Math 301 Integration I - fs2-fanar.comfs2-fanar.com/Math/pdf/Math301-lecturenotes.pdf · Chapter 1 Preliminaries 1.1 Sets and functions Given a set E we denote by P(E) or 2E the set

Math 301

Integration I

Lecture notes of Dr. Hicham [email protected]

Lebanese University, Fanar, Fall 2013-2014www.fs2-fanar.com/Math

2

Introduction and orientation

In previous courses, you studied the Riemann integral. With it, you can integrate many func-tions and you know for example that every piecewise continuous function can be integrated.There are however functions that do not possess a Riemann integral. Consider for instance thefunction f : [0, 1] → IR defined by f(x) = 1 if x is rational and f(x) = 0 if x is irrational. Weshall see that this function is not integrable in the sense of Riemann. It does have however anintegral called the Lebesgue integral. The Lebesgue integral is a generalization of the Riemannintegral: it permits to integrate more functions. One of our targets in this course is to constructthe Lebesgue integral and study its properties.

There are many ways to present the subject. We have chosen the approach based on measuretheory. A measure is a generalization of the familiar notions of length, area, volume andprobability of an event. In this course we shall construct a measure on IR called the Lebesguemeasure which assigns a ”length” to many subsets of IR that are not necessary intervals.

Let me make some remarks about the subject before we start. They will become clear aswe progress.

1. As we said, the Lebesgue integral permits to integrate a wider class of functions than theRiemann integral permits. Also, the sets on which we integrate need not be intervals.

2. The theorems of the Lebesgue theory are stronger and easier to use than those of theRiemann theory.

3. There is an analogy between the completion of the rational numbers by real numbersand the completion of Riemann integrable functions by Lebesgue integrable functions.

4. The Lebesgue theory of measure and integration is fundamental to many fields of Math-ematics like probability, functional analysis, dynamical systems and Fourier series.

References

Paul. R. Halmos, Measure theory (Princeton, Van Nostrand, 1950).

Roger Jean, Mesure et integration (Presses de l’Universite du Quebec, 1982).

Walter Rudin, Real and complex analysis (McGraw-Hill, 1977).

Contents

1 Preliminaries 51.1 Sets and functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Some fundamental properties of the real line and the extended real line . . . . . 8

1.3.1 The real line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.2 The extended real line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3.3 Superior limit and inferior limit . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Axiomatic measure theory 152.1 Measurable spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Measure spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Measurable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.3.1 Measurable real valued functions . . . . . . . . . . . . . . . . . . . . . . . 242.3.2 Simple functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4 Outer measures and Caratheodory’s theorem . . . . . . . . . . . . . . . . . . . . 282.5 Completion of a measure space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 The Lebesgue measure on IR 353.1 Construction and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2.1 A Lebesgue non measurable set . . . . . . . . . . . . . . . . . . . . . . . . 413.2.2 The middle third Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 The Lebesgue integral 454.1 Construction and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.1.1 Simple fonctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.1.2 Nonnegative measurable functions . . . . . . . . . . . . . . . . . . . . . . 484.1.3 Summable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.1.4 Complex valued functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.2 The Lebesgue dominated convergence theorem . . . . . . . . . . . . . . . . . . . 554.3 Relations with the Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . 574.4 Some applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

A The Riemann integral 63A.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63A.2 Criteria of integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64A.3 Classes of integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67A.4 Properties of the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69A.5 Integration and differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69A.6 Limits and integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3

4 CONTENTS

Chapter 1

Preliminaries

1.1 Sets and functions

Given a set E we denote by P(E) or 2E the set of all subsets of E. The reason for the notation2E is that if E is finite and contains n elements, then there are 2n subsets of E.

Let E be a set and let A ⊂ E. The characteristic function of A is the function χA definedby

χA(x) =

1 if x ∈ A

0 if x /∈ A.

It is also denoted by 1A. The complement of A in E is the set of points in E that are not in A.It is denoted by one of the following notations

A, AE , E\A, E −A, Ac.

Given a family of sets (Ai)i∈I , we set⋃i∈I

Ai = x | ∃j ∈ I such that x ∈ Aj,⋂i∈I

Ai = x |x ∈ Ai,∀i ∈ I.

Recall De Morgan’s laws

X\⋃i∈I

Ai =⋂i∈I

(X\Ai)

X\⋂i∈I

Ai =⋃i∈I

(X\Ai),

and the distributivity laws

A ∩⋃i∈I

Ai =⋃i∈I

(A ∩Ai)

A ∪⋂i∈I

Ai =⋂i∈I

(A ∪Ai).

Let f : X → Y be a map between two sets. Let A ⊂ X, the (direct) image of A under f is

f(A) = y ∈ Y | ∃x ∈ X such that y = f(x).

Let B ⊂ Y , the inverse image of B under f is

f−1(B) = x ∈ X | f(x) ∈ B.

5

6 CHAPTER 1. PRELIMINARIES

Regardless of wether f is a bijection or not, f−1 defines a map

f−1 : P(Y ) → P(X)

B 7→ f−1(B).

This map satisfies

1. f−1(∅) = ∅ and f−1(Y ) = X.

2. f−1(Ac) = (f−1(A))c

3. f−1 (∪i∈IBi) = ∪i∈If−1(Bi) and f−1 (∩i∈IBi) = ∩i∈If

−1(Bi).

We use capital letters to denote sets and calligraphic letters to denote sets of sets. If B is acollection of subsets of Y , then we write f−1(B) = f−1(B)|B ∈ B). This is the direct imageunder f−1 of the set B.

Note that if (Ai)i∈I is a family of subsets of X, then

f(⋃i∈I

Ai) =⋃i∈I

f(Ai),

but we only havef(⋂i∈I

Ai) ⊂⋂i∈I

f(Ai),

the equality holds if f is one to one (injective).

A set E is called countable if there is a one to one function f : E → IN∗ where IN∗ is theset of positive integers. This means that we can put the elements of E in a finite or infinitesequence (x1, x2, . . .). So for example the set of all integers Z and the set of all rational numbersQ are countable. A countable union of countable sets is countable, that is, if I is countableand Ai is countable for all i ∈ I, then ∪i∈IAi is countable. Also a finite cartesian product ofcountable sets is countable, so for example IN× IN is countable. However P(IN) and the set ofreal numbers IR are uncountable.

Let X be a set and (An)n∈IN be a sequence of subsets of X. We set

lim inf An = limAn =⋃

n∈IN∗

⋂k≥n

Ak,

andlim supAn = limAn =

⋂n∈IN∗

⋃k≥n

Ak,

Therefore, x ∈ lim inf An if satrting from as certain rank, x belongs to all the An. In probabilitytheory, the event lim inf An is called the event that the An happen almost always (An a.a.).

On the other hand, x ∈ lim supAn if x belongs to An for infintely many indices n. Inprobability theory, the event lim supAn is called the event that the An happen infinitely often(An i.o.).

We claim that ⋂n∈IN∗

An ⊂ limAn ⊂ limAn ⊂⋃

n∈IN

An.

Proof. For the first inclusion, let Bn =⋂

k≥n Ak. Then the first inclusion is equivalent toB1 ⊂

⋃n∈IN∗ Bn which is obviously true. For the third inclusion, let Cn =

⋃k≥n Ak. Then the

1.2. TOPOLOGY 7

third inclusion is equivalent to⋂

n∈IN∗ Cn ⊂ C1 which is obviously true. Now for the secondinclusion observe that

x ∈ lim inf An ⇔ x ∈⋃

n∈IN∗

⋂k≥n

Ak ⇔ ∃n ∀k ≥ n x ∈ Ak ⇔ n ∈ IN∗ |x /∈ An is finite,

and that

x ∈ lim supAn ⇔ x ∈⋂

n∈IN∗

⋃k≥n

Ak ⇔ ∀n ∃k ≥ n x ∈ Ak ⇔ n ∈ IN∗ |x ∈ An is infinite.

Now if n ∈ IN∗ |x /∈ An is finite then its complement n ∈ IN∗ |x ∈ An is infinite becauseIN∗ is infinite. Hence the inclusion.

We say that (An) converges to A if limAn = limAn = A.

We finally state the following axiom.

Axiom of choice. Let (Ai)i∈I be a family of nonempty and pairwise disjoint sets. Thenthere exists a set C such that for all i ∈ I, C ∩ Ai is a singleton. Otherwise stated, given afamily of sets as above, we can select exactly one element form each set.

This axiom seems trivial (and this is why we call it axiom). It has however some far reachingand unexpected consequences like the fact that any set can be well ordered (this is known asthe well ordering theorem). An ordered set (E,≤) is said to be well ordered if every nonemptysubset of E has a smallest element. For example, IN is well ordered in the usual order. HoweverZ and IR are not well ordered in the usual order. The well ordering theorem implies that thereexists a well ordering on IR, a fact that no one can prove without the axiom of choice. Actuallythe axiom of choice is equivalent to the well ordering theorem. The well ordering theoremstartled the mathematical world in the beginning of the twentieth century. A careful analysis ofits proof led to the formulation of the axiom of choice. The axiom of choice asserts the existenceof a set but gives no procedure to construct it, and this is why some mathematicians of thebeginning of the twentieth century refused it. Nowadays, most mathematicians accept it andwe shall use it in chapter 3 to construct a nonmeasurable subset of IR.

1.2 Topology

Let X be a set. A family T of subsets of X (that is, T ⊂ P(X)) is called a topology on Xprovided the following conditions hold.

1. ∅, X ∈ T .

2. If Oi ∈ T for all i ∈ I, then ∪i∈IOi ∈ T .

3. If O1, O2 ∈ T , then O1 ∩O2 ∈ T .

The elements of T are called open sets of X. A set is called closed if its complement is open.

If d is a distance on X, then d generates a topology on X. But a topology can be definednot only by a distance. A general way of defining a topology is by specifying a basis. A subsetB ⊂ P(X) is called a basis provided the following hold.

1. For all x ∈ X, there exists B ∈ B such that x ∈ B.

2. If B1, B2 ∈ B and x ∈ B1 ∩B2, then there exists B3 ∈ B such that x ∈ B3 ⊂ B1 ∩B2.

8 CHAPTER 1. REVIEW AND COMPLEMENTS

The topology T generated by B is defined as follows: O ∈ T if whenever x ∈ O, there existsB ∈ B such that x ∈ B ⊂ O.

A function f : X → Y between two topological spaces is called continuous at a point x0 iffor every neighborhood V of f(x0), there exists a neighborhood of x0 such that f(U) ⊂ V . Afunction f is called continuous on X if it is continuous at every point of X. Then the followingconditions are equivalent.

(i) f : X → Y is continuous.

(ii) The inverse image under f of every open set of Y is an open set of X.

(iii) The inverse image under f of every closed set of Y is a closed set of X.

A subset A of a metric space X is called dense if the closure of A is equal to X. Thefollowing conditions are equivalent.

1. A is dense in X.

2. Every open set of X meets A.

3. For every x ∈ X, there exists a sequence in A that converges to x.

Let A be a subset of a topological space, an open covering of A is a family of open setswhose union contains A. The subset A is called compact if any open covering of A contains afinite subcollection that also covers A. In IRn, a subset is compact if and only if it is closed andbounded.

A subset A of a topological space is connected if the following condition holds. If A ⊂ B∪Cwith B and C open and disjoint, then either A ⊂ B or A ⊂ C. The union of connected setshaving a point in common is connected. The connected subsets of IR are the intervals.

1.3 Some fundamental properties of the real line and the ex-tended real line

1.3.1 The real line

We start by formulating two fundamental properties of the set of real numbers that we shalluse in the sequel. This set is denoted by IR and it is naturally identified to a line. An upperbound of a subset E ⊂ IR is a real number M such that x ≤ M for all x ∈ E. The set E iscalled bounded from above if it has an upper bound. A lower bound of a subset E ⊂ IR is areal number m such that m ≤ x for all x ∈ E. The set E is called bounded from below if it hasa lower bound.

Definition 1.1 Let E ⊂ IR be bounded from above. The number α is called the least upperbound of E if the following hold.

i) α is an upper bound of E.ii) If β < α, then β is not an upper bound of E.

In this case, we write α = supE. If E is not bounded from above, we write supE = +∞.

Proposition 1.1 Let E ⊂ IR be bounded from above. Then α = supE if and only if thefollowing hold.

i) x ≤ α for all x ∈ E.ii) ∀ε > 0 ∃y ∈ E such that α− ε < y.

1.3. SOME FUNDAMENTAL PROPERTIES 9

Definition 1.2 Let E ⊂ IR be bounded from below. The number α is called the greatest lowerbound of E if the following hold.

i) α is a lower bound of E.ii) If β > α, then β is not a lower bound of E.

In this case, we write α = inf E. If E is not bounded from below, we write inf E = −∞.

Proposition 1.2 Let E ⊂ IR be bounded from below. Then α = inf E if and only if thefollowing hold.

i) x ≥ α for all x ∈ E.ii) ∀ε > 0 ∃y ∈ E such that α + ε > y.

Now we can state the two fundamental properties of IR.

Theorem 1.1 Any nonempty subset of IR which is bounded from above has a least upperbound. Any nonempty subset of IR which is bounded from below has a greatest lower bound.

Any proof of this theorem involves going back to the construction of the real numbers fromthe rational numbers. We do not address this issue here.

Recall now that a subset E ⊂ IR is an interval if it is not empty and [x, y] ⊂ E whenever x andy belong to E. An interval has thus one of the following ten forms a, ]a, b[, ]a, b], [a, b[, [a, b],] −∞, a], ] −∞, a[, ]a,+∞[, [a,+∞[ and IR. We shall say that an interval is trivial if it is asingleton.

Theorem 1.2 Any nontrivial interval of the real line contains rational as well as irrationalnumbers.

This theorem is not difficult to prove using the following evident facts: 1- for each realnumber a, there is an integer n bigger than a (this is known as the Archimedean property); 2-every nonempty subset of IN has a smallest element (IN is well ordered).

Proposition 1.3 Let a and b be two real numbers. If a ≤ b + ε for all ε > 0 then a ≤ b.

Theorem 1.3 Every open set O of the real line is the union of a countable family of pairwisedisjoint open intervals.

Proof. Let (Iλ)λ∈L be the collection of connected components of O. We know that each Iλ

is an open interval (recall that in a locally connected space, a connected component of an openset is open). Since the rational numbers are dense in IR, each Iλ contains a rational number.By the axiom of choice, we can choose exactly one rational number in each Iλ. This defines afunction between L and Q which is one to one because two distinct components are disjoint.Therefore L is countable. Finally we have indeed O = ∪λ∈LIλ.

1.3.2 The extended real line

In measure an integration we will have to deal with sets of infinite measure and with unboundedfunctions. So we need to extend the set of real numbers. The extended real line denoted byIR or [−∞,+∞] is obtained from IR by adding two objects +∞ and −∞ (which are not realnumbers).

Rules in IR

I. Order. We extend naturally the usual order on IR by letting −∞ < x < +∞ for every x ∈ IRThis order is total, that is any two numbers are comparable. Also every nonempty subset of IRhas a least upper bound (sup) and greatest lower bound (inf).


II. Arithmetic operations.

1. x +∞ = +∞, for all x ∈]−∞,∞].

2. x−∞ = −∞, for all x ∈ [−∞,∞[.

3. x× (+∞) =

+∞ if x ∈]0,+∞]−∞ if x ∈ [−∞, 0[.

4. 0×±∞ = 0.

Warning. The following operations are not defined: ∞−∞, −∞+∞, ±∞±∞ .

III. The topology of IR. We define a distance on IR by setting

d(x, y) = | arctanx− arctan y|

with the conventions that arctan(+∞) = π2 and arctan (−∞) = −π

2 so that for example

d(x,+∞) = | arctanx− π

2| if x ∈ IR and d(+∞,−∞) = π.

Remark 1.1 Let (xn) be a sequence of IR. Recall that (xn) is said to tend to +∞ if for everyA > 0, we have xn > A for all n large enough. It is proved in calculus that xn →∞ if and onlyif arctanxn → π

2 . It follows that xn → +∞ if and only if lim d(xn,+∞) = 0 in IR, which meansthat (xn) converges to the point +∞ in the topology of IR. Similarly, xn → −∞ if and only iflim d(xn,−∞) = 0.

Here are some properties of the topology of IR.

1. The restriction of d to IR× IR is a distance that generates the usual topology of IR.

Indeed, let first O be open in (IR, d) and let x ∈ O. Then there is r > 0 such thatBd(x, r) := y | d(x, y) < r ⊂ O. But B(x, r) := y | |x − y| < r ⊂ Bd(x, r) sinced(x, y) = | arctanx− arctan y| ≤ |x− y|. Therefore B(x, r) ⊂ O. Hence O is open for theusual topology which therefore contains the topology generated by d.

Conversely, let O be open for the usual topology of IR and let x ∈ O. Then there isε > 0 such that B(x, ε) ⊂ O. Now since the function t 7→ tan t is continuous at arctanx,there is δ > 0 such that | tan arctanx− tan z| < ε whenever |z − arctanx| < δ. We claimthat Bd(x, δ) ⊂ B(x, ε). Indeed, d(x, y) < δ ⇒ | arctanx − arctan y| < δ ⇒ |x − y| =| tan arctanx− tan arctan y| < ε. Therefore, O is open for d.

2. IR is homeomorphic to [−π2 , π

2 ] and therefore to any compact interval [a, b] (a < b) of IR.Therefore IR is compact and connected. Indeed, the map h : IR → [−π

2 , π2 ] defined by

h(x) =

arctanx if x ∈ IRπ2 if x = +∞−π

2 if x = −∞

is a homeomorphism. First, h is clearly a bijection. Next, h is continuous.

• Let x ∈ IR and let xn → x, then arctanxn → arctanx, i.e., h(xn) → h(x).

• Let x = +∞ and let xn → +∞, then h(xn) → π2 = h(+∞).

• Let x = −∞ and let xn → −∞, h(xn) → −π2 = h(−∞).


The inverse map is given by

h−1(x) =

tanx if x ∈]− π

2 , π2 [

+∞ if x = π2

−∞ if x = −π2 .

The proof of the continuity of h−1 is similar to the proof of continuity of h.

3. IR is dense in IR. This follows from the fact that IR is homeomorphic to ] − π2 , π

2 [ whichis dense in [−π

2 , π2 ].

4. The collection of intervals of the form ]a, b[, [−∞, a[, ]a,+∞] with a, b ∈ IR is a basis forthe topology of IR.

Let B denote the collection of such intervals. Observe that (i) every x ∈ IR is containedin some element of B and, (ii) the intersection of two elements of B is again an elementof B (possibly empty). Therefore B is a basis. Let O be open for (IR, d) and let x ∈ O.Then Bd(x, r) ⊂ O for some r > 0. We distinguish between three cases.

Case 1. x ∈ IR. Then ]x− r, x + r[= B(x, r) ⊂ Bd(x, r) ⊂ O. This means that O containsa basis element containing x.

Case 2. x = +∞. Since limy→+∞ arctan y = π2 , then for any r > 0, there exists a ∈ IR

such that | arctan y − π2 | < r for all y > a. It follows that

Bd(+∞, r) = z ∈ IR | d(z,+∞) < r = z ∈ IR | | arctan z − π

2| < r

contains the basis element ]a,+∞] that contains +∞.

Case 3. x = −∞. Similarly to case 2, Bd(−∞, r) contains a basis element [−∞, a[.

Theorem 1.4 Every open set O of IR is the union of a countable family of pairwise disjointopen intervals of IR.

Proof. Observe first that the connected subsets of IR are exactly the intervals of IR. Next,an open interval of IR contains an open interval of IR which therefore contains rational points.Therefore the proof of Theorem 1.3 can be repeated.

We will have to deal later with sequences and series in IR. So it is important to formulatesome rules for using them.

Proposition 1.4 A monotonic sequence (xn) of IR is convergent. Moreover

i) If (xn) is nondecreasing then

lim xn = supxi | i ∈ IN∗.

ii) If (xn) is nonincreasing, then

lim xn = infxi | i ∈ IN∗.

Proof. Let (xn) be a monotonic sequence in IR. Let yn = h(xn) where h : IR → [−π2 , π

2 ] is thehomeomorphism defined above. Then (yn) is also monotonic and therefore convergent (since itis bounded). Continuity of h−1 implies that (xn) is convergent.

i) Let (xn) be nondecreasing. We distinguish between two cases


1) supxi < +∞. We distinguish again between two cases.

a) xn = −∞ for all n ∈ IN∗. Then indeed, lim xn = supxi = −∞.

b) xn0 ∈ IR for some n0. But then xn ∈ IR for all n ≥ n0 because the sequence is non-decreasing. We modify the sequence by letting xn = xn0 for n ≤ n0. This does notchange neither the limit nor the sup and we have xn ∈ IR for all n. Let ε > 0 be given.By a property of the supremum, there exists k ∈ IN∗ such that supxi − ε < xk. Thensupxi − ε < xn for all n ≥ k since the sequence is nondecreasing. But xn ≤ supxi. Itfollows that |xn − supxi| < ε for all n ≥ k and this means that lim xn = supxi.

2) supxi = +∞. We distinguish again between two cases.

a) xn < +∞ for all n ∈ IN∗. The fact that supxi = +∞ means that the set xi|i ∈ IN∗is not bounded from above. This in turn means that for all M > 0, there exists k suchthat xk ≥ M . Then xn ≥ M for all n ≥ k. This means that lim xn = +∞ and solim xn = supxi.

b) xn0 = ∞ for some n0. Then xn = ∞ for all n ≥ n0. In this case, we have lim xn = +∞and so lim xn = supxi.

ii) If (xn) is nonincreasing, then (−xn) is nondecreasing. Therefore lim(−xn) = sup(−xi) =− inf(xi) and so lim xn = inf xi.

Let (xn) be a sequence of [0,+∞]. Then the sequence Sn =∑n

k=1 xk is monotonic. It is thereforeconvergent. We set

∑∞k=1 xk = limn→∞ Sn. This extends the definition of a convergent series

of real numbers. Therefore series of nonnegative terms are always convergent in IR. Note thatthe condition

∑∞k=1 xk < +∞ means that xk ∈ IR for all k and that the series is convergent in

the usual sense.The familiar rules of convergent series still holds. For example, if (xn) and (yn) are sequences

in [0,+∞] then

1.∑∞

n=1(xn + yn) =∑∞

n=1 xn +∑∞

n=1 yn.

2.∑∞

n=1 αxn = α∑∞

n=1 xn for α ≥ 0.

3. (∀n ∈ IN∗, xn ≤ yn) ⇒∑∞

n=1 xn ≤∑∞

n=1 yn.

Lemma 1.1 The sum of a series of nonnegative terms does not depend on the order of sum-mation.

Proof. Let (xn) be a sequence of [0,+∞] and let σ : IN∗ → IN∗ be a bijection. We have toprove that

∑∞n=1 xn =

∑∞n=1 xσ(n). Let n be given and let N = maxσ(1), · · · , σ(n). Then

σ(1), · · · , σ(n) ⊂ 1, · · · , N. It follows that∑n

k=1 xσ(k) ≤∑N

k=1 xk ≤∑∞

k=1 xk becausexk ≥ 0. Letting n → ∞ we get

∑∞k=1 xσ(k) ≤

∑∞k=1 xk. By symmetry

∑∞k=1 xk ≤

∑∞k=1 xσ(k).

Hence the equality.

Remark 1.2 If the sequence (xn) is of variable sign, then the sum of the series∑

xn maydepend on the order of integration. Here is an example. We shall prove later that

1− 12

+13− 1

4+ · · ·+ 1

2k − 1+

12k

+ · · · = ln 2.

However,

1− 12− 1

4+

13− 1

6− 1

8+ · · ·+ 1

2k − 1− 1

4k − 2− 1

4k+ · · · = 1

2ln 2.


Indeed, let Sn be the nth partial sum of the first series and let S′n be the nth partial sum of thesecond series. Then

S′3n =n∑

k=1

(1

2k − 1− 1

4k − 2− 1

4k

)=

n∑k=1

(1

4k − 2− 1

4k

)=

12S2n.

Therefore S′3n → 12 ln 2. Since S′3n−1 = S′3n + 1

4n and S′3n−2 = S′3n−1 + 14n−2 , it follows that

S′3n−1 and S′3n−2 also converge to 12 ln 2. Consequently, S′n converges to 1

2 ln 2.

Let now A be an infinite countable set and let (xa)a∈A be a family of [0,+∞]. Then there existsa bijection ϕ : IN∗ → A. We set ∑

a∈A

xa =∞∑

n=1

xϕ(n).

The previous lemma ensures that this definition makes sense. Using arguments similar to thoseused in the proof of Lemma 1.1, one can prove the following

Proposition 1.5 Let xn,m be a double sequence of [0,+∞]. Then

∑(n,m)∈IN∗×IN∗

xn,m =∞∑

n=1

∞∑m=1

xn,m =∞∑

m=1

∞∑n=1

xn,m.

1.3.3 Superior limit and inferior limit

Let (xn) be an arbitrary sequence in IR. Let yn = supxk | k ≥ n. Then (yn) is nonincreasingand therefore has a limit in IR. This limit is called the superior limit of the sequence xn andit is denoted by limxn or lim supxn. Thus,

lim supxn = infn≥1

supk≥n

xk = limn→∞

supk≥n

xk.

The inferior limit of a sequence xn denoted by limxn or lim inf xn is defined by

lim inf xn = supn≥1

infk≥n

xk = limn→∞

infk≥n

xk.

Examples.1. Let xn = (−1)n. Then yn := infxk | k ≥ n = −1. Hence lim inf xn = lim

n→∞yn = −1. On

the other hand, zn := supxk | k ≥ n = 1. Hence lim supxn = limn→∞

zn = 1.

2. Consider the sequence (1, 2, 1 + 12 , 2 + 1

2 , 1 + 13 , 2 + 1

3 , 1 + 14 , 2 + 1

4 , . . .). Then lim supxn = 2and lim inf xn = 1.

Theorem 1.5 The lim sup and lim inf satisfy the following properties.

(i) xn ≤ yn ⇒ lim inf xn ≤ lim inf yn and lim supxn ≤ lim sup yn.

(ii) lim inf xn ≤ lim supxn.

(iii) lim inf xn = lim supxn = ` if and only if lim xn = `.

(iv) If xn ≥ 0 then lim xn = 0 ⇔ lim supxn = 0.

(v) lim supxn is the biggest limit point of (xn) and lim inf xn is the smallest limit point of(xn).


Proof. (i). We have supxk | k ≥ n ≤ supyk | k ≥ n for all n. Letting n → ∞, we getlim supxn ≤ lim sup yn. The proof for lim inf is similar.

(ii) For all n, we have infxk | k ≥ n ≤ supxk | k ≥ n. Letting n →∞, we get the result.

(iii) Suppose first that lim inf xn = lim supxn = `. If ` = +∞ then the sequence yn := infk≥n xk

tends to ∞. But xn ≥ yn and so xn →∞. If ` = −∞, then zn := supk≥n xk tends to −∞. Butxn ≤ zn and so xn → −∞. Now we assume that ` ∈ IR. Let ε > 0. Then | sup

k≥nxk − `| < ε for

all n large enough. In particular xn ≤ supk≥n

xk < ` + ε. Similarly, | infk≥n

xk − `| < ε for n large

enough. In particular xn ≥ infk≥n xk > `− ε. Thus, `− ε < xn < ` + ε for all n large enough.This proves that lim xn = `.

Conversely, assume that lim xn = `. If ` = +∞, then supk≥n xk →∞ since supk≥n xk ≥ xn.Therefore lim supxn = +∞. Now since xn → ∞, we have for all A > 0, there is n0 such thatxn ≥ A for all n ≥ n0 and so infk≥n xk ≥ infk≥n0 xk ≥ A. Letting n →∞ we get lim inf xn ≥ A.Since A is arbitrary, we conclude that lim inf xn = +∞. The case ` = −∞ is similar. Now weassume that ` ∈ IR. Let ε > 0 be given. Then there is n0 such that ` − ε < xk < ` + ε for allk ≥ n0. Thus

`− ε ≤ infk≥n

xk ≤ supk≥n

xk ≤ ` + ε

for all n ≥ n0. Letting n →∞, we get `− ε ≤ lim supxn ≤ ` + ε and `− ε ≤ lim inf xn ≤ ` + ε.Since ε is arbitrary, we conclude that lim supxn = lim inf xn = `.

(iv) If lim xn = 0 it follows from part (iii) that lim sup xn = 0. Now conversely, suppose thatlim supxn = 0. Since xn ≥ 0, it follows that lim inf xn ≥ 0. But lim inf xn ≤ lim supxn = 0.Therefore lim supxn = lim inf xn = 0. Thus, by part (iii) lim xn = 0.

(v) We prove the result for lim sup, the case of lim inf being similar. We need to show first thatif L := lim supxn, then there is a subsequence (xkp) that converges to L. We consider the caseL ∈ IR, the other cases being left as an exercise. We have the following:

∀ ε > 0 ∃n0 such that L ≤ supk≥n

xk < L + ε ∀n ≥ n0 (by definition of the limsup)

and∀ ε > 0 ∀n ∃m ≥ n such that xm > sup

k≥nxk − ε (by a property of the sup).

Thus, we have the following condition

∀ ε > 0 ∀n ∃m ≥ n such that L + ε > xm > L− ε.

In particular, for ε = 1 and n = 1, ∃k1 ≥ 1 such that L + 1 > xk1 > L− 1. Next take ε = 1/2and n = k1 + 1. Then there exits k2 ≥ k1 + 1 such that L + 1

2 > xk2 > L− 12 . At the pth step,

we get an index kp ≥ kp−1 +1 such that L+ 1p > xkp > L− 1

p . This defines a subsequence (xkp)that converges to L.

Let now (xϕ(n)) be a subsequence that converges to `. We have

xϕ(k)) | k ≥ n ⊂ xk | k ≥ n.

Therefore, supk≥n xϕ(k) ≤ supk≥n xk. Letting n → ∞, we get lim sup xϕ(n) ≤ lim supxn, thatis, lim xϕ(n) ≤ lim supxn.

Chapter 2

Axiomatic measure theory

2.1 Measurable spaces

Definition 2.1 Let X be a set. A σ−algebra on X is a collection A of subsets of X (that is,A ⊂ P(X)) satisfying the following properties.

(i) X ∈ A.

(ii) A ∈ A ⇒ X\A ∈ A. We say that A is closed or stable under complementation.

(iii) If (An) is a sequence of A, then

∞⋃n=1

An ∈ A. We say that A is closed or stable under

countable unions.

Examples. a) Let X be a set. Then P(X) is indeed a σ−algebra on X. It is the biggestσ−algebra on X.

b) ∅, X is a σ−algebra on X. It is the smallest σ−algebra on X.

c) Let A ⊂ X. Then ∅, A, Ac, X is a σ−algebra on X.

d) The collection A of all subsets of IR which are either countable or have a countable comple-ment is a σ−algebra on IR. Indeed, (i) X ∈ A because Xc = ∅ is countable. (ii) If E is countableor Ec is countable, then Ec is countable or E = (Ec)c is countable (this condition is symmetricin E and Ec). (iii) Let An be a sequence of subsets of IR that are either countable or have acountable complement. If all An are countable then

⋃∞n=1 An is also countable. If not, some

Am is uncountable. But then Acm is therefore countable. Now (

⋃∞n=1 An)c =

⋂∞n=1 Ac

n ⊂ Acm is

countable.

e) The collection of all subsets of IN which are either finite or have a finite complement is not aσ−algebra on IN. Indeed, take An = 2n then

⋃∞n=1 An is the set of all even positive integers:

it is infinite and its complement is infinite.

Definition 2.2 A measurable space is a couple (X,A) where X is a set and A is a σ−algebraon X. The elements of A are called the measurable subsets of X.

Proposition 2.1 Let A be a σ− algebra on X. Then

1. ∅ ∈ A.

2. An ∈ A for n = 1, 2, . . . ⇒∞⋂

n=1

An ∈ A. We say that A is closed or stable under countable

intersections.

15

16 CHAPTER 2. AXIOMATIC MEASURE THEORY

3. An ∈ A for n = 1, 2, . . . , N ⇒N⋃

n=1

An ∈ A. We say that A is stable under finite unions.

4. An ∈ A for n = 1, 2, . . . , N ⇒N⋂

n=1

An ∈ A. We say that A is stable under finite intersec-

tions.

Proof. 1. X ∈ A by the first property of a σ−algebra. Therefore, ∅ = X\X ∈ A by thesecond property.

2. Let An ∈ A for n = 1, 2, . . .. Then X\An ∈ A for all n by the second property. There-fore

⋃∞n=1 X\An ∈ A by the third property. But

⋃∞n=1 X\An = X\

⋂∞n=1 An. Therefore

X\⋂∞

n=1 An ∈ A. Thus⋂∞

n=1 An ∈ A by the second property.

3. Set An = ∅ for n > N . Then An ∈ A for all n ∈ IN∗. By property (iii),⋃∞

n=1 An ∈ A. But⋃Nn=1 An =

⋃∞n=1 An.

4. Set An = X for n > N . Then⋂N

n=1 An =⋂∞

n=1 An ∈ A by 2.

Remark 2.1 Let (X,A) be a measurable space and D ⊂ X. Then AD := D ∩E|E ∈ A is aσ−algebra on D. Thus (D,AD) is a measurable space that we may call a subspace of (X,A).Note that AD ⊃ P(D) ∩ A = F ⊂ D|F ∈ A. If in addition D ∈ A (i.e., D is measurable),then AD = P(D) ∩ A.

Lemma 2.1 An intersection of σ−algebras on a set X is a σ−algebra on X.

Proof. Let (Ai)i∈I be a family of σ−algebras. We need to show that A =⋂

i∈I Ai is aσ−algebra.

(i) X ∈ A since X ∈ Ai for all i ∈ I.

(ii) If A ∈ A, then A ∈ Ai for all i ∈ I and so X\A ∈ Ai for all i since Ai is a σ−algebra.Therefore, X\A ∈ A.

(iii) If An ∈ A for all n = 1, 2 . . ., then An ∈ Ai for all n and all i. Therefore⋃∞

n=1 An ∈ Ai

for all i ∈ I since Ai is σ−algebra. Thus,⋃∞

n=1 An ∈ A.

Thanks to this lemma we now can define the notion of a σ−algebra generated by a familyof sets.

Definition 2.3 Let X be a set and S ⊂ P(X) be a collection of subsets of X. The intersectionof all σ−algebras containing S is called the σ−algebra generated by S. It is the smallest (withrespect to inclusion) σ−algebra that contains S. It is denoted by σ(S).

Examples. a) If (X, T ) is a topological space, then the σ−algebra σ(T ) generated by theopen sets of X is called the Borel σ−algebra of (X, T ). It is also denoted by B(X, T ) or justB(X) if no confusion arises. An element B ∈ B(X) is called a Borel set or a Borel measurableset.

b) Let A be a σ−algebra on a set X and B be a σ−algebra on a set Y . We denote by A ⊗ Bthe σ−algebra generated by the following family

A×B ⊂ X × Y such that A ∈ A and B ∈ B .

It is called the product σ−algebra on X × Y .

Remark 2.2 Let X be a set.

2.2. MEASURE SPACES 17

1. If A is a σ−algebra on X, then σ(A) = A. Indeed, first, A ⊂ σ(A). Next, A is aσ−algebra containing A, so A contains the intersection σ(A) of all σ−algebras containingA.

2. If S ⊂ T ⊂ P(X), then σ(S) ⊂ σ(T ).

Remark 2.3 (Methodology) 1. To prove that A = σ(C), we show that A ⊂ σ(C) andthat C ⊂ A (which implies that σ(C) ⊂ σ(A) = A).

2. To prove that σ(C1) = σ(C2), we show that C1 ⊂ σ(C2) and that C2 ⊂ σ(C1).

The following important results will be proved in the exercises.

Proposition 2.2 The Borel σ−algebra on IR, that we denote by B(IR) or B, is generated byanyone of the following collections: (a, b ∈ IR or a, b ∈ Q)

i. the open intervals.

ii. the open intervals of the form ]a, b[.

iii. the closed intervals of the form [a, b].

iv. the intervals of the form ]a,+∞[.

v. the intervals of the form [a,+∞[.

vi. the intervals of the form ]−∞, b[.

vii. the intervals of the form ]−∞, b].

viii. all the intervals.

Proposition 2.3 The Borel σ−algebra of IR that we denote by B(IR) is generated by anyoneof the following collections where (where a, b ∈ Q or a, b ∈ IR).

i. the intervals of the form ]a,+∞].

ii. the intervals of the form [a,+∞].

iii. the intervals of the form [−∞, b[.

iv. the intervals of the form [−∞, b].

2.2 Measure spaces

Let (X,A) be a measurable space (i.e., a set X equipped with a σ−algebra).

Definition 2.4 A measure on (X,A) is a function µ : A → [0,+∞] such that

(i) µ(∅) = 0.

(ii) If An ⊂ A is sequence of pairwise disjoint elements of A then

µ

( ∞⋃n=1

An

)=

∞∑n=1

µ(An).

Condition (ii) is called σ−additivity of the measure.


Remark 2.4 Condition (i) can be replaced by

(i’) There exists A ∈ A such that µ(A) < ∞.

Indeed, if µ(∅) = 0 then µ(∅) < ∞. Conversely, suppose that µ(A) < ∞ for some A ∈ A.Letting A1 = A and An = ∅ for n ≥ 2, we get µ(A) = µ(A) +

∑∞n=2 µ(∅). Since µ(A) is finite,

subtracting it form the equality, we get∑∞

n=2 µ(∅) = 0. This implies that µ(∅) = 0.

Definition 2.5 A measure space is a triple (X,A, µ) such that A is a σ−algebra on X and µis a measure on (X,A).

Examples. In Examples (1) to (4), X is any set and one can take A = P(X).

(1) The trivial measure. µ(A) = 0 for all A ∈ A.

(2) The infinite measure. µ(A) =

0 if A = ∅+∞ if A 6= ∅.

Proof. (i) It is clear that µ(∅) = 0. (ii) Let now (An) be a sequence of pairwise disjointelements of A. If An = ∅ for all n, then

⋃∞n=1 An = ∅ and so µ(

⋃∞n=1 An) = 0. On the

other hand,∑∞

n=1 µ(An) =∑∞

n=1 0 = 0. If An 6= ∅ for some m,⋃∞

n=1 An 6= ∅ and thereforeµ(⋃∞

n=1 An) = ∞. On the other hand the series is equal to ∞ since∑∞

n=1 µ(An) ≥ µ(Am) = ∞.

(3) The Dirac measure or Dirac mass. Let x ∈ X. We set δx(A) =

1 if x ∈ A

0 if x /∈ A.

Proof. (i) Since x /∈ ∅, it follows that δx(∅) = 0. (ii) Let (An) be a sequence of pairwisedisjoint elements of A. We distinguish between two cases. If x belongs to the union, then itbelongs to exactly one Am since the An are pairwise disjoint. Then δx(∪∞n=1An) = 1, δx(Am) = 1and δx(Ak) = 0 for k 6= m. Therefore

∑∞n=1 δx(An) = 1. Next, if x does not belong to the

union, it does not belong to any of the An. So both terms are zero.

(4) The counting measure. Let µ(A) =

card(A) if A is finite+∞ if not.

Proof. (i) µ(∅) = card (∅) = 0. (ii) Let (An) be a sequence of pairwise disjoint elements ofA. We distinguish between two cases.Case 1. ∪∞n=1An is finite. Then there is some N ∈ IN∗ such that An = ∅ for all n ≥ N .Otherwise, cardAn ≥ 1 for infinitely many n. So ∪∞n=1An would be infinite. Then ∪∞n=1An =∪N

n=1An and so

µ

( ∞⋃n=1

An

)= µ

(N⋃

n=1

An

)= card

(N⋃

n=1

An

)=

N∑n=1

cardAn =N∑

n=1

µ(An) =∞∑

n=1

µ(An).

Case 2. ∪∞n=1An is infinite. Then µ(∪∞n=1An) = +∞. Here we also distinguish betweentwo cases. 2a) Some Ak is infinite. Then µ(Ak) = +∞ and so

∑∞n=1 µ(An) = +∞. 2b) All

An are finite. But then An 6= ∅ for infinitely many n because ∪∞n=1An is infinite. But thenµ(An) = card (An) ≥ 1 for infinitely many n and so

∑∞n=1 µ(An) = +∞.

(5) Restriction of a measure. Let (X,A, µ) be a measure space and let D ∈ A. We definea new measure ν on (X,A) by ν(A) = µ(A ∩D). This is indeed a measure which is called therestriction of µ to D. In fact, ν is also a measure on (D,AD) where AD = D ∩A|A ∈ A.

(6) Positive linear combination of measures. Let (µn) be a sequence of measures on ameasurable space (X,A) and let (αn) be a sequence of [0,+∞]. Then µ :=

∑∞n=1 αnµn is a

measure on (X,A).


Proof. (i) µ(∅) =∑

αnµn(∅) = 0. (ii) Let (Ak) be a sequence of pairwise disjoint elementsof A. Then

µ(∪kAk) =∑

n

αnµn(∪kAk) =∑

n

αn

∑k

µn(Ak) =∑

n

∑k

αnµn(Ak)

=∑

k

∑n

αnµn(Ak) we can interchange the order of summation by Proposition 1.5

=∑

k

µ(Ak).

In particular, if (xn) is a sequence of X, then µ :=

∑n αnδxn is a measure. A measure of

this form is called a discrete measure. In particular, if X = x1, . . . , xN is finite thenµ := 1

N

∑Nn=1 δxn is the familiar probability measure

µ(A) =cardA

cardX.

Definition 2.6 Let (X,A, µ) be a measure space. We say that

1. µ is finite if µ(X) < ∞.

2. µ is a probability measure if µ(X) = 1.

3. µ is σ−finite if X =⋃∞

n=1 An with An ∈ A and µ(An) < ∞ for all n ∈ IN.

Example. The counting measure is σ−finite on IN but not on IR.

Proposition 2.4 (Elementary properties of measures) Let (X,A, µ) be a measure space.

(a) If A1, A2, . . . , Ak ∈ A are pairwise disjoint, then

µ

(k⋃

n=1

An

)=

k∑n=1

µ(An).

(b) If B ⊂ A then µ(A\B) + µ(B) = µ(A). In particular, if µ(B) < +∞ then µ(A\B) =µ(A)− µ(B).

(c) If B ⊂ A then µ(B) ≤ µ(A).

(d) µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B). In particular, if µ(A ∩ B) < +∞, then µ(A ∪ B) =µ(A) + µ(B)− µ(A ∩B).

(e) If Ann∈IN∗ ⊂ A then

µ

( ∞⋃n=1

An

)≤

∞∑n=1

µ(An).

This is called the σ−subadditivity (note that the An are not necessarily pairwise disjoint).

Proof. (a). Set An = ∅ for n > k. Then the sequence Ann∈IN∗ is a family of pairwisedisjoint sets. Now, on the one hand, ∪k

n=1An = ∪∞n=1An. On the other hand, µ(An) = 0 forn > k since µ(∅) = 0. Therefore

∞∑n=1

µ(An) =k∑

n=1

µ(An) +∞∑

n=k+1

µ(An) =k∑

n=1

µ(An).


By the σ−additivity of µ, we have

µ

( ∞⋃n=1

An

)=

∞∑n=1

µ(An).

Therefore,

µ

(k⋃

n=1

An

)=

k∑n=1

µ(An).

(b) It follows from (a) that µ(C ∪D) = µ(C) + µ(D) if C and D are disjoint. Now note thatA = (A\B) ∪B and A\B and B are disjoint. Therefore µ(A) = µ(A\B) + µ(B).

(c) follows from (b) and the fact that a measure is non-negative.

(d) Observe first that A ∪ B = A ∪ B\(A ∩ B) and the two sets are disjoint. It follows frompart (a) that µ(A∪B) = µ(A) + µ(B\(A∩B)). But µ(B\(A∩B)) + µ(A∩B) = µ(B) by part(b). Hence the conclusion follows.

(e) Define a sequence Bn in the following way: B1 = A1, B2 = A2\A1, B3 = A3\(A1 ∪ A2)and more generally, Bn = An\(A1 ∪ · · · ∪ An−1). Then the sequence Bn has the followingproperties (see the exercises).

1. The Bn are pairwise disjoint.

2.⋃∞

n=1 An =⋃∞

n=1 Bn.

3. Bn ⊂ An and therefore µ(Bn) ≤ µ(An).

It follows from the above and the σ−additivity of µ, that

µ

( ∞⋃n=1

An

)= µ

( ∞⋃n=1

Bn

)=

∞∑n=1

µ(Bn) ≤∞∑

n=1

µ(An).

Theorem 2.1 (Continuity of measures) Let (X,A, µ) be a measure space and let Ann∈IN∗ ⊂A. Then the following hold.

(i) If An is nondecreasing (i.e, A1 ⊂ A2 · · · ) then

µ

( ∞⋃n=1

An

)= lim

n→∞µ(An).

(ii) If the sequence is nonincreasing (i.e.,A1 ⊃ A2 · · · ) and µ(A1) < ∞, then

µ

( ∞⋂n=1

An

)= lim

n→∞µ(An).

Proof. (i). Define a sequence Bn in the following way: B1 = A1, B2 = A2\A1, B3 = A3\A2

and more generally, Bn = An\An−1. Then the sequence Bn has the following properties

1. The Bn are pairwise disjoint.

2. An = B1 ∪B2 ∪ · · · ∪Bn and therefore µ(An) =∑n

k=1 µ(Bk).


3.⋃∞

n=1 An =⋃∞

n=1 Bn.

It follows from the above and the σ−additivity of µ, that

limn→∞

µ(An) = limn→∞

n∑k=1

µ(Bk) =∞∑

k=1

µ(Bk) = µ

( ∞⋃k=1

Bk

)= µ

( ∞⋃k=1

Ak

).

(ii). Let A = ∩∞n=1An and Cn = A1\An. Then ∪∞n=1Cn = A1\A and therefore

µ

( ∞⋃n=1

Cn

)= µ(A1\A) = µ(A1)− µ(A)

Note that Cn is increasing and therefore by (i) we know that µ(∪∞n=1Cn) = limn→∞ µ(Cn).From the other hand, µ(Cn) = µ(A1\An) = µ(A1)− µ(An) and so

µ(A1)− µ(A) = µ(∪∞n=1Cn) = limn→∞

µ(Cn) = limn→∞

(µ(A1)− µ(An)) = µ(A1)− limn→∞

µ(An).

Whence µ(A) = limn→∞ µ(An).

Remark 2.5 The assumption µ(A1) < ∞ is essential. For example, if X = IN and µ is thecounting measure, then the sequence

Ak = n ∈ IN∗ |n ≥ k

is decreasing but µ(Ak) = ∞ and therefore limk→∞ µ(Ak) = ∞, whereas µ(∞⋂

k=1

Ak) = µ(∅) = 0.

Definition 2.7 Let (X,A, µ) be measure space. A subset A ∈ A is said to be of full measureif µ(Ac) = 0.

Remark 2.6 If µ is finite, then A is of full measure if and only if µ(A) = µ(X). However, ifµ is not a finite measure, then a subset A satisfying µ(A) = µ(X) need not be of full measure.Can you give an example?

Definition 2.8 Let (X,A, µ) be measure space. A subset A ⊂ X is called µ−negligible orsimply negligible, if there exists B ∈ A such that A ⊂ B and µ(B) = 0.

In this definition, the point is that a negligible set need not be measurable. Here is an artificialexample. Let X be a set, A be a proper subset of X and a ∈ A. Let A = ∅, A, Ac, X andconsider the measure space (X,A, δa) where δa is the Dirac measure. Then any proper subsetof Ac is negligible but not measurable.

If every negligible set is measurable (and therefore of measure zero), then the space (X,A, µ)is called complete. Thus, in a complete measure space, a set is negligible if and only if it hasmeasure zero.

Exercise. A subset of a negligible set is negligible and a countable union of negligible sets isnegligible.

Definition 2.9 Let P be a predicate on a measure space (X,A, µ), that is, for each x ∈ X,there is a proposition P (x) which is either true or false depending on x. We say that P holdsµ−almost everywhere (µ−a.e) or just almost everywhere if the set x ∈ X |P (x) is false is µ−negligible.


For example two functions f, g : X → IR are equal almost everywhere if the set x ∈ X | f(x) 6=g(x) is negligible. We shall give more examples in the next chapter. A function from(X,A, µ) → IR is called µ−negligible if it is equal to 0 almost everywhere.

Remark 2.7 A predicate P holds µ−a.e. if and only if there is A ∈ A such that µ(Ac) = 0and P (x) is true ∀x ∈ A. Otherwise stated, P holds µ−a.e. if and only if it holds on a set offull measure. Indeed, suppose first that P holds µ−a.e. This means that there exists B ∈ Asuch that x ∈ X |P (x) is false ⊂ B and µ(B) = 0. Let A = Bc. Then A ∈ A and A =Bc ⊂ x ∈ X |P (x) is true. This means that P (x) is true ∀x ∈ A with µ(Ac) = 0. Conversely,suppose that P (x) is true ∀x ∈ A. This means that A ⊂ x ∈ X |P (x) is true. Thereforex ∈ X |P (x) is false ⊂ Ac with µ(Ac) = 0. This means precisely that x ∈ X |P (x) is falseis negligible, that is, P holds µ−a.e.

Exercise. Let P1 and P2 be two predicates on a measure space. If both P1 and P2 holdalmost everywhere then the predicate P1∧P2 (conjunction) also holds almost everywhere. Moregenerally, if (Pn) is a sequence of predicates that hold almost everywhere then the predicate∧∞n=1Pn also holds almost everywhere. If P ⇒ Q and P holds a.e, then Q holds a.e.

Remark 2.8 Many mathematicians say that a predicate P holds almost everywhere if x ∈X |P (x) is false has measure 0. Our definition is therefore more general. However, due to theprevious remark, this makes little difference in practise.

2.3 Measurable functions

Definition 2.10 Let (X,A) and (Y,B) be two measurable spaces. A function f : X → Y iscalled (A,B)−measurable if f−1(B) ⊂ A, that is if for all B ∈ B we have f−1(B) ∈ A.

Examples. a) Every function f : (X,P(X)) → (Y,B) is measurable.b) A constant function is measurable.

The following lemma gives a useful criterion for the measurability of a function.

Lemma 2.2 Let (X,A) and (Y,B) be two measurable spaces. Suppose that B is generated bya family C ⊂ P(Y ). Then f : X → Y is (A,B)−measurable if and only if f−1(C) ⊂ A.

Proof. Suppose first that f is measurable. Then f−1(C) ⊂ f−1(B) ⊂ A. Conversely, supposethat f−1(C) ⊂ A and consider the family of subsets of Y

F := E ⊂ Y |f−1(E) ∈ A.

Then one can check that F is σ−algebra on Y . But this σ−algebra contains C and so it containsB since B is the smallest σ−algebra containing C. Now B ⊂ F means precisely that f−1(B) ∈ Afor all B ∈ B, that is, f is (A,B)−measurable.

Definition 2.11 Let X and Y be two topological spaces. A function f : X → Y is calledBorel-measurable if it is (B(X),B(Y ))−measurable, that is, if the inverse image under f ofevery Borel subset of Y is a Borel subset of X.

Examples. 1) Every continuous map is Borel-measurable.2) Every monotonic map f : IR → IR is Borel-measurable. Indeed, we claim first that the inverseimage under f of an interval of IR is an interval of IR and hence a Borel subset of IR. Let Ibe an interval of IR and let x, y ∈ f−1(I) with x ≤ y. We need to show that [x, y] ⊂ f−1(I).Let z ∈ [x, y]. If f is increasing then f(x) ≤ f(z) ≤ f(y) and so f(z) ∈ [f(x), f(y)] ⊂ I, thatis, z ∈ f−1(I). If f is decreasing, a similar reasoning yields z ∈ f−1(I) . The claim is proved.Now the Borel σ−algebra of IR is generated by the intervals of IR.

2.3. MEASURABLE FUNCTIONS 23

Remark 2.9 Let (X,A) and (Y,B) be a two measurable space and A ∈ A. We alreadyobserved that (A,P(A) ∩ A) is a measurable space. Therefore it makes sense to say that afunction f : A → Y is measurable. This means that f−1(B) ∈ ∩A, whenever B ∈ B, becausef−1(B) ⊂ A.

Lemma 2.3 (The pasting lemma) Let (X,A) and (Y,B) be a two measurable spaces, andlet A,B ∈ A. Let f : A → Y and g : B → Y be measurable functions that coincide on A ∩ B(this condition is satisfied if A ∩B = ∅). Then the function h : A ∪B → Y defined by

h(x) =

f(x) if x ∈ A

g(x) if x ∈ B

is measurable.

Proof. This follows from the fact that h−1(E) = f−1(E) ∪ g−1(E). Thus if E ∈ B, thenf−1(E) and g−1(E) belong to A (and they are contained in A ∪B).

Corollary 2.1 Let (X,A) a measurable space, A ∈ A, and f : A → IR be measurable. Definef : X → IR by

f(x) =

f(x) if x ∈ A

0 if x /∈ A.

Then f is measurable.

Proof. A constant function is measurable.

This corollary means that we can always extend measurable functions to measurable func-tions defined on the whole space X.

The next corollary shows how one can modify a measurable function and still get a measur-able function.

Corollary 2.2 Let f : X → IR be measurable and let E ⊂ X be a measurable set. Then thefunction h : X → IR defined by

h(x) =

f(x) if x ∈ X\E0 if x ∈ E.

is measurable.

Proof. Note that the restriction of a measurable function to a measurable subset is measurable.

Proposition 2.5 Let f : X → Y be (A,B)−measurable and let µ be a measure on (X,A).Then the formula

ν(B) := µ(f−1(B))

for all B ∈ B defines a measure on (Y,B). This measure is denoted by f∗(µ) and is called theimage measure of µ by f . It is also called the pushforward measure of µ by f .

Proof. (i). ν(∅) = µ(f−1(∅)) = µ(∅) = 0.(ii). Let Bn ⊂ B be sequence of pairwise disjoint sets, then f−1(Bn) ⊂ A is also a

sequence of pairwise disjoint sets and therefore

ν

( ∞⋃n=1

Bn

)= µ

(f−1

( ∞⋃n=1

Bn

))= µ

( ∞⋃n=1

f−1(Bn)

)=

∞∑n=1

µ(f−1(Bn)) =∞∑

n=1

ν(Bn).


Example 2.1 Let X and Y be two arbitrary sets and let A = P(X) and B ⊂ P(Y ) be anarbitrary σ−algebra on Y . Let µ be the counting measure on X. Then every function f : X → Yis (A,B)−measurable and ν = f∗µ is the measure that counts the number of preimages

ν(B) = µ(f−1(B)) =

cardx ∈ X|f(x) ∈ B if this set is finite+∞ if not.

2.3.1 Measurable real valued functions

In the following two subsections we consider functions that take values in IR. It is assumed thatIR is equipped with its Borel σ−algebra B(IR). So given a measurable space (X,A), a functionf : X → IR is called measurable if it is (A,B(IR))-measurable. To simplify, we will also say inthis case that f is A−measurable.

Remark 2.10 Let f : X → IR. To simplify the notation, we write f < a instead off−1([−∞, a[) = x | f(x) < a. This notation is used in probability theory. Most often inprobability theory, a probability space is denoted by (Ω,F , P ) and symbols like X, Y, Z,W areused to denote random variables, i.e, measurable functions on Ω. Then P (X < a) denotes themeasure (probability) of the event X < a.

Lemma 2.4 Let (X,A) be a measurable space and f : X → IR. Then the following conditionsare equivalent.

1. f is measurable.

2. f < a is measurable for every a ∈ IR.

3. f ≤ a is measurable for every a ∈ IR.

4. f < a is measurable for every a ∈ Q.

5. f ≤ a is measurable for every a ∈ Q.

6. f > a is measurable for every a ∈ IR.

7. f ≥ a is measurable for every a ∈ IR.

8. f > a is measurable for every a ∈ Q.

9. f ≥ a is measurable for every a ∈ Q.

Proof. This follows from Lemma 2.2 and the fact that B(IR) is generated by subsets of theform [−∞, a[ etc.

Theorem 2.2 Let (X,A) be a measurable space. Let f, g : X → IR be measurable and letα be a real constant. Then the following functions are defined on measurable subsets and aremeasurable.

f + g, αf, fg, |f |, f

g.

Proof. f +g is not defined when f(x) = +∞ and g(x) = −∞ or vice versa. Otherwise statedf +g is not defined on the set E = f−1(+∞)∩g−1(−∞) ∪ f−1(−∞)∩g−1(+∞). Now observethat +∞ is a closed set in IR. It is therefore a Borel subset of IR. It follows that f−1(+∞)is measurable because f is measurable. Similarly, the sets g−1(−∞), f−1(−∞) g−1(+∞) aremeasurable. Consequently, E is measurable and so the set X\E on which f + g is defined ismeasurable.


For similar reasons, fg is defined on a measurable set. fg is defined everywhere because of

our convention 0×±∞ = 0.

a) Let h = f + g. Let a ∈ Q. We claim that

h < a =⋃f < p ∩ g < q

where the union is taken over all couples (p, q) ∈ Q2 such that p + q = a. Indeed, if f(x) < pand g(x) < q with p + q = a, then h(x) = f(x) + g(x) < p + q = a. Conversely suppose thath(x) < a. Note that this implies that f(x) < ∞ and g(x) < ∞. Let p be a rational numbersuch that f(x) < p < a− g(x). Let q = a− p ∈ Q. Then g(x) < q. This proves the claim. Butthe claim means that h < a is a countable union of measurable sets.

b) αf is measurable because

αf < a =

f < a

α if α > 0∅ if α = 0 and a ≤ 0X if α = 0 and a > 0f > a

α if α ≥ 0.

c) f2 is measurable because

f2 ≤ a =

∅ if a < 0−√

a ≤ f ≤√

a if a ≥ 0.

d) We can write

f(x)g(x) =

+∞ if f(x) = g(x) = +∞ or f(x) = g(x) = −∞−∞ if (f(x) = +∞ and g(x) = −∞) or (f(x) = −∞ and g(x) = +∞)12

((f(x) + g(x))2 − f(x)2 − g(x)2

)if f(x), g(x) ∈ IR.

Since constant functions are measurable, it follows from a), b), c) and the pasting lemma thatfg is measurable.

e) |f | is measurable because

|f | ≤ a =

∅ if a < 0−a ≤ f ≤ a if a ≥ 0.

f) 1g is measurable because

1g

< a =

g > 1

a ∪ g < 0 if a > 0g < 0 if a = 0 1

a < g < 0 if a < 0.

Theorem 2.3 Let (X,A) be a mesurable space and let (fn) be a sequence of measurablefunctions from X to IR, then the following functions

inf fn, sup fn, lim inf fn, lim sup fn

are measurable. In particular, if (fn) converges pointwise, its limit is measurable.


Proof. a) Let h = sup(fn). Then h is measurable because

h ≤ a =∞⋂

n=1

fn ≤ a.

b) Let g = inf(fn). Then g is measurable because

h ≥ a =∞⋂

n=1

fn ≥ a.

c) Consequently,lim inf fn = sup

n≥1infk≥n

fn

is measurable.

d) Similarly,lim sup fn = inf

n≥1supk≥n

fn

is measurable.

Corollary 2.3 If f and g are two measurable functions from X to IR, then max(f, g) andmin(f, g) are measurable. In particular, the functions f+ := max(f, 0) and f− := −min(f, 0) =max(−f, 0) are measurable.

2.3.2 Simple functions

Definition 2.12 Let (X,A) be a measurable space and let f : X → IR be a measurablefunction. We say that f is a simple function if it takes only a finite number of values.

Note that we include the condition of measurability in our definition of simple functions.

Lemma 2.5 Let f : X → IR be a simple function. Then f can be written in the form

f =n∑

i=1

ai1Ai (2.1)

where Ai are measurable sets that form a partition of X.

Proof. Suppose that f takes the values a1, . . . , an with ai 6= aj for i 6= j. Set Ai = f−1(ai)for i = 1, . . . , n. Then first, each Ai is measurable as the inverse image of a measurable set.Second, the sets Ai are pairwise disjoint, for if x ∈ Ai ∩ Aj , then f(x) = ai and f(x) = aj sothat ai = aj and this is impossible if i 6= j. Third, if x ∈ X, then f(x) takes some value ak sothat x ∈ f−1(ak) = Ak. This means that Aii is a partition of X.

We show now that f has the representation (2.1). Let x ∈ X. Then as we already observed,x belongs to exactly one Ak. Now, on the first hand f(x) = ak. On the other hand, 1Ak

(x) = 1and 1Ai(x) = 0 for i 6= k. Therefore

∑ni=1 ai1Ai(x) = ak. Hence the equality.

Remark. If f : X → IR is a simple function then f could be written in the form (2.1) inseveral ways. For example consider the function f defined by f(x) = −1 if x < 0 and f(x) = 1if x ≥ 0. Then f = −χ]−∞,0[ + χ[0,∞[. But also f = −χ]−∞,−2[ − χ[−2,0[ + χ[0,∞[ + 3χ∅. Let ussay that

∑j∈I bjχBj is an admissible representation of f if Bjj∈J form a partition of E.

Theorem 2.4 (Approximation of nonnegative measurable functions by simple func-tions) Let (X,A) be a measurable space and f : X → [0,∞] be measurable. Then there existsa sequence (hn) of simple functions such that


a) hn(x) < +∞ for all n and all x ∈ X.

b) 0 ≤ h1 ≤ h2 ≤ · · · ≤ f .

c) hn(x) → f(x) as n →∞ for all x ∈ X.

d) If f is bounded then the convergence is uniform.

Proof. For each n ∈ IN∗ divide the interval [0, n] into n2n intervals each of length 2−n andset

En,k =

x | k2−n ≤ f(x) < (k + 1)2−n if 0 ≤ k ≤ n2n − 1x | f(x) ≥ n if k = n2n.

Then for each n, the family En,kk=0,...,n2n is a partition of X. Set

hn(x) =

k2−n if x ∈ En,k for some 0 ≤ k ≤ n2n − 1n if x ∈ En,n2n .

Otherwise stated,

hn =n2n∑k=0

k

2n1En,k

is a simple function.

a) It is clear that hn(x) < ∞ for all n ∈ IN∗ and all x ∈ X.b) It is also clear that 0 ≤ hn ≤ f for all n. Let us prove that hn ≤ hn+1. Let x ∈ X. Thereare 2 cases:

(1) k2−n ≤ f(x) < (k + 1)2−n for some 0 ≤ k ≤ n2n − 1. Then hn(x) = k2−n. Also theinequality of this case is equivalent to 2k2−n−1 ≤ f(x) < (2k + 2)2−n−1. So we distinguishbetween two cases.

(1a) 2k2−n−1 ≤ f(x) < (2k + 1)2−n−1, that is, x ∈ En+1,2k. In this case, hn+1(x) =2k2−n−1 = k2−n = hn(x).(1b) (2k+1)2−n−1 ≤ f(x) < (2k+2)2−n−1, that is, x ∈ En+1,2k+1. In this case, hn+1(x) =(2k + 1)2−n−1 > k2−n = hn(x).

(2) f(x) ≥ n. In this case, hn(x) = n. We distinguish between two cases.

(2a) n ≤ f(x) < n + 1. Then k2−n−1 ≤ f(x) < (k + 1)2−n−1 for some k ≤ (n + 1)2n+1 − 1which means that x ∈ En+1,k. But then (k + 1)2−n−1 > n (because f(x) ≥ n) and sok ≥ n2n+1. Therefore, hn+1(x) = k2−n−1 ≥ n = hn(x).

(2b) f(x) ≥ n + 1. Then hn+1(x) = n + 1 > n = hn(x).

c) Let x ∈ X. There are two cases.

1. f(x) < ∞. Let ε > 0 be given. Choose N such that f(x) < N and 2−N < ε. Forn ≥ N we have k2−n ≤ f(x) < (k + 1)2−n for some k, and so hn(x) = k2−n, therefore|f(x)− hn(x)| < 2−n ≤ 2−N < ε. Since ε was arbitrary, it follows that hn(x) → f(x).

2. f(x) = +∞. Then f(x) > n for all n and so hn(x) = n by construction. Therefore,hn(x) → +∞.

d) Let f(x) ≤ M for all x ∈ X. Let ε > 0 be given. Choose N such that N > M and 2−N < ε.Then |f(x)− hn(x)| < 2−n ≤ 2−N < ε for all x ∈ X and all n ≥ N .

Corollary 2.4 (Approximation of measurable functions by simple functions)Let (X,A) be a measurable space and f : X → IR be measurable. Then there exists a sequenceof simple functions that converges pointwise to f .

Proof. See the exercises.


2.4 Outer measures and Caratheodory’s theorem

Definition 2.13 An outer measure on a set X is a function

µ∗ : P(X) → [0,∞]

such that

(i) µ∗(∅) = 0;

(ii) (monotonicity) A ⊂ B ⇒ µ∗(A) ≤ µ∗(B);

(iii) (σ−subadditivity) For any sequence of subset An ⊂ P(Y ) we have

µ∗

( ∞⋃n=1

An

)≤

∞∑n=1

µ∗(An).

Examples. 1) Let X be a set. Any measure on (X,P(X)) is an outer measure on X.

2) Let X be a set. Define m : P(X) → IR+ by

µ∗(A) =

0 if A = ∅1 if not.

Then µ∗ is an outer measure which is not a measure (unless X is a one point set). Indeed,

1. µ∗(∅) = 0.

2. Let A ⊂ B. If B = ∅, then A = ∅ and µ∗(A) = µ∗(B) = 0. If B 6= ∅, then µ∗(B) = 1 ≥µ∗(A).

3. Let (An) be a sequence of P(X). We need to show that µ∗(∪An) ≤∑

µ∗(An).

If An = ∅ for all n, then both sides are zero. If not, Ak 6= ∅ for some k. In this case,µ∗(∪An) = 1 and

∑µ∗(An) ≥ µ∗(Ak) = 1.

We now show that µ∗ is not a measure if X has at least two points. Let a and b be two distinctpoints of X. Then µ∗(a ∪ b) = 1 whereas µ∗(a) + µ∗(b) = 2.

3) Define µ∗ : P(IR) → IR+ by

µ∗(A) =

0 if A is countable1 if not,

Then µ∗ is an outer measure which is not a measure if X is uncountable. Indeed,

1. µ∗(∅) = 0 since ∅ is countable.

2. Let A ⊂ B. If B is countable, then A is also countable and so µ∗(A) = µ∗(B) = 0. If Bis uncountable, then µ∗(B) = 1 ≥ µ∗(A).

3. Let (An) be a sequence of P(IR). We need to show that µ∗(∪An) ≤∑

µ∗(An).

If all An are countable, then ∪An is also countable and both sides are zero. If not, then∪An is uncountable and so µ∗(∪An) = 1. Whereas

∑µ∗(An) ≥ 1.

2.4. OUTER MEASURES AND CARATHEODORY’S THEOREM 29

To show that µ∗ is not a measure, take two uncountable disjoint sets for instance [0,1] and ]2,3].Then the measure of the union is 1 whereas the sum of the measures is 2.

The theorem of Caratheodory below shows that an outer measure is not very far froma measure. More precisely, an outer measure is a measure when restricted to some suitableσ−algebra.

Lemma 2.6 If µ∗ is an outer measure on a set X, then

µ∗(A1 ∪ · · · ∪Ak) ≤k∑

i=1

µ∗(Ai).

Proof. Complete the finite sequence Ai in an infinite one by setting Aj = ∅ for j > k.

Definition 2.14 Let µ∗ be an outer measure on a set X. A subset E ⊂ X is called µ∗−measurablein the sense of Caratheodory if for every A ⊂ X we have

µ∗(A) = µ∗(A ∩ E) + µ∗(A\E).

Remark 2.11 Since A = (A ∩E) ∪ (A\E), it follows from the preceding lemma that µ∗(A) ≤µ∗(A ∩ E) + µ∗(A\E). Thus, a set E ⊂ X is µ∗−measurable if and only of for every A ⊂ X,

µ∗(A) ≥ µ∗(A ∩ E) + µ∗(A\E).

Theorem 2.5 (Caratheodory) Let µ∗ be an outer measure on a set X. Let Mµ∗ denote theset of µ∗−measurable subsets of X. Then

1. Mµ∗ is a σ−algebra.

2. µ := µ∗|Mµ∗ is a measure on (X,Mµ∗).

3. (X,Mµ∗ , µ) is complete.

This theorem associates to each outer measure on a set X a measure space

(X, µ∗) 7→ (X,Mµ∗ , µ).

Proof. We prove 1. and 2. simultaneously. (i) Let A ⊂ X. Then µ∗(A ∩ ∅) +µ∗(A\∅) = µ∗(∅) + µ∗(A) = µ∗(A). Therefore ∅ ∈ Mµ∗ . (ii) It is clear that the condition ofµ∗−measurability is symmetric in E and Ec. Therefore E ∈Mµ∗ implies that Ec ∈Mµ∗ . (iii)To prove the third property of a σ−algebra, we proceed in several steps.

Step 1. Mµ∗ is closed under finite unions. Let E1 and E2 belong to Mµ∗ . We prove thatE1 ∪ E2 ∈Mµ∗ . Let A ⊂ X. Then

µ∗(A) ≤ µ∗(A ∩ (E1 ∪ E2)

)+ µ∗

(A ∩ (E1 ∪ E2)c

)= µ∗

((A ∩ E1) ∪ (A ∩ E2 ∩ Ec

1))

+ µ∗(A ∩ Ec

1 ∩ Ec2

)≤ µ∗(A ∩ E1) + µ∗

(A ∩ Ec

1 ∩ E2))

+ µ∗(A ∩ Ec1 ∩ Ec

2)= µ∗(A ∩ E1) + µ∗

(A ∩ Ec

1

)since E2 is µ∗−measurable

= µ∗(A) since E1 is µ∗−measurable.

Thusµ∗(A) = µ∗(A ∩ (E1 ∪ E2) + µ∗

(A ∩ (E1 ∪ E2)c

),

which means that E1∪E2 ∈Mµ∗ . Now it is easily seen by induction that, if E1, . . . , En ∈Mµ∗ ,then E1 ∪ · · · ∪ En ∈Mµ∗ .


Step 2. Let E1, . . . , En be pairwise disjoint elements of Mµ∗ . We prove that for anysubset A ⊂ X,

µ∗(A ∩ (∪n

i=1Ei))

=n∑

i=1

µ∗(A ∩ Ei).

The claim is indeed true for n = 1. Assume that it is true for some n. Then

µ∗(A ∩ (

n+1⋃i=1

Ei))

= µ∗(A ∩ (

n+1⋃i=1

Ei) ∩ En+1))

+ µ∗(A ∩ (

n+1⋃i=1

Ei) ∩ Ecn+1)

)because En+1 is µ∗−measurable. Now, (∪n+1

i=1 Ej) ∩ En+1 = ∪n+1i=1 (Ei ∩ En+1) = En+1 because

Ei ∩ En+1 = ∅ for i 6= n by assumption. Next, (∪n+1i=1 Ei) ∩ Ec

n+1 = ∪ni=1(Ei ∩ Ec

n+1) = ∪ni=1Ei

because Ei ⊂ Ecn+1 since Ei ∩ En+1 = ∅ for i ≤ n. Therefore

µ∗(A ∩ (∪n+1

i=1 Ei))

= µ∗(A ∩ En+1) + µ∗(A ∩ (∪n

i=1Ei))

= µ∗(A ∩ En+1) +n∑

i=1

µ∗(A ∩ Ei) by the induction assumption

=n+1∑i=1

µ∗(A ∩ Ei).

Therefore, the claim is proved. Observe that if we take A = X, we get

µ∗(∪n

i=1 Ei

)=

n∑i=1

µ∗(Ei).

Step 3. Mµ∗ is closed under countable disjoint unions. Let E1, E2, . . . , be pairwisedisjoint elements of Mµ∗ . Let E = ∪∞i=1Ei. We prove that E ∈Mµ∗ and that for any A ⊂ X,

µ∗(A ∩ (∪∞i=1Ei)

)=

∞∑i=1

µ∗(A ∩ Ei).

Set Fn = ∪ni=1Ei. By Step 1, each Fn is µ∗−measurable and so by Step 2,

µ∗(A) = µ∗(A ∩ Fn) + µ∗(A ∩ F cn) =

n∑i=1

µ∗(A ∩ Ei) + µ∗(A ∩ F cn).

Now Fn ⊂ E so Ec ⊂ F cn and therefore µ∗(A∩Ec) ≤ µ∗(A∩F c

n) by monotonicity of µ∗. Hence,

µ∗(A) ≥n∑

i=1

µ∗(A ∩ Ei) + µ∗(A ∩ Ec).

As this inequality is true for any n ∈ IN∗, we get

µ∗(A) ≥∞∑i=1

µ∗(A ∩ Ei) + µ∗(A ∩ Ec). (2.2)

By the σ−subadditivity, we get

µ∗(A) ≥ µ∗(A ∩ E) + µ∗(A ∩ Ec).

This proves that E ∈Mµ∗ . Now taking in particular A = E in (2.2), we get

µ∗(E) ≥∞∑i=1

µ∗(Ei).

2.5. COMPLETION OF A MEASURE SPACE 31

Since the reverse inequality holds by σ−subadditivity, we have in fact equality.

Step 4. Mµ∗ is closed under arbitrary countable unions. Let An be a sequence of Mµ∗ .Set E1 = A1 and En = An\(A1 ∪ · · · ∪ An−1). We know that the En are pairwise disjoint and∪∞i=1Ei = ∪∞i=1Ai. By Step 3, ∪∞i=1Ei ∈Mµ∗ . Hence the conclusion.

3. Let E be µ−negligible, that is E ⊂ B with µ(B) = 0. Then µ∗(B) = 0 and so µ∗(E) = 0.Therefore µ∗(A ∩ E) = 0. Therefore, µ∗(A ∩ E) + µ∗(A ∩ Ec) = µ∗(A ∩ Ec) ≤ µ∗(A). Thismeans that E is µ∗−measurable.

2.5 Completion of a measure space

You will appreciate the importance of this section in the next chapter when we will comparethe Borel σ−algebra on IR with the Lebesgue σ−algebra.

Theorem 2.6 Let (X,A, µ) be a measure space. Let A denote the set of subsets of X of theform A ∪N where A ∈ A and N is µ−negligible. Then

(1) A is a σ−algebra containing A.

(2) µ extends in a unique way to a measure µ on (X, A).

(3) The measure space (X, A, µ) is complete.

(4) A subset A ⊂ X is µ−negligible if and only if it is µ−negligible.

Proof. (1) We prove first that A is a σ−algebra. Let N denote the set of µ−negligiblesubsets of X.

(i) X = X ∪∅ with X ∈ A and ∅ ∈ N . Therefore X ∈ A.

(ii) Let A ∪N ∈ A. Then (A ∪N)c = Ac ∩N c. Now by definition, there exists C ∈ A suchthat N ⊂ C and µ(C) = 0. Then Cc ⊂ N c and we can write:

Ac ∩N c = (Ac ∩ Cc)⋃

(Ac ∩N c ∩ C) = B⋃

M.

Observe that, B ∈ A and M ⊂ C so that M ∈ N . This means that (A ∪N)c ∈ A.

(iii) Let (Ai∪Ni) be a sequence of A. Then ∪i(Ai∪Ni) = ∪iAi⋃∪iNi. Observe that ∪iAi ∈ A

since A is a σ−algebra and ∪iNi ∈ N for if Ci is such that Ni ⊂ Ci and µ(Ci) = 0 then∪iNi ⊂ ∪iCi with µ(∪Ci) = 0. This means that ∪i(Ai ∪Ni) ∈ A.

Next, A ⊂ A since for every A ∈ A, we have the decomposition A = A ∪ ∅ where A ∈ A and∅ ∈ N . Note also that N ⊂ A since for every N ∈ N , we have the decomposition N = ∅ ∪Nwhere ∅ ∈ A and N ∈ N .

(2) Now we define a measure µ on (X, A) in the following way: let A ∈ A, then there existA ∈ A and N ∈ N such that A = A ∪N . We set µ(A) := µ(A). We need to ensure that thisdefinition makes sense because the representation A ∪N of an element of A is not necessarilyunique. So let B ∪ M be another representation of A so that A = A ∪ N = B ∪ M . Thereexists M1 ∈ A such that M ⊂ M1 and µ(M1) = 0. Then A ⊂ B ∪ M ⊂ B ∪ M1 and soµ(A) ≤ µ(B ∪M1) = µ(B). By symmetry, µ(B) ≤ µ(A). Hence µ is well defined. It is clearthat µ coincides with µ on A because if A ∈ A, the decomposition A = A ∪ ∅ implies thatµ(A) = µ(A).

• Next, we check that µ is a measure on (X, A).


(i) µ(∅) = µ(∅ ∪∅) = µ(∅) = 0.

(ii) Let (An ∪ Nn) be a sequence of pairwise disjoint elements of A. Observe that the sets(An) are also pairwise disjoint. Then

µ

(⋃n

(An ∪Nn)

)= µ (∪nAn ∪ ∪nNn) = µ(∪nAn) =

∑n

µ(An) =∑

n

µ(An ∪Nn).

• We check that µ is the unique extension of µ to A. So let ν be an extension of µ to A.Let B ∈ A. Then B = A ∪ N where A ∈ A and N ∈ N . Then ν(B) ≤ ν(A) + ν(N).Let N ⊂ N1 where N1 ∈ A and µ(N1) = 0. Then ν(N) ≤ ν(N1) = µ(N1) = 0 (becauseν coincides with µ on A). Therefore ν(B) ≤ ν(A) = µ(A) = µ(B). On the other hand,µ(B) = ν(A) ≤ ν(A ∪N) = ν(B). Hence ν(B) = µ(B).

(3) We check that (X, A, µ) is complete. Let C ⊂ D with D ∈ A and µ(D) = 0. ThenD = D1 ∪ D2 where D1 ∈ A and D2 ∈ N . Therefore, µ(D1) = µ(D) = 0. Now D2 ∈ Nimplies that D2 ⊂ D3 where D3 ∈ A and µ(D3) = 0. Consequently, C ⊂ D1 ∪D3 where nowD1 ∪D3 ∈ A and µ(D1 ∪D3) = 0. This means that C ∈ N ⊂ A.

(4) What we just said shows that a µ−negligible set is µ−negligible. Conversely, let A ⊂ Xbe µ−negligible. Then there is M ∈ A such that A ⊂ M and µ(M) = 0. Then M ∈ A andµ(M) = 0. This means that A is µ−negligible.

Definition 2.15 The space (X, A, µ) constructed above is called the completion of the space(X,A, µ). We shall also say that A is the completion of A.

Remark 2.12 In this context, a property holds µ−almost everywhere if and only if it holdsµ−almost everywhere.

Exercise. a) Show that for each B ∈ A, there exist A ∈ A and N ∈ N such that B = A ∪Nand moreover A ∩ N = ∅. Hint. Let M ∈ A be such that N ⊂ M and µ(M) = 0. WriteA ∪N = (A ∪N) ∩M c

⋃(A ∪N) ∩M .

b) Show that B ∈ A if and only if there exist A1, A2 ∈ A such that A1 ⊂ B ⊂ A2 andµ(A2\A1) = 0.

Proposition 2.6 Let (X,A, µ) be a measure space and let (X, A, µ) be its completion. Letf : X → IR be A−measurable. Then there exists a function g : X → IR which is A−measurableand coincides with f almost everywhere.

Proof. We prove this in two steps.

Step 1. Let f =∑

ai1Ai be an A−measurable simple function. Then for each i, there existsBi ∈ A such that Bi ⊂ Ai and Ai\Bi is negligible. Let g =

∑ai1Bi . Then g is A−measurable.

We claim that x /∈ ∪(Ai\Bi) ⇒ f(x) = g(x). Indeed, let x /∈ ∪(Ai\Bi). Since the Ai forma partition of X, there exists a unique k such that x ∈ Ak and so f(x) = ak. Then x ∈ Bk

because otherwise x ∈ Ak\Bk, contrary to our assumption. Then g(x) = ak (because the Bi arealso pairwise disjoint) and so g(x) = f(x). The claim is proved. It follows f 6= g ⊂ ∪(Ai\Bi)and therefore f 6= g is negligible. This means that f = g almost everywhere.

Step 2. Let f be A− measurable. By the approximation theorem, there exists a sequenceof simple functions fn that are A−measurable and converge pointwise to f . Now by step 1,for each n, there exists a simple function gn which is A−measurable and such that fn = gn

almost everywhere. By Remark 2.7, for each n, there exists An ∈ A such that fn(x) = gn(x)

2.5. COMPLETION OF A MEASURE SPACE 33

for all x ∈ An and µ(Acn) = 0. Let A = ∩An. Then µ(Ac) = 0 and fn(x) = gn(x) for all x ∈ A

and for all n. Now for each n, modify gn by setting gn(x) = 0 for x ∈ Ac. Then each gn isstill A−measurable (Corollary 2.2) and moreover (gn) converges to the function g defined byg(x) = f(x) if x ∈ A and g(x) = 0 if x ∈ Ac. Observe that g is A− measurable as a limit of A−measurable functions. Now g(x) = f(x) for all x ∈ A and µ(Ac) = 0. This means that f = galmost everywhere.


Chapter 3

The Lebesgue measure on IR

3.1 Construction and properties

Here is our plan for this section.

1. Construct an outer measure λ∗ on IR and establish its properties.

2. Use the first part of Caratheodory’s theorem to construct a σ−algebra L on IR andestablish its properties.

3. Use the second part of Caratheodory’s theorem to construct a measure λ on (IR, L) calledthe Lebesgue measure. Establish some important properties λ.

The Lebesgue outer measure

The Lebesgue outer measure on IR is the function λ∗ : P(IR) → [0,∞] defined by

λ∗(A) = inf

∞∑n=1

(bn − an)∣∣∣ A ⊂

∞⋃n=1

]an, bn[

. (3.1)

Some comments are in order. First, λ∗ is well defined. Indeed, let

XA =

∞∑n=1

(bn − an)∣∣∣ A ⊂

∞⋃n=1

]an, bn[

.

Then XA ⊂ [0,∞]. Since A ⊂⋃∞

n=1] − n, n[ and∑∞

n=1(2n) = +∞, we have +∞ ∈ XA.Therefore XA is not empty and so it has an infinimum. It is possible that XA = +∞. Forexample, XIR = +∞.

Second, we can write

λ∗(A) = inf

∞∑i=1

`(In)∣∣∣ A ⊂

∞⋃n=1

In, and In is an open interval

where `(In) is the length of the interval In. Indeed, let

YA =

∞∑n=1

`(In)∣∣∣ A ⊂

∞⋃n=1

In, and In is an open interval

.

Then XA ⊂ YA. Conversely, let s ∈ YA then s =∑∞

n=1 `(In) where A ⊂∞⋃

n=1

In. If all intervals

In are bounded, then s ∈ XA. If some interval In is unbounded then s = +∞ ∈ XA.

35

36 CHAPTER 3. THE LEBESGUE MEASURE ON IR

A sequence of intervals In such that A ⊂ ∪In is called a covering of A and∑∞

n=1 `(In) iscalled the total length of the covering.

Third, the word open can be removed from the definition of λ∗(A). Indeed, let

ZA =

∞∑n=1

`(In)∣∣∣ A ⊂

∞⋃n=1

In, and In is an interval

.

Then YA ⊂ ZA and so inf ZA ≤ inf YA. On the other hand, let ε > 0 be agiven and let (In) be acovering of A by intervals. Observe that for each n, there exists an open interval Jn containingIn such that `(Jn) = `(In) + ε

2n ; for example if In = [an, bn], take Jn =]an − ε2n+1 , bn + ε

2n+1 [.It follows that (Jn) is a covering of A by open intervals and

∑∞n=1 `(Jn) =

∑∞n=1 `(In) + ε.

Conssequently,∑∞

n=1 `(In) ≥ inf YA − ε and so inf ZA ≥ inf YA − ε. Since ε was arbotrary, weconclude that inf ZA ≥ inf YA and hence equality.

Proposition 3.1 λ∗ is an outer measure on IR.

Proof.

(i) Let an = bn = 0 for n = 1, 2, 3, . . .. Then ∪∞n=1]an, bn[= ∅ and∑∞

n=1(bn − an) = 0.Therefore, 0 ∈ X∅ and so 0 ≤ λ∗(∅) = inf X∅ ≤ 0.

(ii) We claim that if A ⊂ B ⊂ IR, then XB ⊂ XA. Indeed, let s ∈ XB, then (by the definitionof XB) there exist two sequences an and bn of IR such that s =

∑∞n=1(bn − an) and

B ⊂ ∪∞n=1]an, bn[. Since A ⊂ B, then A ⊂ ∪∞n=1]an, bn[. Thus, s ∈ XA.

It follows that inf XA ≤ inf XB, that is, λ∗(A) ≤ λ∗(B).

(iii) Let An be a sequence of subsets of IR. We need to show that

λ∗(∪nAn) ≤∑

n

λ∗(An).

If λ∗(An) = +∞ for some n then the inequality is of course satisfied. So we assume thatλ∗(An) < +∞ for all n ∈ IN∗. Let ε > 0 be given. By a fundamental property of theinfinimum of a subset of IR, for each n ∈ IN∗, there exists a sequence Inmm∈IN∗ of openintervals such that

An ⊂∞⋃

m=1

Inm and∞∑

m=1

`(Inm) < λ∗(An) +ε

2n.

The countable family Imnn,m∈IN is a covering of⋃∞

n=1 An and therefore

λ∗(∞⋃

n=1

An) ≤∞∑

n,m=1

`(Inm).

We have therefore,

λ∗(∞⋃

n=1

An) ≤∞∑

n=1

∞∑m=1

`(Inm) <

∞∑n=1

(λ∗(An) +

ε

2n

)=

∞∑n=1

λ∗(An) + ε.

Since ε is arbitrary, we conclude that

λ∗

( ∞⋃n=1

An

)≤

∞∑n=1

λ∗(An).

3.1. CONSTRUCTION AND PROPERTIES 37

Definition 3.1 Let E ⊂ IR and a ∈ IR. We set E + a = x + a|x ∈ E. We say that E + a isa translate of E. For example [0,1]+3=[3,4]. We set aE = ax|x ∈ E. aE is the image of Eunder the homothety of center 0 and ratio a. For example if E = [1, 2] then 3E = [3, 6].

Exercise. Prove the following

1. ∪i(Ei + a) = (∪Ei) + a and ∩i(Ei + a) = (∩Ei) + a.

2. (E + a)c = Ec + a and (aE)c = aEc.

3. A ∩ (E + a) = ((A− a) ∩ E) + a and A ∩ (aE) = a( 1aA ∩ E).

4. If a ≥ 0, then inf(aE) = a inf E and sup(aE) = a supE. What if a < 0?

5. If I is an interval, then `(aI) = |a|`(I).

Here are some fundamental properties of λ∗.

Proposition 3.2 The Lebesgue outer measure λ∗ satisfies the following properties.

1. λ∗(p) = 0 for any p ∈ IR.

2. λ∗(E) = 0 for any countable set E ⊂ IR.

3. For any interval I, λ∗(I) = length(I).

4. λ∗ is translation invariant, that is, λ∗(E + a) = λ∗(E) for every E ⊂ IR and every a ∈ IR.

5. λ∗(aE) = |a|λ∗(E) for any E ⊂ IR and any a ∈ IR.

Proof. 1. Let ε > 0 be given. Let I1 =]p − ε, p + ε[ and In = ∅ for n > 1. Then In isa countable covering of p by open intervals whose total length is 2ε. It follows that 2ε ∈ Xpand so λ∗(p) ≤ 2ε. Since ε was arbitrary, we have λ∗(p) = 0.

2. Let E be countable. We can write E = ∪p∈Ep. Since λ∗ is an outer measure, λ∗(E) ≤∑p∈E λ∗(p) = 0.

3. Let I be an interval (possibly unbounded). Set I1 = I and In = ∅ for n >1. Then In isa countable covering of I by intervals whose total length is `(I). Therefore `(I) ∈ XI and soλ∗(I) ≤ `(I).

Conversely, let In be a countable covering of I by open intervals. We already proved inthe exercises that `(I) ≤

∑∞n=1 `(In). Therefore `(I) is a lower bound for XI and consequently,

`(I) ≤ λ∗(I).

Remark. The proof that `(I) ≤∑∞

n=1 `(In) used the compactness of the interval [a, b]. Itseems that the compactness of [a, b] (in terms of open covering) were discovered by Borel andLebesgue in their proof that λ∗([a, b]) = b− a.

4. Let E ⊂ IR and let a ∈ IR. We need to show that λ∗(E + a) = λ∗(E). The claim is indeedtrue if E is an interval (use point 3. above). Let now E be an arbitrary subset of IR. If In isa countable covering of E by open intervals, then In + a is a countable covering of E + a byopen intervals. Hence

λ∗(E + a) ≤ λ∗(∪∞n=1(In + a)) ≤∞∑

n=1

λ∗(In + a) =∞∑

n=1

`(In + a) =∞∑

n=1

`(In).

Therefore λ∗(E + a) is a lower bound of XE and therefore λ∗(E + a) ≤ λ∗(E). Since a isarbitrary, we also have λ∗(E − a) ≤ λ∗(E) (replace a by −a).


On the other hand, E = (E+a)−a and according to what we said λ∗((E+a)−a) ≤ λ∗(E+a).Therefore λ∗(E) ≤ λ∗(E + a). Hence the equality.

5. If a = 0, the result is trivial. If not,

λ∗(aE) = inf∑

`(In)|aE ⊂ ∪In = inf∑

`(In)|E ⊂ ∪1aIn

= inf∑

|a|`(1aIn)|E ⊂ ∪1

aIn

= |a| inf∑

`(1aIn)|E ⊂ ∪1

aIn = |a|λ∗(E).

The Lebesgue σ−algebra L

Since λ∗ is an outer measure on IR, it follows from the first part of Caratheodory’s theoremthat the set of λ∗−measurable sets is a σ−algebra on IR. We call it the Lebesgue σ−algebraand we denote it by L. Therefore E ∈ L if and only if

λ∗(A) = λ∗(A ∩ E) + λ∗(A\E)

for every subset A ⊂ IR. An element in L is called a Lebesgue-measurable set. It turns outthat L is a big set. It contains the Borel σ−algebra of IR, but it is much bigger as we shall see.Recall that the Borel σ−algebra B(IR) is the smallest σ−algebra containing the open subsets ofIR. It is also the smallest σ−algebra containing the closed subsets of IR. An element of B(IR)is called Borel-measurable. We also denote the Borel σ− algebra on IR by B.

Proposition 3.3 B ⊂ L, that is, every Borel subset of the real line is Lebesgue-measurable.

Proof. Recall that B is generated by the family of intervals of the form ]a,∞[. Therefore it isenough to prove that such intervals belong to L. Let A be an arbitrary subset of IR and let a ∈ IR.Set A1 = A∩]a,∞[ and A2 = A∩] −∞, a]. We need to show that λ∗(A1) + λ∗(A2) ≤ λ∗(A).The inequality is satisfied if λ∗(A) = +∞, therefore we assume that λ∗(A) < +∞.

Let ε > 0 be given. Theen, by a fundamental property of the infinimum, there exists asequence In of open intervals such that A ⊂ ∪∞n=1In and

∑∞n=1 `(In) < λ∗(A) + ε. Set

I ′n = In∩]a,∞[ and I ′′n = In∩]−∞, a]. Then, I ′n and I ′′n are disjoint intervals (possibly empty)such that In = I ′n ∪ I ′′n. Therefore,

`(In) = `(I ′n) + `(I ′′n).

Now, since A1 ⊂ ∪∞n=1I′n, we have

λ∗(A1) ≤∞∑

n=1

`(I ′n).

Similarly,

λ∗(A2) ≤∞∑

n=1

`(I ′′n).

Therefore,

λ∗(A1) + λ∗(A2) ≤∞∑

n=1

(`(I ′n) + `(I ′′n)

)=

∞∑n=1

`(In) < λ∗(A) + ε.

Since ε was arbitrary, we conclude that

λ∗(A1) + λ∗(A2) ≤ λ∗(A).


Corollary 3.1 Countable sets, intervals, open sets and closed sets are Lebesgue measurable.

In practise, all the real sets that we deal with are Lebesgue measurable. Constructing a non-measurable set is not a trivial matter. See below.

Remark 3.1 You could ask if the inclusion B ⊂ L, is strict. It is. There are two ways to seethis. First, we can construct explicitly a Borel set which is not Lebesgue measurable. But thisis not trivial. The known examples use the middle third Cantor set that we introduce in thenext section. The second way is to show that there is no bijection form B to L. This result usesthe theory of cardinals that you probably do not know. In fact it can be shown that B is inbijection with IR whereas L is in bijection with P(IR). To prove this is is also a nontrivial task.

Proposition 3.4 If E ∈ L and a ∈ IR, then E + a ∈ L and aE ∈ L.

Proof. Let E ∈ L. We first show that E + a ∈ L, that is, we show that

λ∗(A) = λ∗(A ∩ (E + a)) + λ∗(A ∩ (E + a)c)

for every A ⊂ IR.Let δ = λ∗(A ∩ (E + a)) + λ∗(A ∩ (E + a)c). Using the reults of a previous exerxise, we get

δ = λ∗(((A− a) ∩ E) + a)

)+ λ∗

(A ∩ (Ec + a)

)= λ∗

(((A− a) ∩ E) + a)

)+ λ∗

(((A− a) ∩ Ec) + a)

)= λ∗

((A− a) ∩ E)

)+ λ∗

((A− a) ∩ Ec)

)by the translation invariance of λ∗

= λ∗(A− a) since E ∈ L= λ∗(A) by the translation invariance of λ∗.

Next, we show that aE ∈ L. Let A ⊂ IR, then

λ∗(A ∩ aE) + λ∗(A ∩ (aE)c) = λ∗[a(1aA ∩ E)] + λ∗[a(

1aA ∩ Ec)]

= |a|λ∗[ 1aA ∩ E] + |a|λ∗[ 1

aA ∩ Ec]

= |a|λ∗(1aA) = λ∗(A).

Recall that f : (X,A) → (Y,B) is called (A,B)−measurable if f−1(B) ⊂ A. Since on IR,

there are two main σ−algebras: the Lebesgue σ−algebra L and the Borel σ−algebra B, wedistinguish between two types of measurable functions. Recall that a function f : IR → IR iscalled Borel-measurable if it is (B,B(IR))−measurable, that is, if the inverse image under f ofevery Borel set is a Borel set. On the other hand we have the following definition.

Definition 3.2 A function f : IR → IR is called Lebesgue-measurable (sometimes just measur-able) if it is (L,B(IR))−measurable, that is, if the inverse image under f of every Borel set isLebesgue measurable.

Remark 3.2 Every Borel-measurable function is Lebesgue-measurable. The converse is nottrue. Why?

In general, if f : (X,A) → (Y,B) is (A,B)−measurable and g : (Y,B) → (Z, C) is(B, C)−measurable, then g f is (A, C)−measurable. In particular let f : IR → IR andg : IR → IR, then


1. If f and g are Borel-measurable then g f is Borel-measurable.

2. If f is Lebesgue-measurable and g is Borel-measurable, then gf is Lebesgue-measurable.

3. Warning. If f is Borel-measurable and g is Lebesgue-measurable, then g f is not nec-essarily Lebesgue-measurable. In particular, the composition of two Lebesgue-measurablefunctions need not be Lebesgue-measurable.

The Lebesgue measure λ

The second part of Caratheodory’s theorem ensures that λ∗ restricted to L is a measure. Wecall it the Lebesgue measure on IR and we denote by λ. Therefore (IR, L, λ) is a measure space.In addition, to the properties satisfied by any measure, the Lebesgue measure satisfies somespecial properties.

Proposition 3.5 The Lebesgue measure λ satisfies the following properties.

1. The measure of a countable set is zero.

2. The measure of an interval is equal to its length.

3. λ is invariant under translations and symmetries.

4. The measure space (IR, L, λ) is complete.

5. λ is σ−finite.

Proof. The first three points follow from the same properties of λ∗ after recalling thatcountable sets and intervals are measurable, that L is invariant under translations and sym-metries. Point 4. follows from Caratheodory’s theorem. Point 5. follows from the fact thatIR = ∪∞n=1[−n, n] and λ([−n, n]) = 2n < ∞.

Here is another important relation between the Lebesgue σ−algebra and the Borel σ−algebraon IR.

Proposition 3.6 The Lebesgue σ−algebra is the completion of the Borel σ−algebra. Thismeans that a A is Lebesgue measurable if and only if there exists a Borel set B and a negligibleset N , such that A = B ∪N .

Proof. See the exercises.

Remark 3.3 It follows from Theorem 2.6 that a set N ⊂ IR is negligible if and only if it iscontained in a Borel set of measure 0.

Corollary 3.2 Let f : IR → IR be Lebesgue measurable. Then there exists a Borel measurablefunction g : IR → IR that coincides with f almost everywhere.

Proof. This follows from the previous proposition and Proposition 2.6.

We end this section with a characterization of the sets of measure zero. The proof isstraightforward if you recall the definition of the measure in terms of covering by open intervals.We can think of a set of measure zero as a ”thin” set.

Proposition 3.7 A set E ⊂ IR has measure zero if and only if for every ε > 0 there exists acountable covering of E by open intervals whose total length is less that ε.

3.2. COUNTEREXAMPLES 41

3.2 Counterexamples

3.2.1 A Lebesgue non measurable set

This example is due to Vitali.

Construction.

1. Define on ]0,1[ the equivalence relation: x ∼ y if x− y is rational. This defines a partitionof ]0,1[ into equivalence classes. The class containing x is [x] = x + q | q ∈ Q∩]− 1, 1[.

2. By the axiom of choice, there exists a set P that contains exactly one element from eachof the equivalence classes. It is clear that P ⊂]0, 1[.

Proposition 3.8 The set P defined above is not Lebesgue-measurable.

Proof.

1. Let rn be a sequence that counts the rational numbers of ]-1,1[. We assume that thereis no redundancy in the sequence, that is rn 6= rm if n 6= m. Set Pn = P + rn. We claimthat

]0, 1[⊂∞⋃

n=1

Pn ⊂]− 1, 2[. (3.2)

Indeed, first note that Pn = P +rn ⊂]0, 1[+]−1, 1[=]−1, 2[. Therefore, ∪∞n=1Pn ⊂]−1, 2[.Next, let y ∈]0, 1[. There is exactly one element x ∈ [y] such that x ∈ P . Then x ∼ yand so q := y − x is rational. Note that q ∈]− 1, 1[ since x, y ∈]0, 1[. Therefore q = rj forsome j ∈ IN∗. Conclusion: y = x + rj ∈ P + rj = Pj .

2. The Pn are pairwise disjoint. Otherwise there are two integers n 6= m such that Pn∩Pm 6=∅. Let t ∈ Pn∩Pm. Then there are two numbers x and z in P such that t = x+rn = z+rm.It follows that x − z = rm − rn is rational and so x ∼ z. But this contradicts theconstruction of P (P contains only one element from each equivalence class).

3. Suppose that P is measurable. Then each Pn is measurable and λ(Pn) = λ(P ) by thetranslation invariance of the Lebesgue measure. It follows from the σ−additivity of λ andthe previous step that

λ(∪∞n=1Pn) =∞∑

n=1

λ(Pn) =∞∑

n=1

λ(P ).

It follows from relation (3.2) and the monotonicity of λ that 1 ≤∑∞

n=1 λ(P ) ≤ 3. But thisis impossible because

∑∞n=1 λ(P ) is either 0 (if λ(P ) = 0) or +∞ (if λ(P ) > 0). Therefore

we have to conclude that P is not measurable.

3.2.2 The middle third Cantor set

We know that every countable set has Lebesgue measure 0. Is the converse true? The followingexample shows that this not the case, that is, there is an uncountable set of measure 0. I believethat every educated mathematician should know about this beautiful set which is the source ofmany examples and counterexamples in Analysis. It is the simplest example of a fractal set.

Construction. Start with the interval [0,1].

Step 1: Remove the open middle third of [0,1] to obtain

K1 =[0,

13

]∪[23, 1]

.


Step 2: Remove the open middle third of each of the intervals of K1, we get

K2 =[0,

19

]∪[29,13

]∪[0,

23

]∪[79, 1]

...

Step n: Remove the open middle third of each of the intervals of Kn−1, to get

Kn =⋃

i∈Jn

[a

(n)i , b

(n)i

].

Note that Kn is a union of 2n pairwise disjoint intervals each of length13n

.

This defines a decreasing sequence (Kn) of closed subsets of [0,1]. We set

K =∞⋂

n=1

Kn.

This set is called the middle third Cantor set. It is clear that K is a compact set. It is nonempty because it is the intersection of a decreasing sequence of nonempty closed sets in thecompact space [0,1]. In fact, it should be clear from the construction that the endpoints of theintervals removed at each step belong to K, so for example 0, 1, 1

3 , 23 , 1

9 , . . . belong to K, that is,the sequences a

(n)i and b

(n)i belong to K.

Proposition 3.9 The middle third Cantor set satisfies the following properties.

1. Its Lebesgue measure is 0.

2. It has an empty interior.

3. It has no isolated points.

4. It is uncountable.

Proof. 1. We already observed that Kn is a disjoint union of 2n intervals of length 13n .

Therefore λ(Kn) =(

23

)n so that λ(Kn) → 0 as n →∞. By the continuity property of measuresand since λ(K1) < ∞ we, get

λ(K) = λ

( ∞⋂n=1

Kn

)= lim

n→∞λ(Kn) = 0.

2. Any set E of Lebesgue measure zero has an empty interior. For otherwise, E would containan nonempty open interval I. But then λ(E) ≥ λ(I) > 0.

3. Let x ∈ K. We need to show that any neighborhood of x meets K at a point y 6= x. Lettherefore ε > 0 be given. Let n be an integer satisfying

(13

)n< ε. Since x ∈ Kn, there is some

i ∈ Jn such that x ∈[a

(n)i , b

(n)i

]. The length of this interval being

(13

)n. We already observed

that a(n)i and b

(n)i belong to K. Now, if x = a

(n)i , let y = b

(n)i . If x = b

(n)i , let y = a

(n)i . Finally,

if a(n)i < x < b

(n)i , let y = a

(n)i . In all cases, y ∈ K, y 6= x and |y − x| ≤

(13

)n< ε. Since ε was

arbitrary, this means that any neighborhood of meets K at a point different from x.

4. A theorem of topology states that a compact Hausdorff space with no isolated points isuncountable.

3.2. COUNTEREXAMPLES 43

Remark 3.4 Let I denote the collection of open intervals of IR and let J denote the collectionof all intervals of IR. Then we have

I ⊂ J ⊂ B ⊂ L ⊂ P(IR).

The first inclusion is clear. The second inclusion follows from Proposition 2.2. The thirdinclusion follows from Proposition 3.3. The last inclusion is clear. In fact all the inclusionsare strict. This is clear for the first two. That the third inclusion is strict was pointed out inRemark 3.1. The last inclusion is strict because of the existence of a Lebesgue non measurableset.


Chapter 4

The Lebesgue integral

4.1 Construction and properties

We shall define the Lebesgue integral successively for

1. nonnegative simple functions,

2. nonnegative measurable functions,

3. a large class of measurable functions called summable functions,

4. summable complex valued functions.

Remark 4.1 The Riemann integral is defined by approximation from the integrals of stepfunctions, whereas the Lebesgue integral is constructed from the integral of simple functions.In the first case, we divide the domain of the function (that is the interval on which it is defined)into small parts. In the second case, we divide the range of the function. This is a fundamentaldifference.

4.1.1 Simple fonctions

Lemma 4.1 Let (X,A, µ) be a measure space and f : X → IR be a nonnegative simplefunction. If

∑ni=1 bi1Bi and

∑mj=1 cj1Cj are two admissible representations of f , then

n∑i=1

biµ(Bi) =m∑

j=1

cjµ(Cj).

Proof. Note that by assumption, Cjj=1,...,m form a partition of E. Therefore, Bi =∪m

j=1Bi ∩ Cj (disjoint union) and so 1Bi =∑m

j=1 1Bi∩Cj . Thus

n∑i=1

bi1Bi =∑i,j

bi1Bi∩Cj .

Similarly,m∑

j=1

cj1Bj =∑j,i

cj1Cj∩Bi .

Therefore ∑i,j

bi1Bi∩Cj =∑j,i

cj1Cj∩Bi

45

46 CHAPTER 4. THE LEBESGUE INTEGRAL

Thus, bi = cj , for all i, j such that Bi ∩ Cj 6= ∅. Let Λ be the set of indices (i, j) for whichBi ∩ Cj 6= ∅. Now

n∑i=1

biµ(Bi) =n∑

i=1

bi

m∑j=1

µ(Bi ∩ Cj) =∑

(i,j)∈Λ

biµ(Bi ∩ Cj).

Similarly,m∑

j=1

cjµ(Cj) =m∑

j=1

cj

n∑i=1

µ(Bi ∩ Cj) =∑

(i,j)∈Λ

cjµ(Bi ∩ Cj).

Hence the equality.

Thanks to this lemma we can now give the following definition.

Definition 4.1 Let (X,A, µ) be a measure space, E ∈ A and f : X → IR be a nonnegativesimple function. The Lebesgue integral of f on E denoted by

∫E f dµ or just

∫E f is∫

Ef dµ =

n∑i=1

aiµ(Ai ∩ E),

where∑n

i=1 ai1Ai is any admissible representation of f .

Note that the integral is a value in [0,∞].

Example 4.1 Let (X,A, µ) = (IR, L, λ). Consider f : IR → IR defined by

f(x) =

1 if x ∈ [0, 1] ∪ [3, 4]3 if x ∈]1, 3[0 elsewhere.

Then f = 1× 1[3,4] + 3× 1]1,3[ + 0× 1IR\[0,4]. Then,∫IR

f dλ = 1× λ([0, 1] ∪ [3, 4]) + 3× λ(]1, 3[) + 0× λ(IR\[0, 4]) = 8,

where we have used the convention 0×∞ = 0.

Example 4.2 Let A ⊂ IR be a Lebesgue measurable set. Then∫IR 1A dλ = λ(A). More

generally, if (X,A, µ) is a measure space and A ∈ A, then,∫X

1A dµ = µ(A).

Example 4.3 Consider the measure space (X,P(X), δa) where δa is the Dirac measure ata ∈ X. Let f : X → [0,∞] be a simple function. Then∫

Xf dδa = f(a).

Indeed, let f =∑

αi1Ai be a simple nonnegative function (this implies that Aii is apartition of X). Then,

∫X f dδa =

∑αiδa(Ai) = αk where k is the index of the unique set Ak

to which a belongs. But if a ∈ Ak then f(a) =∑

αiδa(Ai) = αk. Hence the equality.


Example 4.4 Consider the measure space (IN,P(IN), µ) where µ is the counting measure. Letf : IN → [0,∞] be a simple function. Then∫

INf dµ =

∑n∈IN

f(n).

Indeed, note first that for E ⊂ IN, µ(E) =∑

n∈E 1 =∑

n∈IN 1E(n). Next, let f =∑mi=1 αi1Ai be a simple nonnegative function. Then,

∫IN

f dµ =m∑

i=1

αiµ(Ai) =m∑

i=1

αi

∑n∈IN

1Ai(n) =m∑

i=1

∑n∈IN

αi1Ai(n) =∑n∈IN

m∑i=1

αi1Ai(n) =∑n∈IN

f(n).

Lemma 4.2 Let (X,A, µ) be a measure space, E,F ∈ A and let f : X → [0,∞] be simple.Then the following hold.

(i)∫E 0 dµ = 0.

(ii) E ⊂ F ⇒∫E f dµ ≤

∫F f dµ.

(iii) µ(E) = 0 ⇒∫E f dµ = 0.

Proof. (i) is trivial. (ii) Let f =∑n

i=1 ai1Ai . Then∫E f dµ =

∑ni=1 aiµ(Ai ∩ E) and∫

F f dµ =∑n

i=1 aiµ(Ai ∩ F ). Since µ(Ai ∩ E) ≤ µ(Ai ∩ F ) and ai ≥ 0, the result follows. (iii)follows from the definition and the fact that µ(Ai ∩ E) = 0.

Lemma 4.3 Let f be a simple nonnegative function defined on a measure space (X,A, µ).Then A 7→

∫A f dµ is a measure on (X,A).

Proof. Set µ(A) =∫A f dµ with f =

∑ni=1 ai1Ai .(i) It should be clear from the definition

and the convention 0 ×∞ = 0 that µ(∅) = 0. (ii) Let (En) be a sequence of pairwise disjointmeasurable sets. Then∫

∪∞n=1En

f dµ =n∑

i=1

aiµ(∪∞n=1 (Ai ∩ En)

)=

n∑i=1

ai

∞∑n=1

µ(Ai ∩ En) =∞∑

n=1

n∑i=1

aiµ(Ai ∩ En)

=∞∑

n=1

∫En

f dµ.

Proposition 4.1 The Lebesgue integral satisfies the following properties (f and g are twononnegative simple functions).

(i) Additivity:

∫E(f + g) dµ =

∫E

f dµ +∫

Ef dµ.

(ii) Positive homogeneity:

∫E(αf) dµ = α

∫E

f dµ for any constant α ≥ 0.

(iii) Monotonicity: f ≤ g on E ⇒∫

Ef dµ ≤

∫E

g dµ.


Proof. Let f =∑

i∈I ai1Ai and g =∑

j∈J bj1Bj be admissible representations of f and g.(i) Note that (Ai ∩Bj)(i,j)∈I×J form a finite partition of E. It is also easy to see that

f + g =∑

(i,j)∈I×J

(ai + bj)1Ai∩Bj

is an admissible representation of f + g. Then∫E(f + g) dµ =

∑(i,j)∈I×J

(ai + bj)µ(Ai ∩Bj) =∑

(i,j)∈I×J

aiµ(Ai ∩Bj) +∑

(i,j)∈I×J

bjµ(Ai ∩Bj)

=∑i∈I

ai

∑j∈J

µ(Ai ∩Bj) +∑j∈J

bj

∑i∈I

µ(Ai ∩Bj)

=∑i∈I

aiµ(Ai ∩ ∪j∈JBj) +∑j∈J

bjµ(Bj ∩ ∪i∈IAi)

=∑i∈I

aiµ(Ai) +∑j∈J

bjµ(Bj)

=∫

Ef dµ +

∫E

g dµ.

(ii) For every α ≥ 0, αf =∑

i∈I αaiχAi , hence∫E(αf) dµ =

∑i∈I

αaiµ(E ∩Ai) = α∑i∈I

aiµ(E ∩Ai) = α

∫E

f dµ.

(iii) Write g = (g − f) + f where g − f is nonnegative and simple. It follows from (i) that∫E

g dµ =∫

E(g − f) dµ +

∫E

f dµ ≥∫

Ef dµ

since the integral of a nonnegative function is nonnegative.

4.1.2 Nonnegative measurable functions

Let (X,A, µ) be a measure space. The set nonnegative measurable functions defined on X isdenoted by M+(X,A, µ) or just M+(X) i.e.,

M+(X) = f : X → [0,∞] | f is measurable.

Definition 4.2 Let E ⊂ X be a measurable set and f ∈ M+(X). The Lebesgue integral of fon E (with respect to the measure µ) is∫

Ef dµ = sup

∫E

h dµ |h : X → [0,∞] is simple and 0 ≤ h ≤ f.

We need first to justify that this definition of the integral is an extension of the previousone, i.e., that both definitions coincide when f is simple. Indeed, let f be simple and let usdenote its integral according to the first definition by SE(f). Also let

If = SE(h) |h : X → [0,∞] is simple and 0 ≤ h ≤ f.

Then of course SE(f) ∈ If and therefore SE(f) ≤ sup If =∫E f dµ. On the other hand, for

any simple function h satisfying 0 ≤ h ≤ f , we have SE(h) ≤ SE(f) since h ≤ f . It followsthat

∫E f = sup If ≤ SE(f). Hence the equality.


Proposition 4.2 For every f, g ∈ M+(X), the following hold

(i) f ≤ g ⇒∫E f ≤

∫E g.

(ii) E ⊂ F ⇒∫E f ≤

∫F f .

(iii)∫E(αf) = α

∫E f for every constant α ≥ 0.

Proof. (i) Let f ≤ g. If h is a simple function such that 0 ≤ h ≤ f then 0 ≤ h ≤ g. Thereforesup0≤h≤f

∫E h ≤ sup0≤h≤g

∫E g. Hence the result.

(ii) follows similarly by appealing to Lemma 4.2 (ii).(iii) follows from the same property for nonnegative simple functions:∫

E(αf) = sup

0≤h≤f

∫E(αh) = sup

0≤h≤fα

∫E

h = α sup0≤h≤f

∫E

h = α

∫E

f.

Proposition 4.3 Let f ∈ M+(X,A, µ) and E ∈ A. Then∫E f dµ = 0 if and only if f = 0

µ−almost everywhere on E.

Proof. Suppose first that f = 0 on E\A where A has measure zero. Then every simplefunction h such that 0 ≤ h ≤ f is zero on E\A. Therefore, by Lemma 4.3 and Lemma 4.2 (iii)∫

Eh dµ =

∫E\A

h dµ +∫

Ah dµ = 0,

and so∫E f = sup0≤h≤f

∫E h = 0. Conversely, suppose that

∫E f dµ = 0. Let A = x ∈

E | f(x) > 0 and An = x ∈ E | f(x) > 1n so that A = ∪∞n=1An. The functions hn := 1

nχAn

are simple functions such that 0 ≤ hn ≤ f . Consequently,

1n

µ(An) =∫

Ehn dµ ≤

∫E

f dµ = 0

which implies that µ(An) = 0 for all n. Therefore, µ(A) ≤∑∞

n=1 µ(An) = 0.

Corollary 4.1 Let f, g ∈ M+(X). If f = g µ−a.e, then∫X f dµ =

∫X g dµ.

Proof. Let

h =

max(f, g)−min(f, g) if min(f, g) < ∞0 otherwise.

Then h ∈ M+(X) and max(f, g) = min(f, g) + h. Furthermore, h = 0 on the set f = g.Therefore, h = 0 µ−a.e.and so

∫X h dµ = 0 by he previous proposition. This implies that∫

Xmax(f, g) dµ =

∫X

min(f, g) dµ.

But both∫X f dµ and

∫X g dµ lie between theses equal value by monotonicity of the integral,

and therefore they are equal to each other.

Lemma 4.4 (Markov’s inequality) For every f ∈ M+(X) and every constant a ≥ 0, wehave ∫

Ef dµ ≥ aµ(f ≥ a).


Proof. Let A = x | f(x) ≥ a. Then h := a1A is a simple function such that 0 ≤ h ≤ f .Consequently, ∫

Ef dµ ≥

∫E

h dµ = aµ(A) = aµ(f ≥ a).

Corollary 4.2 Let f ∈ M+(X) satisfy∫X f dµ < ∞. Then µ(f = ∞) = 0.

Proof. Let An = x|f(x) ≥ n. Then f = ∞ = ∩An. By Markov’s inequality,∫X

f dµ ≥ nµ(An).

It follows that µ(An) →0. But (An) is decreasing and µ(A1) < ∞, therefore by the continuityproperty of a measure, µ(f = ∞) = µ(∩An) = lim µ(An) = 0.

Theorem 4.1 (Monotone convergence theorem or Beppo Levi theorem) Let (X,A, µ)be a measure space and let (fn) be a sequence of M+(X) satisfying

1. fn(x) ≤ fn+1(x) for all n and all x ∈ X (that is, fn is nondecreasing).

2. fn(x) → f(x) for every x ∈ X.

Then,

limn→∞

∫X

fn dµ =∫

Xf dµ.

Otherwise stated limn→∞

∫X

fn dµ =∫

Elim

n→∞fn dµ, that is, we can interchange lim and

∫.

Proof. Observe first that fn(x) ≤ f(x) for all x ∈ E and n ∈ IN∗. It follows from Proposition4.2 (i) that

∫X fn dµ ≤

∫X f dµ for all n and therefore

limn→∞

∫X

fn dµ ≤∫

Xf dµ

(the limit exists in [0,∞] since the sequence ∫X fn dµ is nondecreasing). So it remains to

prove the reverse inequality.Let α ∈]0, 1[ and let h be a simple function such that 0 ≤ h ≤ f . Set

An = x ∈ X | fn(x) ≥ αh(x).

Then An is an increasing sequence of sets and we claim that X = ∪∞n=1An. Indeed, let x ∈ X.If h(x) = 0, then fn(x) ≥ αh(x) = 0 and so x ∈ An for all n. If not, choose ε < (1 − α)h(x).Since fn(x) → f(x), there exists m such that f(x) − fn(x) < ε for all n ≥ m. In particular,f(x)− fm(x) < (1− α)h(x) and so f(x)− fm(x) < f(x)− αh(x) since h(x) ≤ f(x). Thereforefm(x) ≥ αh(x) and so x ∈ Am.

Now we already know that A 7→∫A h is a measure ν, and as any measure it satisfies

ν(∪∞n=1An) = limn→∞ ν(An), that is∫X

hdµ = limn→∞

∫An

h dµ.

Note also that∫An

αh dµ ≤∫An

fn dµ ≤∫X fn dµ. Hence

α limn→∞

∫An

h dµ ≤ limn→∞

∫X

fn dµ


and thusα

∫X


∫E

h dµ.

Letting α → 1, we get ∫X


∫X

fn dµ.

Since this inequality holds for any simple function h such that 0 ≤ h ≤ f , by taking thesupremum over such h, we finally get∫

Xf dµ ≤ lim

n→∞

∫X

f dµ.

Example 4.5 Consider the measure space (X,P(X), δa) where δa is the Dirac measure ata ∈ X. Let f : X → [0,∞] be a nonnegative measurable function. Then∫

Xf dδa = f(a).

The result is true for simple functions. Let now f ∈ M+(X). Then there exists an increasingsequence (hn) of simple nonnegative functions that converge to f . By Beppo Levi’s theorem,∫

Xf dδa = lim

∫X

hn dδa = lim hn(a) = f(a).

Example 4.6 Consider the measure space (IN,P(IN), µ) where µ is the counting measure. Letf : IN → [0,∞] be a nonnegative measurable function. Then∫

Xf dµ =

∑n∈IN

f(n).

The result is true for simple nonnegative functions. Set

sk(n) =

f(n) if n ≤ k

0 otherwise.

Then, (sk)k is an increasing sequence of simple functions that converges to f . By Beppo Levi’stheorem ∫

INf dµ = lim

k→∞

∫IN

sk dµ = limk→∞

∑n∈IN

sk(n) = limk→∞

k∑n=0

f(n) =∞∑

n=0

f(n).

Proposition 4.4 Let f, g ∈ M+(X) and let E ⊂ X be measurable. Then∫E(f + g) =

∫E

f +∫

Eg.

Proof. By Theorem 2.4, there exist two nondecreasing sequences (fn) and (gn) of simplenonnegative functions that converge respectively to f and g. Therefore (fn + gn) is a nonde-creasing sequence of simple nonnegative functions that converges to (f + g). By the previoustheorem

limn→∞

∫E

fn =∫

Ef, lim

n→∞

∫E

gn →∫

Eg, and lim

n→∞

∫E(fn + gn) =

∫E(f + g)


By the additivity of the integral of simple functions, we have∫E(fn + gn) =

∫E

fn +∫

Egn,

and thus,∫E(f + g) = lim

n→∞

(∫E

fn +∫

Egn

)= lim

n→∞

∫E

fn + limn→∞

∫E

gn =∫

Ef +

∫E

g.

Lemma 4.5 (Additivity of domains) Let f ∈ M+(X). If E1 and E2 are disjoint measurablesets then ∫

E1∪E2

f =∫

E1

f +∫

E2

f.

Proof. The result follows from the additivity of domains for simple functions and theapproximation of functions in M+(X) by simple functions (reason as above).

Corollary 4.3 Let f ∈ M+(X). If E and A are measurable sets then

∫E

f1A =∫

A∩Ef .

Proof. By the previous lemma,∫E

f1A =∫

E∩Af1A +

∫E\A

f1A =∫

E∩Af +

∫E\A

0 =∫

E∩Af.

Proposition 4.5 (Integration term by term) Let (fn) be a sequence of M+(X). Then∫X

( ∞∑i=1

fi

)=

∞∑i=1

∫X

fi.

Proof. Set gn =∑n

i=1 fi. Then gn is nondecreasing sequence of M+(X) that converges to∑∞i=1 fi. By the monotone convergence theorem

limn→∞

∫X

gn =∫

X

∞∑i=1

fi.

On the other hand, it follows by induction from the previous proposition that∫X

gn =∫

X

(n∑

i=1

fi

)=

n∑i=1

∫X

fi.

Letting n →∞ we get the result.

Corollary 4.4 Let f ∈ M+(X,A, µ). Then E 7→∫

Ef dµ is a measure on (X,A).

Proof. (i) It is clear that∫

∅ f dµ = 0. (ii). Let (En) be sequence of pairwise disjointelements of A. We proved in the exercises that 1∪∞n=1En =

∑∞n=1 1En and therefore f1∪∞n=1En =∑∞

n=1 f1En . By Corollary 4.3 and the previous proposition∫∪∞n=1En

f dµ =∫

Xf1∪∞n=1En dµ =

∫X

( ∞∑n=1

f1En

)dµ =

∞∑n=1

∫X

f1En dµ =∞∑i=1

∫En

f dµ.


4.1.3 Summable functions

Definition 4.3 Let (X,A, µ) be a measure space, E ∈ A and f : E → IR be measurable. f iscalled integrable or summable on E if ∫

E|f |dµ < ∞.

We denote by L1(E) the set of summable functions on E. Other notations: L1IR

(E), L1(µ),L1(E,µ). The set of summable functions f : E → IR is denoted by LIR(E). So we haveLIR(E) ⊂ LIR(E). But as we shall see, there is little difference between these sets.

Recall that for a function f we defined f+ = max(f, 0) and f− = −min(f, 0) and if f ismeasurable if and only if both f+ and f− are measurable. Recall also that

f = f+ − f− and |f | = f+ + f−.

Therefore, f is summable if and only if both f+ and f− are summable. It is natural to definethe integral of a summable function by∫

Ef =

∫E

f+ −∫

Ef−.

Example 4.7 Consider the measure space (X,P(X), δa) where δa is the Dirac measure ata ∈ X. Let f : X → IR be a measurable function. Then f is summable if and only if|f(a)| < +∞, that is, if and only if f(a) ∈ IR. In this case∫

Xf dδa =

∫X

f+ dδa −∫

Xf− dδa = f+(a)− f−(a) = f(a).

Example 4.8 Consider the measure space (IN,P(IN), µ) where µ is the counting measure. Letf : IN → IR be a function (it is necessary measurable). Then f is summable if and only ifthe series

∑n∈IN |f(n)| is convergent, that is, if and only if the series

∑n∈IN f(n) is absolutely

convergent. In this case we have∫IN

f dµ =∫

INf+ dµ−

∫X

f− dµ =∑n∈IN

f+(n)−∑n∈IN

f−(n) =∑n∈IN

(f+(n)− f−(n)) =∑n∈IN

f(n).

Proposition 4.6 Let (X,A, µ) be a measure space and E ∈ A. Then the following hold.

1. L1(E) is real vector space.

2. The map f 7→∫

Ef is linear, i.e.,

∫E(αf + βg) = α

∫E

f + β

∫E

g.

Proof. 1. Let f, g ∈ L1(E) and α, β ∈ IR. Then |αf + βg| ≤ |α||f |+ |β||g|. Consequently,∫E|αf + βg| ≤

∫E|α||f |+ |β||g| = |α|

∫E|f |+ |β|

∫E|g| < +∞,

and so αf + βg ∈ L1(E).

2. We check only that∫

(f + g) =∫

f +∫

g. The homogeneity∫E αf = α

∫E f is left as an

exercise (distinguish between the case α ≥ 0 and α < 0).From the identity

f + g = (f + g)+ − (f + g)− = (f+ − f−) + (g+ − g−),


we deduce that (f + g)+ + f− + g− = (f + g)− + f+ + g+, and therefore∫E(f + g)+ + f− + g− =

∫E(f + g)− + f+ + g+

Now all the functions belong to M+(E) and so by the linearity of the integral for functions inM+(E) we deduce that∫

E(f + g)+ +

∫E

f− +∫

Eg− =

∫E(f + g)− +

∫E

f+ +∫

Eg+

and so ∫E(f + g)+ −

∫E(f + g)− =

∫E

f+ −∫

Ef− +

∫E

g+ −∫

Eg−

which means that ∫E(f + g) =

∫E

f +∫

Eg.

Proposition 4.7 Let f ∈ L1(E), and let (En) be a sequence of pairwise disjoint measurablesubsets of E. Then ∫

∪En

f =∑∫

En

f

Proof. Using the additivity of domains for nonnegative measurable functions, we have∫∪En

f =∫∪En

f+ −∫∪En

f− =∑∫

En

f+ −∑∫

En

f− =∑

(∫

En

f− −∫

En

f−) =∑∫

En

f.

Proposition 4.8 For every f ∈ L1(E) we have∣∣∣∣∫E

f

∣∣∣∣ ≤ ∫E|f |.

Proof. We have∣∣∣∣∫E

f

∣∣∣∣ = ∣∣∣∣∫E

f+ −∫

Ef−∣∣∣∣ ≤ ∫

Ef+ +

∫E

f− =∫

E(f+ + f−) =

∫E|f |.

Proposition 4.9 Let f, g ∈ L1(X,A, µ). If f = g µ−a.e, then∫X f dµ =

∫X g dµ.

Proof. It is easy to check that if f = g µ−a.e, then f+ = g+ a.e. and f− = g− a.e. ByCorollary 4.1,

∫f+ =

∫g+ and

∫f− =

∫g−. Hence the conclusion follows.

4.1.4 Complex valued functions

Let (X,A, µ) be a measure space and let f : X → C. We say that f is summable or integrableif its components Re f and Im f are summable and in this case we set∫

Xf dµ =

∫X

Re f dµ + i

∫X

Im f dµ.

The space of summable complex valued functions is denoted by L1C(X). Other notations: L1

C(µ),L1(X, C), L1(X, µ, C).

It is not difficult to see that L1C(X) is a vector space on C and that the map f 7→

∫X f dµ

is a linear functional on L1C(X). The additivity on domains also holds.

4.2. THE LEBESGUE DOMINATED CONVERGENCE THEOREM 55

Proposition 4.10 For every f ∈ L1C(E) we have∣∣∣∣∫

Ef

∣∣∣∣ ≤ ∫E|f |.

Proof. If∫

f = 0, the inequality is satisfied. Assume therefore that∫

f 6= 0 and let α = |∫

f |∫f

.Then

|∫

f | = α

∫f =

∫αf =

∫Re(αf) + i

∫Im(αf) =

∫Re (αf) ≤

∫|αf | =

∫|f |

4.2 The Lebesgue dominated convergence theorem

Suppose that we have a sequence of functions fn that converges pointwise to a limit f , do wehave

limn→∞

∫X

fn dµ =∫

Xlim

n→∞fn dµ?

Otherwise stated, can we interchange the∫

and lim? The answer in general is no as showed bythe following examples.

Counter-examples. 1) Consider the sequence defined by fn(x) = nx(1− x2)n for x ∈ [0, 1].This sequence converges to the zero function. Note however that

∫ 10 fn(x) dx = n

2n+2 → 12

whereas∫ 10 lim fn(x) dx = 0.

2) Consider the sequence fn = 1[−n,n]. For each n, fn ∈ L1(IR). This sequence convergespointwise to the constant function 1 which is not even in L1(IR)

There are of course many cases in which the interchange of∫

and lim is possible. Inthe Riemann theory, a sufficient condition is uniform convergence. However this a restrictiverequirement. A better criterion for the interchange of

∫and lim is the monotone convergence

theorem. Now we give another criterion known as the Lebesgue dominated convergence theorem.But first, a lemma.

Lemma 4.6 (Fatou) Let fn be a sequence of M+(X), then∫X

lim inf fn ≤ lim inf∫

Xfn.

Proof. Set gn = infk≥n

fk. Then gn ≤ fn, gn ≤ gn+1 and lim inf fn = lim gn. Therefore∫gn ≤

∫fn and so lim

∫gn ≤ lim inf

∫fn. It follows from this and the monotone convergence

theorem that ∫lim inf fn =

∫lim gn = lim

∫gn ≤ lim inf

∫fn.

Theorem 4.2 (The Lebesgue dominated convergence theorem) Let (X,A, µ) be a mea-sure space and E ∈ A. Let (fn) be a sequence of measurable functions from E to K whereK = IR or C. Suppose that

(i) There exists a measurable function f such that fn → f a.e. on E.

(ii) There exist g ∈ L1(E) such that |fn| ≤ g a.e. on E.


Then f ∈ L1K(E) and

(a) limn→∞

∫E|fn − f |dµ = 0.

(b) limn→∞

∫E

fn dµ =∫

Ef dµ.

Proof. Let A be a set on which the assumptions hold and such that µ(Ac) = 0. Modify f, fn

and g by setting f(x) = fn(x) = g(x) = 0 for x /∈ A. This does not modify the measurabilityand summability properties but (i) and (ii) now hold everywhere.

Now the fn are summable since |fn| ≤ g and g is summable. Also it follows by lettingn → ∞ in (ii) that |f | ≤ g and therefore f is also summable. Next, |f − fn| ≤ |f | + |fn| ≤ 2gand so setting ϕn := 2g−|f−fn| we have that ϕn is summable and nonnegative. Since fn → f ,it follows that lim inf ϕn = lim ϕn = 2g and by Fatou’s lemma∫

E2g =

∫E

lim inf ϕn ≤ lim inf∫

Eϕn =

∫E

2g + lim inf∫

E(−|f − fn|).

It follows that lim inf∫E(−|f − fn|) ≥ 0 and so lim sup

∫E |f − fn| ≤ 0. Therefore

limn→∞

∫|f − fn| = 0.

This proves (a). Part (b) follows from the estimate∣∣∣∣∫E

f −∫

Efn

∣∣∣∣ = ∣∣∣∣∫E(f − fn)

∣∣∣∣ ≤ ∫E|f − fn| → 0.

Corollary 4.5 (Integration term by term) Let (fn) be a sequence of measurable functionsfrom X to IR or C. Suppose that ∑

n≥1

∫X|fn|dµ < +∞.

Then∑

n≥1 fn is µ− integrable and∫X

∑n≥1

fn dµ =∑n≥1

∫X

fn dµ.

Proof. Let g =∑

n≥1 |fn|. Using Proposition 4.5, we get∫X

g dµ =∫

X

∑n≥1

|fn|dµ =∑n≥1

∫X|fn|dµ < ∞.

Therefore g ∈ L1(X). It follows that g is finite µ−a.e. and so the series∑

n≥1 fn is absolutelyconvergent (and hence convergent) a.e.

Let gn =∑n

k=1 fk. Then (gn) converges to∑

k≥1 fk a.e. and |gn| ≤ g a.e. By the dominatedconvergence theorem,

∑k≥1 fk is µ− integrable and

limn→∞

∫X

gn dµ =∫

X

∑k≥1

fk dµ,

4.3. RELATIONS WITH THE RIEMANN INTEGRAL 57

that is

limn→∞

∫X

n∑k=1

fk =∫

X

∑k≥1

fk dµ.

But limn→∞∫X(∑n

k=1 fk) dµ = limn→∞∑n

k=1

∫X fk dµ =

∑k≥1

∫X fk dµ. Therefore,

∑k≥1

∫X

fk dµ =∫

X

∑k≥1

fk dµ.

4.3 Relations with the Riemann integral

Theorem 4.3 Let f : [a, b] → IR be bounded. If f is Riemann integrable then f is Lebesgueintegrable on [a, b] and the integrals coincide, i.e.,∫

[a,b]f dλ =

∫ b

af(x) dx.

Proof. Let (Pn) be an increasing1 sequence of partitions of [a, b] such that ||Pn|| → 0. Forexample one can take

Pn = a +k

2n(b− a) | k = 0, . . . , 2n,

which divides [a, b] into 2n equal subintervals. Then form the Darboux upper and lower sumscorresponding to each Pn

U(f, Pn) =∑i∈In

Mn,i∆xn,i, L(f, Pn) =∑i∈In

mn,i∆xn,i.

We claim that

limn→∞

U(f, Pn) = limn→∞

L(f, Pn) =∫ b

af(x) dx.

Indeed, let ε > 0 be given, then by Theorem A.3, there exists n0 such that U(f, Pn)−L(f, Pn) <

ε for all n ≥ n0. Therefore∫ ba f(x) dx ≤ U(f, Pn) ≤

∫ ba f(x) dx + ε. Similarly,

∫ ba f(x) dx− ε ≤

L(f, Pn) ≤∫ ba f(x) dx.

Now define two sequence of step functions Gn and gn by

Gn(x) = Mn,i if xi−1 ≤ x < xi; gn(x) = mn,i if xi−1 ≤ x < xi; Gn(b) = gn(b) = f(b).

Otherwise stated,

Gn =∑i∈In

Mn,i1[xi−1,xi[ + f(b)1b, gn =∑i∈In

gn,i1[xi−1,xi[ + f(b)1b.

Then clearly, ∫[a,b]

Gn dλ = U(f, Pn) and∫

[a,b]gn dλ = L(f, Pn).

Moreover,G1(x) ≥ G2(x) ≥ · · · ≥ f(x) and g1(x) ≤ g2(x) ≤ · · · ≤ f(x).

HenceG(x) := lim

n→∞Gn(x) ≥ f(x) and g(x) := lim

n→∞gn(x) ≤ f(x).

1By increasing we mean that Pn+1 is a refinement of Pn


Note that G and g and measurable as limits of measurable functions. Also G and g are summableon [a, b] since they are bounded. It follows from the monotone convergence theorem2 that∫

[a,b]G dλ = lim

n→∞

∫[a,b]

Gn dλ = limn→∞

U(f, Pn) =∫ b

af(x) dx,

and ∫[a,b]

g dλ = limn→∞

∫[a,b]

gn dλ = limn→∞

L(f, Pn) =∫ b

af(x) dx.

Therefore, ∫[a,b]

G dλ =∫

[a,b]g dλ

and so G = g almost everywhere. Hence f = G = g a.e. This implies first that f is measurable(because G is measurable as a limit of step functions), and second that∫

[a,b]f dλ =

∫[a,b]

G dλ =∫ b

af(x) dx.

Next we show the relation between absolutely convergent Riemann integrals and Lebesgue

integrals. Recall that the integral∫∞a f(x) dx is called absolutely convergent if limb→∞

∫ ba |f(x)|dx

exists.

Theorem 4.4 Let f : [a,∞[→ IR be Riemann integrable on any compact interval [a, b]. Thenthe following conditions are equivalent.

(i)

∫ ∞

af(x) dx is absolutely convergent.

(ii) f ∈ L1([a,∞[).

If one of these conditions holds, then∫[a,∞[

f dλ =∫ ∞

af(x) dx.

Proof. (i)⇒(ii). Let fn = f1[a,n]. Then (fn) is a sequence of measurable functions thatconverges to f and so f is measurable. Now the sequence (|fn|) is nondecreasing and convergesto |f |. By the monotone convergence theorem,∫

[a,∞[|f |dλ = lim

n→∞

∫[a,∞[

|fn|dλ = limn→∞

∫[a,n]

|f |dλ.

But according to the previous theorem∫[a,n]

|f |dλ =∫ n

a|f(x)|dx.

But this last integral has a finite limit by assumption and therefore∫[a,∞[ |f |dλ < +∞, that is

f is summable on [a,∞[.

2(Gn−G) is a decreasing sequence in M+([a, b]) that converges to 0 and such that∫

(G1−G) <∞. Therefore,∫(Gn − G) → 0, i.e

∫Gn →

∫G. Also, (gn − g1) is an increasing sequence M+([a, b]) that converges to g − g1.

Therefore∫

(gn − g1)→∫

(g − g1) and so∫

gn →∫

g

4.3. RELATIONS WITH THE RIEMANN INTEGRAL 59

Now by the dominated convergence theorem∫[a,∞[ fn →

∫[a,∞[ f . But∫

[a,∞[fn dλ =

∫[a,n]

f dλ =∫ n

af(x) dx

and this last integral converges to∫∞a f(x) dx. Hence the equality.

(ii)⇒(i). It is enough to prove that for any sequence tn ⊂ [a,∞[ converging to +∞, theintegral

∫ tna |f(x)|dx has a finite limit independent of (tn). So let (tn) be such sequence and

set fn = fχ[a,tn]. Then (|fn|) converges to |f | and |fn| ≤ |f | with |f | ∈ L1. By the dominatedconvergence theorem

∫[a,∞[ |fn|dλ =

∫[a,tn] |f |dλ converges to

∫[a,∞[ |f |dλ. But

∫[a,tn] |fn|dλ =∫ tn

a |f(x)|dx by the previous theorem. This means that∫ tna |f(x)|dx →

∫[a,∞[ |f |dλ. It follows

that∫∞a |f(x)|dx is convergent.

Theorem 4.5 Let f : [a, b[→ IR be Riemann integrable on any compact interval [a, c] ⊂ [a, b[.Then the following conditions are equivalent.

(i)

∫ b

af(x) dx is absolutely convergent.

(ii) f ∈ L1([a, b])

If one of these conditions holds, then∫[a,b]

f dλ =∫ b

af(x) dx.

Proof. Exercise. Proceed as above.

Remark 4.2∫∞a f(x) dx may be semi convergent and so f /∈ L1([a,∞[). For example,∫ ∞

0

sinx

xdx =

π

2but

∫ ∞

0

∣∣∣∣sin x

x

∣∣∣∣ dx = +∞.

In this case∫[0,∞[ f dλ does not exist, although

∫∞0 f(x) dx exists.

Theorem 4.6 [Lebesgue’s theorem on Riemann’s integral] Let f : [a, b] → IR be a boundedfunction. Then f is Riemann integrable if and only if it is continuous almost everywhere, i.e.,the set of its discontinuity points is of Lebesgue measure zero.

Proof. Suppose first that f is Riemann integrable. We proceed as in the proof of Theorem4.3: that is, we consider an increasing sequence (Pn) of partitions of [a, b] whose norms tend tozero, and then define two monotone sequences of step functions Gn and gn converging to G andg respectively so that g(x) ≤ f(x) ≤ G(x) and G = g a.e. Now let

E := x ∈ [a, b] | g(x) 6= G(x) ∪⋃n

Pn.

Then E has measure zero since⋃

n Pn is countable. We claim that f is continuous on [a, b]\E.Indeed, let x0 ∈ [a, b]\E and let ε > 0 be given. Then g(x0) = G(x0) and since gn(x0) → g(x0)and Gn(x0) → G(x0) we deduce that Gk(x0) − gk(x0) < ε for all k large enough. Choose andfix such k. Now x0 /∈ Pk and so it must be an interior point of some subinterval of the partitionPk where gk and Gk are constant. Hence there exists δ > 0 such that Gk(x) = Gk(x0) and


gk(x) = gk(x0) for all |x − x0| < δ. From the above and the inequalities g(x) ≤ f(x) ≤ G(x),we conclude that

−ε < g(x0)−G(x0) ≤ f(x)− f(x0) ≤ G(x0)− g(x0) < ε.

This proves that f is continuous at x0.

Suppose now conversely, that f is continuous on some [a, b]\D where λ(D) = 0. Let ε > 0 begiven and M be such that |f(x)| ≤ M . Then |f(x)−f(y)| ≤ 2M for all x, y ∈ [a, b]. Since D hasmeasure zero there exists a countable cover of D by open intervals In such that

∑n `(In) < ε

4M .Now for all x /∈ D there exists (by the continuity of f at x) an open interval Jx containing xand such that |f(z)− f(y)| < ε

2(b−a) for all z, y ∈ Jx ∩ [a, b]. Now Inn ∪ Jxx/∈D is an opencover of [a, b] and so by compactness, there exists a finite subcover Ink

nk=1 ∪ Jxim

i=1. LetP = t0, . . . tN be the partition of [a, b] determined by the endpoints of Ink

and Jxi (which areinside [a, b]). Then each interval ]tj−1, tj [ is contained either in some Ink

or in some Jxi . LetJ = j | ]tj−1, tj [⊂ Ink

for some k and ∆j = tj − tj−1. Then

U(f, P )− L(f, P ) =N∑

j=1

∆j sup|f(y)− f(z)| | y, z ∈ [tj−1, tj ]

≤∑j∈J

∆j2M +∑j /∈J

∆jε

2(b− a)

≤ ε

4M2M + (b− a)

ε

2(b− a)= ε

Hence f is Riemann integrable.

Remark 4.3 The characteristic function f of Q∩ [0, 1] is not Riemann integrable because it isdiscontinuous everywhere (this is because every interval, no matter how small, contains rationaland irrational numbers). However it is Lebesgue integrable because it is Lebesgue measurable(as Q ∩ [0, 1] is Lebesgue measurable) and its integral, by definition, is λ(Q ∩ [0, 1]) = 0.

On the other hand the characteristic function of the middle third Cantor set (restricted to[0,1]) is Riemann integrable.

4.4 Some applications

We give some applications of the power of the convergence theorems.

Example. Compute limn→∞

∫ n

0

(1 +

x

n

)ne−2x dx. The classical theorems of Riemann integra-

tion do not apply since we work on an unbounded domain and there is no uniform convergence.However, we can work with the Lebesgue integral instead. First write

In =∫ n

0

(1 +

x

n

)ne−2x dx =

∫[0,n]

(1 +

x

n

)ne−2x dλ =

∫[0,∞[

(1 +

x

n

)ne−2xχ[0,n] dλ.

Let fn(x) =(1 + x

n

)ne−2xχ[0,n](x). Then fn(x) → e−x and |fn(x)| ≤ e−x. But the function

x → e−x is summable on [0,∞[. By the dominated convergence theorem

limn→∞

In =∫

[0,∞[e−x dx =

∫ ∞

0e−x dx = 1.

Remark. Note that the sequence (1 + x/n)n is nondecreasing for x ≥ 0 and therefore we canalso use the monotone convergence theorem.

4.4. SOME APPLICATIONS 61

Integrals depending on a parameter

Theorem 4.7 Let (X,A, µ) be a measure space, J be metric space and f : X × J → IR (or C)be a function satisfying

(i) For every s ∈ J , the function x 7→ f(x, s) is measurable.

(ii) For almost every x ∈ X, the function s 7→ f(x, s) is continuous at some point s0.

(iii) There exists a function g ∈ L1(X) such that |f(x, s)| ≤ g(x) for all s ∈ I and almost everyx ∈ X.

Then the function I defined by

I(s) =∫

Xf(x, s) dµ(x)

is continuous at s0.

Proof. We need to show that for every sequence tn ⊂ J converging to s0, we haveI(tn) → I(s0). Let fn(x) = f(x, tn) and h(x) = f(x, s0). Then fn → h a.e. The dominatedconvergence theorem gives the result.

Theorem 4.8 (Derivation under the integral) Let (X,A, µ) be a measure space and J bean open interval of IR and f : X × J → IR (or C) be a function satisfying

(i) For every s ∈ J , the function x 7→ f(x, s) is µ−integrable.

(ii) For almost every x ∈ X, the function s 7→ f(x, s) is differentiable on J .

(iii) There exists a function g ∈ L1(X) such that |∂f(x,s)∂s | ≤ g(x) for all s ∈ J and almost every

x ∈ X.

Then the function I defined by

I(s) =∫

Xf(x, s) dµ(x)

is differentiable on J anddI

ds=∫

X

∂f(x, s)∂s

dµ(x).

Proof. Let t ∈ J and let tn ⊂ J be a sequence converging to t with tn 6= t. Setϕn(x) = f(x,tn)−f(x,t)

tn−t . By the mean value theorem and assumption (iii),

|ϕn(t)| ≤ sups|∂f(x, s)

∂s| ≤ g(x).

By the dominated convergence theorem

dI

ds= lim

n→∞

I(tn)− I(t)tn − t

= limn→∞

∫X

f(x, tn)− f(x, t)tn − t

dµ(x) =∫

Xlim

n→∞

f(x, tn)− f(x, t)tn − t

dµ(x)

=∫

X

∂f(x, s)∂s

dµ(x).


Appendix A

The Riemann integral

The integral of a function f over an interval [a, b] is thought of as the area of the planar regionbounded by the graph of f , and the lines y = 0, x = a and x = b. But how to define this area?The Riemann approach is to divide the interval [a, b] into many small subintervals [xi−1, xi],next, to choose a point ci in each subinterval and then consider the rectangles with base [xi−1, xi]and height f(ci). The sum of the areas of all these small rectangles is then an approximationof the ”area” under the graph of f . It is natural to expect that the more we have rectangles,or the finer is the decomposition of the interval [a, b], the better will be the approximation ofthe area under the graph of f . In what follows, we shall elaborate these intuitive ideas andconstruct a theory of the integral known as the Riemann integral.

A.1 Definitions

Definition A.1 Let [a, b] be a given interval. A partition P of [a, b] is a finite set of pointsx0, x1, . . . , xn such that a = x0 < x1 < · · · < xn−1 < xn = b. We write ∆xi = xi − xi−1 fori = 1, . . . , n. The biggest of the numbers ∆xi is called the norm of the partition and is denotedby ‖P‖.

Definition A.2 Let P1 and P2 be two partitions of [a, b]. We say that P2 is a refinement of P1

if P1 ⊂ P2.

Let f : [a, b] → IR be a bounded function. In order to define the integral of f over [a, b], weproceed as follows. Let P = x0, . . . , xn be a partition of [a, b] and let c1, . . . , cn be a sequenceof points in [a, b] such that ci ∈ [xi−1, xi] for all i = 1, . . . , n. Then we form the finite sum

σ(f, P, c1, . . . , cn) =n∑

i=1

f(ci)∆xi

called a Riemann sum of f corresponding to the the partition P . This Riemann sum is said tohave a limit I as the partition becomes finer and finer, or as ‖P‖ → 0 if for every ε > 0, thereexists a number δ > 0 such that

|σ(f, P )− I| < ε

for every partition P of [a, b] such that ‖P‖ < δ and any choice of the points c1, . . . , cn. In thiscase we write

lim‖P‖→0

σ(f, P ) = I.

Definition A.3 A function f : [a, b] → IR is called integrable in the sense of Riemann orRiemann integrable if the Riemann sums of f have a limit. In this case, this limit is called the

63

64 APPENDIX A. THE RIEMANN INTEGRAL

integral of f over [a, b] and is denoted by ∫ b

af(x) dx.

Otherwise stated, the integral of f is defined by∫ b

af(x) dx = lim

‖P‖→0σ(f, P )

provided that the limit exists.

Theorem A.1 Let f : [a, b] → IR be integrable, then

limn→∞

b− a

n

n∑k=1

f

(a + k

b− a

n

)=∫ b

af(x) dx.

Proof. Take a uniform partition of [a, b], that is, divide [a, b] into equal intervals each oflength b−a

n and choose ck = a + k b−an , we obtain the Riemann sum

n∑k=1

f

(a + k

b− a

n

)b− a

n.

According to the above, this sum tends to∫ b

af(x) dx as n →∞.

Definition A.4 If a > b and f is integrable on [b, a] we set∫ ba f(x) dx = −

∫ ab f(x) dx. Also

we set∫ aa f(x) dx = 0.

A.2 Criteria of integrability

Let f : [a, b] → IR be a bounded function and let P = x0, . . . , xn be a partition of [a, b]. Weset

Mi = supxi−1≤x≤xi

f(x)

mi = infxi−1≤x≤xi

f(x).

We need the following quantities called respectively the Darboux upper sum and the Darbouxlower sum of f corresponding to the partition P

U(f, P ) =n∑

i=1

Mi∆xi

L(f, P ) =n∑

i=1

mi∆xi.

Lemma A.1 Let P be a partition of [a, b] and let f be a bounded function. Then,

L(f, P ) ≤ σ(f, P, c1, . . . , cn) ≤ U(f, P ),

for any choice of the intermediate points ci.

A.2. CRITERIA OF INTEGRABILITY 65

Proof. Straightforward.

Theorem A.2 Let P ∗ be a refinement of a partition P of [a, b]. Then

(i) L(P, f) ≤ L(P ∗, f).

(ii) U(P ∗, f) ≤ U(P, f).

Otherwise stated, inserting an extra point into a partition increases the lower sum and decreasesthe upper sum.

Proof. We prove (i). Let P = x0, . . . , xn. It is enough to prove the claim when P ∗

contains just one point more than P . Let this point be x∗. Then there is i = 1, . . . , n such thatxi−1 < x∗ < xi. Let

w1 = infxi−1≤x≤x∗

f(x) and w2 = infx∗≤x≤xi

f(x).

Clearly w1 ≥ mi and w2 ≥ mi where

mi = infxi−1≤x≤xi

f(x).

Now

L(f, P ∗)− L(f, P ) = w1(x∗ − xi−1) + w2(xi − x∗)−mi(xi − xi−1)= (w1 −mi)(x∗ − xi−1) + (w2 −mi)(xi − x∗).

Therefore the difference is nonnegative, and so our claim is true.

The proof of (ii) is similar.

Corollary A.1 For any partition P and any partition Q of [a, b],

L(f, P ) ≤ U(f,Q).

Therefore, supL(f, P ) ≤ inf U(f,Q) where the sup and inf are taken over all possible partitionsof [a, b].

Proof. Let P and Q be two partitions of [a, b] and let P ∗ = P ∪Q so that P ∗ is a refinementof both P and Q. It follows from the theorem above that L(P, f) ≤ L(P ∗, f) ≤ U(P ∗, f) ≤U(Q, f).

Lemma A.2 Let P ∗ be a refinement of a partition P of [a, b]. Then

U(f, P )− U(f, P ∗) ≤ 2m||f ||||P || and L(f, P ∗)− L(f, P ) ≤ 2m||f ||||P ||

where m is the number of points in P ∗\P and ||f || = supx∈[a,b] |f(x)|.

Proof. It is enough to prove the lemma when P ∗ has just one more point than P . Let thispoint be x∗. Then there is i = 1, . . . , n such that xi−1 < x∗ < xi. Then,

U(f, P )− U(f, P ∗) ≤ ||f ||(x∗ − xi−1) + ||f ||(xi − x∗) + ||f ||(xi − xi−1) = 2||f ||(xi − xi−1)≤ 2||f ||||P ||.

The proof for L(U, f) is similar.


Theorem A.3 (Darboux) Let f : [a, b] → IR be bounded. Then

lim||P ||→0

U(f, P ) = inf U(f, P ) and lim||P ||→0

L(f, P ) = supL(f, P ),

where the sup is taken over all partitions of [a, b].

Proof. Let I∗ = inf U(f, P ) and let ε > 0 be given. Then there exists a partition P1 =x0, . . . , xm such that U(f, P1) < I∗ + ε

2 . Let δ = ε4m||f || where ||f || is the supremum of f on

[a, b]. Let P be a partition such that ||P || < δ and finally let P ∗ = P1 ∪ P . According to theprevious lemma U(f, P ) − U(f, P ∗) ≤ 2m||f ||||P || < 2m||f ||δ = ε

2 . Now since P ∗ is finer thatP1, we have U(f, P ∗) ≤ U(f, P1) < I∗ + ε

2 . Thus,

I∗ ≤ U(f, P ) < U(f, P ∗) +ε

2< I∗ + ε.

The proof for I∗ = supL(f, P ) is similar or can be deduced from the preceding by noting thatinf(A) = − sup(−A) and L(−f, P ) = −U(f, P ).

Lemma A.3 By a choice of the intermediate points ci, the Riemann sum σ(f, P, c1, . . . , cn) canbe made arbitrarily close to the upper Darboux sum U(f, P ) as well as to the lower Darbouxsum L(f, P ).

Proof. By the property of the supremum, for any ε > 0, there exists a point ci in [xi−1, xi]such that

Mi −ε

b− a< f(ci) ≤ Mi.

Multiplying both inequalities by ∆xi and summing from i = 1 to i = n, we get

U(f, P )− ε < σ(f, P, c1, . . . , cn) ≤ U(f, P ).

This proves the first assertion of the lemma. The second assertion is proved similarly by usingthe property of the infinimum.

Theorem A.4 Let f : [a, b] → IR be a bounded function. Then the following conditions areequivalent.

(i) f is Riemann integrable on [a, b].

(ii) For every ε > 0 there exits a number δ > 0 such that U(f, P ) − L(f, P ) < ε, for anypartition P of [a, b] such that ‖P‖ < δ.

(iii) supL(f, P ) = inf U(f, P ).

(iv) For every ε > 0 there exits a partition P of [a, b] such that U(f, P )− L(f, P ) < ε.

In addition, if one of the above conditions hold then∫ ba f(x) dx = supL(f, P ) = inf U(f, P ).

Proof. (i)⇒(ii). Let f be integrable and set I =∫ ba f(x) dx. Let ε > 0 be given. Then by

definition, there exists a number δ > 0 such that

|σ(f, P, c1, . . . , cn)− I| < ε

2, or I − ε

2< σ(f, P, c1, . . . , cn) < I +

ε

2,

for every partition P of [a, b] such that ||P || < δ and any choice of the intermediate points ci.By the previous lemma, the lower Darboux sum L(f, P ) belongs to this interval for some choiceof ci. The same is true for the upper Darboux sum U(f, P ). This means that both sums belongto the same interval of length ε.

A.3. CLASSES OF INTEGRABLE FUNCTIONS 67

(ii)⇒(iii). For every partition P of [a, b], we have

L(f, P ) ≤ supL(f, P ) ≤ inf U(f, P ) ≤ U(f, P ),

where the inf and sup are taken over all possible partitions of [a, b]. Condition (ii) then impliesthat

0 ≤ inf U(f, P )− supL(f, P ) < ε.

for all ε > 0. This means that inf U(P, f) = supL(P, f).

(iii)⇒(i). Let I be this common number. It follows from Darboux’s theorem that

lim||P ||→0

U(f, P ) = lim||P ||→0

L(f, P ) = I.

But L(f, P ) ≤ σ(f, P, c1, . . . , cn) ≤ U(f, P ) and so lim||P ||→0

σ(f, P, c1, . . . , cn) = I. This means

that f is integrable (and its integral is I).

(iii)⇒(iv). Denote this common number by I. By the property of the sup, there exists apartition P such that

I = supL(P, f) < L(f, P ) +ε

2.

By the property of the inf, there exists a partition Q such that

I = inf U(P, f) > U(f, P )− ε

2.

Let P ∗ = P ∪Q. ThenU(f, P ∗)− ε

2< I < L(f, P ∗) +

ε

2.

Hence the conclusion.

(iv)⇒(iii). It follows that inf U(P, f) ≤ U(f, P ) < L(f, P )+ ε ≤ supL(f, P )+ ε for every ε > 0and therefore, inf U(P, f) ≤ supL(f, P ). But the reverse inequality always holds, hence theequality.

Remark A.1 Observe that U(f, P )−L(f, P ) =∑

(Mi−mi)∆xi. The difference ωi = Mi−mi

is called the oscillation of f on [xi−1, xi]. Thus condition (ii) of the previous theorem can beformulated as follows: For every ε > 0, there is δ > 0 such that

∑ωi∆xi < ε, for every partition

P such that ‖P‖ < δ.

A.3 Classes of integrable functions

Not all functions are integrable in the sense of Riemann (this is why we sought another criterionof integrability). Consider for example the Dirichlet function f : [0, 1] → IR defined by f(x) = 1if x is rational and f(x) = 0 if x is irrational. Let P = x0, . . . , xn be an arbitrary partitionof [0,1]. Since every subinterval [xk−1, xk] contains rational as well as irrational points, theoscillation of f on every subinterval is 1. Therefore

∑ωi∆xi = 1. Therefore this sum cannot

be made arbitrarily small for some choice of a fine partition. This means that the function isnot integrable in the sense of Riemann.

However, there are many functions which are integrable in the sense of Riemann. In fact allelementary functions are integrable. We shall prove this shortly, and to this end we start witha lemma that is proved in the exercises.

Lemma A.4 Let f : [α, β] → IR be a bounded function. Then the oscillation of f is also equalto sup|f(x)− f(y)| |x, y ∈ [α, β].


Theorem A.5 A continuous function is Riemann integrable.

Proof. Let f : [a, b] → IR be continuous and let ε > 0 be given. Since f is uniformlycontinuous on [a, b], there exists δ > 0 such that |f(x)−f(t)| < ε

b−a for all x, t ∈ [a, b] such that|x− t| < δ. Let P = x0, . . . xn be a partition such that ‖P‖ < δ. Let ωi be the oscillation off on the interval [xi−1, xi]. According to the previous lemma ωi ≤ ε

b−a and therefore

n∑i=1

ωi∆xi ≤ε

b− a

n∑i=1

∆xi =ε

b− a(b− a) = ε.

By Theorem A.4, f is integrable.

Theorem A.6 A monotonic function is Riemann integrable.

Proof. We prove the theorem for an increasing function f . We distinguish between two cases(i) f(a) = f(b) and (ii) f(a) < f(b). In the first case f is constant and therefore integrable (sinceit is continuous). So we consider the second case. Let ε > 0 be given, and let P = x0, . . . , xnbe a partition of [a, b] with ‖P‖ < δ = ε

f(b)−f(a) . Let as before, Mi = supxi−1≤x≤xif(x) and

mi = infxi−1≤x≤xi f(x). Then mi = f(xi−1) and Mi = f(xi) (because f is increasing) so that

U(f, P )− L(f, P ) =n∑

i=1

[f(xi)− f(xi−1)

]∆xi ≤ δ

n∑i=1

[f(xi)− f(xi−1)

]= δ[f(b)− f(a)] = ε.

By Theorem A.4, f is integrable.

Theorem A.7 Let f : [a, b] → IR be a function and let c ∈]a, b[. If f is integrable on [a, c] andon [c, b], then f is integrable on [a, b] and∫ b

af(x) dx =

∫ c

af(x) dx +

∫ b

cf(x)dx.

Proof. Consider the sum∑

ωi∆xi for some partition. If the point c belongs to the partition,then this sum consists of two similar sums for the intervals [a, c] and [c, b], each of which tendsto zero as the norm of the partition tends to zero. The conclusion remains true also in the casewhere c does not belong to the partition: adding c to the partition we would change only oneterm in the sum which itself tends to zero.

Theorem A.8 A piecewise continuous function is Riemann integrable.

Proof. It is enough to prove the theorem when f has only one point of discontinuity c ∈]a, b[because the general case follows by induction. Now f is integrable on [a, c] because it continuousthere. Similarly f is integrable on [c, b]. By the previous theorem f is integrable on [a, b].

In fact we have a stronger theorem.

Theorem A.9 A bounded function having finitely many points of discontinuities is Riemannintegrable.

Example. Consider the function x 7→ sin( 1x). This function is discontinuous at x = 0 but it is

not piecewise continuous. According to to the previous theorem it is integrable on any compactinterval.

The above theorem and the next one can be proved inside the Riemann theory but they can beeasily deduced from Theorem 4.6.

Theorem A.10 Let f be integrable and ϕ be continuous. Then ϕ f is integrable.

In particular, if f is integrable then so are |f |, f2 and any positive power of f .

A.4. PROPERTIES OF THE INTEGRAL 69

A.4 Properties of the integral

Proposition A.1 Suppose that f and g are integrable functions on [a, b] and that k is aconstant. Then

a) kf is integrable and

∫ b

akf(x) dx = k

∫ b

af(x) dx.

b) f + g is integrable and

∫ b

af(x) + g(x) dx =

∫ b

af(x) dx +

∫ b

ag(x) dx.

c) fg is integrable.

Proof. a) and b) are proved by taking Riemann sums and going to the limit.

c) is proved in the exercises .

Proposition A.2 If f is integrable on an interval containing the points a, b, and c, then∫ b

af(x) dx =

∫ c

af(x) dx +

∫ b

cf(x) dx.

Proposition A.3 If an integrable function f : [a, b] → IR satisfies f(x) ≥ 0 for every x ∈ [a, b],

then

∫ b

af(x) dx ≥ 0.

Proof. If f(x) ≥ 0 then any Riemann sum is nonnegative. Going to the limit, we obtain theresult.

Corollary A.2 If f and g are two integrable functions satisfying f(x) ≤ g(x) for all x ∈ [a, b],

then

∫ b

af(x) dx ≤

∫ b

ag(x) dx.

Corollary A.3 If f is integrable on an interval containing a and b and f satisfies |f(x)| ≤ M

then

∣∣∣∣∫ b

af(x) dx

∣∣∣∣ ≤ M |b− a|.

A.5 Integration and differentiation

Theorem A.11 (First fundamental theorem of Calculus) Let f : [a, b] → IR be inte-

grable. Then the function F defined by F (x) =∫ x

af(t) dt is continuous and differentiable at

each point where f is continuous. At these points, F ′(x) = f(x).

Proof. Since f is bounded, there is a constant M such that |f(t)| ≤ M for all t ∈ [a, b]. Notethat F (x)− F (y) =

∫ xy f(t) dt and so

|F (x)− F (y)| ≤ M |x− y|.

This means that F is Lipschitz continuous and therefore continuous.Let now x be a point at which f is continuous. Then, given ε > 0, there exists δ > 0 such

that |f(t)− f(x)| < ε for all t such that |t− x| < δ.∣∣∣∣F (x + h)− F (x)h

− f(x)∣∣∣∣ = ∣∣∣∣1h

∫ x+h

x(f(t)− f(x)) dt

∣∣∣∣ ≤ ε for |h| < δ.

This means that limh→0F (x + h)− F (x)

h= f(x) and so F is differentiable at x with derivative

f(x).


Corollary A.4 Every continuous function defined on a interval has an antiderivative.

We rarely compute integrals by going to the definition. The most practical way to computethe integral of elementary functions is given by the following theorem which connects integrationand the search for antiderivatives.

Theorem A.12 (Second fundamental theorem of Calculus) Let f be an integrable func-tion defined on an interval I and let a, b ∈ I. If G is an antiderivative of f , then∫ b

af(x) dx = G(b)−G(a).

Proof. It is enough to prove the result for a < b. Note that for any partition P = x0, . . . , xnof [a, b] we have G(b) − G(a) =

∑nk=1 G(xk) − G(xk−1). Now by the mean value theorem

there exists ck ∈]xk−1, xk[ such that G(xk) − G(xk−1) = G′(ck)∆xk = f(ck)∆xk. Therefore,G(b) − G(a) =

∑nk=1 f(ck)∆xk. But the right hand side of this equality is a Riemann sum of

f . Since f is integrable this sum converges to∫ ba f(x) dx as ‖P‖ → 0.

Remark. The quantity G(b)−G(a) is usually denoted by G(x)|ba.

Thanks to the above theorem we can write for example∫ 1

0x2 dx =

x3

3

∣∣∣∣10

=13.

Example of computing Riemann sums. Find limn→∞

1n√

n

(√1 +

√2 + · · ·+

√n). The sum

can be written as

1n

(√1n

+

√2n

+ · · ·+√

n

n

)=

b− a

n

n∑k=1

f

(a + k

b− a

n

)

where a = 0, b = 1 and f(x) =√

x. Therefore the sum is a Riemann sum of f and convergestherefore to

∫ 10

√xdx = 2

3 .

There are of course several techniques for integration. The most important are integrationby parts, substitutions or change of variables and partial fractions. These are usually studiedin elementary Calculus courses. We mention the first two.

Theorem A.13 (Integration by parts) Let f and g be differentiable functions defined on[a, b]. If f ′ and g′ are integrable on [a, b], then∫ b

af ′(x)g(x) = f(b)g(b)− f(a)g(a)−

∫ b

ag′(x)f(x) dx.

Proof. The result follows from the derivative of a product and the second fundamentaltheorem of Calculus.

Theorem A.14 (Integration by substitution) Let ϕ : [a, b] → [c, d] be differentiable andlet f : [c, d] → IR be continuous. If ϕ is integrable, then∫ b

af(ϕ(t))ϕ′(t) dt =

∫ ϕ(b)

ϕ(a)f(x) dx.

Proof. By the chain rule and the second fundamental theorem of Calculus, both sides areequal to F (ϕ(b))− F (ϕ(a)) where F is an antiderivative of f .

A.6. LIMITS AND INTEGRATION 71

A.6 Limits and integration

Example A.1 Consider the sequence defined by fn(x) = nx(1 − x2)n for x ∈ [0, 1]. Thissequence converges to the zero function. Note however that

∫ 10 fn(x) dx = n

2n+2 →12 whereas∫ 1

0 lim fn(x) dx = 0.

The above example show that one should be careful before interchanging limits and integrals.However, there is a stronger form of convergence that permits this interchange and also preservesthe properties of continuity and integrability.

Example A.2 The sequence of functions defined by fn(x) = nx(1 − x2)n for x ∈ [0, 1] doesnot converge uniformly because

supx∈[0,1]

fn(x) = fn

(1√

2n + 1

)=

n√2n + 1

[1− 1

2n + 1

]n

→∞ as n →∞.

Theorem A.15 Let fn : [a, b] → IR be a sequence of Riemann integrable functions whichconverges uniformly to a function f . Then f is Riemann integrable and

limn→∞

∫ b

afn(t) dt =

∫ b

af(t) dt.

Otherwise stated,

limn→∞

∫ b

afn(t) dt =

∫ b

alim

n→∞fn(t) dt.

Proof. The first step is to show that the sequence ∫ ba fn converges. Let ε > 0. Since fn

converges uniformly, there exists an integer N such that |fn(x) − fm(x)| < ε for all x ∈ [a, b]and all n, m ≥ N . It follows that∣∣∣∣∫ b

afn(x) dx−

∫ b

afm(x) dx

∣∣∣∣ = ∣∣∣∣∫ b

a[fn(x)− fm(x)] dx

∣∣∣∣ ≤ ∫ b

a|fn(x)− fm(x)|dx ≤ ε(b− a)

for all n, m ≥ N . This means that the sequence of integrals is a Cauchy sequence and thereforeis convergent to some limit L.

Chose n ≥ N . Since fn is integrable, there exists δ > 0 such that |σ(fn, P, c1, . . . , cn) −∫ ba fn(x) dx| ≤ ε for any partition P with ‖P‖ ≤ δ. From the other hand, it is easily seen that|σ(f, P, c1, . . . , ck)− σ(fn, P, c1, . . . , ck)| ≤ ε(b− a).

The result now follows from

|σ(f, P )− L| ≤ |σ(f, P )− σ(fn, P )|+∣∣∣∣σ(fn, P )−

∫ b

afn(x) dx

∣∣∣∣+ ∣∣∣∣∫ b

afn(x) dx− L

∣∣∣∣ .

Theorem A.16 Let fn : [a, b] → IR be a sequence of continuously differentiable functionswhich satisfies the following conditions.

(i) There is some point x0 ∈ [a, b] such that fn(x0) converges.

(ii) The sequence of derivatives f ′n converges uniformly to some function g.

Then fn converges uniformly to a C1 function f which satisfies f ′ = g.


Proof. By the second fundamental theorem of calculus we can write

fn(x) = fn(x0) +∫ x

x0

f ′n(t) dt.

Let ` = limn→∞ fn(x0) and define now the function f by

f(x) = ` +∫ x

x0

g(t) dt.

Then f ′(x) = g(x). We shall prove that fn converges uniformly to f . Taking the differencebetween the last two identities, it follows from the triangle inequality that

|fn(x)− f(x)| ≤ |fn(x0)− `|+∣∣∣∣∫ x

x0

|f ′n(t)− g(t)|dt

∣∣∣∣ .Now given ε > 0, we can make |fn(x0)−`| < ε for all n large enough. Also we can make |f ′n(t)−g(t)| < ε for all n large enough and all t ∈ [a, b]. Thus integrating we get

∣∣∣∫ xx0|f ′n(t)− g(t)|dt

∣∣∣ ≤(b− a)ε for all x ∈ [a, b]. This means that

|fn(x)− f(x)| ≤ ε + (b− a)ε,

for all n large enough and all x ∈ [a, b]. Since ε is arbitrary, the theorem is proved.

Now we give the analog of the above theorems for series. The following theorem can beeasily proved by taking partial sums.

Theorem A.17 Let fn : I → IR be sequence of functions such that the series∑

fn convergesuniformly to a function f .

a) If each fn is continuous then f is continuous.

b) If each fn is integrable on an interval [a, b] ⊂ I, then f is integrable on [a, b] and∫ ba f(x) dx =∑∫ b

a fn(x). This means that we can integrate term by term a uniformly convergent series.

c) If each fn is continuously differentiable and∑

f ′n is uniformly convergent on I then f iscontinuously differentiable and f ′(x) =

∑f ′n(x) for all x ∈ I.

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Math 301 Integration I - fs2-fanar.comfs2-fanar.com/Math/pdf/Math301-lecturenotes.pdf · Chapter 1 Preliminaries 1.1 Sets and functions Given a set E we denote by P(E) or 2E the set