∈ σ(∂J) .

We know that ∂J is negligible. Let us prove that v ∈ J

: σ(v) ∈ σ(∂I) isnegligible, as well.


Let v0 ∈ J

be such that σ(v0) ∈ σ(∂I). Then, there is a u0 ∈ ∂I such thatσ(u0) = σ(v0). We argue as above, and define Ψ : W0 → V0 × Ω0. Let U0 be anopen neighborhood of u0 such that σ(U0 ∩ I) ⊆W0. Let us see that

J∩ σ−1(σ(U0 ∩ ∂I)) ⊆ (Ψ1 σ)(U0 ∩ ∂I) .

Indeed, taking v ∈ J∩σ−1(σ(U0∩∂I)), we have that σ(v) ∈ σ(U0∩∂I). Then, being

Φ(v,0) = σ(v), we have that Ψ(σ(v)) = (v,0) ∈ V0×Ω0, hence v ∈ Ψ1(σ(U0∩∂I)),and the inclusion is thus proved. Now, since Ψ1 σ is of class C1, by Lemma 2.36

we have that (Ψ1 σ)(U0 ∩ ∂I) is negligible. Finally, the conclusion that v ∈ J

:σ(v) ∈ σ(∂I) is negligible follows from the fact that ∂I is compact, so that it canbe covered by a finite number of such open sets as U0.

It remains to be proved that v ∈ J

: σ(v) ∈ σ(∂J) is negligible. Let v0 ∈ J

be such that σ(v0) ∈ σ(∂J). Then, there is a v0 ∈ ∂J such that σ(v0) = σ(v0). LetV0 be an open neighborhood of v0 such that σ(V0 ∩ J) ⊆ W0. As above, one sees

that J∩ σ−1(σ(V0 ∩ ∂J)) ⊆ (Ψ1 σ)(V0 ∩ ∂J), showing that J

∩ σ−1(σ(V0 ∩ ∂J))

is negligible. The conclusion is obtained as above, covering ∂J by a finite numberof such open sets V0.

3.7 The integral of a differential form

We want to define the notion of integral of a M -differential form on a M -surface.Let

ω(x) =∑


a M -differential form defined on a subset U of RN containing the support of a M -surface σ : I → RN, with 1 ≤M ≤ N. We consider, when the indices i1, . . . , iM varyin the set 1, . . . , N, the functions σ(i1,...,iM ) : I → RM defined by

σ(i1,...,iM ) :

u1...uM

7→ σi1(u1, . . . , uM )

...σiM (u1, . . . , uM )

.

Definition 3.9 We say that the M -differential form ω : U → ΩM (RN ) is inte-grable on the M -surface σ : I → U if, for every choice of the indices i1, . . . , iM inthe set 1, . . . , N, the function (fi1,...,iM σ) detσ′(i1,...,iM ) is integrable on I. In thatcase, we set ∫

σω =

∑1≤i1<...<iM≤N

∫Ifi1,...,iM (σ(u)) detσ′(i1,...,iM )(u) du .

3.7. THE INTEGRAL OF A DIFFERENTIAL FORM 127

For example, ω is surely integrable on σ when all its components are continuousfunctions. Notice that

σ′(i1,...,iM )(u) =∂(σi1 , . . . , σiM )

∂(u1, . . . , uM )(u) =

∂σi1∂u1

(u) . . .∂σi1∂uM

(u)... · · ·

...∂σiM∂u1

(u) . . .∂σiM∂uM

(u)

.

If we define, for every x ∈ U and every u ∈ I, the(NM

)−dimensional vectors

F (x) = (fi1,...,iM (x))1≤i1<...<iM≤N ,

Σ(u) =(

detσ′(i1,...,iM )(u))

1≤i1<...<iM≤N,

we have that ∫σω =

∫I〈F (σ(u)) ,Σ(u)〉 du ,

where 〈· , ·〉 denotes here the euclidean scalar product in R(NM).

It is important to analyze how the integral of a differential form ω changes ontwo equivalent M -surfaces having the same orientations, or opposite orientations.

Theorem 3.10 Let σ : I → RN and σ : J → RN be two equivalent M -surfaces. Ifthey have the same orientations, then∫

σω =

∫σω ;

if they have opposite orientations, then∫σω = −

∫σω .

Proof We have a M -differential form of the type

ω(x) =∑

1≤i1<...<iM≤Nfi1,...,iM (x) dxi1 ∧ . . . ∧ dxiM .

Let ϕ : A→ B, be as in the definition of equivalent M -surfaces, such that σ = σ ϕ.By the Theorem on the Change of Variables in the integral, it is∫

σω =

∑1≤i1<...<iM≤N

∫Afi1,...,iM (σ(ϕ(u))) det(σ ϕ)′(i1,...,iM )(u) du

=∑

1≤i1<...<iM≤N

∫Afi1,...,iM (σ(ϕ(u))) det σ′(i1,...,iM )(ϕ(u)) detϕ′(u) du


= ±∑

1≤i1<...<iM≤N

∫Bfi1,...,iM (σ(v)) det σ′(i1,...,iM )(v) dv

= ±∫σω ,

with positive sign if detϕ′ > 0, negative if detϕ′ < 0.

Remark. In general, if σ and σ are equivalent, we do not necessarily have the equality |∫σω| =

|∫σω|. It is not guaranteed, indeed that they have the same or opposite orientations. For example,

if we consider the two surfaces σ, σ : [1, 2]× [0, 2π]→ R3, defined by

σ(u, v) =

((3

2+

(u− 3

2

)cos

v

2

)cos v,

(3

2+

(u− 3

2

)cos

v

2

)sin v,

(u− 3

2

)sin

v

2

),

σ(u, v) = σ(u, v +

π

2

),

it is possible to see that they are both parametrizations of the same set (a Mobius strip), andtherefore they are equivalent (the reader is invited to explicitly find a diffeomorphism ϕ : A → Bwith the properties required by the definition). On the other hand, if we consider the 2-differentialform ω(x1, x2, x3) = dx12, determined by the constant vector field (0, 0, 1), computation yields∫

σ

ω = 0 ,

∫σ

ω = −3√

2 .

We now consider the important case when M = N.

Theorem 3.11 Let M = N ; if σ is regular and injective on I

with detσ′ > 0, andω is of the type

ω(x) = f(x) dx1 ∧ . . . ∧ dxN ,

then ∫σω =

∫σ(I)

f .

Proof Using the theorem of local inversion it is seen that σ induces a diffeomorphism

between I

and σ(I). Being both the boundary of I and its image through σ negligible

(see Lemma 2.36), by the Theorem on the Change of Variables in the integral, wehave ∫

σω=

∫If(σ(u)) det(σ′(u)) du

=

∫If(σ(u)) det(σ′(u)) du

=

∫σ(I

)f =

∫σ(I)

f .


3.7. THE INTEGRAL OF A DIFFERENTIAL FORM 129

If σ is the identity function, then σ(I) = I, and instead of∫σ ω one usually writes∫

I ω. Hence, we have that∫If(x) dx1 ∧ . . . ∧ dxN =

∫If .

Let us see the meaning of the given definition in the case N = 3. If M = 1,σ : [a, b]→ R3 is a curve and ω is a 1-differential form:

ω(x) = F1(x) dx1 + F2(x) dx2 + F3(x) dx3 .

Hence, ∫σω=

∫ b

a[F1(σ(t))σ′1(t) + F2(σ(t))σ′2(t) + F3(σ(t))σ′3(t)] dt

=

∫ b

a〈F (σ(t)) , σ′(t)〉 dt .

This quantity will be called line integral2 of the vector field F = (F1, F2, F3) alongthe curve σ, and will be denoted by ∫

σ〈F, d`〉 .

Example. Let us compute the line integral of the vector field F (x, y, z) = (−y, x, z2) along thecurve σ : [0, 2π]→ R3, defined by σ(t) = (cos t, sin t, t) :∫

σ

〈F, d`〉 =

∫ 2π

0

[(sin t)2 + (cos t)2 + t2] dt = 2π +8π3

3.

If M = 2, σ : [a1, b1]× [a2, b2]→ R3 is a surface and ω is a 2-differential form:

ω(x) = F1(x) dx2 ∧ dx3 + F2(x) dx3 ∧ dx1 + F3(x) dx1 ∧ dx2 .

Hence,

∫σω=

∫ b2

a2

∫ b1

a1

F1(σ(u, v)) det

∂σ2∂u (u, v) ∂σ2∂v (u, v)

∂σ3∂u (u, v) ∂σ3∂v (u, v)

+

+F2(σ(u, v)) det

∂σ3∂u (u, v) ∂σ3∂v (u, v)

∂σ1∂u (u, v) ∂σ1∂v (u, v)

+

2In mechanics this concept is used, for example, to define the work done by a field of forces ona particle moving along a curve.


+F3(σ(u, v)) det

∂σ1∂u (u, v) ∂σ1∂v (u, v)

∂σ2∂u (u, v) ∂σ2∂v (u, v)

du dv=

∫ b2

a2

∫ b1

a1

⟨F (σ(u, v)) ,

∂σ

∂u(u, v)× ∂σ

∂v(u, v)

⟩du dv .

This quantity is called the surface integral or flux3 of the vector field F =(F1, F2, F3) through the surface σ, and will be denoted by∫

σ〈F, dS〉 .

Example. Let us compute the flux of the vector field F (x, y, z) = (−y, x, z2) through the surfaceσ : [0, 1]× [0, 1]→ R3, defined by σ(u, v) = (u2, v, u+ v) :∫

σ

〈F, dS〉 =

∫ 1

0

∫ 1

0

[(−v)(−1) + u2(−2u) + (u+ v)2(2u)] du dv =3

2.

3.8 Scalar functions and M-superficial measure

We recall that, if ω is a M -differential form defined on a subset U of RN, with1 ≤M ≤ N,

ω(x) =∑


and σ : I → RN is a M -surface whose support is contained in U , then∫σω =

∫I〈F (σ(u)) ,Σ(u)〉 du ,

where

F (x) = (fi1,...,iM (x))1≤i1<...<iM≤N ,

Σ(u) =(

detσ′(i1,...,iM )(u))

1≤i1<...<iM≤N.

In view of the applications, besides the integral of a M -differential form, it is usefulto define also the integral of a scalar function f : U → R on a M -surface.

3In fluidodynamics this concept is used, for instance, to define the amount of fluid crossing agiven surface in the unit time.

3.8. SCALAR FUNCTIONS AND M -SUPERFICIAL MEASURE 131

Definition 3.12 The function f : U → R is integrable on the M -surface σ : I →RN if (f σ)‖Σ‖ is integrable on I. In that case, we set∫

σf =

∫If(σ(u)) ‖Σ(u)‖ du

=

∫If(σ(u))

[ ∑1≤i1<...<iM≤N

(detσ′(i1,...,iM )(u)

)2] 1

2

du .

In this context, the integral does not differ for equivalent M -surfaces.

Theorem 3.13 If σ and σ are two equivalent M -surfaces, then∫σf =

∫σf .

Proof With the notations introduced previously, since σ = σ ϕ, with ϕ : A → B,we have

Σ(u) =(

detσ′(i1,...,iM )(u))

1≤i1<...<iM≤N

=(

det(σ′(i1,...,iM )(ϕ(u))ϕ′(u)

))1≤i1<...<iM≤N

=(

det σ′(i1,...,iM )(ϕ(u)))

1≤i1<...<iM≤Ndetϕ′(u)

= Σ(ϕ(u)) detϕ′(u) .

Therefore, by the Theorem on the Change of Variables in the integral, being I \ Aand J \B negligible, we have that∫

σf =

∫Af(σ(u)) ‖Σ(u)‖ du

=

∫Af(σ(ϕ(u))) ‖Σ(ϕ(u))‖ |detϕ′(u)| du

=

∫Bf(σ(v)) ‖Σ(v)‖ dv

=

∫σf ,

thus proving the claim.

In the case M = 1, we have a curve σ : [a, b]→ RN and, given a scalar functionf defined on the support of σ,∫

σf =

∫ b

af(σ(t)) ‖σ′(t)‖ dt .


Consider the interesting case when f is constantly equal to 1 : having in mind thephysical situation of a particle in motion along the curve described by σ, in this casethe line integral is called the length4 (or curvilinear measure) of the curve σ, andwe write

ι1(σ) =

∫ b

a‖σ′(t)‖ dt .

Example. Let σ : [0, b]→ R3 be defined by σ(t) = (t, t2, 0). Its support is an arc of parabola, andits length is given by

ι1(σ) =

∫ b

0

√1 + (2t)2 dt

=

∫ sinh−1(2b)

sinh−1(0)

1

2(coshu)2 du

=1

2

[u+ sinhu coshu

2

]sinh−1(2b)

0

=1

4

(sinh−1(2b) + 2b

√1 + 4b2

)=

1

4ln(

2b+√

1 + 4b2)

+b

2

√1 + 4b2 .

If M = 2 and N = 3, we have the surface σ : [a1, b1] × [a2, b2] → R3 and, givena scalar function f , defined on the support of σ,∫

σf =

∫ b2

a2

∫ b1

a1

f(σ(u, v))

∥∥∥∥∂σ∂u (u, v)× ∂σ

∂v(u, v)

∥∥∥∥ du dv .Again it is interesting to consider the case when f is constantly equal to 1 : in thiscase we call area (or surface measure) of the surface σ the following integral:

ι2(σ) =

∫ b2

a2

∫ b1

a1

∥∥∥∥∂σ∂u (u, v)× ∂σ

∂v(u, v)

∥∥∥∥ du dv .In the case when, for example, the surface happens to be a 2-parametrization of aset in R3, this integral is the flux of a vector field which at every point coincideswith the normal versor to the surface itself.5

Example. Let σ : [0, π]× [0, 2π]→ R3 be defined by

σ(φ, θ) = (R sinφ cos θ,R sinφ sin θ,R cosφ) .

4This definition could also be justified by geometrical considerations, which we omit here for thesake of briefness.

5Also the definition of the area of a surface can be justified by geometrical considerations, evenif the procedure is much more delicate than in the case of a curve.

3.8. SCALAR FUNCTIONS AND M -SUPERFICIAL MEASURE 133

Its support is a sphere of radius R, and its area is given by

ι2(σ) =

∫ 2π

0

∫ π

0

√(R2 sin2 φ cos θ)2 + (R2 sin2 φ sin θ)2 + (R2 sinφ cos θ)2 dφ dθ

=

∫ 2π

0

∫ π

0

R2 sinφdφ dθ

= 4πR2 .

In general, the case when f is constantly equal to 1 gives∫σ

1 =

∫I‖Σ(u)‖ du ,

and leads to the following.

Definition 3.14 We call M-superficial measure of a M -surface σ : I → RN thefollowing integral:

ιM (σ) =

∫I

[ ∑1≤i1<...<iM≤N

(detσ′(i1,...,iM )(u)

)2] 1

2

du .

As reasonably one expects, as a direct consequence of Theorem 3.13 and Theo-rem 3.8 one has the following.

Corollary 3.15 Two equivalent M -surfaces always have the same M -superficialmeasure. In particular, this is true for any two M -parametrizations of a given set.

Example. Consider the two curves σ, σ : [0, 2π]→ R2, defined by

σ(t) = (cos(t), sin(t)) , σ(t) = (cos(2t), sin(2t)) .

Notice that, even if they have the same support, these curves are not equivalent. Indeed, as is easily

seen, ι1(σ) = 2π while ι1(σ) = 4π.

The above considerations naturally lead to the following.

Definition 3.16 We call M-dimensional measure of a M -parametrizable setM⊆ RN the M -superficial measure of any of its M -parametrizations.

In the cases when M = 1, 2, the M -dimensional measure of M is often calledlength or area of M, respectively. We may thus consider, for example, the lengthof a circle or the area of a sphere.6

If M = N, it can be verified that the N−dimensional measure of the set M isthe same as the usual measure which has been treated in Chapter 2.

6We emphasize here the fact that the area of a sphere of radius R, which we found to be equalto 4πR2 by taking a particular parametrization, will always be the same when computed with anyparametrization. This fact is not always proved in other textbooks.


Exercises

1. Find the length of the elicoidal curve γ : [0, 2π]→ R3 defined as

γ(t) = (cos t, sin t, t) .

2. Compute the integral ∫γxyz d` ,

where γ : [0, 1]→ R3 is defined as γ(t) = (t, t2, t3).

3. Find the area of the ellipsoid(x, y, z) ∈ R3 :

x2

a2+y2

b2+z2

c2= 1

.

4. Compute the integral ∫σ(x+ y + z) dS ,

where σ : [1, 2]× [0, 1]→ R3 is the surface defined as

σ(u, v) = (u sin v, v sinu, cos(uv)) .

5. Find a parametrization σ : I → R3 of the set

M = (x, y, z) ∈ R3 : x22 + 4y2 + 9z2 = 1 ,

where I ⊆ R2 is some rectangle. Then, compute the integral∫σ f , where f : R3 → R

is the function defined by the formula

f(x, y, z) = xyz .

6. Find a parametrization σ : I → R3 of the set

M = (x, y, z) ∈ R3 : |z| ≤ 4x2 + 9y2 ≤ 1 ,

taking as I ⊆ R3 a rectangle (parallelepiped). Then, compute the volume of such asolid.

3.9. THE ORIENTED BOUNDARY OF A RECTANGLE 135

3.9 The oriented boundary of a rectangle

Assume that σ1 : I1 → RN, . . . , σn : In → RN are some M -surfaces. We can easilyfind some equivalent M -surfaces σ1 : J1 → RN,. . . , σn : Jn → RN, with the sameorientation, such that the rectangles J1, . . . , Jn be non overlapping and whose unionbe a rectangle I.

Definition 3.17 We call glueing of the M -surfaces σ1, . . . , σn any function σ :

I → RN whose restrictions to J1

, . . . , Jn

coincide with σ1 , . . . , σn, respectively; it

is almost everywhere differentiable, and we can define∫σ ω by the same formula we

have used for the M -surfaces of class C1. Hence,∫σω =

∫σ1

ω + . . .+

∫σn

ω .

We have thus “glued” together the M -surfaces σ1, . . . , σn and defined an integralof this glueing which does not depend on the particular choice of the equivalent M -surfaces, since anyway they conserve the orientation. In practice, however, we willnever need to construct explicitly the glueing; what will actually be important is theformula for the integral.

Assume now that I be a rectangle7 of RM+1, with M ≥ 1:

I = [a1, b1]× . . .× [aM+1, bM+1] .

We denote by Ik the rectangle of RM obtained from I by suppression of the k−thcomponent:

Ik = [a1, b1]× . . .× [ak−1, bk−1]× [ak+1, bk+1]× . . .× [aM+1, bM+1] .

Consider, for every k, the M -surfaces α+k , β

+k : Ik → RM+1 defined by

α+k (u1, . . . , uk, .., uM+1) = (u1, . . . , uk−1, ak, uk+1, . . . , uM+1) ,

β+k (u1, . . . , uk, .., uM+1) = (u1, . . . , uk−1, bk, uk+1, . . . , uM+1) ,

where the meaning of the symbol is to suppress the underlying variable. Considermoreover some M -surfaces α−k , β

−k : Ik → RM+1, equivalent to α+

k , β+k , respectively,

with opposite orientation. For example, we can take

α−k (u1, . . . , uk, .., uM+1) = (u1, . . . , uk−1, ak + bk − uk, uk+1, . . . , uM+1) ,

β−k (u1, . . . , uk, .., uM+1) = (u1, . . . , uk−1, ak + bk − uk, uk+1, . . . , uM+1) ,

7We are thus considering here the situation when N = M + 1. This setting will be maintainedalso in the next section.


Definition 3.18 We call oriented boundary of the rectangle I a function ∂Iwhich is a glueing of the following M -surfaces:

(a) α−k and β+k if k is odd;

(b) α+k and β−k if k is even.

If ω is a M -differential form defined on a subset U of RM+1 containing the imageof ∂I, we will then have∫

∂Iω =

M+1∑k=1

(−1)k∫α+k

ω +M+1∑k=1

(−1)k−1

∫β+k

ω

=

M+1∑k=1

(−1)k−1

(∫β+k

ω −∫α+k

ω

).

Let M = 1, and consider the rectangle [a1, b1]× [a2, b2]. Then, for example,

α−1 : [a2, b2]→ R2, v 7→ (a1, a2 + b2 − v) ,

β+1 : [a2, b2]→ R2, v 7→ (b1, v) ,

α+2 : [a1, b1]→ R2, u 7→ (u, a2) ,

β−2 : [a1, b1]→ R2, u 7→ (a1 + b1 − u, b2) .

We can visualize geometrically ∂I as the glueing of the sides of the rectangle Ioriented in such a way that the perimeter be described in counter-clockwise direction.

a1 b1

α2+

b2

a2

β2−

β1+

α1−

If M = 2, we have, for example,

α−1 : [a2, b2]× [a3, b3]→ R3, (v, w) 7→ (a1, a2 + b2 − v, w) ,

β+1 : [a2, b2]× [a3, b3]→ R3, (v, w) 7→ (b1, v, w) ,

α+2 : [a1, b1]× [a3, b3]→ R3, (u,w) 7→ (u, a2, w) ,

3.10. THE GAUSS FORMULA 137

β−2 : [a1, b1]× [a3, b3]→ R3, (u,w) 7→ (u, b2, a3 + b3 − w) ,

α−3 : [a1, b1]× [a2, b2]→ R3, (u, v) 7→ (a1 + b1 − u, v, a3) ,

β+3 : [a1, b1]× [a2, b2]→ R3, (u, v) 7→ (u, v, b3) .

In this case, we can visualize ∂I as the glueing of the six faces of the parallelepipedI, each oriented in such a way that the normal versor be alway directed towards theexterior.

α2+ β2−

α1−

α3−

β3+

β1+

3.10 The Gauss formula

In this section, I will be a rectangle in RN, with N ≥ 2. In the following theorem,the elegant Gauss formula is obtained.8

Theorem 3.19 If ω is a (N − 1)-differential form of class C1 defined on an openset containing the rectangle I in RN, then∫

Idω =

∫∂Iω .

Proof We can write ω as

ω(x) =N∑j=1

Fj(x) dx1 ∧ . . . ∧ dxj ∧ . . . ∧ dxN .

8As mentioned in the previous footnote, referring to the previous section, we have N = M + 1.


Then,

dω(x) =N∑j=1

N∑m=1

∂Fj∂xm

(x) dxm ∧ dx1 ∧ . . . ∧ dxj ∧ . . . ∧ dxN

=N∑j=1

(−1)j−1∂Fj∂xj

(x) dx1 ∧ . . . ∧ dxN .

Being the partial derivatives of each Fj continuous, they are integrable on the rect-angle I, and we can use Fubini Theorem:

∫Idω=

N∑j=1

(−1)j−1

∫I

∂Fj∂xj

(x) dx1 . . . dxN

=N∑j=1

(−1)j−1

∫Ij

(∫ bj

aj

∂Fj∂xj

(x1, . . . , xN ) dxj

)dx1 . . . dxj . . . dxN

=N∑j=1

(−1)j−1

∫Ij

[Fj(x1, . . . , xj−1, bj , xj+1, . . . , xN )−

−Fj(x1, . . . , xj−1, aj , xj+1, . . . , xN )] dx1 . . . dxj . . . dxN ,

by the Fundamental Theorem. On the other hand, we have

∫α+k

ω=N∑j=1

∫α+k

Fj dx1 ∧ . . . ∧ dxj ∧ . . . ∧ dxN

=

N∑j=1

∫Ik

(Fj α+k ) det(α+

k )′(1,...,j,...,N)

dx1 . . . dxj . . . dxN

=

∫Ik

Fk(x1, . . . , xk−1, ak, xk+1, . . . , xN ) dx1 . . . dxk . . . dxN ,

being

det(α+k )′

(1,...,j,...,N)=

0 if j 6= k ,

1 if j = k .

Similarly,∫β+

ω =

∫Ik

Fk(x1, . . . , xk−1, bk, xk+1, . . . , xN ) dx1 . . . dxk . . . dxN ,

3.11. ORIENTED BOUNDARY OF A M -SURFACE 139

so that∫∂Iω=

N∑k=1

(−1)k−1

(∫β+k

ω −∫α+k

ω

)

=N∑k=1

(−1)k−1

∫Ik

[Fk(x1, . . . , xk−1, bk, xk+1, . . . , xN )−

−Fk(x1, . . . , xk−1, ak, xk+1, . . . , xN )] dx1 . . . dxk . . . dxN ,

and the proof is completed.

Remark. The regularity assumption on the differential form ω could be consider-ably weakened. However, for briefness, we will not enter into this discussion. Theinterested reader is referred to [19].

3.11 Oriented boundary of a M-surface

In this section, I will be a rectangle in RM+1 and σ : I → RN a (M + 1)-surface.

Definition 3.20 For 1 ≤M ≤ N −1, we call oriented boundary of σ a function∂σ = σ ∂I, which is a glueing of the following M -surfaces:

(a) σ α−k and σ β+k if k is odd;

(b) σ α+k and σ β−k if k is even.

Given a M -differential form ω whose domain contains the support of ∂σ, we willthen have ∫

∂σω =

M+1∑k=1

(−1)k∫σα+

k

ω +

M+1∑k=1

(−1)k−1

∫σβ+

k

ω

=M+1∑k=1

(−1)k−1

(∫σβ+

k

ω −∫σα+

k

ω

).

Remark. It is useful to extend the meaning of∫∂σ ω to the case when σ : [a, b]→ RN

is a curve, with N ≥ 1, and ω = f : U → R is a 0-differential form; in this case, weset ∫

∂σω = f(σ(b))− f(σ(a)) .

Examples. As an illustration, consider as usual the case N = 3. We begin with three examples oforiented boundaries of surfaces.


1. Let σ : [r,R]× [0, 2π]→ R3, with 0 ≤ r < R, be given by

σ(u, v) = (u cos v, u sin v, 0) .

Its support is a disk if r = 0, an annulus if r > 0. The oriented boundary ∂σ is given by a glueingof the following four curves:

σ α−1 (v) = (r cos v,−r sin v, 0) ,

σ β+1 (v) = (R cos v,R sin v, 0) ,

σ α+2 (u) = (u, 0, 0) ,

σ β−2 (u) = (r +R− u, 0, 0) .

The first curve has as support a circle with radius r, which degenerates in the origin in the casewhen r = 0. The second has as support a circle with radius R. Notice however that these two circlesare described by the two curves in opposite directions. The last two curves are equivalent withopposite orientations.

Consider, for example, the vector field F (x, y, z) = (−y, x, xyez). Then,∫∂σ

〈F, d`〉=

∫σα−

1

〈F, d`〉+

∫σβ+

1

〈F, d`〉

=

∫ 2π

0

[−r2 sin2 v − r2 cos2 v] dv +

∫ 2π

0

[R2 sin2 v +R2 cos2 v] dv

= 2π(R2 − r2) .

2. Consider the surface σ : [r,R]× [0, 2π]→ R3, with 0 < r < R, defined by

σ(u, v) =

((r +R

2+

(u− r +R

2

)cos(v

2

))cos v,(

r +R

2+

(u− r +R

2

)cos(v

2

))sin v,(

u− r +R

2

)sin(v

2

)),

whose support is a Mobius strip. In this case, the oriented boundary is given by a glueing of

σ α−1 (v) =

((r +R

2+R− r

2cos(v

2

))cos v,

−(r +R

2+R− r

2cos(v

2

))sin v,

−R− r2

sin(v

2

)),

3.11. ORIENTED BOUNDARY OF A M -SURFACE 141

σ β+1 (v) =

((r +R

2+R− r

2cos(v

2

))cos v,(

r +R

2+R− r

2cos(v

2

))sin v,

R− r2

sin(v

2

)),

σ α+2 (u) = (u, 0, 0) ,

σ β−2 (u) = (u, 0, 0) .

Notice that in this case the two last curves are the same.

3. Consider the surface σ : [0, π]× [0, 2π]→ R3 defined by

σ(φ, θ) = (R sinφ cos θ,R sinφ sin θ,R cosφ) ,

whose support is the sphere with radius R > 0, centered at the origin. In this case, the orientedboundary is a glueing of

σ α−1 (θ) = (0, 0, R) ,

σ β+1 (θ) = (0, 0,−R) ,

σ α+2 (φ) = (R sinφ, 0, R cosφ) ,

σ β−2 (φ) = (R sinφ, 0,−R cosφ) .

Notice that the first two curves are degenerated in one point, while the two others are equivalentwith opposite orientations. Hence, for any choice of a vector field F, it will be

∫∂σ〈F, d`〉 = 0.

Let us see now an example of oriented boundary of a volume in R3. Let σ : [0, R] × [0, π] ×[0, 2π]→ R3 be the volume defined by

σ(ρ, φ, θ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ,

whose support is the closed ball, centered at the origin, with radius R > 0. The oriented boundary∂σ is a glueing of the following six surfaces:

σ α−1 (φ, θ) = (0, 0, 0) ,

σ β+1 (φ, θ) = (R sinφ cos θ,R sinφ sin θ,R cosφ) ,

σ α+2 (ρ, θ) = (0, 0, ρ) ,

σ β−2 (ρ, θ) = (0, 0,−ρ) ,

σ α−3 (ρ, φ) = ((R− ρ) sinφ, 0, (R− ρ) cosφ) ,

σ β+3 (ρ, φ) = (ρ sinφ, 0, ρ cosφ) .


Notice that the first surface is degenerated in a point (the origin), the second has as support theentire sphere, the third and the fourth are degenerated in two lines, while the remaining two areequivalent with opposite orientations. Hence, given a vector field F, we will have∫

∂σ

〈F, dS〉 =

∫σβ+

1

〈F, dS〉 .

3.12 The Stokes–Cartan formula

Let us state the following generalization of the Gauss theorem, where the importantStokes–Cartan formula is obtained.

Theorem 3.21 Let 0 ≤M ≤ N − 1. If ω : U → ΩM (RN ) is a M -differential formof class C1 and σ : I → RN is a (M + 1)-surface whose support is contained in U,then ∫

σdω =

∫∂σω .

Notice that the case M = 0, N = 1 and σ(u) = u is a version of the FundamentalTheorem, with the further assumption on the derivative of ω to be continuous.

The general proof of the above theorem is given in Appendix B. We will con-centrate here on some corollaries obtained, in the case N = 3, when M takes thevalues 0, 1 and 2. It is interesting to prove directly these corollaries, showing howthe general proof adapts to them.

The case M = 0. We consider a 0-differential form ω = f : U → R, and we obtainthe following.

Theorem 3.22 Let f : U → R be a scalar function of class C1 and σ : [a, b] → R3

a curve whose support is contained in U. Then,∫σ〈grad f, d`〉 = f(σ(b))− f(σ(a)) .

Proof Consider the function G : [a, b] → R defined by G(t) = f(σ(t)). It is of classC1, and by the Fundamental Theorem we have∫ b

aG′(t) dt = G(b)−G(a) .

Since G′(t) = 〈grad f(σ(t)), σ′(t)〉, the conclusion follows.

Remark. The line integral of the gradient of a function f does not depend on thechosen curve itself, but only on the values of the function at the two extrema σ(b)and σ(a).

3.12. THE STOKES–CARTAN FORMULA 143

Example. Given

F (x, y, z) = −(

x

[x2 + y2 + z2]3/2,

y

[x2 + y2 + z2]3/2,

z

[x2 + y2 + z2]3/2

)and the curve σ : [0, 4π]→ R3 defined by σ(t) = (cos t, sin t, t), we want to compute the line integral∫σ〈F, d`〉. Observe that F = grad f, with

f(x, y, z) =1√

x2 + y2 + z2.

Hence, ∫σ

〈F, d`〉 = f(σ(4π))− f(σ(0)) =1√

1 + 16π2− 1 .

The case M = 1. We consider a 1-differential form

ω(x) = F1(x) dx1 + F2(x) dx2 + F3(x) dx3 ,

and we obtain the Stokes–Ampere formula.

Theorem 3.23 Let F : U → R3 be a C1-vector field and σ : [a1, b1]× [a2, b2]→ R3

be a surface whose support is contained in U. Then,∫σ〈curlF, dS〉 =

∫∂σ〈F, d`〉 .

Verbally. The flux of the curl of the vector field F through the surface σ is equalto the line integral of F along the oriented boundary of σ.

Proof Let I = [a1, b1]× [a2, b2], and define the following 1-differential form ω : I →Ω1(R2) :

ω(u, v) =

⟨F (σ(u, v)) ,

∂σ

∂u(u, v)

⟩du+

⟨F (σ(u, v)) ,

∂σ

∂v(u, v)

⟩dv .

We first consider its integral on α−1 :∫α−1

ω=

∫ b2

a2

⟨F (σ(a1, a2 + b2 − v)) ,−∂σ

∂v(a1, a2 + b2 − v)

⟩dv

=

∫σα−1〈F, d`〉 .

The integrals on β+1 , α

+2 and β−2 are then treated analogously, so that∫

∂Iω =

∫∂σ〈F, d`〉 .


Assume now that σ is of class C2. Then, ω is of class C1 and, with some computations,we get

dω(u, v) =

[∂

∂u

⟨F (σ(u, v)) ,

∂σ

∂v(u, v)

⟩− ∂

∂v

⟨F (σ(u, v)) ,

∂σ

∂u(u, v)

⟩]du ∧ dv

=

⟨curlF (σ(u, v)) ,

∂σ

∂u(u, v)× ∂σ

∂v(u, v)

⟩du ∧ dv ,

so that ∫Idω =

∫σ〈curlF, dS〉 .

The Gauss formula applied to ω thus yields the conclusion in this case.

The assumption that σ be of class C2 can eventually be eliminated by an approx-imation procedure: it is possible to construct a sequence (σn)n of surfaces of classC2 which converge to σ together with all their partial derivatives of the first order.The Stokes–Ampere formula holds then for those surfaces, and passing to the limit,by the dominated convergence theorem, we have the conclusion.

Example. Let F (x, y, z) = (−y, x, 0) and γ : [0, 2π] → R3 be the curve defined by γ(t) =(R cos t, R sin t, 0); we want to compute the line integral

∫γ〈F, d`〉. We have already seen how to

compute this integral by the direct use of the definition. We now proceed in a different way:consider the surface σ : [0, R] × [0, 2π] → R3 given by σ(ρ, θ) = (ρ cos θ, ρ sin θ, 0). Observe thatγ = σ β+

1 , so ∫γ

〈F, d`〉 =

∫σβ+

1

〈F, d`〉 =

∫∂σ

〈F, d`〉 =

∫σ

〈curlF, dS〉 .

Being curlF (x, y, z) = (0, 0, 2) and

∂σ

∂ρ(ρ, θ)× ∂σ

∂θ(ρ, θ) = (0, 0, ρ) ,

we then have ∫γ

〈F, d`〉 =

∫ R

0

∫ 2π

0

〈(0, 0, 2) , (0, 0, ρ)〉 dθ dρ = 2πR2 .

The case M = 2. We consider a 2-differential form

ω(x) = F1(x) dx2 ∧ dx3 + F2(x) dx3 ∧ dx1 + F3(x) dx1 ∧ dx2 ,

and we obtain the Gauss–Ostrogradski formula.

Theorem 3.24 Let F : U → R3 be a C1-vector field and σ : I = [a1, b1]× [a2, b2]×[a3, b3]→ R3 be a volume whose support is contained in U . Then,∫

σdivF dx1 ∧ dx2 ∧ dx3 =

∫∂σ〈F, dS〉 .

3.12. THE STOKES–CARTAN FORMULA 145

Hence, if σ is regular and injective on I, with detσ′ > 0, then∫

σ(I)divF =

∫∂σ〈F, dS〉 .

In intuitive terms. The integral of the divergence of the vector field F on theset V = σ(I) is equal to the flux of F which exits from V.

Proof Let ω : I → Ω2(R3) be the 2-differential form defined by

ω(u) =

⟨F (σ(u)) ,

∂σ

∂u2(u)× ∂σ

∂u3(u)

⟩du2 ∧ du3 +

+

⟨F (σ(u)) ,

∂σ

∂u3(u)× ∂σ

∂u1(u)

⟩du3 ∧ du1 +

+

⟨F (σ(u)) ,

∂σ

∂u1(u)× ∂σ

∂u2(u)

⟩du1 ∧ du2 .

Considering β+1 , we have:∫

β+1

ω =

∫ b2

a2

∫ b3

a3

⟨F (b1, u2, u3) ,

∂σ

∂u2(b1, u2, u3)× ∂σ

∂u3(b1, u2, u3)

⟩du2 du3

=

∫β+1

〈F, dS〉 .

With the analogous computations on the remaining five surfaces which determine∂I we can say that ∫

∂Iω =

∫∂σ〈F, dS〉 .

Assume now that σ be of class C2. Then ω is of class C1 and, carrying over thecomputations, with some tenacity, we have

dω(u) =

[∂

∂u1

⟨F (σ(u)) ,

∂σ

∂u2(u)× ∂σ

∂u3(u)

⟩+

+∂

∂u2

⟨F (σ(u)) ,

∂σ

∂u3(u)× ∂σ

∂u1(u)

⟩+

+∂

∂u3

⟨F (σ(u)) ,

∂σ

∂u1(u)× ∂σ

∂u2(u)

⟩]du1 ∧ du2 ∧ du3

= divF (σ(u)) detσ′(u) du1 ∧ du2 ∧ du3 .

Hence, ∫Idω =

∫I

divF (σ(u)) detσ′(u) du1 ∧ du2 ∧ du3 =

∫σ

divF .


The Gauss formula applied to ω thus yields the first conclusion. On the other hand,

if σ is regular and injective on I, with detσ′ > 0, Theorem 3.11 yields the second

formula, and the proof is completed. The assumption that σ be of class C2 caneventually be eliminated, as in the previous theorem.

Example. We want to compute the flux of the vector field

F (x, y, z) = ([x2 + y2 + z2]x, [x2 + y2 + z2]y, [x2 + y2 + z2]z)

through a spherical surface parametrized by η : [0, π]× [0, 2π]→ R3, defined as

η(φ, θ) = (R sinφ cos θ,R sinφ sin θ,R cosφ) .

Recall that η = σ β+1 , where σ : I = [0, R]× [0, π]× [0, 2π]→ R3 is the volume given by

σ(ρ, φ, θ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) .

Hence, ∫η

〈F, dS〉 =

∫∂σ

〈F, dS〉 =

∫σ(I)

divF .

Being divF (x, y, z) = 5(x2 + y2 + z2), passing to spherical coordinates we have∫σ(I)

divF =

∫ 2π

0

∫ π

0

∫ R

0

(5ρ2)(ρ2 sinφ) dρ dφ dθ = 4πR5 .

3.13 Analogous results in R2

Assuming that U be a subset of R2 we find two interesting corollaries of the Stokes–Cartan theorem. Like in the case N = 3, the line integral of a vector field F =(F1, F2) along a curve σ : [a, b]→ R2 is defined as∫

σ〈F, d`〉 =

∫ b

a[F1(σ(t))σ′1(t) + F2(σ(t))σ′2(t)] dt .

We have the following result, analogous to the one obtained in the previous sectionfor the case N = 3.

Theorem 3.25 Let f : U → R be a scalar function of class C1 and σ : [a, b] → R2

be a curve with support contained in U. Then,∫σ

⟨(∂f

∂x1,∂f

∂x2

), d`

⟩= f(σ(b))− f(σ(a)) .

Taking now M = 1, we obtain the Gauss–Green formula.

3.13. ANALOGOUS RESULTS IN R2 147

Theorem 3.26 Let F = (F1, F2) : U → R2 be a C1-vector field and σ : I =[a1, b1]× [a2, b2]→ R2 be a surface whose support is contained in U . Then,∫

σ

(∂F2

∂x1− ∂F1

∂x2

)dx1 ∧ dx2 ∧ dx3 =

∫∂σ〈F, dS〉 .

Hence, if σ is regular and injective on I, with detσ′ > 0, then∫

σ(I)

(∂F2

∂x1− ∂F1

∂x2

)=

∫∂σ〈F, d`〉 .

Proof As for the Stokes–Ampere theorem, we consider the auxiliary differentialform ω : I → Ω1(R2), defined by

ω(u, v) =

⟨F (σ(u, v)) ,

∂σ

∂u(u, v)

⟩du+

⟨F (σ(u, v)) ,

∂σ

∂v(u, v)

⟩dv ,

and see that ∫∂Iω =

∫∂σ〈F, d`〉 .

If σ is of class C2, then ω is of class C1 and computation shows that

dω(u, v) =

[∂

∂u

⟨F (σ(u, v)) ,

∂σ

∂v(u, v)

⟩−

− ∂

∂v

⟨F (σ(u, v)) ,

∂σ

∂u(u, v)

⟩]du ∧ dv

=

(∂F2

∂x1(σ(u, v))− ∂F1

∂x2(σ(u, v))

)detσ′(u, v) du ∧ dv .

Hence,∫Idω =

∫I

(∂F2

∂x1(σ(u, v))− ∂F1

∂x2(σ(u, v))

)detσ′(u, v) du∧ dv =

∫σ

(∂F2

∂x1− ∂F1

∂x2

).

The Gauss formula applied to ω gives us the first conclusion. If σ is regular and

injective on I, with detσ′ > 0, then Theorem 3.11 yields the conclusion. As men-

tioned in the case N = 3, the assumption that σ is of class C2 can be eliminated byan approximation procedure.

Example. Consider the surface σ : I = [0, 1]× [0, 2π]→ R2 defined by σ(ρ, θ) = (Aρ cos θ,Bρ sin θ),whose support is an elliptical surface with semi-axes having lengths A > 0 and B > 0. Take thevector field F (x, y) = (−y, x). Being

∂F2

∂x(x, y)− ∂F1

∂y(x, y) = 2 ,


and (as for the disk) ∫∂σ

〈F, d`〉 =

∫σβ+

1

〈F, d`〉 ,

the Gauss–Green formula gives us∫σ(I)

2 dx dy =

∫ 2π

0

〈(−B sin θ,A cos θ) , (−A sin θ,B cos θ)〉 dθ = 2πAB.

We then find the area of the elliptic surface: µ(σ(I)) = πAB .

Exercises

1. Let γ : [0, 4]→ R2 be the curve defined as γ(t) = (t,√t ). Compute the integral∫

γ〈(xy2 , yx2), d`〉 ,

first directly, then by the use of a scalar potential.

2. Let σ : [0, 1]× [0, 1]→ R2 be the surface defined by

σ(u, v) = (u2, v2) .

Compute∫∂σ〈F, d`〉 with F : R2 → R2 given by

F (x, y) = (x2 − y2, 2xy) .

3. Let σ : [0, 1]× [0, 1]→ R3 be the surface defined by

σ(u, v) = (u2, v2, uv) .

Compute∫σ〈curlF, dS〉, both directly and by the use of the Stokes–Ampere formula,

whenF (x, y, z) = (x− y, y − z, z − x) .

4. Let σ : [0, 1]× [0, 1]× [−1, 1]→ R3 be the volume defined as

σ(u, v, w) = (u2 + v2, u2 + v2, w) ,

and consider the vector field F : R3 → R3 given by

F (x, y, z) = (x2, y3, z4) .

Compute∫σ divF dx∧dy∧dz, both directly and by the use of the Gauss–Ostrogradski

formula.

5. Let σ : [0, 1] × [0, π] × [0, 2π] → R3 be the parametrization of the unit ball inspherical coordinates, given by

σ(ρ, φ, θ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) .

Compute∫∂σ〈F, dS〉, where F : R3 → R3 is the vector field defined as

F (x, y, z) = (x− y + z , y − z + x , z − x+ y) .

3.14. EXACT DIFFERENTIAL FORMS 149

3.14 Exact differential forms

We are now interested in the following problem. Given a differential form ω, whenis it possible to write it as the external differential of another differential form ω, tobe determined? In the following, let M ≥ 1.

Definition 3.27 A M -differential form ω is said to be closed if dω = 0; it is saidto be exact if there is a (M − 1)-differential form ω such that dω = ω.

Every exact differential form is closed: if ω = dω, then dω = d(dω) = 0. Thecontrary is not always true.

Example. The 1-differential form defined on R2 \ (0, 0) by

ω(x, y) =−y

x2 + y2dx+

x

x2 + y2dy

is closed, as easily verified: setting

F1(x, y) =−y

x2 + y2, F2(x, y) =

x

x2 + y2,

for every (x, y) 6= (0, 0), it is∂F2

∂x(x, y) =

∂F1

∂y(x, y) .

Let us compute the line integral of its vector field F = (F1, F2) on the curve σ : [0, 2π] → R2,defined by σ(t) = (cos t, sin t) :∫

σ

〈F, d`〉=

∫ 2π

0

〈F (σ(t) , σ′(t)〉 dt

=

∫ 2π

0

〈(− sin t, cos t) , (− sin t, cos t)〉 dt

= 2π .

Assume by contradiction that ω be exact, i.e., that there exists a C1-function f : R2 \ (0, 0) → Rsuch that ∂f

∂x= F1 and ∂f

∂y= F2. In that case, being σ(0) = σ(2π), we should have∫

σ

〈F, d`〉 =

∫σ

⟨(∂f

∂x,∂f

∂y

), d`

⟩= f(σ(2π))− f(σ(0)) = 0 ,

which contradicts the above.

The Poincare theorem says that the situation of the preceding example cannever happen if, for example, the open set U on which the differential form is definedis star-shaped.

Definition 3.28 A set U is star-shaped with respect to a point x if, with each ofits points x, the set U contains the whole segment joining x to x, i.e.,

x + t(x− x) : t ∈ [0, 1] ⊆ U .


For example, every convex set is star-shaped (with respect to any of its points).In particular, a ball, or a rectangle, or even the whole space RN are star-shaped.Clearly, the set R2 \ (0, 0) considered above is not star-shaped.

Theorem 3.29 Let U be an open subset of RN , star-shaped with respect to a pointx. For 1 ≤ M ≤ N, a M -differential form ω : U → ΩM (RN ) of class C1 is exact ifand only if it is closed. In that case, if ω is of the type

ω(x) =∑


a (M − 1)-differential form ω such that dω = ω is given by

ω(x) =∑

1≤i1<...<iM≤N

M∑s=1

(−1)s+1(xis − xis) ·

·(∫ 1

0tM−1fi1,...,iM (x + t(x− x)) dt

)dxi1 ∧ . . . ∧ dxis ∧ . . . ∧ dxiM .

The proof of this theorem will be given in Appendix B. As for the Stokes–Cartantheorem, we consider here some corollaries which hold true for the case N = 3, givingfor each of them a direct proof. In order to simplify the notations, we will assumewithout loss of generality that x = (0, 0, 0).

The case M = 1. A C1-vector field F = (F1, F2, F3), defined on an open subset U ofR3, determines a 1-differential form

ω(x) = F1(x) dx1 + F2(x) dx2 + F3(x) dx3 .

This is closed if and only if curlF = 0. In this case, the vector field is said to beirrotational. On the other hand, the vector field is said to be F conservative ifthere is a function f : U → R such that F = grad f. In that case, f is a scalarpotential of the vector field F .9

Theorem 3.30 If U ⊆ R3 is star-shaped with respect to the origin, then the vectorfield F : U → R3 is conservative if and only if it is irrotational, and in that case afunction f : U → R such that F = grad f is given by

f(x) =

∫ 1

0〈F (tx),x〉 dt .

Any other function f : U → R which is such that F = grad f is obtained from f byadding a constant.

9Beware that in Mechanics it is often the function −f which is called “the potential”.


Proof Set ω = f : U → R. Let us verify that dω = ω. Using the fact that curlF = 0and the Leibniz rule, we have

∂ω

∂xj(x) =

∫ 1

0

∂

∂xj〈F (tx),x〉 dt

=

∫ 1

0

(3∑i=1

(∂Fi∂xj

(tx)txi

)+ Fj(tx)

)dt

=

∫ 1

0

(3∑i=1

(∂Fj∂xi

(tx)txi

)+ Fj(tx)

)dt .

Defining φ(t) = tFj(tx), since

φ′(t) =3∑i=1

(∂Fj∂xi

(tx)txi

)+ Fj(tx) ,

by the Fundamental Theorem we have

∂ω

∂xj(x) =

[tFj(tx)

]t=1

t=0= Fj(x) .

This proves that F = grad f. The second part of the theorem follows directly fromthe fact that, if grad f = grad f , then f − f has to be constant on U.

Example. Consider the vector field F (x, y, z) = (2xz + y, x, x2) which, as easily verified, is irrota-tional. A scalar potential is then given by

f(x, y, z) =

∫ 1

0

((2t2x2z + txy) + txy + t2x2z) dt = xy + x2z .

The case M = 2. A C1-vector field F = (F1, F2, F3), defined on an open subset U ofR3, determines a 2-differential form

ω(x) = F1(x) dx2 ∧ dx3 + F2(x) dx3 ∧ dx1 + F3(x) dx1 ∧ dx2 .

This is closed if and only if divF = 0. In this case, the vector field is said to besolenoidal. One says that F has a vector potential if there is a vector fieldV = (V1, V2, V3) such that F = curlV.

Theorem 3.31 If U ⊆ R3 is star-shaped with respect to the origin, then the vectorfield F : U → R3 has a vector potential if and only if it is solenoidal, and in that


case a vector field V : U → R3 for which F = curlV is given by

V (x) =(∫ 1

0t(F2(tx)x3 − F3(tx)x2) dt ,∫ 1

0t(F3(tx)x1 − F1(tx)x3) dt ,∫ 1

0t(F1(tx)x2 − F2(tx)x1) dt

),

which we will briefly write as

V (x) =

∫ 1

0t(F (tx)× x) dt .

Any other vector field V : U → R3 such that F = curl V is obtained from V byadding the gradient of an arbitrary scalar function.

Proof Consider the 1-differential form determined by the vector field V ,

ω(x) =

(∫ 1


)dx1 +

+

(∫ 1


)dx2 +

+

(∫ 1


)dx3 .

We have to prove that dω = ω. By the Leibniz rule, taking into account the factthat ω is closed, we find

∂

∂x2

∫ 1

0t(F1(tx)x2 − F2(tx)x1) dt− ∂

∂x3

∫ 1

0t(F3(tx)x1 − F1(tx)x3) dt =

=

∫ 1

0

(t2(∂F1

∂x1(tx)x1 +

∂F1

∂x2(tx)x2 +

∂F1

∂x3(tx)x3

)+ 2tF1(tx)

)dt

= F1(x) ,

by applying the Fundamental Theorem to the function φ(t) = t2F1(tx). Analogouslyone obtains the remaining two identities, thus proving the formula. The second partof the theorem follows from the fact that, if curlV = curl V , then, by the previoustheorem, V − V is a conservative vector field.


Example. Consider the solenoidal vector field F (x, y, z) = (y, z, x). A vector potential is then givenby

V (x, y, z) =

∫ 1

0

t(ty, tz, tx)× (x, y, z) dt =1

3(z2 − xy, x2 − yz, y2 − xz) .

The case M = 3. A C1-scalar function f, defined on an open subset U of R3, deter-mines a 3-differential form

ω(x) = f(x) dx1 ∧ dx2 ∧ dx3 .

This is necessarily always closed, since dω is a 4-differential form defined on a subsetof R3.

Theorem 3.32 If U ⊆ R3 is star-shaped with respect to the origin, the functionf : U → R is always of the type f = divW, where W : U → R3 is the vector fielddefined by

W (x) =

(∫ 1

0t2f(tx) dt

)x .

Any other vector field W : U → R3 such that F = div W is obtained from W byadding the curl of an arbitrary vector field.

Proof Using the Leibniz rule, we have

∂

∂x1

∫ 1

0t2f(tx)x1 dt+

∂

∂x2

∫ 1

0t2f(tx)x2 dt+

∂

∂x3

∫ 1

0t2f(tx)x3 dt

=

∫ 1

0

(t3(∂f

∂x1(tx) +

∂f

∂x2(tx) +

∂f

∂x3(tx)

)+ 3t2f(tx)

)dt

= f(x) ,

by applying the Fundamental Theorem to the function φ(t) = t3f(tx). The secondpart of the theorem follows from the fact that, if divW = div W , then, by thepreceding theorem, W − W has a vector potential.

Example. Consider the scalar function f(x, y, z) = xyz. Then, a vector field whose divergence is fis given by

W (x, y, z) =

(∫ 1

0

t5xyz dt

)(x, y, z) =

1

6(x2yz, xy2z, xyz2) .


Exercises

1. Let F : R3 → R3 be the vector field defined by the formula

F (x, y, z) = (exyz + eyz + ezy , exz + eyxz + ezx , exy + eyx+ ezxy) ,

and let ω be the associated 1-differential form. Prove that ω is exact, and find a0-differential form ω such that dω = ω. Compute then the integral∫

γω ,

where γ : [0, 1]→ R3 is the curve defined as

γ(t) = (t, t2, t3) .

2. Let F : R3 \ z = 0 → R3 be the vector field defined as

F (x, y, z) =(2xy2

z2,

2x2y

z2,−2x2y2

z3

).

Prove that it is conservative, and find a scalar potential.

3. Let F : R3 → R3 be defined as

F (x, y, z) =(− 2xyz ,

xz

2, yz2

).

Prove that it is solenoidal, and find a vector potential. Finally, compute the flux∫σ F · dS, where σ : [0, 2π]× [0, π]→ R3 is defined by

σ(u, v) = (cosu sin v , 2 sinu sin v , 3 cos v) .

4. Let ω : R4 → Ω3(R4) be the 3-differential form defined by

ω(x1, x2, x3, x4) = x1x2 dx2 ∧ dx3 ∧ dx4 + x22 dx1 ∧ dx3 ∧ dx4 +

+x2x3 dx1 ∧ dx2 ∧ dx4 + x21x2 dx1 ∧ dx2 ∧ dx3 .

Prove that ω is exact, and find a 2-differential form ω such that dω = ω.

5. Prove that the differential form

ω : R3 \ (x = 0 ∪ y = 0 ∪ z = 0)→ Ω1(R3) ,

defined as

ω(x, y, z) =1

x2yzdx+

1

xy2zdy +

1

xyz2dz,

is exact, and find a 0-differential form ω such that dω = ω.

Appendix A

Differential calculus in RN

Let Ω ⊂ RN be an open set, x0 a point of Ω, and f : Ω→ RM be a given function.We want to extend the notion of derivative already known in the case M = N = 1.Here is the definition.

Definition A.1 We say that f is differentiable at x0 if there exists a linear function` : RN → RM for which one can write

f(x) = f(x0) + `(x− x0) + r(x) ,

where r is a function satisfying

limx→x0

r(x)

‖x− x0‖= 0 .

If f is differentiable at x0, the linear function ` is called differential of f at x0,and is denoted by

df(x0) .

Following the tradition for linear functions, taking h ∈ RN we will often writedf(x0)h instead of df(x0)(h).

We will now review the main results needed in this book concerning the differ-ential calculus.

A.1 The differential of a scalar-valued function

Assume first, for simplicity, that M = 1. We start by fixing a direction i.e., avector v ∈ RN with ‖v‖ = 1, also called a versor. Whenever it exists, we calldirectional derivative of f at x0 in the direction v the limit

limt→0

f(x0 + tv)− f(x0)

t,

155

156 APPENDIX A. DIFFERENTIAL CALCULUS IN RN

which will be denoted by∂f

∂v(x0) .

If v coincides with an element ek of the canonical basis (e1,e2, . . . ,eN ) of RN, thedirectional derivative is called k-th partial derivative of f at x0 and is denotedby

∂f

∂xk(x0) .

If x0 = (x01, x

02, . . . , x

0N ), then

∂f

∂xk(x0) = lim

t→0

f(x0 + tek)− f(x0)

t

= limt→0

f(x01, x

02, . . . , x

0k + t, . . . , x0

N )− f(x01, x

02, . . . , x

0k, . . . , x

0N )

t,

so that it is frequent to call it “partial derivative with respect to the k-th variable”.

Theorem A.2 If f is differentiable at x0, then f is continuous at x0. Moreover,all the directional derivatives of f at x0 exist: for every direction v ∈ RN one has

∂f

∂v(x0) = df(x0)v .

Proof We know that the function ` = df(x0), being linear, is continuous, and`(0) = 0. Then,

limx→x0

f(x) = limx→x0

[f(x0) + `(x− x0) + r(x)]

= f(x0) + `(0) + limx→x0

r(x)

= f(x0) + limx→x0

r(x)

‖x− x0‖lim

x→x0

‖x− x0‖

= f(x0) ,

showing that f is continuous at x0. Concerning the directional derivatives, we have

limt→0

f(x0 + tv)− f(x0)

t= lim

t→0

df(x0)(tv) + r(x0 + tv)

t

= limt→0

t df(x0)v + r(x0 + tv)

t

= df(x0)v + limt→0

r(x0 + tv)

t.

A.1. THE DIFFERENTIAL OF A SCALAR-VALUED FUNCTION 157

On the other hand, being ‖v‖ = 1, it is

limt→0

∣∣∣∣r(x0 + tv)

t

∣∣∣∣ = limx→x0

|r(x)|‖x− x0‖

= 0 ,


In particular, if v coincides with an element ek of the canonical basis (e1,e2,. . . ,eN ) , then

∂f

∂xk(x0) = df(x0)ek .

Writing the vector h ∈ RN as h = h1e1 + h2e2 + . . .+ hNeN , by linearity we have

df(x0)h = h1df(x0)e1 + h2df(x0)e2 + . . .+ hNdf(x0)eN

= h1∂f

∂x1(x0) + h2

∂f

∂x2(x0) + . . .+ hN

∂f

∂xN(x0) ,

i.e.,

df(x0)h =N∑k=1

∂f

∂xk(x0)hk .

Theorem A.3 If f has partial derivatives defined in a neighborhood of x0, and theyare continuous at x0, then f is differentiable at x0.

Proof In order to simplify the notations, we will assume that N = 2. We define thefunction ` : R2 → R associating to every vector h = (h1, h2) the real number

`(h) =∂f

∂x1(x0)h1 +

∂f

∂x2(x0)h2 .

We will prove that ` is indeed the differential of f at x0. First of all, it is readilyverified that it is linear. Moreover, writing x0 = (x0

1, x02) and x = (x1, x2), by the

Lagrange Mean Value Theorem one has

f(x)− f(x0) = (f(x1, x2)− f(x01, x2)) + (f(x0

1, x2)− f(x01, x

02))

=∂f

∂x1(ξ1, x2)(x1 − x0

1) +∂f

∂x2(x0

1, ξ2)(x2 − x02) ,

for some ξ1 ∈ ]x01, x1[ and ξ2 ∈ ]x0

2, x2[ . Hence,

r(x) = f(x)− f(x0)− `(x− x0)

=

[∂f

∂x1(ξ1, x2)− ∂f

∂x1(x0

1, x02)

](x1 − x0

1) +

[∂f

∂x2(x0

1, ξ2)− ∂f

∂x2(x0

1, x02)

](x2 − x0

2) .


Then, being |x1 − x01| ≤ ‖x− x0‖ and |x2 − x0

2| ≤ ‖x− x0‖,

|r(x)|‖x− x0‖

≤∣∣∣∣ ∂f∂x1

(ξ1, x2)− ∂f

∂x1(x0

1, x02)

∣∣∣∣+

∣∣∣∣ ∂f∂x2(x0

1, ξ2)− ∂f

∂x2(x0

1, x02)

∣∣∣∣ .Letting x tend to x0, we have that (ξ1, x2) → (x0

1, x02) and (x0

1, ξ2) → (x01, x

02) so

that, being ∂f∂x1

and ∂f∂x2

continuous at x0 = (x01, x

02), it has to be

limx→x0

|r(x)|‖x− x0‖

= 0 ,


We say that f : Ω→ R is of class C1, or a C1-function, if f has partial derivativeswhich are continuous on the whole domain Ω. From the previous theorem we havethat a function of class C1 is differentiable, i.e., differentiable at every point of Ω.

A.2 Twice differentiable scalar-valued functions

Let Ω be an open subset of RN , and f : Ω → R be a differentiable function. Wewant to extend the notion of “second derivative”, which is well-known in the caseN = 1. For simplicity, let us deal with the case N = 2. If the partial derivatives∂f∂x1

, ∂f∂x2

: Ω → R have themselves the partial derivatives at a point x0, these aresaid to be “second order partial derivatives” of f at x0 and are denoted by

∂2f

∂x21

(x0) =∂

∂x1

∂f

∂x1(x0) ,

∂2f

∂x2∂x1(x0) =

∂

∂x2

∂f

∂x1(x0) ,

∂2f

∂x1∂x2(x0) =

∂

∂x1

∂f

∂x2(x0) ,

∂2f

∂x22

(x0) =∂

∂x2

∂f

∂x2(x0) .

We now prove the Schwarz Theorem.

Theorem A.4 If the second order partial derivatives ∂2f∂x2∂x1

, ∂2f∂x1∂x2

exist in a neigh-borhood of x0 and they are continuous at x0, then

∂2f

∂x2∂x1(x0) =

∂2f

∂x1∂x2(x0) .

Proof Let ρ > 0 be such that B(x0, ρ) ⊆ Ω.1 We write x0 = (x01, x

02) and we take

an x = (x1, x2) ∈ B(x0, ρ) such that x1 6= x01 and x2 6= x0

2. It is then possible todefine

g(x1, x2) =f(x1, x2)− f(x1, x

02)

x2 − x02

, h(x1, x2) =f(x1, x2)− f(x0

1, x2)

x1 − x01

.

1Let us recall the notation B(x0, ρ) for the open ball centered at x0 with radius ρ > 0, andB(x0, ρ) for the closed ball.

A.2. TWICE DIFFERENTIABLE SCALAR-VALUED FUNCTIONS 159

One can verify that

g(x1, x2)− g(x01, x2)

x1 − x01

=h(x1, x2)− h(x1, x

02)

x2 − x02

.

By the the Lagrange Mean Value Theorem, there is a ξ1 ∈ ]x01, x1[ such that

g(x1, x2)− g(x01, x2)

x1 − x01

=∂g

∂x1(ξ1, x2) =

∂f∂x1

(ξ1, x2)− ∂f∂x1

(ξ1, x02)

x2 − x02

,

and there is a ξ2 ∈ ]x02, x2[ such that

h(x1, x2)− h(x1, x02)

x2 − x02

=∂h

∂x2(x1, ξ2) =

∂f∂x2

(x1, ξ2)− ∂f∂x2

(x01, ξ2)

x1 − x01

.

Again by the Lagrange Mean Value Theorem, there is a η2 ∈ ]x02, x2[ such that

∂f∂x1

(ξ1, x2)− ∂f∂x1

(ξ1, x02)

x2 − x02

=∂2f

∂x2∂x1(ξ1, η2) ,

and there is a η1 ∈ ]x01, x1[ such that

∂f∂x2

(x1, ξ2)− ∂f∂x2

(x01, ξ2)

x1 − x01

=∂2f

∂x1∂x2(η1, ξ2) .

Hence,∂2f

∂x2∂x1(ξ1, η2) =

∂2f

∂x1∂x2(η1, ξ2) .

Taking the limit, as x = (x1, x2) tends to x0 = (x01, x

02), we have that both (ξ1, η2)

and (η1, ξ2) converge to x0, and the continuity of the second order partial derivativesleads to the conclusion.

We say that f : Ω → R is of class C2, or a C2-function if all its second orderpartial derivatives exist and are continuous on Ω.

It could be useful to consider the Hessian matrix of f at x0:

Hf(x0) =

∂2f∂x21

(x0) ∂2f∂x2∂x1

(x0)

∂2f∂x1∂x2

(x0) ∂2f∂x22

(x0)

;

if f is of class C2, this is a symmetric matrix.

What has been said above extends without difficulties for any N ≥ 2. If f is ofclass C2, the Hessian matrix is an N ×N symmetric matrix.

One can further define by induction the n−th order partial derivatives. It is saidthat f : Ω → R is of class Cn if all its n−th order partial derivatives exist and arecontinuous on Ω.


A.3 The differential of a vector-valued function

When M ≥ 2, let f1, f2, . . . , fM be the components of the function f : Ω→ RM , sothat

f(x) = (f1(x), f2(x), . . . , fM (x)) .

Theorem A.5 The function f is differentiable at x0 if and only if such are all itscomponents. In this case, for any vector h ∈ RN it is

df(x0)h = (df1(x0)h, df2(x0)h, . . . , dfM (x0)h) .

Proof Considering the components in the equation

f(x) = f(x0) + `(x− x0) + r(x) ,

we can writefj(x) = fj(x0) + `j(x− x0) + rj(x) ,

with j = 1, 2, . . . ,M , and we know that

limx→x0

r(x)

‖x− x0‖= 0 ⇐⇒ lim

x→x0

rj(x)

‖x− x0‖= 0 for every j = 1, 2, . . . ,M ,


It is useful to consider the matrix associated to the linear function ` = df(x0), givenby

`1(e1) `1(e2) . . . `1(eN )`2(e1) `2(e2) . . . `2(eN )

......

...`M (e1) `M (e2) . . . `M (eN )

,

where e1,e2, . . .eN are the vectors of the canonical basis of RN . This matrix iscalled Jacobian matrix associated to the function f at x0, and is denoted byJf(x0). Recalling that

∂fj∂xk

(x0) = dfj(x0)ek ,

with j = 1, 2, . . . ,M and k = 1, 2, . . . , N , we obtain the matrix

Jf(x0) =

∂f1∂x1

(x0) ∂f1∂x2

(x0) · · · ∂f1∂xN(x0)

∂f2∂x1

(x0) ∂f2∂x2

(x0) · · · ∂f2∂xN(x0)

......

...∂fM∂x1

(x0) ∂fM∂x2 (x0) · · · ∂fM∂xN(x0)

.

The function f : Ω → RM is said to be of class C1, or C2, if all its components aresuch. This definition naturally extends to functions of class Cn.

A.4. SOME COMPUTATIONAL RULES 161

A.4 Some computational rules

Let us start with some easy propositions.

1. If f : Ω→ Y is constant, then df(x0) = 0, for every x0 ∈ Ω.

2. If A : Ω→ Y is linear and continuous, then dA(x0) = A, for every x0 ∈ Ω.

3. If RN = RN1 × RN2 and B : Ω → Y is bilinear and continuous, writing x0 =(x0

1, x02) and h = (h1, h2), with x0

1, h1 ∈ RN1 and x02, h2 ∈ RN2 , one has

dB(x0)(h) = B(x01, h2) + B(h1, x

02) .

All this can be generalized to n-linear continuous functions.

We now recall the usual laws of calculus.

Theorem A.6 If f, g : Ω → Y are differentiable at x0 and α, β are two real num-bers, then

d(αf + βg)(x0) = αdf(x0) + βdg(x0) .

Proof Writing

f(x) = f(x0) + df(x0)(x− x0) + r1(x) , g(x) = g(x0) + dg(x0)(x− x0) + r2(x) ,

we have that

(αf + βg)(x) = (αf + βg)(x0) + (αdf(x0) + βdg(x0))(x− x0) + r(x) ,

with r(x) = αr1(x) + βr2(x), and

limx→x0

r(x)

‖x− x0‖= α lim

x→x0

r1(x)

‖x− x0‖+ β lim

x→x0

r2(x)

‖x− x0‖= 0 .

Hence, αf + βg is differentiable at x0 with differential αdf(x0) + βdg(x0).

We now study the differentiability of a composite function.

Theorem A.7 If f : Ω→ RM is differentiable at x0, while Ω′ ⊆ RM is an open setcontaining f(Ω) and g : Ω′ → RL is differentiable at f(x0), then gf is differentiableat x0, and

d(g f)(x0) = dg(f(x0)) df(x0) .


Proof Setting y0 = f(x0), we have

f(x) = f(x0) + df(x0)(x− x0) + r1(x) , g(y) = g(y0) + dg(y0)(y− y0) + r2(y) ,

with

limx→x0

r1(x)

‖x− x0‖= 0 , lim

y→y0

r2(y)

‖y− y0‖= 0 .

Let us introduce the function R2 : Ω′ → RL, defined as

R2(y) =

r2(y)

‖y− y0‖se y 6= y0 ,

0 se y = y0 .

Notice that R2 is continuous at y0. Then,

g(f(x)) = g(f(x0)) + dg(f(x0))[f(x)− f(x0)] + r2(f(x))

= g(f(x0)) + dg(f(x0))[df(x0)(x− x0) + r1(x)] + r2(f(x))

= g(f(x0)) + [dg(f(x0)) df(x0)](x− x0) + r3(x) ,

where

r3(x) = dg(f(x0))(r1(x)) + r2(f(x))

= dg(f(x0))(r1(x)) + ‖f(x)− f(x0)‖R2(f(x))

= dg(f(x0))(r1(x)) + ‖df(x0)(x− x0) + r1(x)‖R2(f(x)) .

Hence,

‖r3(x)‖‖x− x0‖

≤∥∥∥∥dg(f(x0))

(r1(x)

‖x− x0‖

)∥∥∥∥+

+

(∥∥∥∥df(x0)

(x− x0

‖x− x0‖

)∥∥∥∥+‖r1(x)‖‖x− x0‖

)‖R2(f(x))‖ .

If x → x0, the first summand tends to 0, since dg(f(x0)) is continuous; f is con-tinuous at x0 and R2 is continuous at y0 = f(x0), with R2(y0)) = 0, so that‖R2(f(x))‖ tends to 0; df(x0), being continuous, is bounded on the compact setB(0, 1). Therefore,

limx→x0

‖r3(x)‖‖x− x0‖

= 0 .

We conclude that g f is differentiable at x0, with differential dg(f(x0)) df(x0).

A.5. THE IMPLICIT FUNCTION THEOREM 163

It is well-known that the matrix associated to the composite of two linear func-tions is the product of the two respective matrices. From the above theorem we thenhave the following formula for the Jacobian matrices:

J(g f)(x0) = Jg(f(x0)) · Jf(x0) ,

i.e., the matrix ∂(gf)1∂x1

(x0) · · · ∂(gf)1∂xN

(x0)... · · ·

...∂(gf)L∂x1

(x0) · · · ∂(gf)L∂xN

(x0)

is equal to the product

∂g1∂y1

(f(x0)) · · · ∂g1∂yM(f(x0))

... · · ·...

∂gL∂y1

(f(x0)) · · · ∂gL∂yM(f(x0))

∂f1∂x1

(x0) · · · ∂f1∂xN(x0)

... · · ·...

∂fM∂x1

(x0) · · · ∂fM∂xN(x0)

.

We can thus derive the formula for the partial derivatives, usually called chain rule:

∂(g f)i∂xk

(x0) =

=∂gi∂y1

(f(x0))∂f1

∂xk(x0) +

∂gi∂y2

(f(x0))∂f2

∂xk(x0) + . . .+

∂gi∂yM

(f(x0))∂fM∂xk

(x0)

=M∑j=1

∂gi∂yj

(f(x0))∂fj∂xk

(x0) ,

where i = 1, 2, . . . , L and k = 1, 2, . . . , N .

A.5 The implicit function theorem

Let Ω be an open subset of RM × RN , and g : Ω→ RN be a C1-function. Hence, ghas N components

g(x,y) = (g1(x,y), . . . , gN (x,y)) .

Here x = (x1, . . . , xM ) ∈ RM , and y = (y1, . . . , yN ) ∈ RN . We will use the followingnotation for the Jacobian matrices:

∂g

∂x(x,y)=

∂g1∂x1

(x,y) · · · ∂g1∂xM(x,y)

... · · ·...

∂gN∂x1

(x,y) · · · ∂gN∂xM(x,y)

,∂g

∂y(x,y)=

∂g1∂y1

(x,y) · · · ∂g1∂yN(x,y)

... · · ·...

∂gN∂y1

(x,y) · · · ∂gN∂yN(x,y)

.


Theorem A.8 Let Ω ⊆ RM ×RN be open, g : Ω→ RN a C1-function, and (x0,y0)a point in Ω for which

g(x0,y0) = 0 , and det∂g

∂y(x0,y0) 6= 0 .

Then, there exist an open neighborhood U of x0, an open neighborhood V of y0, anda C1-function η : U → V such that U × V ⊆ Ω and, taking x ∈ U and y ∈ V, onehas that

g(x,y) = 0 ⇐⇒ y = η(x) .

Moreover, the function η is of class C1, and the following formula holds true:

Jη(x) = −(∂g

∂y(x, η(x))

)−1 ∂g

∂x(x, η(x)) .

Proof 2 It will be carried out by induction. We first prove the case N = 1.

Assume for instance that ∂g∂y (x0, y0) > 0. By the continuity of ∂g

∂y , there is a

δ > 0 such that, if ‖x − x0‖ ≤ δ and |y − y0| ≤ δ, then ∂g∂y (x, y) > 0. Hence, for

every x ∈ B(x0, δ), the function g(x, ·) is strictly increasing on [y0−δ, y0 +δ]. Beingg(x0, y0) = 0, we have that

g(x0, y0 − δ) < 0 < g(x0, y0 + δ) .

By continuity again, there is a δ′ > 0 such that, if x ∈ B(x0, δ′), then

g(x, y0 − δ) < 0 < g(x, y0 + δ) .

We define U = B(x0, δ′), and V = ]y0−δ, y0+δ[ . Hence, for every x ∈ U, since g(x, ·)

is strictly increasing, there is exactly one y ∈ ]y0 − δ, y0 + δ[ for which g(x, y) = 0;we call η(x) such a y. We have thus defined a function η : U → V such that, takingx ∈ U and y ∈ V,

g(x, y) = 0 ⇐⇒ y = η(x) .

In order to verify the continuity of η, let us fix a x ∈ U and prove that η is continuousat x. Taken x ∈ U and considered the function γ : [0, 1]→ U × V , defined as

γ(t) = (x + t(x− x), η(x) + t(η(x)− η(x))),

2The proof reported here is most probably due to Giuseppe Peano, and can be found in the Italianbook “Calcolo differenziale e principii di calcolo integrale”, published in 1884. This important workwas written by Peano himself, who at that time was only 25 years old, but the official author isAngelo Genocchi, the professor who was in care of Peano as an assistant at the University of Torino.Genocchi got indeed very angry when he was told that the book had been published, and publiclydeclared that he was not aware of what had been written in the volume, recalling however the factthat Peano had followed his lessons in infinitesimal analysis.


the Lagrange Mean Value Theorem applied to g γ tells us that there is a ξ ∈ ]0, 1[for which

g(x, η(x))− g(x, η(x)) =∂g

∂x(γ(ξ))(x− x) +

∂g

∂y(γ(ξ))(η(x)− η(x)) .

Being g(x, η(x)) = g(x, η(x)) = 0, we have that

|η(x)− η(x)| = 1

|∂g∂y (γ(ξ))|

∥∥∥∥ ∂g∂x(γ(ξ))(x− x)

∥∥∥∥ .Since the partial derivatives of g are continuous and ∂g

∂y in not zero on the compact

set U × V , we have that there is a constant c > 0 for which

1

|∂g∂y (γ(ξ))|

∥∥∥∥ ∂g∂x(γ(ξ))(x− x)

∥∥∥∥ ≤ c‖x− x‖ .

As a consequence, η is continuous at x.

We now prove the differentiability. Taken x = (x1, x2, . . . , xM ), let x = (x1 +h, x2, . . . , xM ); proceeding as above, for h small enough we have

η(x1 + h, x2, . . . , xM )− η(x1, x2, . . . , xM )

h= −

∂g∂x1

(γ(ξh))∂g∂y (γ(ξh))

,

with γ(ξh) belonging to the segment joining (x, η(x)) to (x, η(x)). If h tends to 0,we have that γ(ξh) tends to (x, η(x)), and hence

∂η

∂x1(x) = lim

h→0

η(x1 + h, x2, . . . , xM )− η(x1, x2, . . . , xM )

h= −

∂g∂x1

(x, η(x))∂g∂y (x, η(x))

.

The partial derivatives with respect to x2, . . . , xM are computed similarly, thus yield-ing that η is of class C1, and

Jη(x) = − 1∂g∂y (x, η(x))

∂g

∂x(x, η(x)) .

We now assume that the statement holds till N − 1, for some N ≥ 2 (and anyM ≥ 1), and prove that it then also holds for N. We will use the notation

y1 = (y1, . . . , yN−1) ,


and we will write y = (y1, yN ). Since

det

∂g1∂y1

(x0,y0) · · · ∂g1∂yN(x0,y0)

... · · ·...

∂gN∂y1

(x0,y0) · · · ∂gN∂yN(x0,y0)

6= 0 ,

at least one of the elements in the last column is different from zero. We can assumewithout loss of generality, possibly changing the rows, that ∂gN

∂yN(x0,y0) 6= 0. Writing

y0 = (y01, y

0N ), with y0

1 = (y01, . . . , y

0N−1), we then have

gN (x0,y01, y

0N ) = 0 , and

∂gN∂yN

(x0,y01, y

0N ) 6= 0 .

Then, by the already proved one-dimensional case, there are an open neighborhoodU1 of (x0,y0

1), an open neighborhood VN of y0N , and a C1-function η1 : U1 → VN

such that U1× VN ⊆ Ω, with the following properties: if (x,y1) ∈ U1 and yN ∈ VN ,

gN (x,y1, yN ) = 0 ⇐⇒ yN = η1(x,y1) ,

and

Jη1(x,y1) = − 1∂gN∂yN

(x,y1, η1(x,y1)))

∂gN∂(x,y1)

(x,y1, η1(x,y1)) .

We may assume U1 to be of the type U × V1, with U being an open neighborhoodof x0 and V1 an open neighborhood of y0

1.

Let us define the function φ : U × V1 → RN−1 by setting

φ(x,y1) = (g1(x,y1, η1(x,y1)), . . . , gN−1(x,y1, η1(x,y1))) .

For briefness, we will write

g(1,...,N−1)(x,y) = (g1(x,y), . . . , gN−1(x,y)) ,

so thatφ(x,y1) = g(1,...,N−1)(x,y1, η1(x,y1)) .

Notice that, being η1(x0,y01) = y0

N , we have that

φ(x0,y01) = g(1,...,N−1)(x0,y0) = 0 ,

and

∂φ

∂y1

(x0,y01) =

∂g(1,...,N−1)

∂y1

(x0,y0) +∂g(1,...,N−1)

∂yN(x0,y0)

∂η1

∂y1

(x0,y01) . (∗)


Moreover, since gN (x,y1, η1(x,y1)) = 0, for every (x,y1) ∈ U1, differentiating withrespect to y1 we see that

0 =∂gN∂y1

(x0,y0) +∂gN∂yN

(x0,y0)∂η1

∂y1

(x0,y01) . (∗∗)

Let us write the identity

det∂φ

∂y1

(x0,y01) =

1∂gN∂yN

(x0,y0)det

∂φ

∂y1

(x0,y01)

∂g(1,...,N−1)

∂yN(x0,y0)

0∂gN∂yN

(x0,y0)

,

Substituting the two equalities (∗), (∗∗), we have that

det

∂φ

∂y1

(x0,y01)

∂g(1,...,N−1)

∂yN(x0,y0)

0∂gN∂yN

(x0,y0)

=

=det

∂g(1,...,N−1)

∂y1

(x0,y0)+∂g(1,...,N−1)

∂yN(x0,y0)

∂η1

∂y1

(x0,y01)∂g(1,...,N−1)

∂yN(x0,y0)

∂gN∂y1

(x0,y0) +∂gN∂yN

(x0,y0)∂η1

∂y1

(x0,y01)

∂gN∂yN

(x0,y0)

= det

(∂g

∂y1

(x0,y0) +∂g

∂yN(x0,y0)

∂η1

∂y1

(x0,y01)

∂g

∂yN(x0,y0)

)

= det

[∂g

∂y(x0,y0) +

(∂g

∂yN(x0,y0)

∂η1

∂y1

(x0,y01)

∂g

∂yN(x0,y0)

)].

We now recall that, adding a scalar multiple of one column to another column of amatrix does not change the value of its determinant. Hence, being(

∂g

∂yN(x0,y0)

∂η1

∂y1

(x0,y01)

∂g

∂yN(x0,y0)

)=

=

∂g1

∂yN(x0,y0)

∂η1

∂y1(x0,y

01) · · · ∂g1

∂yN(x0,y0)

∂η1

∂yN−1(x0,y

01)

∂g1

∂yN(x0,y0)

... · · ·...

...

∂gN∂yN

(x0,y0)∂η1

∂y1(x0,y

01) · · · ∂gN

∂yN(x0,y0)

∂η1

∂yN−1(x0,y

01)

∂g1

∂yN(x0,y0)

,


it has to be

det

[∂g

∂y(x0,y0) +

(∂g

∂yN(x0,y0)

∂η1

∂y1

(x0,y01)

∂g

∂yN(x0,y0)

)]= det

∂g

∂y(x0,y0) .

So, finally we have

φ(x0,y01) = 0 , and det

∂φ

∂y1

(x0,y01) 6= 0 .

By the inductive assumption, there are an open neighborhood U of x0, an openneighborhood V1 of y0

1 and a C1-function η2 : U → V1 such that U × V1 ⊆ U × V1,and the following holds: for every x ∈ U and y1 ∈ V1,

φ(x,y1) = 0 ⇐⇒ y1 = η2(x) .

In conclusion, for x ∈ U and y = (y1, yN ) ∈ V1 × V2, we have that

g(x,y) = 0 ⇐⇒

g(1,...,N−1)(x,y1, yN ) = 0

gN (x,y1, yN ) = 0

⇐⇒

g(1,...,N−1)(x,y1, yN ) = 0

yN = η1(x,y1)

⇐⇒

φ(x,y1) = 0

yN = η1(x,y1)

⇐⇒

y1 = η2(x)

yN = η1(x,y1)

⇐⇒ y = (η2(x), η1(x, η2(x))) .

Setting V = V1 × V2, we may then define the function η : U → V as

η(x) = (η2(x), η1(x, η2(x))) .

This function is of class C1, since both η1 and η2 are such. Since g(x, η(x)) = 0 forevery x ∈ U, one easily deduces that

∂g

∂x(x, η(x)) +

∂g

∂y(x, η(x))Jη(x) = 0 ,

whence the formula for Jη(x).

Clearly, the following analogous statement holds true, where the roles of x andy are interchanged.

A.6. LOCAL DIFFEOMORPHISMS 169

Theorem A.9 Let Ω ⊆ RM ×RN be open, g : Ω→ RN a C1-function, and (x0,y0)a point in Ω for which


∂x(x0,y0) 6= 0 .

Then, there exist an open neighborhood U of x0, an open neighborhood V of y0, anda C1-function η : V → U such that U × V ⊆ Ω and, taking x ∈ U and y ∈ V, onehas that

g(x,y) = 0 ⇐⇒ x = η(y) .

Moreover, the function η is of class C1, and the following formula holds true:

Jη(y) = −(∂g

∂x(y, η(y))

)−1 ∂g

∂y(y, η(y)) .

A.6 Local diffeomorphisms

Let us introduce the notion of diffeomorphism.

Definition A.10 Given A and B, two open subsets of RN , a function ϕ : A → Bis said to be a diffeomorphism if it is of class C1, it is a bijection, and its inverseϕ−1 : B → A is also of class C1.

Let us state the important Local Diffeomorphism Theorem.

Theorem A.11 Let A and B be open subsets of RN , and ϕ : A → B be a C1-function. If, for some x0 ∈ A, one has that det Jϕ(x0) 6= 0, then there exist anopen neighborhood U of x0 contained in A, and an open neighborhood V of ϕ(x0)contained in B, such that the restricted function ϕ|U : U → V is a diffeomorfism.

Proof We consider the function g : A×B → RN defined as

g(x,y) = y− ϕ(x) .

Setting y0 = ϕ(x0), we have that


∂x(x0,y0) = det Jϕ(x0) 6= 0 .

By the Implicit Function Theorem, there exist an open neighborhood V of y0, anopen neighborhood U of x0 and a C1-function η : V → U such that, taken y ∈ Vand x ∈ U,

ϕ(x) = y ⇐⇒ g(x,y) = 0 ⇐⇒ x = η(y) .

Hence, η = ϕ−1|U , and the proof is thus completed.


Appendix B

Stokes–Cartan and Poincaretheorems

In this appendix we will give a complete proof of the Stokes–Cartan and Poincaretheorems, of which only particular cases have been proved in Chapter 3.

Let U be an open set in RN, V an open set in1 RP and φ : V → U a function ofclass C1 :

φ(y) = (φ1(y), . . . , φN (y)) ,

with y = (y1, . . . , yP ) ∈ V. Given a M -differential form ω : U → ΩM (RN ),

ω(x) =∑


we can define a M -differential form Tφω : V → ΩM (RP ), which we will call thetransformation through φ of ω, in the following way2:

Tφω(y) =∑

1≤i1<...<iM≤Nfi1,...,iM (φ(y)) dφi1(y) ∧ . . . ∧ dφiM (y) .

Notice that

dφi1(y) ∧ . . .∧ dφiM (y) =

=

P∑j=1

∂φi1∂yj

(y)dyj

∧ . . . ∧ P∑j=1

∂φiM∂yj

(y)dyj

=

P∑j1,...,jM=1

∂φi1∂yj1

(y) · · · ∂φiM∂yjM

(y) dyj1 ∧ . . . ∧ dyjM

1Whenever the sets would not be open, see the footnote in section 3.6.2This is usually called pull-back and denoted by φ ∗ ω.

171

172 APPENDIX B. STOKES–CARTAN AND POINCARE THEOREMS

(being aware that here the indices j1, . . . , jM are not in an increasing order). It isreadily verified that, taken a c ∈ R, it is

Tφ(cω) = c Tφω ;

if ω is a M−differential form defined on U, then

Tφ(ω ∧ ω) = Tφω ∧ Tφω ,

and if moreover M = M, then

Tφ(ω + ω) = Tφω + Tφω .

Let us prove now the following properties.

Proposition 1. If ψ : W → V and φ : V → U, then

Tψ(Tφω) = Tφψω .

Proof By the linearity properties seen above, it will be sufficient to consider thecase of a differential form of the type

ω(x) = fi1,...,iM (x) dxi1 ∧ . . . ∧ dxiM .

Then,

Tψ(Tφω) =

(fi1,...,iM φ)

P∑j1,...,jM=1

∂φi1∂yj1

· · · ∂φiM∂yjM

ψ dψj1 ∧ . . . ∧ dψjM .

On the other hand,

Tφψω = (fi1,...,iM φ ψ) d(φ ψ)i1 ∧ . . . ∧ d(φ ψ)iM ,

and being

d(φ ψ)ik = d(φik ψ) =P∑j=1

(∂φik∂yj

ψ)dψj ,

equality then holds.

173

Proposition 2. Assume that φ be of class C2. If ω is of class C1, then also Tφω issuch, and

d(Tφω) = Tφ(dω) .

Proof Here, too, it is sufficient to consider the case ω = fi1,...,iM dxi1 ∧ . . . ∧ dxiM .We have

d(Tφω) = d(fi1,...,iM φ) ∧ dφi1 ∧ . . . ∧ dφiM + (fi1,...,iM φ) d(dφi1 ∧ . . . ∧ dφiM )

= d(fi1,...,iM φ) ∧ dφi1 ∧ . . . ∧ dφiM

=

[N∑m=1

(∂fi1,...,iM∂xm

φ)dφm

]∧ dφi1 ∧ . . . ∧ dφiM .

On the other hand,

dω(x) =N∑m=1

∂fi1,...,iM∂xm

(x) dxm ∧ dxi1 ∧ . . . ∧ dxiM ,

hence

Tφ(dω) =

N∑m=1

(∂fi1,...,iM∂xm

φ)dφm ∧ dφi1 ∧ . . . ∧ dφiM ,

and the formula is thus proved.

Proposition 3. If σ : I → RN is a M - surface whose support is contained in U,then ∫

σω =

∫ITσω .

Proof As above, we just consider the case ω = fi1,...,iMdxi1 ∧ . . . ∧ dxiM . We have

∫ITσω=

∫Ifi1,...,iM (σ(u))

M∑j1,...,jM=1

∂σi1∂uj1

(u) . . .∂σiM∂ujM

(u) duj1 ∧ . . . ∧ dujM

=

∫Ifi1,...,iM (σ(u)) detσ′(i1,...,iM )(u) du

=

∫σω .



We are now ready to give the proof of the Stokes–Cartan theorem, whosestatement, we recall, is the following.

Theorem B.1 Let 0 ≤ M ≤ N − 1. If ω : U → ΩM (RN ) is a M -differential formof class C1 and σ : I → RN is a (M + 1)-surface whose support is contained in U,then ∫

σdω =

∫∂σω .

Proof The case M = 0 follows from the Fundamental Theorem applied to thefunction ω σ : [a, b]→ R. Assume now 1 ≤M ≤ N − 1. Being∫

σα+k

ω =

∫Ik

Tσα+kω =

∫Ik

Tα+k

(Tσω) =

∫α+k

Tσω ,

and analogously for β+k we have∫

∂σω=

M+1∑k=1

(−1)k∫σα+

k

ω +M+1∑k=1

(−1)k−1

∫σβ+

k

ω

=

M+1∑k=1

(−1)k∫α+k

Tσω +

M+1∑k=1

(−1)k−1

∫β+k

Tσω

=

∫∂ITσω .

If σ is of class C2, then Tσω is of class C1 and, applying the Gauss formula to Tσω,we have ∫

∂ITσω =

∫Id(Tσω) .

But ∫Id(Tσω) =

∫ITσ(dω) =

∫σdω .

Hence, we have see that∫σdω =

∫Id(Tσω) =

∫∂ITσω =

∫∂σω ,

and the theorem is proved in this case.

The assumption that σ : I → RN be of class C2 can be eliminated by an ap-proximation procedure: it is possible to construct a sequence (σn)n of M -surfacesof class C2 which converge to σ together with all first order partial derivatives. TheStokes–Cartan formula then holds for such surfaces, and the dominated convergencetheorem permits to pass to the limit and conclude.

175

Consider now the set [0, 1] × U, whose elements will be denoted by (t,x) =(t, x1, . . . , xN ). Let us define the linear operator K which transforms a generic M -differential form α : [0, 1] × U → ΩM (RN+1) in a (M − 1)-differential form K(α) :U → ΩM−1(RN ) in the following way:

a) if α(t,x) = f(t,x) dt ∧ dxi1 ∧ . . . ∧ dxiM−1 (notice that here the term dtappears), then

K(α)(x) =

(∫ 1

0f(t,x) dt

)dxi1 ∧ . . . ∧ dxiM−1 ;

b) if α(t,x) = f(t,x) dxi1 ∧ . . .∧ dxiM (here the term dt does not appear), thenK(α) = 0;

c) in all the other cases, K is defined by linearity (for each component of ageneric M -differential form α, the term dt may appear or not, and the previous twodefinitions apply).

We moreover define the functions ψ, ξ : U → [0, 1]× U as follows:

ψ(x1, . . . , xN ) = (0, x1, . . . , xN ) , ξ(x1, . . . , xN ) = (1, x1, . . . , xN ) .

Lemma B.2 If α : [0, 1] × U → ΩM (RN+1) is a M -differential form of class C1,then

d(K(α)) +K(dα) = Tξα− Tψα .

Proof Because of the linearity, it will be sufficient to consider the two cases whenthe differential form α is of one of the two kinds considered in a) and b).

a) If α(t,x) = f(t,x) dt ∧ dxi1 ∧ . . . ∧ dxiM−1 , by the Leibniz rule we have

d(K(α))(x) =N∑m=1

(∫ 1

0

∂f

∂xm(t,x) dt

)dxm ∧ dxi1 ∧ . . . ∧ dxiM−1 ;

on the other hand,

dα(t,x) =∂f

∂t(t,x) dt ∧ dt ∧ dxi1 ∧ . . . ∧ dxiM−1 +

+N∑m=1

∂f

∂xm(t,x) dxm ∧ dt ∧ dxi1 ∧ . . . ∧ dxiM−1

=−N∑m=1

∂f

∂xm(t,x) dt ∧ dxm ∧ dxi1 ∧ . . . ∧ dxiM−1 ,


and hence

K(dα)(x) =−N∑m=1

(∫ 1

0

∂f

∂xm(t,x) dt

)dxm ∧ dxi1 ∧ . . . ∧ dxiM−1

=−d(K(α))(x) .

Moreover, since the first component of ψ and of ξ is constant, it is Tψα = Tξα = 0.Hence, the identity is proved in this case.

b) If α(t,x) = f(t,x) dxi1 ∧ . . . ∧ dxiM , it is K(α) = 0 and hence d(K(α)) = 0;on the other hand,

dα(t,x) =∂f

∂t(t,x) dt ∧ dxi1 ∧ . . . ∧ dxiM +

+N∑m=1

∂f

∂xm(t,x) dxm ∧ dxi1 ∧ . . . ∧ dxiM ,

and hence

K(dα)(x) =

(∫ 1

0

∂f

∂t(t,x) dt

)dxi1 ∧ . . . ∧ dxiM

= (f(1,x)− f(0,x)) dxi1 ∧ . . . ∧ dxiM .

Moreover,

Tξα(x) = f(1,x) dξi1(x) ∧ . . . ∧ dξiM (x)

= f(1,x) dxi1 ∧ . . . ∧ dxiM ,

Tψα(x) = f(0,x) dψi1(x) ∧ . . . ∧ dψiM (x)

= f(0,x) dxi1 ∧ . . . ∧ dxiM .

The formula is thus proved in this case, as well.

We can now give the proof of the Poincare theorem, whose statement is re-called below.

Theorem B.3 Let U be an open subset of RN , star-shaped with respect to a pointx. For 1 ≤ M ≤ N, a M -differential form ω : U → ΩM (RN ) of class C1 is exact ifand only if it is closed. In that case, if ω is of the type

ω(x) =∑


177

a (M − 1)-differential form ω such that dω = ω is given by

ω(x) =∑

1≤i1<...<iM≤N

M∑s=1

(−1)s+1(xis − xis) ·

·(∫ 1

0tM−1fi1,...,iM (x + t(x− x)) dt

)dxi1 ∧ . . . ∧ dxis ∧ . . . ∧ dxiM .

Proof To simplify the notations, we can assume without loss of generality thatx = (0, 0, . . . , 0); let φ : [0, 1]× U → U be defined by

φ(t, x1, . . . , xN ) = (tx1, . . . , txN ) .

Moreover, by the linearity, we may assume, for simplicity, that

ω(x) = fi1,...,iM (x) dxi1 ∧ . . . ∧ dxiM .

Consider Tφω, the transformation through φ of ω. It is the differential form of degreeM defined on [0, 1]× U as follows:

Tφω(t,x) = fi1,...,iM (tx)(xi1dt+ t dxi1) ∧ . . . ∧ (xiMdt+ t dxiM )

= fi1,...,iM (tx)[tMdxi1 ∧ . . . ∧ dxiM +

+tM−1M∑s=1

(−1)s−1xisdt ∧ dxi1 ∧ . . . ∧ dxis ∧ . . . ∧ dxiM ] .

Observe that

K(Tφω)(x) =

=

(∫ 1

0fi1,...,iM (tx)tM−1

M∑s=1

(−1)s+1xis dt

)dxi1 ∧ . . . ∧ dxis ∧ . . . ∧ dxiM ,

so that ω = K(Tφω). We want to prove that dω = ω. Being ω closed, we have

K(d(Tφω)) = K(Tφ(dω)) = K(Tφ(0)) = K(0) = 0 .

By the preceding lemma,

dω= d(K(Tφω))

= Tξ(Tφω)− Tψ(Tφω)−K(d(Tφω))

= Tξ(Tφω)− Tψ(Tφω)

= Tφξω − Tφψω .

Being φξ the identity function and φψ the null function, we have that Tφξω = ωand Tφψω = 0, which concludes the proof.


We have thus concluded the proof of the two main theorems of the theory, andmaybe something should be said about the need for such a theory. Of course, itsmathematical beauty would alone justify its existence and development. Neverthe-less, such a nice theory also finds a lot of applications in the physical world. Thereason of this probably lies in the fact that it was motivated by the classical theoremsin Electromagnetism and Fluidodynamics, hence the abstract construction lies onsome very concrete bases.

So we find an example here of how Physics and Mathematics interact to helpand motivate each other, leading to wonderful successful theories which may have alot of pratical implications in our lifes.

Appendix C

On differentiable manifolds

We would like to show here how the theory on differential forms developed in Chap-ter 3, and in particular the Stokes–Cartan theorem, can be adapted to the context ofdifferentiable manifolds. However, contrary to our habit, we will not give the com-plete proofs of all the results of this section; the interested reader will find useful torefer, e.g., to [20]. We consider a subset M of RN .

Definition C.1 The set M is a M-dimensional differentiable manifold, with1 ≤M ≤ N (or briefly a M -manifold) if, taken a point x in M, there are an openneighborhood A of x, an open neighborhood B of 0 in RN and a diffeomorphismϕ : A→ B such that ϕ(x) = 0 and, either

(a) ϕ(A ∩M) = y = (y1, . . . , yN ) ∈ B : yM+1 = . . . = yN = 0 ,or

(b) ϕ(A ∩M) = y = (y1, . . . , yN ) ∈ B : yM+1 = . . . = yN = 0 and yM ≥ 0 .

It can be seen that (a) and (b) cannot hold at the same time. The points xfor which (b) is verified make up the boundary of M, which we denote by ∂M. If∂M is empty, we are speaking of a M -manifold without boundary; otherwise,M issometimes said to be a M -manifold with boundary.

First of all we notice that the boundary of a differentiable manifold is itself adifferentiable manifold, with a lower dimension.

Theorem C.2 The set ∂M is a (M − 1)-manifold without boundary:

∂(∂M) = Ø .

Proof Taken a point x in ∂M, there are an open neighborhood A of x, an openneighborhood B of 0 in RN and a diffeomorphism ϕ : A → B such that ϕ(x) = 0and

ϕ(A ∩M) = y=(y1, . . . , yN )∈B : yM+1 = . . . = yN = 0 and yM ≥ 0.

179

180 APPENDIX C. ON DIFFERENTIABLE MANIFOLDS

Reasoning on the fact that the conditions (a) an (b) of the definition can not holdsimultaneously for any point of M, it is possible to prove that

ϕ(A ∩ ∂M) = y = (y1, . . . , yN ) ∈ B : yM = yM+1 = . . . = yN = 0 .


Let us now see that, given a M -manifoldM, correspondingly to each of its pointx it is possible to find a local M -parametrization.

Theorem C.3 For every x ∈M, there is a neighborhood A′ of x such that A′∩Mcan be M -parametrized with a function σ : I → RN, where I is a rectangle of RM ofthe type

I =

[−α, α]M if x 6∈ ∂M ,

[−α, α]M−1 × [0, α] if x ∈ ∂M ,

and σ(0) = x.

Proof Consider the diffeomorphism ϕ : A → B given by the above definition andtake an α > 0 such that the rectangle B′ = [−α, α]N be contained in B. SettingA′ = ϕ−1(B′), we have that A′ is a neighborhood of x (indeed, the set B′′ =]−α, α[N

is open and hence also A′′ = ϕ−1(B′′) is open, and x ∈ A′′ ⊆ A′). We can thentake the rectangle I as in the statement and define σ(u) = ϕ−1(u,0). It is readilyseen that σ is injective and σ(I) = A′ ∩M. Moreover, ϕ(1,...,M)(σ(u)) = u for everyu ∈ I; hence, ϕ′(1,...,M)(σ(u)) · σ′(u) is the identity matrix, so that σ′(u) has rankM, for every u ∈ I.

Remark. In the proof we have seen that M can be covered by a family of opensets of the type A′′, so that for each of them there is a local M -parametrization σ,defined on an open set containing I and injective there, such that A′′ ∩M ⊆ σ(I).Restricting if necessary the sets A′′, this property still holds if instead of A′′ wetake an open ball B(x, ρx). Moreover, if x is a point of the boundary ∂M, theM -parametrization σ is such that the interior points of a single face of the rectangleI are sent on ∂M.

We want to define an orientation for M, which will automatically induce onealso for ∂M. Given x ∈ M, let σ : I → RN be a local M -parametrization, withσ(0) = x. Since σ′(u) has rank M, for every u ∈ I, we have that the vectors[

∂σ

∂u1(u) , . . . ,

∂σ

∂uM(u)

]form a basis for a vector space of dimension M which will be called the tangentspace to M at the point σ(u) and will be denoted by Tσ(u)M (in particular, ifu = 0, we have the tangent space TxM).

181

Now, once u ∈ I is considered, the point σ(u) will belong also to the images ofother M -parametrizations. There can be a σ : J → RN such that σ(u) = σ(v), forsome v ∈ J. We know from Chapter 3 how it is possible to change the orientationto such a σ with a simple change of variable. Hence, we can choose these local M -parametrizations so that the bases of the tangent space Tσ(u)M = Tσ(v)M associatedto them all be coherently oriented; this means that the matrix which permits to passfrom one basis to the other has a positive determinant. We will call coherent sucha choice.

A coherent choice of the local M -parametrizations is therefore always possible,remaining in a neighborhood of x. But we are interested in the possibility of makinga global coherent choice, i.e., for all possible local M -parametrizations of M. Thisis not always possible: for example, it can be seen that this can not be done for aMobius strip, which is a 2-manifold.

Whenever all the local M -parametrizations of M can be chosen coherently, wesay thatM is orientable. From now on we will always assume thatM is orientableand that a coherent choice of all the local M -parametrizations has been made. Wethen say that M has been oriented.

Once we have oriented M, let us see how it is possible to define, from that, anorientation on ∂M. Given x ∈ ∂M, let σ : I → RN be a local M -parametrizationwith σ(0) = x; recall that in this case I is the rectangle [−α, α]M−1 × [0, α]. Being∂M a (M − 1)-manifold, the tangent vector space Tx∂M has dimension M − 1and is a subspace of TxM, which has dimension M. Hence, there are two versors inTxM which are orthogonal to Tx∂M. We denote by ν(x) the one which is obtainedas a directional derivative ∂σ

∂v (0) = dσ(0)v, for some v = (v1, . . . , vM ) with vM <

0. At this point, we choose a basis [v(1)(x), . . . , v(M−1)(x)] in Tx∂M such that[ν(x), v(1)(x), . . . , v(M−1)(x)] be a basis of TxM oriented coherently with the onealready chosen in this space. Proceeding in this way for every x, it can be seen that∂M is thus oriented: we have assigned to it the induced orientation from that ofM.

Assume now thatM, besides being oriented, be compact. Given a M -differen-tial form ω : U → ΩM (RN ), with U containing M, we would like to define what wemean by integral of ω on M.

In the case when ω|M, the restriction of ω to the set M, be zero outside the

support of a single local M -parametrization σ : I → RN, we simply set∫Mω =

∫σω .


In general, we have seen thatM can be covered by some open sets A′′ of RN, whichwe can assume to be open balls, for each of which there is a local M -parametrizationσ : I → RN with A′′ ∩M ⊆ σ(I). Being M compact, there is a finite sub-covering:let it be given by A′′1, . . . , A

′′n. The open set V = A′′1 ∪ . . .∪A′′n contains thenM. We

now need the following result.

Theorem C.4 There exist some functions φ1, . . . , φn : V → R, of class C∞, suchthat, for every x and every k ∈ 1, . . . , n, the following properties hold:

(i) 0 ≤ φk(x) ≤ 1 ,

(ii) x 6∈ A′′k ⇒ φk(x) = 0 ,

and, for x ∈M,

(iii)∑n

k=1 φk(x) = 1.

Proof Let A′′k = B(xk, ρxk), with k = 1, . . . , n. Consider the function f : R → Rdefined by

f(u) =

exp

(1

u2−1

)if |u| < 1 ,

0 if |u| ≥ 1 ,

and set

ψk(x) = f

(‖x− xk‖

ρk

).

Then, for every x ∈ V, it is ψ1(x)+ . . .+ψn(x) > 0, and we can define the functions

φk(x) =ψk(x)

ψ1(x) + . . .+ ψn(x).

The required properties are now easily verified.

The functions φ1, . . . , φn are said to be a partition of unity. Since each φk ·ω|Mis zero outside the support of a single local M -parametrization, we can define theintegral of ω on M in this way:∫

Mω =

n∑k=0

∫Mφk · ω .

It is possible to prove that such a definition does not depend neither on the (coherent)choice of the local M -parametrizations, nor on the particular partition of unitydefined above.

We can now state the analogue of the Stokes–Cartan theorem.

183

Theorem C.5 If ω : U → ΩM (RN ) is a M -differential form of class C1 and M isa compact, oriented (M + 1)-manifold contained in U, then∫

Mdω =

∫∂M

ω

(provided the orientation on ∂M is the induced one).

Proof Let us first assume that there is a local M -parametrization σ : I → RN suchthat

σ(I) ∩ ∂M = Ø ,

and ω|M is equal to zero outside σ(I). By the injectivity of σ and the continuity ofω, we have that ω has to be zero on all points of the support of ∂σ, so that∫

Mdω =

∫σdω =

∫∂σω = 0 .

On the other hand, since ω is zero on ∂M,∫∂M

ω = 0 .

Hence, the identity is verified in this case.

Assume now that there is a local M -parametrization σ : I → RN which sendsthe interior points of a single face Ij of I on the boundary of M and that ω|M isequal to zero outside σ(I). Then,∫

Mdω =

∫σdω =

∫∂σω ,

and since ω is zero outside the support of ∂σ except for the points coming from Ij ,which belong to ∂M, we have ∫

∂σω =

∫∂M

ω .

Hence, even in this case the identity holds.

Consider now the general case. With the above found partition of unity, each ofthe φk · ω is of one of the two kinds just considered. Being

n∑k=1

dφk ∧ ω = d

(n∑k=1

φk

)∧ ω = d(1) ∧ ω = 0 ,


we then have ∫Mdω=

n∑k=1

∫Mφk · dω

=n∑k=1

∫Mdφk ∧ ω +

n∑k=1

∫Mφk · dω

=

n∑k=1

∫Md(φk · ω)

=

n∑k=1

∫∂M

φk · ω

=

∫∂M

ω ,

and the proof is completed.

A final remark about orientation. In Chapter 1, when dealing with functionsf defined on an interval [a, b], the natural orientation of R was implicitly used,

suggesting also the introduction of the notation∫ ab f = −

∫ ba f .

In Chapter 2 this issue was not emphasized, even if some care was needed in theChange of Variables theorem, where only diffeomorphisms ϕ are allowed, and thefactor | detϕ′| appears in the main formula.

In Chapter 3 we did not really define an orientation for a M -surface σ, butnevertheless introduced the concept of equivalent M -surfaces which could have thesame or the opposite orientation. Moreover, each M -surface σ induces an orientationon its boundary ∂σ, in such a way that the Stokes-Cartan theorem appears as thetheory’s most natural conclusion.

Finally, in this appendix, we have seen that the orientation of a differentiablemanifold plays a crucial role. It is a delicate question, and it could be not so easy toverify in practice. But, again, the whole theory is motivated by the Stokes-Cartantheorem, which can be stated in this framework in its full elegance.

Appendix D

The Banach–Tarski paradox

Let us start with the following

Definition D.1 Two subsets A,B of R3 are said to be equi-decomposable if thereare some sets A1, . . . , An, which are pairwise disjoint, and B1, . . . , Bn, also pairwisedisjoint, such that

A = A1 ∪ . . . ∪An , B = B1 ∪ . . . ∪Bn ,

and each Bi happens to be a roto-translation of Ai . In that case, we will write A ∼ B.

We recall that a roto-translation is just the composition of a rotation with atranslation. It is not difficult to prove that ∼ is an equivalence relation. Let usintroduce the following notations:

B = x ∈ R3 : ‖x‖ ≤ 1 ; S = x ∈ R3 : ‖x‖ = 1 .

Moreover, once we have fixed a vector v ∈ R3 such that ‖v‖ > 2, for every subsetE of R3 we denote by ET the translation of E by the vector v :

ET = x ∈ R3 : x− v ∈ E .

Let us state and prove the astonishing Banach–Tarski theorem.

Theorem D.2 It truly happens that B ∼ B ∪BT .

How is this possible? Everybody knows that a roto-translation of a body pre-serves its volume, while the Banach–Tarski theorem affirms that two sets may beequidecomposable even if they have different volumes. The point is that, among the“pieces” A1, . . . , An, there clearly are some which are non-measurable, hence thereis no volume to be preserved! This is a serious task, which shows us the importanceof the concept of measurable set.

185

186 APPENDIX D. THE BANACH–TARSKI PARADOX

Before starting the proof, we need an important property of the rotations in R3.We will say that two rotations ρ1, ρ2 of R3 are independent if, by a finite numberof compositions of the elements in ρ1, ρ

−11 , ρ2, ρ

−12 , it is not possible to obtain

the identity function, unless allowing the appearance of couples of the type ρ1ρ−11 ,

ρ−11 ρ1 , ρ2ρ

−12 , ρ−1

2 ρ2 . We will call simplified the compositions where these couplesare not allowed. In the following, we will consider only simplified compositions.

Lemma D.3 There exist two independent rotations of R3.

Proof Let ρ be the rotation with angle arccos 13 in counter-clockwise direction

around the x-axis and φ be the analogous rotation around the z-axis. We havethat ρ, ρ−1, φ, φ−1 are represented by the following matrices:

ρ±1 =

1 0 0

0 13 ∓2

√2

3

0±2√

23

13

, φ±1 =

13 ∓2

√2

3 0

±2√

23

13 0

0 0 1

.

We will show that any simplified composition f of elements in ρ, ρ−1, φ, φ−1 cannot be the identity, proving by induction the following proposition, for n ≥ 1 :

(Pn) For every f having n components, one has that f(1, 0, 0) is of the form

1

3n(a, b√

2, c) ,

where a, b, c are integer numbers and b is not a multiple of 3.

Without loss of generality, let us assume that the last component to the right of f

be φ or φ−1. If n = 1, we have f = φ±1 and φ±1(1, 0, 0) = (13 ,±

2√

23 , 0); hence, (P1)

holds with a = 1, b = ±2 and c = 0. Assume now that (Pk) holds for k = 1, . . . , nand consider f having n+ 1 components. Then, f = φ±1f ′ or f = ρ±1f ′, where f ′

has n components and f ′(1, 0, 0) = (a′, b′√

2, c′)/3n, with b′ not being a multiple of3. In the first case,

f(1, 0, 0) = φ±1f ′(1, 0, 0) = (a′ ∓ 4b′, (b′ ± 2a′)√

2, 3c′)/3n+1 ,

while in the second case,

f(1, 0, 0) = ρ±1f ′(1, 0, 0) = (3a′, (b′ ∓ 2c′)√

2, c′ ± 4b′)/3n+1 .

It still remains to prove that neither (b′ ± 2a′) nor (b′ ∓ 2c′) are multiples of 3. Wewill do it considering the four possible cases when f is one of the following:

φ±1ρ±1f ′′ , ρ±1φ±1f ′′′ , φ±1φ±1f ′′′ , ρ±1ρ±1f ′′ ,

where f ′′′ does not appear at all if n = 1, while, if n ≥ 2, f ′′ has n− 1 componentsand f ′′(1, 0, 0) = (a′′, b′′

√2, c′′)/3n−1, with b′′ not being a multiple of 3.

187

In the first case, b = b′ ± 2a′ and a′ = 3a′′, so that b is not a multiple of 3.

In the second case, b = b′ ∓ 2c′ and c′ = 3c′′, similarly as in the first case.

In the third case, b = b′ ± 2a′ = b′ ± 2(a′′ ∓ 4b′′) = b′ + (b′′ ± 2a′′)− 9b′′ = 2b′ − 9b′′,hence b is not a multiple of 3.

In the fourth case, b = b′ ∓ 2c′ = b′ ∓ 2(c′′ ± 4b′′) = b′ + (b′′ ∓ 2c′′)− 9b′′ = 2b′ − 9b′′,similarly as in the third case.

In any case, (Pn+1) holds, and the proof is thus completed.

Proof of the Theorem. We denote by ρ, φ two independent rotations of R3, whoseexistence is guaranteed by Lemma D.3. Let F be the set of all the rotations whichcan be obtained as the composition of a finite number of elements in ρ, ρ−1, φ, φ−1,to which we add the identity. Let D be the subset of S made by the fixed points ofthe rotations in F , except for the identity. Being F countable, Lemma D.3 tells usthat D is countable, too. The sequel of the proof is divided into three steps.

Step 1. We want to prove that S \D ∼ (S \D) ∪ (S \D)T . Given x ∈ S \D, wedefine the orbit of x through F :

σ(x) = f(x) : f ∈ F .

It is easily seen that σ(x) is contained in S \D and that two orbits either coincideor are disjoint. Hence, the set of all orbits makes up a partition of S \ D. By theaxiom of choice, we can construct a set M taking a single point from each of theseorbits.

Let us prove now that, varying f ∈ F , the sets f(M) generate a partition of S\D.First observe that, taken x ∈ S \D, there is a u ∈ σ(x)∩M ; it will be u = g(x), forsome g ∈ F . Then, setting f = g−1, we have that f ∈ F and x = f(u) ∈ f(M). Thisproves that the sets f(M) cover S \D. Secondly, if there is a x ∈ f1(M) ∩ f2(M),with f1, f2 ∈ F , then both f−1

1 (x) and f−12 (x) belong to σ(x) ∩M, and therefore

coincide, by the way M has been defined. Then, x = f2(f−11 (x)), which means that

x is a fixed point for f2f−11 . Since x 6∈ D, it has to be f1 = f2 . this proves that the

sets f(M) are pairwise disjoint.

We now define the following sets:

A1 =⋃f(M) : f ∈ F starts on the left side with ρ ,

A2 =⋃f(M) : f ∈ F starts on the left side with ρ−1 ,

A3 =⋃f(M) : f ∈ F starts on the left side with φ ,

A4 =⋃f(M) : f ∈ F starts on the left side with φ−1 .

188 APPENDIX D. THE BANACH–TARSKI PARADOX

They form a partition of (S \ D) \M, since the identity is excluded from those fstarting on the left side with ρ, ρ−1, φ, φ−1. Let us prove that

A1 ∪ ρ(A2) = S \D , A1 ∩ ρ(A2) = Ø ,

A3 ∪ φ(A4) = S \D , A3 ∩ φ(A4) = Ø .

If x ∈ S \ D, it has to belong to one and only one of the sets f(M), with f ∈ F .If f begins with ρ, then x ∈ A1 ; otherwise, since we are only considering simplifiedcompositions, ρ−1f begins with ρ−1, so that ρ−1(x) ∈ A2, that is x ∈ ρ(A2). Thetwo first equalities then follow. Analogously one proves the two second ones.

If we define B1 = A1 , B2 = ρ(A2) , B3 = (A3)T , B4 = (φ(A4))T , we thus haveseen that

(S \D) \M ∼ (S \D) ∪ (S \D)T .

Consider now the bijective function

β : (S \D) \M → (S \D) ∪ (S \D)T ,

which is obtained using the respective roto-translations of the sets A1, A2, A3 andA4: we will have that β coincides with the identity on A1, with the rotation ρ onA2, with the translation by the vector v on A3, with the rotation φ followed by thetranslation by v on A4. We set α = β−1 and define the set

C =∞⋃n=0

αn((S \D)T ) = (S \D)T ∪ α((S \D)T ) ∪ α2((S \D)T ) ∪ . . . ,

where αn denotes the function α iterated n times.

We now show that

(S \D) \ α(C) = ((S \D) ∪ (S \D)T ) \ C .

Indeed, if x ∈ (S \D) \ α(C), then x does not belong to

α(C) = α((S \D)T ) ∪ α2((S \D)T ) ∪ . . . ,

and since surely it does not belong to (S \ D)T , then x is not an element of C.Consequently, x ∈ ((S \D)∪ (S \D)T )\C. Vice versa, if x ∈ (S \D)∪ (S \D)T \C,then, not being an element of C, x does not belong to (S \D)T , so that x ∈ S \D;moreover, since α(C) ⊆ C, x can not belong to α(C), hence x ∈ (S \D) \ α(C).

In conclusion, being C ∼ α(C), we have

S \D = α(C)∪((S \D)\α(C)) ∼ C∪(((S \D)∪(S \D)T )\C) = (S \D)∪(S \D)T ,

which is what we wanted to prove.

189

Step 2. We will now prove that S ∼ S ∪ ST . Let ` be a line, passing through theorigin 0 of R3, which does not intersect the countable set D. Let us see that there isa rotation ψ, having ` as axis, such that

∀n ≥ 1 ψn(D) ∩D = Ø .

Consider first the set Θ1 made of those angles which are rational multiples of 2π.These angles determine all the possible rotations f around the axis ` for whicha point of D comes back to itself by applying an iterate fn. Consider now twopoints x1,x2 ∈ D which lie on the same orthogonal plane to ` : the point x1 willbe moved onto the point x2 by a rotation around the axis ` by a certain angleθ12 ∈ [0, 2π[ . Define the set Θ2 made of the angles of the type (θ12 + 2πm)/n; theseangles determine all the possible rotations f around the axis ` for which the pointx1 is moved on x2 by applying an iterate fn. Since D is countable, the sets Θn

can be ordered in a sequence. Moreover, each of the sets Θn is countable and hencetheir union is countable, too. It is therefore sufficient to take an angle which doesnot depend on any Θn to find a rotation ψ with the required property.

Consider now the set

D =

∞⋃n=0

ψn(D) = D ∪ ψ(D) ∪ ψ2(D) ∪ . . . .

Then,S = D ∪ (S \ D) ∼ ψ(D) ∪ (S \ D) = S \D .

Analogously one sees that ST ∼ (S\D)T . Recalling the equi-decomposability provedin Step 1, using the symmetry and the transitivity of the relation ∼, we obtain theequi-decomposability we are looking for.

Step 3. Associating to each x ∈ S the radius without its origin λx : λ ∈ ]0, 1], wecan proceed exactly as above to prove that B \ 0 ∼ (B \ 0) ∪ (B \ 0)T . Letnow η be a rotation by an angle which is an irrational multiple of 2π around theline (x, 0, 1/2) : x ∈ R. Consider the set O = ηn(0) : n ≥ 0. Then,

B = O ∪ (B \ O) ∼ η(O) ∪ (B \ O) = B \ 0 .

In an analogous way one sees that BT ∼ (B \ 0)T ; by the symmetry and thetransitivity of the relation ∼, we conclude that B ∼ B ∪ BT , and the theorem isthus proved.

As we said above, the Banach–Tarski theorem should make anyone aware of thestrange consequences one can face when dealing with non-measurable sets. Thosesets are constructed by the use of the axiom of choice. Nevertheless, this axiom isstill usually accepted for its remarkable usefulness in modern mathematics.


Appendix E

A brief historical note

In this appendix, I would like to take a brief a look at the historical evolution ofthe concept of integral, without laying any claims to completeness. In particular, Iwould like to stress the role played by the Riemann sums, in different stages.

The primary motive for the integral calculus stems from geometrical problemssuch as the computation of the length of a curve, the area of a surface, planar or not,and the volume of a solid. Some of these were already computed since Ancient Greektimes, in particular by Eudoxus (4th century B.C.) and Archimedes (3rd centuryB.C.). The method used at that time was based on two steps: first, a candidatefor the integral was found by the use of approximations which could resemble theRiemann sums; then, the rigorous proof was given by the so-called “exhaustionmethod”. Obviously, the notations were completely different from ours, and theprocedure followed was mainly geometrical rather than analytical.

The main significant change in the setting of the problem was made in the 17thcentury, when Descartes discovered Analytic Geometry. In particular, differentialcalculus started to be developed as the method of determining the tangents to agiven curve. A fundamental step was then made by Leibniz (1682) and Newton(1687). They independently understood the link between differential and integralcalculus, finding that, if F is a function whose derivative coincides with f, then∫ b

af(x) dx = F (b)− F (a) .

This is what we have called the Fundamental Theorem.

Thanks to the contributions of Euler (1768) and others, the theory of primitivablefunctions was developed in the set of those functions defined by a single analyticalformula or by a power series, which were about the only functions deemed worthyof interest at that time. In this framework, the above formula was sometimes used

191

192 APPENDIX E. A BRIEF HISTORICAL NOTE

as the very definition of integral, and the the Riemann sums approach took on asecondary role.

However, a fundamental research by Fourier (1811) put in evidence that sucha theory was too restrictive: it was time to develop a theory which could dealwith some discontinuous, not primitivable functions. Cauchy (1823) was the first toprovide a rigorous basis for the theory, but it was Riemann (1854) who introducedthe definition of integrable function, the one we have called R-integrable in thisbook, which we recall here.

Definition E.1 A function f : [a, b] → R is R-integrable and its integral is somereal number A if the following property holds: for every ε > 0 there is a real numberδ > 0 for which one has ∣∣∣∣∣∣

m∑j=1


∣∣∣∣∣∣ ≤ ε ,for every choice of points aj and xj such that

a = a0 < a1 < . . . < am−1 < am = b ,

withaj − aj−1 ≤ δ , and aj−1 ≤ xj ≤ aj .

Nevertheless, the Cauchy-Riemann theory was not the final solution to the prob-lem of integration: Volterra (1881) gave an example of a primitivable and boundedfunction which is not R-integrable on [a, b]. The main problem was that this proce-dure did not take into account the particular properties of the functions involved.According to Borel, it was like “a ready-made outfit which doesn’t suit to each andeveryone”. The problem of finding a theory where both R-integrable and primi-tivable functions were integrable persisted.

Lebesgue (1902) faced this problem introducing the following integration proce-dure.

Given f : [a, b] → R, bounded with non-negative values, and C > 0 such that0 ≤ f(x) < C for every x ∈ [a, b], let us consider the subdivision of the interval [0, C[

(and not of the interval [a, b]!) in n parts[j−1n C, jnC

[, and consider the sets

Ejn =

x ∈ [a, b] :

j − 1

nC ≤ f(x) <

j

nC

.

The problem is how to “measure” these sets. In general, given a set E ⊆ [a, b],Lebesgue calls “outer measure” of E the infimum of the set of all sums

∑k(dk− ck),

193

finite or countable, obtained by considering a covering of E with the intervals [ck, dk].Denoting by µ∗(E) this exterior measure, the “interior measure” of E is given byµ∗(E) = (b − a) − µ∗([a, b] \ E). The set E is measurable if µ∗(E) = µ∗(E), inwhich case this number is called the “measure of E” and is denoted by µ(E). Atthis point, Lebesgue defines the integrable functions, which in this book have beencalled L-integrable.

Definition E.2 A function f : [a, b] → R, such that 0 ≤ f(x) < C for everyx ∈ [a, b], is L-integrable and its integral is a real number A if the following propertyholds: for every ε > 0 there is a natural number n such that, taken n ≥ n, the setsEjn are measurable and ∣∣∣∣∣

n∑j=1

j

nCµ(Ejn)−A

∣∣∣∣∣ ≤ ε .The definition is then extended to functions which are not bounded and with

arbitrary real values in the following way.

Definition E.3 Assume that the function f : [a, b]→ R has non-negative values butis not bounded above. In this case, f is said to be L-integrable on [a, b] if, for everypositive integer k, the function fk(x) = minf(x), k is L-integrable on [a, b] and the

sequence of the integrals (∫ ba fk)k has a finite limit as k tends to +∞. Such a limit

is said to be the “integral of f on [a, b]”. Whenever f not only has non-negativevalues, it is said to be L-integrable if both its positive part f+ and its negative partf− are L-integrable; the integral of f is then the difference of the respective integralsof f+ and f−.

Again according to Borel, this procedure “is custom made, and perfectly adaptsto the properties of each function”. Lebesgue’s theory was completely satisfactory forseveral aspects. It was shown that every R-integrable function is also L-integrableand the two integrals have the same value, and if a function is primitivable andbounded, then it is L-integrable.

However, the problem of the integrability of not bounded primitivable functionsremained unsolved at the time. It finally found a solution some years later by Denjoy(1912) and Perron (1914), who extended Lebesgue’s theory with two different butequivalent approaches. While Denjoy used a transfinite induction method startingfrom Lebesgue definition, Perron’s method is more in line with the formula for theintegral of primitivable functions. Let us take a simplified and brief look at Perron’smethod.

We call lower-primitive of f a function F− such that F−(a) = 0 and, for everyx ∈ [a, b], it is F ′−(x) ≤ f(x); similarly, we define an upper-primitive F+ of f, by

194 APPENDIX E. A BRIEF HISTORICAL NOTE

changing the inequality sign. We say that f is almost primitivable if there are a lower-primitive and an upper-primitive. In the following definition, we call P-integrablethe functions which are integrable according to Perron.

Definition E.4 A function f : [a, b]→ R is P-integrable if it is almost primitivableand

supF−(b) : F− is a lower primitive = infF+(b) : F+ is an upper primitive .

In such a case, this value is called “integral of f on [a, b]”.

While it is clear that every primitivable function is P-integrable, to prove thatevery L-integrable function is P-integrable (with the same value for the integral) israther complicated. It comes out that a function f is L-integrable if and only if bothf and |f | are P-integrable.

The approach to the definition of integral had been set aside for a long timewhen Kurzweil (1957) and Henstock (1961) proposed the following modification toRiemann’s definition: after observing that the two conditions aj − aj−1 ≤ δ andaj−1 ≤ xj ≤ aj can be replaced by

xj − δ ≤ aj−1 ≤ xj ≤ aj ≤ xj + δ ,

without modifying the definition, in order to adapt the procedure to each functionthey decided to allow greater freedom of choice for the δ, in correspondence with thepoints xj where the function is calculated. In other words, the δ no longer neededto be constant, but could vary in the interval [a, b], according to the needs of thefunction. This is how they reached the definition that we have adopted in this book:

Definition E.5 A function f : [a, b] → R is integrable (according to Kurzweil andHenstock) and its integral is a real number A if the following property holds: forevery ε > 0 there is a function δ : [a, b]→ R, with positive values, for which one has∣∣∣∣∣∣

m∑j=1


∣∣∣∣∣∣ ≤ ε ,for every choice of the points aj and xj in such a way that

a = a0 < a1 < . . . < am−1 < am = b ,

andxj − δ(xj) ≤ aj−1 ≤ xj ≤ aj ≤ xj + δ(xj) .

Surprisingly, a function is integrable (according to Kurzweil and Henstock) ifand only if it is P-integrable, and in that case the value of the integral is the same.

195

Some years later Mac Shane (1969) proved that, by modifying in the abovedefinition the condition aj−1 ≤ xj ≤ aj with the less restrictive one xj ∈ [a, b], analternative definition of L-integrable functions is obtained.

In conclusion, it can be said that the Riemann sums have played and still playa fundamental role in the theory of integration, even if with alternate fortunes.Intuitively used by the ancient Greeks, they were placed on the back burner once thetheory of Leibniz and Newton was introduced, only to be back in the limelight thanksto Cauchy and Riemann. Overshadowed once again by the theories of Lebesgue,Denjoy and Perron, they proved to be important again in the work of Kurzweil andHenstock, who re-introduced them to unify these theories in an easy and intuitiveway. And they are the subject of interesting research developments even today.


Bibliography

[1] R.G. Bartle, A Modern Theory of Integration, American Mathematical Society,Providence, 2001.

[2] Z. Buczolich, The g-integral is not rotation invariant, Real Analysis Exchange18 (1992/93), 437–447.

[3] A. Fonda, Lezioni sulla Teoria dell’Integrale, Ed. Goliardiche, Roma, 2001.

[4] R.A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, Amer-ican Mathematical Society, Providence, 1994.

[5] R. Henstock, Definitions of Riemann type of the variational integrals, Proceed-ings of the London Mathematical Society 11 (1961), 402–418.

[6] R. Henstock, Theory of Integration, Butterworths, London, 1963.

[7] R. Henstock, The Generalized Theory of Integration, Clarendon Press, Oxford,1991.

[8] J. Kurzweil, Generalized ordinary differential equations and continuous depen-dence on a parameter, Czechoslovak Mathematical Journal 7 (1957), 418–449.

[9] J. Kurzweil, Nichtabsolut Konvergente Integrale, Teubner, Leipzig, 1980.

[10] J. Kurzweil, Henstock–Kurzweil Integration: Its Relation to Topological VectorSpaces, World Scientific, Singapore, 2000.

[11] S. Leader, The Kurzweil–Henstock Integral and its Differentials, Marcel Dekker,New York, 2001.

[12] P.Y. Lee, Lanzhou Lectures on Henstock Integration, World Scientific, Singa-pore, 1989.

[13] P.Y. Lee and R. Vyborny, The Integral. An Easy Approach after Kurzweil andHenstock, Cambridge University Press, Cambridge, 2000.

197

198 BIBLIOGRAPHY

[14] T.Y. Lee, Henstock–Kurzweil integration on Euclidean spaces, World Scientific,Singapore, 2011.

[15] J. Mawhin, Analyse: Fondements, Techniques, Evolution, De Boeck, Bruxelles,1979–1992.

[16] R.M. McLeod, The Generalized Riemann Integral, Mathematical Associationof America, Washington, 1980.

[17] E.J. McShane, Unified Integration, Academic Press, New York, 1983.

[18] W.F. Pfeffer, The Riemann Approach to Integration, Cambridge UniversityPress, 1993.

[19] W.F. Pfeffer, Derivation and Integration, Cambridge University Press, Cam-bridge, 2001.

[20] M. Spivak, Calculus on Manifolds, Benjamin, Amsterdam, 1965.

[21] Ch. Swartz, Introduction to Gauge Integrals, World Scientific, Singapore, 2001.

Index

additivity of the integral, 22, 56, 71, 103almost everywhere, 65area, 60

of a surface, 132, 133

boundary of a manifold, 179

Cebicev inequality, 63chain rule, 163cilindrical coordinates, 98curl, 119curve, 120

diffeomorphism, 169differentiable, 155, 158differentiable manifold, 179

orientable, 181oriented, 181

differential, 155differential form, 113

closed, 149exact, 149

direction, 155directional derivative, 155divergence, 119

equi-decomposable sets, 185equivalent M -surfaces, 123external differential, 116external product, 114

flux, 130Formulas:

Gauss, 137

Gauss–Green, 146Gauss–Ostrogradski, 144Stokes–Ampere, 143Stokes–Cartan, 142

functions:differentiable, 155integrable, 5L-integrable, 31of class C1, or C1-functions, 158, 160of class C2, or C2-functions, 159, 160of class Cn, or Cn-functions, 159, 160R-integrable, 24

gauge, 3, 54glueing, 135gradient, 119

Hessian matrix, 159

induced orientation, 181integrable function:

Kurzweil–Henstock, 5, 55Lebesgue, 31, 56, 100Riemann, 24, 56

integralimproper, 42indefinite, 12line integral, 129of a differential form, 126of a function, 6, 56, 100surface integral, 130

integrationby parts, 16by substitution, 18

199

200 INDEX

irrotational, 150

Jacobian matrix, 160

L-integrable, 31, 56Leibniz rule, 73, 103length of a curve, 132, 133line integral, 129local diffeomorphism, 169

M-surface, 120measurable, 60, 101measure, 60, 101

M -dimensional, 133M -superficial, 133

negligible, 63non-overlapping, 53, 70normal versor, 121

orientationinduced, 181

oriented boundaryof a M -surface, 139of a rectangle, 136

P-partition, 1, 53δ-fine, 3, 55

parametrizable, 124parametrization, 124partial derivative, 156partition of unity, 182polar coordinates, 97potential

scalar, 150vector, 151

primitivable function, 11primitive of a function, 11projection of a set, 81, 104pull-back, 171

R-integrable, 24, 56rectangle, 53

reflection, 95

regular M -surface, 120

Riemann sum, 2, 54

rotations, 95

independent, 186

scalar product, 119

section of a set, 81, 104

sets

M -parametrizable, 124

equi-decomposable, 185

measurable, 60, 101

negligible, 63

non-overlapping, 70

star-shaped, 149

solenoidal, 151

spherical coordinates, 98

star-shaped set, 149

support of a M -surface, 120

surface, 120

surface integral, 130

tangent

plane, 121

space, 180

versor, 121

Theorems:

Banach–Tarski, 185

Beppo Levi, 36, 101

Change of Variables, 85, 93, 106

Cousin, 3, 55

Fubini, 78, 80, 81, 104

Fundamental Theorem, 10, 11

Hake, 46

Implicit Function, 164, 169

Lebesgue, 39, 102

Local Diffeomorphism, 169

Poincare, 150, 176

Saks–Henstock, 28

Schwarz, 158

Stokes–Cartan, 174, 182

INDEX 201

translation, 95trasformation of a differential form, 171

vector fieldirrotational, 150solenoidal, 151

vector product, 119versor, 155volume, 123

of a set, 60

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The Kurzweil{Henstock integral for undergraduatesfonda/p2017_book_KH.pdf · The integral in question is indeed more general than Lebesgue’s in RN, but its construction is rather