Atmnotes 2 Hay

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Measure and Integration:

Concepts, Examples and Exercises

INDER K. RANA

Indian Institute of Technology Bombay

India

Department of Mathematics, Indian Institute of Technol-

ogy, Bombay, Powai, Mumbai 400076, India

Current address: Department of Mathematics, Indian Institute of Tech-nology, Powai, Mumbai 400076, India

E-mail address: [email protected]


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These notes were specially prepared for the participants of the first Annual

Foundation School, May 2004, held at IIT Bombay, India.

Abstract. These notes present a quick overview of the theory of Mea-

sure and Integration. For a more detailed and motivated text, the reader

may refer authors book:

An Introduction to Measure and Integration,

Narosa Publishers, Delhi, 1997

or,

An Introduction to Measure and Integration,Second Edition,

Graduate Text in Mathematics, Volume 45,

American Mathematical Society, 2002.

May, 2004 Mumbai 400076 Inder K. Rana


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Contents

Chapter 1. Classes of sets 1

1.1. Semi-algebra and algebra of sets 1

1.2. Sigma algebra and monotone class 5

Chapter 2. Measure 9

2.1. Set functions 9

2.2. Countably additive set functions on intervals 14

2.3. Set functions on algebras 15

2.4. Uniqueness problem for measures 17

Chapter 3. Construction of measures 19

3.1. Extension from semi-algebra to the generated algebra 19

3.2. Extension from algebra to the generated-algebra 20

3.3. Choosing nice sets: Measurable sets 22

3.4. Completion of a measure space 24

3.5. The Lebesgue measure 27

Chapter 4. Integration 35

4.1. Integral of nonnegative simple measurable functions 35

4.2. Integral of nonnegative measurable functions 38

4.3. Intrinsic characterization of nonnegative measurable functions 42

4.4. Integrable functions 50

4.5. The Lebesgue integral and its relation with the Riemannintegral 55

vii


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viii Contents

4.6. L1[a, b] as the completion ofR[a, b] 59

Chapter 5. Measure and integration on product spaces 63

5.1. Introduction 63

5.2. Product of measure spaces 65

5.3. Integration on product spaces: Fubinis theorems 69

5.4. Lebesgue measure on R2 and its properties 75

Chapter 6. Lp-spaces 79

6.1. Integration of complex-valued functions 79

6.2. Lp-spaces 82

6.3. L(X, S, ) 866.4. L2(X, S, ) 87

6.5. L2-convergence of Fourier series 93

Appendix A. Extended real numbers 97

Appendix B. Axiom of choice 101

Appendix C. Continuum hypothesis 103

Appendix. References 105

Appendix. Index 107


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Chapter 1

Classes of sets

1.1. Semi-algebra and algebra of sets

Concepts and examples:

1.1.1. Definition:LetXbe a nonempty set and let Cbe a collection of subsets ofX. We sayC is a semi-algebra of subsets ofXif it has the following properties:

(i) , X C.

(ii) A B C for every A, B C.

(iii) For everyA Cthere existn Nand sets C1, C2, . . . , C n Csuchthat Ci Cj = for i =j and Ac =

ni=1 Ci.

1.1.2. Definition:LetXbe a nonempty set and Fa collection of subsets ofX. The collectionF is called analgebra of subsets ofX ifFhas the following properties:

(i) , X F.

(ii) A B F, whenever A, B F.

(iii) Ac

F, whenever A F.

1.1.3. Examples:

(i) Let Xbe any nonempty set. The collections {, X} and P(X) :={E| EX} are trivial examples of algebras of subsets ofX. The collectionP(X)is called the power set ofX.

1


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2 1. Classes of sets

(ii) The collectionIof all intervals forms a semi-algebra of subsets ofR.For

a, b R with a < b, consider the collection I of all intervals of the form(a, b], (, b], (a, ), (, +).We call Ithe collection of all left-open,right-closed intervals ofR.It is easy to check that Iis also a semi-algebraof subsets ofR.

(iii) The collection

F(I) :=

E R E=

nk=1

Ik, Ik I, Ik I= for k =, n N

.

is an algebra of subsets ofR. So is the class

F(I) :=

E R | E=n

k=1

Ik, Ik I, Ik I = for k =, n N

.

(iv) Let Xbe any nonempty set. Let

C :={EX| either Eor Ec is finite}.

ThenC is an algebra of subsets ofX.

(v) LetXand Ybe two nonempty sets, and Fand Gsemi-algebras of subsetsofX and Y, respectively. Let

FG= {FG| F F, G G}.

Then,F Gis a semi-algebra of subsets ofXY.

Exercises:

(1.1) Let Fbe any collection of subsets of a set X. Show that F is analgebra if and only if the following hold:

(i) , X F.

(ii) Ac

F whenever A F.(iii) A B F whenever A, B F.

(1.2) LetFbe an algebra of subsets ofX. Show that(i) IfA, B F thenAB := (A \ B) (B\ A) F.

(ii) If E1, E2, . . . , E n F then F1, F2, . . . , F n F such thatFi Eifor eachi,FiFj = fori=j and

ni=1 Ei=

nj=1 Fj.

The next set of exercise describes some methods of constructing algebrasand semi-algebras.


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1.1. Semi-algebra and algebra of sets 3

(1.3) Let X be a nonempty set. Let = E X and let Cbe a semi-

algebra (algebra) of subsets ofX. Let

C E :={A E| A C}.

Show that C E is a semi-algebra (algebra) of subsets ofE. NotethatC Eis the collection of those subsets ofEwhich are elementsofCwhenE C.

(1.4) LetX, Ybe two nonempty sets and f :XY be any map. ForE Y, we write f1(E) := {x X| f(x) E}. Let C be anysemi-algebra (algebra) of subsets ofY . Show that

f1(C) :={f1(E)| E C }

is a semi-algebra (algebra) of subsets ofX.

(1.5) Give examples of two nonempty sets X, Y and algebras F, G ofsubsets ofX and Y , respectively such that F G :={A B | AF, B G} is not an algebra. (It will of course be a semi-algebra,as shown in example 1.1.3(v).)

(1.6) Let {F}I be a family of algebras of subsets of a set X. LetF :=

IF. Show that F is also an algebra of subsets ofX.

(1.7) Let{Fn}n1 be a sequence of algebras of subsets of a set X. Underwhat conditions onFn can you conclude thatF :=

n=1 Fn is also

an algebra?(1.8) LetCbe a semi-algebra of subsets of a set X. A setA X is called

a-setif there exist setsCi C, i= 1, 2, . . . ,such thatCi Cj =for i =j and

i=1 Ci= A. Prove the following:

(i) For any finite number of sets C, C1, C2, . . . , C nin C, C\(n

i=1 Ci)is a finite union of pairwise disjoint sets from Cand hence is a-set.

(ii) For any sequence {Cn}n1 of sets in C,

n=1 Cn is a -set.(iii) A finite intersection and a countable union of-sets is a -set.

(1.9) Let C be any collection of subsets of a set X. Then there exists aunique algebraFof subsets ofX such thatC F and ifA is any

other algebra such thatC A, thenF A. This unique algebragiven is called the algebra generated by C and is denoted byF(C).

(1.10) Show that he algebra generated by I, the class of all intervals, is{E R |E=

nk=1 Ik, Ik I , Ik I= if 1 k = n}.

(1.11) Let Cbe any semi-algebra of subsets of a set X.Show that F(C),thealgebra generated by C, is given by {EX |E=

ni=1Ci, Ci C

and Ci Cj = for i =j, n N}.


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Remark:

Exercise 1.11 gives a description ofF(C),the algebra generated bya semi-algebra C. In general, no description is possible for F(C)whenCis not a semi-algebra.

(1.12) Let Xbe any nonempty set and C={{x} |x X}

{, X}. IsCa semi-algebra of subsets ofX? What is the algebra generated byC? Does your answer depend upon whether X is finite or not?

(1.13) Let Ybe any nonempty set and let Xbe the set of all sequenceswith elements from Y , i.e.,

X={x= {xn}n1| xn Y, n= 1, 2, . . .}.

For any positive integerk let A Yk,the k-fold Cartesian productofYwith itself, and let i1< i2< < ik be positive integers. Let

C(i1, i2, . . . , ik; A) :={x= (xn)n1 X |(xi1 , . . . , xik) A}.

We call C(i1, i2, . . . , ik; A) a k-dimensional cylinder set in Xwith base A. Prove the following assertions:(a) Every k-dimensional cylinder can be regarded as a n-dimensional

cylinder also for n k.(b) Let

A= {EX|E is ann-dimensional cylinder set for some n}.

Then,A {, X} is an algebra of subsets ofX.

(1.14) LetCbe any collection of subsets of a set Xand let EX. Let

C E:= {C E|C C }.

Then the following hold:(a)

C E F(C) E :={A E|A F(C)}.

Deduce that

F(C E) F(C) E.

(b) Let

A= {A X|A E F(C E)}.

Then,A is an algebra of subsets ofX,C Aand

A E= F(C E).

(c) Using (a) and (b), deduce thatF(C) E= F(C E).


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1.2. Sigma algebra and monotone class 5

1.2. Sigma algebra and monotone class


1.2.1. Definition:Let Xbe any nonempty set and let Sbe a class of subsets ofX with thefollowing properties:

(i) andX S.

(ii) Ac S whenever A S.

(iii)i=1 Ai S whenever Ai S, i= 1, 2, . . . .Such a class S is called a sigma algebra (written as -algebra) of subsetsofX.

1.2.2. Examples:

(i) Let X be any set. Then {, X} and P(X) are obvious examples of -algebras of subsets ofX.

(ii) Let Xbe any uncountable set and let

S={A X |A orAc is countable}.

ThenS is a -algebra of subsets ofX.(iii) LetXbe any set and let Cbe any class of subsets ofX. Let S(C) :=

S,

where the intersection is taken over all -algebrasSof subsets ofX suchthat S C(note that P(X) is one such -algebra). It is easy to see thatS(C) is also a-algebra of subsets ofX andS(C) C. In fact, ifSis any-algebra of subsets ofX such that S C, then clearly S S(C). ThusS(C) is the smallest -algebra of subsets ofXcontainingC, and is calledthe -algebra generated byC.In general it is not possible to representan element ofS(C) explicitly in terms of elements ofC.

1.2.3. Definition:LetXbe a nonempty set and M be a class of subsets ofX. We sayM is amonotone class if

(i)

n=1 An M,wheneverAn Mand An An+1for n = 1, 2, . . . ,

(ii)

n=1 An M,wheneverAn Mand An An+1for n = 1, 2, . . . .

1.2.4. Examples:

(i) Clearly, every -algebra is also a monotone class.


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(ii) Let X be any uncountable set. Let M := {A X | A is countable}.

ThenM is a monotone class but not a -algebra.

(iii) LetXbe any nonempty set and letCbe any collection of subsets ofX.Clearly P(X) is a monotone class of subsets ofX such that C P(X).Let M(C) :=

M, where the intersection is over all those monotone

classes M of subsets of X such that C M. Clearly, M(C) is itself amonotone class, and ifM is any monotone class such that C M, thenM(C) M. Thus M(C) is the smallest monotone class of subsets ofX such that C M(C). The class M(C) is called the monotone classgenerated byC.

Exercises:

(1.15) Let S be a -algebra of subsets of X and let Y X. Show thatS Y :={E Y |E S}is a -algebra of subsets ofY .

(1.16) Let f :XYbe a function and Ca nonempty family of subsetsof Y. Let f1(C) := {f1(C) | C C}. Show that S(f1(C)) =f1(S(C)).

(1.17) Let Xbe an uncountable set and C={{x} |x X}. Identify the-algebra generated by C.

(1.18) Let Cbe any class of subsets of a set X and let Y X. LetA(C)be the algebra generated byC.

(i) Show thatS(C) =S(A(C)).(ii) LetC Y :={E Y |E C}.Show thatS(C Y) S(C) Y.

(iii) Let

S :={E (B Yc)| E S(C Y), B C}.

Show that S is a -algebra of subsets of X such that C Sand S Y =S(C Y).

(iv) Using (i), (ii) and (iii), conclude that S(C Y) =S(C) Y.

(1.19) LetCbe a class of subsets of a setXsuch that C.ThenE S(C)iff setsC1, C2, . . . inCsuch thatE S({C1, C2, . . .}).

Hint:The technique used to prove of exercise 1.19 is very useful, andis often used to prove various properties of-algebras under con-sideration, is as follows: The sets satisfying the required propertyare collected together. One shows that this collection itself is a-algebra and includes a subfamily of the original -algebra which


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(iv) Using (iii), deduce that M(A) L(E) for every E M(A),i.e.,

M(A) is closed under unions also. Now use exercise (1.22) todeduce that S(A) M(A).


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Chapter 2

Measure

2.1. Set functions


2.1.1. Definition:LetCbe a class of subsets of a set X. A function : C [0, +] is called

a set function. Further,

(i) is said to be monotoneif(A) (B) whenever A, B C andA B.

(ii) is said to be finitely additive if

ni=1

Ai

=

ni=1

(Ai).

whenever A1, A2, . . . , An C are such that Ai Aj = for i = jandni=1 Ai C.

(iii) is said to be countably additive if

n=1

An

=

n=1

(An)

whenever A1, A2, . . . in Cwith AiAj = for i=j and

n=1 An C.

9


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10 2. Measure

(iv) is said to be countably subadditive if

(A)

n=1

(An).

whenever A C, A=

n=1 An with An C for every n.

(v) is called ameasure on Cif Cwith() = 0 andis countablyadditive onC.

Here are some more examples of finitely/countably additive set func-tions:

2.1.2.Example:

Let X be any infinite set and let xn X, n = 1, 2, . . . . Let {pn}n1 be asequence of nonnegative real numbers. For anyA X, define

(A) :=

{i|xiA}

pi.

It is easy to show that is a countably additive set function on the algebraP(X). We say is a discrete measurewith mass pi at xi. The measureis finite (i.e.,(X)< +) iff

i=1pi< +. If

i=1pi= 1,the measure

is called a discrete probability measure/distribution. Note that({xi}) = pi i and ({x}) = 0 ifx = xi. So, one can regard as a setfunction defined on the subsets of the set Y := {xn : n 1}. Some of the

special cases whenX={0, 1, 2, . . .} are:(a) Binomial distribution: Y :={0, 1, 2, . . . , n}and, for 0< p 0.

(c) Uniform distribution: Y :={1, 2, . . . , n},

pk := 1/k k.

An important example of a set function on the collection of intervals inR.

2.1.3 Example:We denote the set of real numbers by R.Let R denote the set ofextendedreal numbers.

LetIdenote the collection of all intervals ofR. If an interval I I hasend points a and b we write it as I(a, b). By convention, the open interval


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(a, a) = a R. Let [0, +] := {x R|x 0} = [0, +) {+}.

Define the function : I [0, ] by

(I(a, b)) :=

|b a| ifa, b R,+ if eithera = or b = +or both.

The function , as defined above, is called the length function and hasthe following properties:

Property (1): () = 0.

Property (2): (I) (J) ifIJ.

This is called the monotonicity property of (or one says that is

monotone).Property (3): LetI Ibe such thatI=

ni=1 Ji, whereJi Jj = for

i=j. Then

(I) =n

i=1

(Ji).

This property of is called the finite additivity of , or one says that is finitely additive.

Property (4): LetI Ibe a finite interval such thatI

i=1 Ii, whereIi I. Then

(I)

i=1

(Ii).

Property (5): LetI Ibe a finite interval such thatI=

n=1 In,whereIn I andIn Im= forn=m. Then

(I) =

n=1

(In).

Property (6): LetI I be any interval. Then

(I) =

n=

(I [n, n + 1)).

Property (7): Let I I be any interval such thatI=

n=1 In, In IandIn Im= forn=m. Then

(I) =

n=1

(In).


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12 2. Measure

This property of is called the countable additivity of , or one says

that is countably additive.

Property (8): LetI I andI

n=1 In, In I. Then

(I)

n=1

(In).

This property of is called the countable subadditivity of, or onesays that is countably subadditive.

Property (9): (I) = (I +x), for every I I and x R, where

I+ x:= {y+ x| y I}.

This property of the length function is calledtranslation invariance, orone says that is translation invariant.

Exercises:

(2.1) LetXbe any countably infinite set and let

C= {{x} |x X}.

Show that the algebra generated by C is

F(C) :={A X |A or Ac is finite}.

Let :F(C)[0, ) be defined by

(A) :=

0 ifA is finite,1 ifAcis finite.

Show that is finitely additive but not countably additive. IfX isan uncountable set, show that is also countably additive.

(2.2) LetX= N,the set of natural numbers. For every finite setA X,let #Adenote the number of elements in A. Define for A X,

n(A) := #{m: 1 m n, m A}n

.

Show thatnis countably additive for everynon P(X).In a sense,n is the proportion of integers between 1 to n which are in A. Let

C= {A X | limn

n(A) exists}.

Show that C is closed under taking complements, finite disjointunions and proper differences. Is it an algebra?


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(2.3) Let : I (0, 1][0, ] be defined by

(a, b] :=

b a ifa = 0, 0< a < b 1,+ otherwise.

(Recall that I (0, 1] is the class of all left-open right-closed inter-vals in (0, 1].)Show that is finitely additive. Is countably additive also?

(2.4) LetXbe a nonempty set.(a) Let : P(X) [0, ) be a finitely additive set function

such that (A) = 0 or 1 for every A P(X). Let

U={A P(X)| (A) = 1}.

Show thatUhas the following properties:(i) U.

(ii) IfA X and B A, then B U.(iii) IfA, B U, thenA B U.(iv) For every A P(X), either A U or Ac U.

(Any U P(X) satisfying (i) to (iv) is called an ultrafilterin X.)

(b) LetUbe any ultrafilter in X. Define :P(X)[0, ) by

(A) :=

1 ifA U,0 ifA U.

Show that is finitely additive.(2.5) LetA be an algebra of subsets of a set X.

(i) Let1, 2 be measures onA, and let and be nonnegativereal numbers. Show that 1+ 2 is also a measure on A.

(ii) For any two measures 1, 2 onA, we say

1 2 if 1(E) 2(E), E A.

Let{n}n1 be a sequence of measures on A such that

n n+1, n 1.

Define E A,

(E) := limn n(E).

Show that is also a measure on A and E B,

(E) = sup {n(E)| n 1}.

(2.6) Let Xbe a compact topological space and A be the collection ofall those subsets ofXwhich are both open and closed. Show thatAis an algebra of subsets ofX. Further, every finitely additive setfunction onA is also countably additive.


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14 2. Measure

2.2. Countably additive set functions on

intervals


We saw in the previous section that the length function is a countably ad-ditive set function on the class of all intervals. One can ask the question:do there exist countably additive set functions on intervals, other than thelength function? The answer is given by the following:

2.2.1. Proposition:Let F : R R be a monotonically increasing function. Let F : I

[0, ] be defined by

F(a, b ] := F(b) F(a),

F(, b ] := limx

[F(b) F(x)],

F(a, ) := limx

[F(x) F(a)],

F(, ) := limx

[F(x) F(x)].

Then, F is a well-defined finitely additive set function on I. Further, Fis countably additive ifF is right continuous.

One calls Fthe set function induced by F.

The converse of proposition 2.2.1 is also true.

2.2.2. Proposition:Let : I [0, ]be a finitely additive set function such that(a, b]< +

for every a, b R. Then there exists a monotonically increasing functionF : R R such that (a, b] = F(b) F(a) a, b R. If is alsocountably additive, thenF is right-continuous.

(Hint: Define F as follows:

F(x) :=

(0, x] ifx >0,0 ifx = 0,

(x, 0] if x


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2.3. Set functions on algebras 15

(ii) Propositions 2.2.1 and 2.2.2 completely characterize the non-trivial count-

ably additive set functions on intervals in terms of functions F : R Rwhich are monotonically increasing and right continuous. Such functionsare called distribution functions on R. The set function F induced bythe distribution function F is non-trivial in the sense that it assigns finitenon-zero values to bounded intervals.

Exercises:

(2.7) Let F(x) = [x], the integral part of x, x R. Describe the setfunction F.

(2.8) R. Show that F1 :=F+ is also a distribution function andF =F1 . Is the converse true?

(2.9) (i) Let Cbe a collection of subsets of a set X and : C [0, ]be a set function. If is a measure on C, show that is finitelyadditive. Is monotone? Countably subadditive?

(ii) IfCbe a semi-algebra, thenis countably subadditive iff A Cwith A

i=1 Ai, Ai C implies

(A)

i=1(Ai).

2.3. Set functions on algebras


In this section, we give some general properties of a set function definedon an algebra A of subsets of an arbitrary set X.

2.3.1. Theorem:LetA be an algebra of subsets of a setXand let : A [0, ] be a set

function. Then the following hold:

(i) If is finitely additive and(B)< +, then(B A) =(B) (A)for every A, B A with A B. In particular, () = 0 if is finitelyadditive and(B)< + for someB A.

(ii) If is finitely additive, then is also monotone.

(iii) Let() = 0. Then is countably additive iff is both finitely additiveand countably subadditive.


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16 2. Measure

Another characterization of countable additivity of set functions defined

on algebras is given in the next theorem.

2.3.2. Theorem:LetA be an algebra of subsets of a setX and let : A [0, ] be suchthat() = 0.

(a) If is countably additive then the following hold:(i) For any A A, if A=

n=1 An, where An A and An

An+1 n, then

(A) = limn

(An).

This is called the continuity from below of atA.(ii) For any A A, if A=

n=1 An, where An A with An An+1 n and(An)< + for somen, then

limn

(An) =(A).

This is called the continuity from above of at A.

Conversely,

(b) If is finitely additive and(i) holds, then is countably additive.

(c) If(X)< +, is finitely additive and(ii)holds, then is count-ably additive.

Exercises:

(2.10 ) (i) In the proofs of part (ii) and part (c) of theorem 2.3.2, where doyou think we used the hypothesis that (X)< +? Do you thinkthis condition is necessary?

(ii) LetA be an algebra of subsets of a set Xand : A [0, ] bea finitely additive set function such that (X) < +. Show thatthe following statements are equivalent:

(a) limk (Ak) = 0, whenever {Ak}k1 is a sequence in A withAk Ak+1 k, and

k=1 Ak =.

(b) is countably additive.

(2.11 ) Extend the claim of theorem 2.3.2 when A is only a semi-algebraof subsets ofX.(Hint: Use exercise 1.8)

(2.12 ) Let A be a -algebra and : A [0, ] be a measure. For anysequence{En}n1 inA, show that


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2.4. Uniqueness problem for measures 17

(i) (lim infn En) lim infn (En).

(ii) (lim supn En) lim supn (En).

(Hint: For a sequence {En}n1 of subsets of a set X,

lim infn

En:=

n=1

k=n

Ek limsupn

En:=

n=1

k=n

Ek. )

2.4. Uniqueness problem for measures


The problem that we want to analyze is the following: Let be a -finitemeasure on an algebra A of subsets ofX. Let 1 and 2 be two measureson S(A), the -algebra generated by the algebra A, such that 1(A) =2(A) A A. Is 1= 2?

2.4.1. Definition:Let Cbe a collection of subsets ofXand let: C [0, ] be a set function.We say is totally finite (or just finite) if (A) < + A C. Theset function is said to be sigma finite(written as -finite) if there existpairwise disjoint setsXn C, n= 1, 2, . . . ,such that(Xn)< +for every

n and X=

n=1 Xn.

2.4.2. Examples:

(i) The length function on the class of intervals is -finite.

(ii) Let A = A(I), the algebra generated by left-open right-closed intervalsin R. For A A, let (A) = + if A = and () = 0. Then is ameasure on A and it is not -finite. Let R be chosen arbitrarily andfixed. LetA denote the algebra of subsets ofRgenerated by A and {}.Define for A A,

(A) := + ifA \ {} =,

0 if eitherA = or A = {}.

It is easy to check that is also a measure on A and is not -finite.

(iii) Let Xdenote the set of rationals in (0,1] and let A be as in (ii) above.Show that the-algebraXS(A) =P(X) and that every nonempty set inthe algebraXAhas an infinite number of points. For anyEXS(A)and c >0, define c(E) =c times the number of points in E. Show thatc is a measure on X S(A) =S(X A) and is not -finite.


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18 2. Measure

2.4.3. Proposition:

Let 1 and 2 be totally-finite measures on a -algebra S. Then the classM= {E S |1(E) =2(E)} has the following properties:

(i) M is a monotone class.

(ii) IfS =S(A), and 1(A) = 2(A) A A, then1(A) = 2(A)AS(A).

Exercises:

(2.13 ) LetA be an algebra of subsets of a set X. Let 1and 2be-finitemeasures on a -algebraS(A) such that 1(A) =2(A) A A.Then, 1(A) =2(A)A S(A).

(2.14) Show that a measure defined on an algebra A of subsets of a setXis finite if and only if(X)< +.


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Chapter 3

Construction of

measures


3.1. Extension from semi-algebra to the

generated algebra

3.1.1. Definition:LetCi, i= 1, 2 be classes of subsets of a set X, with C1 C2. Let function1 : C1 [0, +] and 2 : C2 [0, +] be set functions. The setfunction2 is called anextensionof1 if1(E) =2(E) for every E 1.

3.1.2. Examples:In example 2.4.2(ii), each is an extension of the measure . Similarly, inexample 2.4.2(iii) each c is an extension of.

Above examples show that in general a measure on an algebra Acanhave more than one extension to S(A), the -algebra generated by A.

Our next theorem describes a method of uniquely extending a measurefrom a semi-algebra to the algebra generated by it.

3.1.3. Theorem:Given a measure on a semi-algebraC, there exists a unique measure onF(C) such that(E) =(E) for everyE C.

The measure is called the extensionof.

19


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20 3. Construction of measures

[Hint: For E F(C), with E= ni=1 Eifor pairwise disjoint sets E1, . . . , E nC, define

(E) :=

i=1

(Ei).]

Exercises:

(3.1) LetCbe a collection of subsets of a set X and : C [0, ] be aset function. If is a measure onC,show that is finitely additive.Is monotone? Countably subadditive?

(3.2) IfCbe a semi-algebra, then is countably subadditive iff A Cwith A

i=1 Ai, Ai C implies

(A)

i=1

(Ai).

(3.3) Using theorem 3.1.3, show that length function, which is initiallydefined on the semi-algebra Iof all intervals, can be uniquely ex-tended to a set function on F(I), the algebra generated. It isworth mentioning a result due to S.M. Ulam (1930) which, underthe assumption of the continuum hypothesis, implies that it isnot possible to extend the length function to all subsets ofR.

Theorem (Ulam): Let be a measure defined on all subsets ofR such that((n, n+ 1])< n Z and ({x}) = 0 for everyx R. Then(E) = 0 for everyE R.

3.1.3. Remark:Ulams theorem shows the impossibility of extending the length functionfrom intervals to all subsets ofR, assuming the continuum hypothesis. Weshall see later that similar results can be proved if one assumes the axiomof choice. Ulams result uses the property of that ({x}) = 0 x R,and the fact that ([n, n+ 1]) < + for every n Z. In the later resultswe shall use the translation invariance property of the length function .

3.2. Extension from algebra to the generated

-algebra

Concepts and Examples:

Given an arbitrary measure on an algebra A of subsets of a set X, ouraim is to try to extend to a class of subsets of Xwhich is larger than


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3.2. Extension from algebra to the generated-algebra 21

A. Intuitively, sets A in A are those whose size (A) can be measured

accurately. The approximate size of any set E X is given by the outermeasure as defined next. Recall that, for any nonempty set A [0, +],we write inf(A) := infA [0, +) ifA [0, +) = , and inf(A) := +otherwise.

3.2.1. Definition:LetA be an algebra of subsets of a set Xand : A [0, ] be a measureonA. ForE X, define

(E) := inf

i=1(Ai) Ai A,

i=1Ai E

.

The set function is called the outer measure induced by .

3.2.2. Proposition (Properties of outer measure):The set function :P(X)[0, ] has the following properties:

(i) () = 0 and(A) 0 A X.

(ii) is monotone, i.e.,

(A) (B) whenever A B X.

(iii) is countably subadditive, i.e.,

(A)

i=1

(Ai) whenever A=

i=1

Ai.

(iv) is an extension of , i.e.,

(A) =(A) if A A.

3.2.3. Remarks:(i) A set function defined on all subsets of a set X is called an outermeasureifhas properties (i), (ii) and (iii) in proposition 3.2.2. The outermeasure induced by is characterized by the property that if is anyouter measure on X such that (A) = (A) A A, then (A) (A).In other words, is the largest of all the outer measures which agree with

on A.

(ii) In the definition of (E) the infimum is taken over the all possiblecountable coverings ofE. To see that finite coverings will not suffice, con-sider E := Q (0, 1), the set of all rationals in (0, 1), and let I1, I2, . . . , I nbe any finite collection of open intervals such that E

ni=1 Ii. Then it is

easy to see thatn

i=1 (Ii) 1. This will imply (E) 1 if only finite

coverings are considered in the definition of, which contradicts the factthat (E) = 0, Ebeing a countable set.


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3.2.4. Example:

LetA:= {A R | EitherA or Ac is countable}.

It is easy to see that A is a -algebra. For A A, let (A) = 0 if A iscountable and (A) = 1 ifAc is countable. Then, is a measure on A. Let be the outer measure induced by on P(R). It follows from proposition3.2.2 that is countably subadditive on P(R).

IfA R is countable, then clearlyA A,and hence(A) =(A) = 0.Further,(A) = 1 iffA is uncountable. Since, R = (, 0] (0, ) and

(R) = 1 < 2 =(, 0] + (0, ).

This shows that need not be even finitely additive on all subsets.

Exercises:

(3.4) Show that (E), as in definition 3.2.1, is well-defined.

(3.5) The set function(E) can take the value +for some sets E .

(3.6) Show that

(E) = inf

i=1

(Ai) Ai A, Ai Aj = for i=j and

i=1

AiE

.

(3.7) LetXbe any nonempty set and letA be any algebra of subsets ofX. Let x0 X be fixed. For A A, define

(A) :=

0 ifx0A,1 ifx0 A.

Show that is countably additive. Let be the outer measureinduced by . Show that (A) is either 0 or 1 for every A X,and(A) = 1 ifx0 A. Can you conclude that(A) = 1 impliesx0 A? Show that this is possible if{x0} A.

3.3. Choosing nice sets: Measurable sets


In the previous section we defined the notion of, the outer measure in-duced by on all subsets ofX. We saw that (A) = (A), A A, butin general need not be even finitely additive on P(X).Let us try to iden-tify some subclassS ofP(X) such that restricted toSwill be countableadditive. This is the classS which we call the class of nice subsets ofX.


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3.3. Choosing nice sets: Measurable sets 23

But the problem is how to pick these nice sets? This motivates our next

definition.

3.3.1. Definition:A subset EX is said to be -measurable if for every Y X,

(Y) =(Y E) + (Y Ec). (3.1)

We denote by S the class of all -measurable subsets ofX. Note thatE S iffEc S, due to the symmetry in equation (3.1).

Thus, a setEXis a nice set if we use it as a knife to cut any subsetY ofX into two parts,Y EandY Ec, so that their sizes (Y E) and

(Y E

c

) add up to give the size

(Y) ofY.Thus a nice set is in a sensea sharp knife.

3.3.2. Theorem:LetE X. Show that the following statements are equivalent:

(i) E S.

(ii) For everyY X,

(Y) (Y E) + (Y Ec).

(iii) For everyY X, with(Y)< +,

(Y)

(Y E) +

(Y Ec

).(iv) For everyA A,

(A) (A E) + (A Ec).

We give an equivalent definition of measurable sets when (X)< +.

3.3.3. Theorem:Let(X)< +. ThenE X is-measurable iff

(X) =(E) + (Ec)

Next, we check that S is indeed the required collection of nice sets.

3.3.4. Proposition:The collectionS has the following properties:

(i) A S.

(ii) S is an algebra of subsets ofX, S(A) S, and restricted toS is finitely additive.


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(iii) If An S, n = 1, 2, . . . , thenn=1 An S and restricted toS is countably additive.

(iv) LetN :={E X |(E) = 0}. ThenN S.

This together with proposition 2.4.3 gives us the following:

3.3.5. Theorem:Let be a measure on an algebraA of subsets of a setA. If is-finite, thenthere exists a unique extension of to a measure onS(A), the-algebragenerated byA.

3.3.6. Remark:Theorems 3.2.2.(iv) and 3.3.4 together give us a method of constructing anextension of a measure defined on an algebraA to a classS S(A) A.

Exercises:

(3.8) Identify the collection of -measurable sets for as in example3.2.4.

(3.9) LetX= [a, b] and letSbe the-algebra of subsets ofXgeneratedby all subintervals of [a, b]. Let , be finite measures on Ssuch that([a, c]) =([a, c]), c [a, b]. Show that (E) =(E) E S.

(3.10) LetFbe the measure on the algebra A(I) as given in proposition

2.2.1. LetF itself denote the unique extension ofF to LF, the-algebra ofF-measurable sets, as given by theorem 3.3.4. Showthat

(i) BR LF.(ii) F({x}) =F(x) lim

yxF(y). Deduce that the function F is

continuous atx iffF({x}) = 0.(iii) Let F be differentiable with bounded derivative. IfA R is

a null set, then F(A) = 0.The measure F is called the Lebesgue-Stieltjes measure in-duced by the distribution function F.

3.4. Completion of a measure space


Theorem 3.3.4 showed that, given a -finite measure on an algebra A ofsubsets of a set X, can be extended to a unique measure on the -algebra S of-measurable subsets ofX, and S S(A). In this sectionwe describe the relation between S and the sets in S(A).


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3.4. Completion of a measure space 25

We first give an equivalent ways of describing (E) for any set E

X, being the outer measure induced by . Let A denote the collectionof sets of the form

i=1 Ai, Ai A.

3.4.1. Proposition:For every setE X,

(E) = inf {(A)| A A, EA}

= inf {(A)| A S(A), EA}

= inf {(A)| A S, E A}.

3.4.2. Proposition:For every E X, there exists a set F S(A) such that E F, (E) =(F) and(F\ E) = 0.

The set F is called a measurable cover ofE .

3.4.3. Corollary:LetEX. Then there exists a setKE, K S(A), such that(A) = 0

for every setA E\ K.

The set K is called a measurable kernel ofE.

3.4.4. Definition:LetXbe a nonempty set,Sa-algebra of subsets ofXand a measure onS.The pair (X, S) is called ameasurable spaceand the triple (X, S, ) iscalled a measure space. Elements ofSare normally called measurablesets.

Till now what we have done is that, given a measure on an algebra A ofsubsets of a set X, we have constructed the measure spaces (X, S(A), ),(X, S, ) and exhibited the relations between them. The measure space(X, S, ) has the property that ifE Xand(E) = 0,thenE S.Thisproperty is called the completeness of the measure space (X, S, ).

The measure space (X, S(A),

) need not be complete in general. However,S is obtainable from S(A) andN := {E X|(E) = 0} by

S = S(A) N := {E N |E S(A), N N }.

One calls (X, S, ) thecompletionof (X, S(A), ).This construction canbe put in a general context as follows.

3.4.5. Definition:Let (X, S, ) be a measure space and let N :={E X |EN for some


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N Swith(N) = 0}.One says (X, S, ) iscompleteifN S.Elements

ofNare called the -null subsets ofX.

The abstraction of the relation between the measure spaces (X, S(A), )and (X, S, ) is described in the next theorem.

3.4.6. Theorem:Let(X, S, ) be a measure space and letNbe the class of-null sets (as indefinition 3.4.5). Let S N := {E N | E S, N N } and S N :={E N| E S, N N }. ThenS N =S N is a-algebra of subsets ofX. Let(E N) =(E), E S, N N. Then is a measure onS Nand(X, S N, )is a complete measure space,called thecompletionof the

measure space (X, S, ). (The measure space (X, S N, ) is also denotedby (X, S, ).

Finally we describe the relation between onP(X) and onA.

3.4.7. Proposition:Let be a measure on an algebraA of subsets of a setX and let be theinduced outer measure. LetE S be such that(E)< + and let >0be arbitrary. Then there exists a setF A such that(E F)< .

3.4.8. Note:Whenever (X, S, ) is a finite measure space with (X) = 1, it is called a

probability spaceand the measure is called aprobability. The reasonfor this terminology is that the triple (X, S, ) plays a fundamental rolein the axiomatic theory of probability. It gives a mathematical model foranalyzing statistical experiments. The setXrepresents the set of all possibleoutcomes of the experiment, the -algebra S represents the collection ofevents of interest in that experiment, and for every E S, the nonnegativenumber (E) is the probability that the event E occurs. For more detailssee Kolmogorov [9] and Parthasarathy [10].

Exercises:

(3.11 ) Let E X, and let G1, G2 be two measurable covers ofE. Showthat (G1G2) = 0.

(3.12) LetE1 E2 E3 . . . be subsets ofX. Then

n=1

En

= lim

n(En).

(3.13 ) Let EX, and let K1, K2 be two measurable kernels ofE. Showthat (K1K2) = 0.


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(3.14 ) Let N := {E X | (E) = 0}. Show that N is closed under

countable unions and

S =S(A) N :={E N |E S(A), N N },

where S is the-algebra of-measurable sets. Further, A S

(A) =(E), ifA = E N, with E S(A) and N N.

3.5. The Lebesgue measure


We now apply the extension theory of measures, developed in previous sec-tions, to the particular case whenX= R, A= A(I), the algebra generatedby all intervals, and on A is the length function as described in section

3.1. The outer measure

, induced by the length function , on all sub-sets ofR is called the Lebesgue outer measure and can be described asfollows: for E R,

(E) := inf

i=1

(Ii) Ii I i, Ii Ij = for i =j and E

i=1

Ii

.

The-algebra of

-measurable sets, as obtained in section 3.4, is called the-algebra ofLebesgue measurable sets and is denoted by LR, or simply

byL.The -algebraS(I) =S(A) :=BR, generated by all intervals, is calledthe-algebra ofBorel subsetsofR.We denote the restriction of

toL or

BR by itself. The measure space (R, L, ) is called the Lebesgue mea-sure space and is called the Lebesgue measure. We note that since on I is -finite (e.g., R =

+n=(n, n+ 1]), the extension of to BR is

unique. It is natural to ask the question:What is the relation between the classes BR, L and P(R)?

As a special case of theorem 3.4.6, we have L= BR N, where

N :={N R |NE BR, (E) = 0}.

Thus,BR L P(R).

The question arises:

Is BR a proper subset ofL?

That is, are there sets in L which are not Borel sets?. First of all, we note

that Cantors ternary set C N L. Further, E C, then

(E) = 0and hence E L. In other words, P(C) L. Thus, the cardinality of L


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is at least 2c (here c denotes the cardinality of the real line, also called the

cardinality of the continuum). SinceL P(R), we get the cardinalityof L to be 2c. On the other hand, BR is the -algebra generated by allopen intervals of R with rational endpoints. One can show that the -algebra BR of Borel subsets ofR has cardinality c, that of the continuum.Thus, there exist sets which are Lebesgue measurable but are not Borel sets.The actual construction of such sets is not easy. One such class of sets iscalled analytic sets. An analytic setis a set which can be represented as acontinuous image of a Borel set. For a detailed discussion on analytic sets,see Srivastava [38], Parthasarathy [29].

Since L, the class of all Lebesgue measurable subsets ofR, has 2c ele-ments, i.e., same as that ofP(R), the natural question arises:

Is L = P(R)?

We stated earlier that, if we assume the continuum hypothesis, it is notpossible to define a countably additive set function on P(R) such that({x}) = 0 x R. In particular, if we assume the continuum hypothesis,we cannot extend to all subsets ofR. HenceL =P(R). What can be saidif one does not assume the continuum hypothesis? To answer this question,one can either try to construct a set ER such that E L ,or, assumingthat such a set exists, try to see whether one can reach a contradiction.G. Vitali (1905), F. Bernstein (1908), H. Rademacher (1916) and othersconstructed such sets assuming the axiom of choice (see appendix B). The

example of Vitali used the translation invariance property of the Lebesguemeasure, and that of Bernstein used the regularity properties of the Lebesguemeasure. Rademacher proved that every set of positive outer Lebesguemeasure includes a Lebesgue nonmeasurable set. Even today, more andmore nonmeasurable sets with additional properties are being constructed.For example, one can construct nonmeasurable subsets A of R such that

(A I) =

(I) for every interval I R. Of course, all these constructionsare under the assumption of the axiom of choice. Lebesgue himself didnot accept such constructions. In 1970, R. Solovay [37] proved that if oneincludes the statement all subsets of R are Lebesgue measurable as anaxiom in set theory, then it is consistent with the other axioms of set theoryif the axiom of choice is not assumed. Construction of a nonmeasurableset(due to Vitali), assuming the axiom of choice, is given in exercise (3.26).

We recall that the -algebra BR includes all topologically nice subsetsofR, such as open sets, closed sets and compact sets. Also, forE BR, ifwe transform E with respect to the group operation on R, e.g., for x R,considerE+ x:= {y+ x| y E}, thenE+ x BR.For this, note that themap y x+y is a homeomorphism ofR onto R, and hence E+ x Bfor every open set E. We leave it for the reader to verify (using -algebra


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techniques) that this is true for all sets E BR. The relation of on L

with on topologically nice subsets ofR and the question as to whetherE+ x Lfor E L, x R, i.e., do the group operations on Rpreserve theclass of Lebesgue measurable sets, will be analyzed in this section. We give

below some properties of

which are also of interest.

3.5.1. Theorem:LetE R and

(E)< +. Then, given >0, there exists a setF which

is a finite disjoint union of open intervals and is such that

(E F)< .

We give next some more characterizations of Lebesgue measurable sets.

3.5.2. Theorem:For any setE R the following statements are equivalent:

(i) E L, i.e., E is Lebesgue measurable.

(ii) For every >0, there exists an open setG such that

E G and

(G\ E)< .

(iii) For every >0, there exists a closed setF such that

F E and

(E\ F)< .

(iv) There exists aG-setG such that

E G and

(G \ E) = 0.

(v) There exists anF-setF such that

F E and

(E\ F) = 0.

[Hint: Prove the following implications:

(i) = (ii) = (iv) = (i)and

(i) = (iii) = (v) = (i).]

3.5.3. Note:Theorem 3.5.2 tells us the relation between L, the class of Lebesgue mea-surable sets, and the topologically nice sets, e.g., open sets and closed sets.The property that for E L and > 0, there exists an open set G Ewith(G \ E)< can be stated equivalently as:

(E) = inf{(U)| U open, UE}.


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This is called the outer regularity of. Other examples of outer regular

measures on R (in fact any metric space) are given in the exercise 3.25.

Another topologically nice class of subsets ofR is that of compact subsetsofR. It is natural to ask the question: does there exist a relation betweenL and the class of compact subsets ofR? Let Kbe any compact subset ofR. Since Kis closed (and bounded), clearly K BR L and (K)< +.It is natural to ask the question: can one obtain(E) for a set E BR, if(E)< +, from the knowledge of(K), Kcompact in R? The answer isgiven by the next proposition.

3.5.4. Proposition:

LetE L with0< (E)< + and let >0 be given. Then there exists acompact setKE such that(E\ K)< .

On the set R,we have the group structure given by the binary operation ofthe addition of two real numbers. We analyze the behavior of on L underthe map y y + x, y R and x R fixed. We saw that A+ x is aLebesgue measurable set whenever A is Lebesgue measurable andx R. Itis natural to ask the question: forA L and x R, is (A+x) = (A)?The answer is given by the following:

3.5.5. Theorem (Translation invariance property):LetE L. ThenE+ x L for everyx R, and(E+ x) =(E).

We saw that Lebesgue measure is the unique extension of the length functionfrom the classIof intervals to BR,the -algebra of Borel subsets ofR.Thisgave us a measure on BR with the following properties:

(i) For every nonempty open set U, (U)> 0.

(ii) For every compact set K, (K)< +.

(iii) For every E BR,

(E) = inf {(U)| U open, UE},= sup{(C)| CE, Cclosed}.

If(E)< +, then we also have

(E) = sup{(K)| KE, K compact}.

(iv) For every E BR and x R, E+ x BR and (E+ x) =(E).

Thus the Lebesgue measure is a translation invariant -finite regularmeasure on BR. The question arises: are there other -finite measures onBRwith these properties? Obviously, ifc >0 thenc defined by (c)(E) :=c(E), E BR, is also a -finite measure and is translation invariant. Infact the following hold:


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3.5.6. Theorem:

Let be a measure onBR such that

(i) (U)> 0 for every nonempty open setU R.

(ii) (K)< + for every compact setK R.

(iii) (E+ x) =(E), E BR and x R.

Then there exists a positive real number c such that (E) = c(E)E BR.

3.5.7. Note:In fact the above theorem has a far-reaching generalization to abstract topo-logical groups. Let us recall that the set of real numbers R is a group underthe binary operation +, the addition of real numbers. Also, there is a topol-ogy on R which respects the group structure, i.e., the maps (t, s)t + sand t t from R R R and R R, respectively, are continuouswhen R R is given the product topology. In an abstract setting, ifG is aset with a binary operation and a topology T such that (G, ) is a groupand the maps G G G, (g, h) g.h and G G, g g1 arecontinuous with respect to the product topology on G G, one calls G atopological group. Given a topological group, let BGdenote the-algebragenerated by open subsets ofG, called the -algebra of Borel subsets ofG.The question arises: does there exist a-finite measure on G such that ithas the properties as given in theorem 3.5.5? A celebrated theorem due to A.

Haar states that such a measure exists and is unique up to a multiplicative(positive) constant ifGis locally-compact. Such a measure is called a (right)Haar measure on G. Theorem 3.5.5 then states that for the topologicalgroup R, the Lebesgue measure is a Haar measure. Consider the group(R \ {0}, ), where R \ {0}= {t R|t= 0}and is the usual multiplicationof real numbers. LetR \ {0} be given the subspace topology from R. It iseasy to show that R \ {0} is a topological group and, E BR\{0},

(E) :=

E

1

|x|d(x) (3.2)

is a Haar measure on R \ {0}.

3.5.8. Note:In the previous sections we have seen how the general extension theory,as developed earlier, can be applied to the particular situation when thesemi-algebra is that of intervals and the set function is the length function.More generally, if we consider the semi-algebra Iof left-open right-closedintervals in R and consider F : R R as a monotonically increasingright continuous function, then we can construct a countably additive setfunction F on the semi-algebra I, as in example 1.1.2. Using theorem


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3.4.6, we can construct a complete measure F on a -algebra of subsets

ofR which includes BR.This measureF is called the Lebesgue-Stieltjesmeasure induced by the function F. Note that F has the property thatF(a, b] < + a, b R, a < b. Conversely, given a measure on BRsuch that (a, b] < + a, b R, a < b , we can restrict it to I and,using proposition 2.2.2, define a monotonically increasing right continuousfunction F : R R such that the unique Lebesgue-Stieltjes measure Finduced by F is nothing but (by the uniqueness of the extension). Thusmeasures on BR which have the property that (a, b] < + a < bcan be looked upon as a Lebesgue-Stieltjes measure F for some F. Wepoint out that it is possible to find different F1, F2 : R R such thatboth are monotonically increasing and right continuous and F1 = F2 . If

is finite measure, i.e., (R) < +, then it is easy to see that F(x) :=(, x], x R, is a monotonically increasing right continuous functionsuch that = F. ThisFis calledthe distribution function of. When(R) = 1, is called aprobabilityand its distribution functionF, which ismonotonically increasing and is right continuous with lim

x[F(x) F(x)]

=F(R) = 1, is called aprobability distribution function.

Exercises:

(3.15) LetI0denote the collection of allopen intervalsofR.ForE X,show that

(E) = inf

i=1

(Ii) Ii I0 i, for i =j and E

i=1

Ii

.

(3.16) Let E R and let > 0 be arbitrary. Show that there exists an

open setU Esuch that(U)

(E)+.Can you also concludethat (U\ E) ?

(3.17) For E R, let

diameter(E) := sup{|x y| |x, y E}.

Show that

(E) diameter(E).

(3.18) Show that for E R,

(E) = 0 if and only if for every > 0,there exist a sequence{In}n1of intervals such thatE n=1and

(n=1\ E) < . Such sets are called Lebesgue null sets.Provethe following:

(i) Every singleton set {x}, x R, is a null set. Also every finiteset is a null set.

(ii) Any countably infinite set S= {x1, x2, x3, . . .} is a null set.(iii) Q, the set of rational numbers, is a null subset ofR.


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(iv) Every subset of a null set is also a null set.

(iv) Let A1, A2, . . . , An, . . . be null sets. Then

n=1 An is a nullset.

(v) Let E [a, b] be any set which has only a finite number oflimit points. Can Ebe uncountable? Can you say E is a nullset?

(vi) LetEbe a null subset ofR andx R.What can you say aboutthe sets E+ x:= {y+ x| y E} and xE:= {xy | y E}?

(vii) LetIbe an interval having at least two distinct points. Showthat Iis not a null set.

(viii) IfEcontains an interval of positive length, show that it is nota null set. Is the converse true, i.e., ifE

R is not a null set,

then does Econtain an interval of positive length?(ix) Show that Cantors ternary set is an uncountable null set.

(3.19) LetE[0, 1] be such that

([0, 1] \ E) = 0. Show that Eis densein [0, 1]

(3.20) LetE R be such that

(E) = 0. Show thatEhas empty interior.

(3.21) Let{En}n1 be any increasing sequence of subsets (not necessarilymeasurable) ofR. Then,

n=1 En

= limn

(En).

(3.22) LetE R. Show that the following statements are equivalent:(i) E L.

(ii)

(I) =

(E I) +

(Ec I) for every interval I .(iii) E [n, n + 1) L for every n Z.

(iv)

(E [n, n + 1)) +

(Ec [n, n + 1)) = 1 for every n Z.

(3.23) LetA Land x R. Using theorem 4.2.2, show that(i) A + x L, where A + x:= {y+ x| y A}.(ii) A L, where A:= {y| y A}.

(3.24) Let (X, d) be any metric space and let be a measure on BX

, the-algebra generated by open subsets ofX, called the -algebra ofBorel subsets of X. The measure is called outer regular ifE BX,

(E) = inf {(U)| U open, UE},

= sup{(C)| Cclosed, CE}. (3.3)

(i) If(X)< +, show that is outer regular iff for every EBX and > 0 given, there exist an open set U and a closed


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setC such that

U EC and (U C)< .

(ii) For A X, let

d(x, A) := inf{d(x, y)| y A}.

Show that for every A X, xd(x, A) is a uniformly con-tinuous function.(Hint: |d(x, A) d(y, A)| d(x, y) x, y.)

(iii) Let(X)< +and

S :={E BX |(3.3) holds for E}.

Show thatS is a -algebra of subsets ofX.(iv) LetCbe any closed set in X. Show that C S.

(Hint: C=

n=1{x X :d(x, C)< 1/n}.)(v) Show that is outer regular on BX.

(3.25) LetE BR. Show that E+ x BR for every x R.

(3.26) LetE L and x R. Let

xE:= {xy| y E} and E:= {x| x E}.

Show that E,xE L for every x E. Compute (xE) and(E) in terms of(E).

(3.27) Example (Vitali)(Existence of nonmeasurable sets:

Define a relation on [0,1] as follows: forx, y [0, 1], we say x isrelated to y, written as x y, if x y is a rational. Prove thefollowing:

(i) Show that is an equivalence relation on [0, 1].(ii) Let {E}I denote the set of equivalence classes of elements

of [0, 1]. Using the axiom of choice, choose exactly one elementx E for every Iand construct the set E :={x| I}.Let r1, r2, . . . , rn, . . . denote an enumeration of the rationals in[1, 1]. Let

En:= rn+ E, n= 1, 2, . . . .

Show that En Em= for n =m and En [1, 2] for every n.Deuce that

[0, 1]

n=1

En [1, 2].

(iii) Show that Eis not Lebesgue measurable.


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Chapter 4

Integration

Unless stated otherwise, we shall work on a fixed -finite measure space(X, S, ).

4.1. Integral of nonnegative simple measurable

functions


4.1.1. Definition:

Lets : X[0, ] be defined by

s(x) =n

i=1

aiAi (x), x X,

wherenis some positive integer;a1, a2, . . . , anare nonnegative extended realnumbers; Ai Sfor everyi; Ai Aj = for i =j; and

ni=1 Ai= X. Such a

function s is called a nonnegative simple measurable function on (X, S)and

ni=1 aiAi (x) is called a representation ofs. We say

ni=1 aiAi is

the standard representationofs ifa1, a2, . . . , an are all distinct.

We denote by L+0 the class of all nonnegative simple measurable func-

tions on (X, S).Note that s L+0 iffs takes only a finite number of distinct values, say

a1, a2, . . . , an, the value ai being taken on the set Ai S, i = 1, 2, . . . , n .And in that case its standard representation is

ni=1 aiAi . Also note that

the class L+0 depends only upon the set Xand the-algebraS; the measure plays no part in the definition of functions in L+0.

35


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36 4. Integration

4.1.2. Examples:

(i) Clearly, ifs(x) c for some c [0, +], thens L+0.

(ii) For A X, consider A : X[0, +], the indicator function of theset A, i.e., A(x) = 1 ifx A and A (x) = 0 ifxA. ThenA L

+0 iff

A S, for A =aA+ bAc witha = 1 and b = 0.

(iii) Let A, B S. Thens = A B L+0, since s = AB .

(iv) LetA, B S. IfA B = , then clearly A+ B =AB L+0.

4.1.3. Definition:For s L+0 with a representation s =

ni=1 aiAi , we define

s(x)d(x),

the integral ofs with respect to , by s(x)d(x) :=

ni=1

ai(Ai).

The integral

s(x)d(x) is also denoted by

sd. It is well-defined (seeexercise (4.1)).

4.1.4. Proposition:

Fors, s1, s2 L+0 and R with 0, the following hold:

(i) 0

sd +.

(ii) s L+0 and (s)d=

sd.

(iii) s1+ s2 L+0 and

(s1+ s2)d =

s1d +

s2d.

(iv) ForE Swe havesE L+0, and the set function

E(E) :=

sEd

is a measure onS. Further, (E) = 0 whenever(E) = 0, E S.

Before we proceed further, we observe that

s(x)d(x) is well-defined,(see

Exercise 4.1). The properties of

sd, for s L+0 , are given by the nextproposition.


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4.1. Integral of nonnegative simple measurable functions 37

4.1.5. Proposition:

Lets L+0. Then the following hold:

(i) If {sn}n1 is any increasing sequence in L+0 such that limn sn(x)

=s(x), x X, then sd= lim

n

snd.

(ii)

sd= sup

s

d| 0 s

s, s

L+0

.

Exercises:

(4.1) Show that for s L+0,

s(x)d(x) is well-defined by proving thefollowing: let

s=n

i=1

aiAi =m

j=1

bjBj ,

where {A1, . . . , An} and {B1, . . . , Bm} are partitions ofX by ele-ments ofS, then(ii)

s =n

i=1

ai

m

j=1

AiBj =m

j=1

bj

n

i=1

AiBj .

(ii)n

i=1

ai(Ai) =m

j=1

bj(Bj).

(iii)

s(x)d(x) is independent of the representation of the func-tion s(x) =

ni=1 aiAi .

(4.2) LetA, B S. Express the functions |A B | andA +B ABas indicator functions of sets in S and hence deduce that theybelong to L+0.

(4.3) Lets1, s2 L+0. Prove the following:(i) Ifs1 s2, then

s1d

s2d.

(ii) Let x X,

(s1 s2)(x) := max{s1(x), s2(x)} and (s1 s2)(x) := min{s1(x), s2(x)}.

Thens1 s2 and s1 s2 L+0 with

(s1 s2)d

sid

(s1 s2)d, i= 1, 2.


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38 4. Integration

(4.4) Express the functionsA B andA B ,for A, B S,in terms

of the functions A and B .

(4.5) Let X = (0, 1], S = B(0,1], the -algebra of Borel subsets of (0, 1]and = , the Lebesgue measure restricted to S. Forx (0, 1], ifx has non-terminating dyadic expansion x =

n=1 xn/2

n. Let

fi(x) :=

+1 if xi= 1,1 if xi= 0, i= 1, 2, . . . .

Show that for every i, there exists simple function si L+0 such

that fi= si 1. Compute

sid.

(4.6) Let s : X R be any nonnegative function such that the rangeofs is a finite set. Show thats L+

0

iffs1{t} Sfor everyt R.

(4.7) For s1, s2 L+0 show that {x | s1(x) s2(x)} S. Can you say

that the sets {x X | s1(x)> s2(x)}, {x X | s1(x) s2(x)} and{x X| s1(x) =s2(x)} are also elements ofS?

(4.8) Let s1, s2 L+0 be real valued and s1 s2. Let = s1 s2. Show

that L+0. Can you say that d=

s1d

s2d?

(4.9) Let {sn}n1 and {s

n}n1 be sequences in L+0 such that for each

x X, both{sn(x)}n1 and{s

n(x)}n1 are increasing and

limn sn(x) = limn s

n(x).

Show that

limn

snd= lim

n

s

nd.

(Hint: Apply exercise (4.3) and proposition 4.1.5 to {sn s

m}n forall fixedm to deduce that

s

md limn

snd.)

(4.10) Show that in general L+0 need not be closed under limiting oper-ations. For example, consider the Lebesgue measure space (R, L, )and construct a sequence {sn}n1in L

+0 such that limn

sn(x) =f(x)

exists butfL+

0.

4.2. Integral of nonnegative measurable

functions


Having defined the integral for functions s L+0, i.e., nonnegative simplemeasurable functions, we would like to extend it to a larger class.


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4.2.1. Definition:

(i) A nonnegative function f : X R is said to be S-measurable ifthere exists an increasing sequence of functions {sn}n1 in L

+0 such that

f(x) = limn

sn(x) x X.

If the underlying -algebra is clear from the context, a S-measurablefunction is also called measurable. We denote the set of all nonnegativemeasurable functions by L+.

(ii) For a function f L+, we define the integral offwith respect to by f(x)d(x) := lim

n

sn(x)d(x).

It follows from exercise (4.9) that forf L+,

f dis well-defined. Clearly,L+0 L

+ and

sdfor an element s L+0 is the same as

sd, for s as anelement ofL+. The next proposition gives a characterization of functions inL+ and the integrals of its elements. Another (intrinsic) characterization ofL+ will be given in the next section.

4.2.2. Proposition:Letf :X R be a nonnegative function. Then the following hold:

(i) f L+ iff there exist functionssn L+0, n 1,such that0 sn f n

andf(x) = limn

sn(x) x X.

(ii) Iff L+

ands L+0 is such that0 s f, then

sd

f d and

f d= sup

sd 0 s f, s L+0

.

4.2.3. Definition:Let (X, S, ) be a measure space and Y S. We say a property P holdsalmost everywhere on Ywith respect to the measure if the set E= {xY |Pdoes not hold at x} S and (E) = 0. We write this as P for a.e.x()or Pfor a.e. ()x Y.If the setY and are clear from the context,we shall simply write P a.e. For example if f : X R is a function,then f(x) = 0 for a.e. x() means that E = {x X | f(x) = 0} S and

(E) = 0.

We describe next the properties of

fd, for f L+.

4.2.4. Proposition:Letf, f1, f2 L+. Then the following hold:

(i)

f d 0 and forf1 f2 f1d

f2d.


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40 4. Integration

(ii) For, 0 we have(f1+ f2) L+ and (f1+ f2)d =

f1d +

f2d.

(iii) For everyE S we haveEf L+. If

(E) :=

Efd, E S,

then is a measure onS and(E) = 0 whenever(E) = 0.

The integral

f Ed is also denoted by

Ef d and is called theintegraloff over E.

(iv) Iff1(x) =f2(x) for a.e. x(), then f1d=

f2d.

Since the class L+0 is not closed under limiting operations (exercise 4.10),we defined the class L+ by taking limits of sequences in L+0. Naturally, weexpect L+ to be closed under limits. The next theorem discusses this andthe behavior of

f d under increasing limits, extending proposition 4.1.5

to functions in L+.

4.2.5. Theorem (Monotone convergence):Let {fn}n1 be an increasing sequence of functions in L+, and f(x) :=

limn

fn(x), x X. Thenf L+ and f d= lim

n

fnd.

4.2.6. Remark:If{fn}n1 is a sequence in L+ decreasing to a function f L+, then theequality

f d = limn

fnd need not hold. For example, let X = R,

S=L and = , the Lebesgue measure. Let fn= [n,) .Thenfn L+0

L+, and {fn}n1 decreases to f0.Clearly,

fnd= +for every n and

f d= 0. In fact, at this stage it is not clear whether f L+ whenever{fn}n1 decreases to f, with each fn L+. That this is true will be shownas a consequence of the characterization ofL+ proved in the next section.


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Exercises:

(4.11) Let f L+ and let {sn}n1 be in L+0 and such that {sn(x)}n1

is decreasing and xX, limn

sn(x) =f(x). Can you conclude

that f d= lim

n

snd?

(4.12) Letf L+. Show that f d= sup

sd 0 s(x) f(x) for a.e. x(), s L+0

.

(4.13) Let{fn}n1be an increasing sequence of functions in L+ such thatf(x) := lim

nfn(x) exists for a.e. x(). Show that f L+ and

f d= limn

fnd,

where f(x) is defined as an arbitrary constant for all those x forwhich lim

nfn(x) does not converge.

(4.14) Let (X)< , and let f L+ be a bounded function. LetP :={E1, E2, . . . , E n} be such that

ni=1 Ei= X, Ei Ej = for i=j

andEi S i. Such a Pis called a measurable partitionofX.Given a measurable partitionP ={E

1, . . . , E

n}, define

Mi:= sup{f(x) | x Ei} and mi:= inf{f(x) | x Ei}.

Let

P :=n

i=1

miEi and P :=n

i=1

MiEi .

Prove the following:(i) For every partition P, show that P, P L

+0 and P f

P.(ii)

f d = sup

Pd | P is a measurable partition ofX

,

= inf Pd | Pis a measurable partition ofX .(This gives an equivalent way of defining

fd,in a way sim-ilar to that for the Riemann integral.)

(iii) Let

= sup

0

sd |s L+0, s f

and

= inf

sd |s L+0, fs

.


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42 4. Integration

Show that

= sup

pd| Pis a measurable partition ofX

and

= inf

pd| Pis a measurable partition ofX

.

(iv) Deduce thatf L+ implies =

f d= .

Note:Exercise 4.14 tells us that forf L+,in defining

f dit is enough

to consider approximations off from below, as the approximations

from above will also give the same value for

fd.This is becausef L+, i.e., f is nonnegative measurable. The converse is alsotrue, i.e., if= , for f : X [0, ], then f L+. To see this,first note that

= < M((X)) < ,

where M is such that |f(x)| M x X. Thus for everyn wecan choose functions n, n L

+0 such that

n f n and

(n n)d .

(ii) Letf :X R be measurable, and R. Thenfis also measurable,since

{x X |(x)> c}=

{x X|f(x)< c/} if >0,X if = 0, c 0, if = 0, c > 0,

{x X|f(x)< c/} if


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g(x)}, {x X| {f(x) g(x)}, {x X| f(x)< g(x)}, {x X| f(x) g(x)}

and{x X| f(x) =g(x)} S.

4.3.8. Proposition:Letf, g: X R be measurable functions and let R be arbitrary. Let

A:= {x X| f(x) = +, g(X) =}{x X|f(x) =, g(x) = +}.

Define x X

(f+ g)(x) :=

f(x) + g(x) ifxA,

ifx A.

Thenf+ g: X R is a well-defined measurable function.

4.3.9. Proposition:Letf :X R be measurable and let : R R be such thatR {xR | (x) } BR, R. Then f is also measurable.

4.3.10. Proposition:Let fn : X R, n = 1, 2, . . ., be measurable functions. Then each of the

functions supn

fn, infn

fn,lim supn

fn and lim infn

fn is a measurable function.

In particular, if{fn}n1 converges to f, thenfis a measurable function.

4.3.11. Corollary:Let{fn}n1 be a sequence inL+. Then each of the functionssup

nfn, inf

nfn,

limsupn

fn andlim infn

fn is inL+. In particular, if limn

fn=: fexists, then

f L+.

Recall that in theorem 4.2.5 we analyzed the limit of

fnd for an in-creasing sequence of nonnegative measurable functions. We analyze next thebehavior of

fnd when {fn}n1 is not necessarily an increasing sequence

of nonnegative measurable functions.

4.3.12. Theorem (Fatous lemma):Let{fn}n1 be a sequence of nonnegative measurable functions. Then

lim infn

fn

d liminfn

fnd.

4.3.13. Proposition:Let(X, S)be a measurable space and letf :X R beS-measurable. Let be a measure on (X, S). Let g : X R be such that {x X| f(x) =g(x)}is a-null set. If(X, S, )is a complete measure space, theng is alsoS-measurable.


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46 4. Integration

Exercises:

(4.15)(i) Let f L+ and E Sbe such that f(x) > 0 for every x Eand (E)> 0. Show that

Efd >0.

(ii) Let f , g L+ be such that f d=

gd


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(iv) Let , , R be arbitrary. Define for x R,

(1/f)(x) :=

1/f(x) if f(x) {0, +, }, if f(x) = 0, if f(x) =, if f(x) = +.

Then 1/fis a measurable function.(v) Let R be arbitrary and A be as in proposition 4.3.8.

Define for x R,

(f g)(x) := f(x)g(x) if xA,

if x A.

Thenf g is a measurable function.

(4.19) Let f : X R be S-measurable. Show that |f| is also S-measurable. Give an example to show that the converse need notbe true.

(4.20) Let (X, S) be a measurable space such that for every function f :X R, f is S-measurable iff |f| is S-measurable. Show thatS=P(X).

(4.21) Letfn L, n= 1, 2, . . . .Show that the sets

{x X | {fn(x)}n is convergent}

and

{x X| {fn(x)}n1is Cauchy}

belong toS.

(4.22) Give an example to show that strict inequality can occur in Fatouslemma.

(4.23) Let {fn}n1be a sequence of functions in L+ and let

n=1 fn(x) =:f(x), x X. Show that f L+ and

f d =

n=1

fnd.

(4.24) Show that each of the functions f : R R defined below is L-measurable, and compute

f d:

(i) f(x) :=

0 if x 0,1/x if x >0.

(ii) f(x) :=Q(x),the indicator function ofQ,the set of rationals.


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48 4. Integration

(iii) f(x) :=

0 if x >1 or x


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(4.29) Letf L+ be a bounded function, sayf(x) N x Xand for

some N N. Show that f d = lim

n

N2nk=1

k 1

2n

x|

k 1

2n f(x) 0, there exists >0 such that

(E)< whenever, forE S, (E)< .

(iv) If(E) = 0 E, thenf(x) = 0 for a.e. x() onE.

4.4.4. Remark:It is easy to see that (iii) in the above proposition implies (ii). In fact (ii)also implies (iii). To see this, suppose (iii) does not hold. Then there existan >0 and sets En S, n 1, such that (En)< 2n but(En) . LetAn=

k=n Ek. Then{An}n1 is a decreasing sequence in S and (An)

(En) 2n

. Thus by theorem 2.3.2,

n=1

An

= lim

n(An) = 0.

On the other hand, (An) (En) , , contradicting (ii).

We prove next the most frequently used theorem which allows us tointerchange the operations of integration and limits.

4.4.5. Theorem (Lebesgues dominated convergence theorem):Let {fn}n1 be a sequence of measurable functions and let g L1() be

such that n, |fn(x)| g(x) for a.e. x(). Let {fn(x)}n1 converge tof(x) for a.e. x(). Then the following hold:

(i) fL1().

(ii)

f d = limn

fn d.

(iii) limn

|fn f|d= 0.

We state another version of this theorem, which is applicable to seriesof functions.


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52 4. Integration

4.4.6. Corollary:

Let {fn}n1 be a sequence of functions in L1() such that

n=1

|fn|d 0 such that |fn(x)| M a.e. x() andfn(x) f(x) a.e. x(). Thenf, fn L1(X, S, ) and

f d = limn

fnd.

4.4.8. Notes:

(i) The monotone convergence theorem and the dominated convergence theo-rem (along with its variations and versions) are the most important theoremsused for the interchange of integrals and limits.

(ii) Simple function technique: This is an important technique (similarto the-algebra technique) used very often to prove results about integrableand nonnegative measurable functions. Suppose we want to show that acertain claim () holds for all integrable functions. Then technique is thefollowing:

(1) Show that () holds for nonnegative simple measurable functions.

(2) Show that () holds for nonnegative measurable / integrable func-tions by approximating them by nonnegative simple measurablefunctions and using (1).

(3) Show that () holds for integrable functions f,by using (2) and the

fact that for fL1, f =f+ f and both f+, f L1.

The proof of the next proposition is an illustration of this technique.

4.4.9. Proposition:Let(X, S, )be a-finite measure space andfL1(X, S, )be nonnegative.For everyE S, let

(E) :=

E

fd.


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4.4. Integrable functions 53

Then is a finite measure on S. Further, f g L1(X, S, ) for every g

L1(X, S, ), and f d =

fgd.

We shall see some more applications of the dominated convergence the-orem in the remaining sections. In the next section we look at some specialproperties of L1(X, S, ) in the particular case when X = R, S = L, the-algebra of Lebesgue measurable sets, and = , the Lebesgue measure.

As an application of the dominated convergence theorem, we exhibit the

possibility of interchanging the order of integration and differentiation inthe next theorem.

4.4.10. Theorem:Let ft L1() for every t (a, b) R. Let t0 (a, b) be such that fora.e. x(), t ft(x) is differentiable in a neighborhood U of t0 and there

exists a functiong L1() such that dft

dt(x) g(x) for a.e. x() and for

everyt U. Then(t) :=

ft(x)d(x) is differentiable att0 and

(t0) =

dftdt

(x)

t0

d(x).

Exercises:

(4.32) For f L, prove the following:

(i) fL1() iff|f| L1(). Further, in either case

f d

|f|d.

(ii) IffL1(), then|f(x)|< +for a.e. x().

(4.33) Let(X)< + and letf L be such that |f(x)| Mfor a.e.x()and for some M . Show that fL1().

(4.34) LetfL1() and E S. Show that EfL1(), whereE

f d :=

Efd.

Further, ifE , F Sare disjoint sets, show thatEF

f d =

E

f d +

F

fd.


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54 4. Integration

(4.35) Let f L1() and Ei S, i 1, be such that Ei Ej = for

i=j . Show that the series

i=1

Ei f d is absolutely convergent,and ifE:=

i=1 Ei, then

i=1

Ei

f d =

E

fd.

(4.36) (i) For every >0 andfL1(), show that

{x X | |f(x)| } 1

|f|d < .

This is called Chebyshevs inequality.

(ii) LetfL1

(), and let there exist M >0 such that 1(E)

E

f d M

for every E S with 0 < (E) < . Show that |f(x)| M fora.e. x().

(4.37) Let be the Lebesgue measure on R, and let f L1(R, L, ) besuch that

(,x)f(t)d(t) = 0, x R.

Show that f(x) = 0 for a.e. ()x R.

(4.38) Let I R be an interval and t I, let ft L. Let g L1()be such that t, |ft(x)| g (x) for a.e. x(). Let t0 R be anyaccumulation point ofIand letf(x) := lim

tt0ft(x) exist for a.e.x().

ThenfL1() and f d = lim

tt0

ft(x)d(x).

Further, if x the function t ft(x) is continuous, then so isthe functionh(t) :=

ft(x) d, t I .

(Hint: Apply theorem 4.4.5 to every sequence tn t0.)

(4.39) Let {fn}n1and {gn}n1be sequences of measurable functions such

that|fn| gn n. Let f andg be measurable functions such thatlim

nfn(x) =f(x) for a.e. x() and lim

ngn(x) =g(x) for a.e. x().

If

limn

gn d =

g d < +,

show that

limn

fn d=

fd.

(Hint: Apply Fatous lemma to (gn fn) and (gn+ fn).)


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4.5. The Lebesgue integral and its relation with the Riemann integral 55

(4.40) Let {fn}n0be a sequence inL1(X, S, ).Show that |fn|dn1converges to

|f0|d iff

|fn f0|d

n1

converges to zero.

(Hint: Use exercise 4.38.)

(4.41) Let (X, S) be a measurable space and f :X R be S-measurable.Prove the following:

(i) S0 := {f1(E) | E BR}is the -algebra of subsets ofX, andS0 S.

(ii) If : R R is Borel measurable, i.e., 1(E) BR E BR, then f is anS0-measurable function onX.

(iii) If : X R is any S0-measurable function, then there existsa Borel measurable function : R R such that = f.

(Hint: Use the simple function technique and note that if isa simple S0-measurable function, then

=n

i=1

anf1(Ei)

for some positive integer n, ai R for each i, and Ei BR,then

=n

i=1

ai(Eif).

(4.42) Let {fn}n1 be a decreasing sequence of nonnegative functions inL1() such that fn(x) f(x). Show that

limx

fnd = 0 iff f(x) = 0 a.e. x().

(4.43) Let , be as in proposition 4.4.9. Let Sdenote the -algebra ofall -measurable subsets ofX. Prove the following:

(i) S S.(ii) There exist examples such that S is a proper subclass ofS.

Show thatS=S if{x X| f(x) = 0}= 0.

(4.44) Let (X, S, ) be a finite measure space and {fn}n1 be a sequencein L1() such that fn f uniformly. Show that fL1() and

limn

|fn f| d = 0.

Can the condition of(X)< +be dropped?

4.5. The Lebesgue integral and its relation with

the Riemann integral



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56 4. Integration

In this section we analyze the integral, as constructed in the previous section,

for the particular situation when X= R,S=L (the -algebra of Lebesguemeasurable sets) and= , the Lebesgue measure. The space L1(R, L, ),also denoted byL1(R) or L1(),is called the space ofLebesgue integrablefunctions on R, and

f d is called the Lebesgue integral of f. For any

set E L, we write L1(E) for the space of integrable functions on themeasure space (E, L E, ), where is restricted to L E. In the specialcase when E= [a, b], we would like to show that the new notion of integralforfL1[a, b] indeed extends the notion of Riemann integral. To be precise,we have the following theorem:

4.5.1. Theorem:

Letf : [a, b] R be a Riemann integrable function. ThenfL1[a, b] and f d=

ba

f(x)dx.

4.5.2. Remark:In fact, the proof of the above theorem includes a proof of the following: Iff R[a, b], then f is continuous a.e. x(). This is because

f(x) = limn

n(x) = limn

n(x) a.e. x().

Thus, if we put

E:= {x [a, b]| f(x) = limn

n(x) = limn

n(x)},

thenEis a Lebesgue measurable set and ([a, b] \ E) = 0. Forx E, givenan arbitrary >0, we can choose n0 such that

n0(x) n0(x)< . (4.2)

Further, if x is not a point in any partition Pn, then we can choose >0 such that whenever y [a, b] and |x y| < , then y belongs to thesame subinterval of the partition Pn0 to which x belongs. Thus by (4.2),|f(x) f(y)| < , showing that x is a point of continuity of f. Thus the

set of discontinuity points of f forms a subset of ([a, b])\ E) P, whereP is the set of partition points of Pn, n = 1, 2, . . . . Hence f is continuousalmost everywhere. Another proof of the converse is given in the one of theexercises.

Exercises:

(4.45) Letf : [a, b] R be bounded and continuous for a.e. x().


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4.5. The Lebesgue integral and its relation with the Riemann integral 57

(i) Let {Pn}n1 be any sequence of partitions of [a, b] such that

each Pn+1 is a refinement of Pn and Pn 0 as n .Let n, n be as constructed in theorem 4.5.1. Letx (a, b)be a point of continuity off. Show that

limn

n(x) = f(x) = limn

n(x).

(ii) Using (i) and the dominated convergence theorem, deduce thatfL1([a, b]) and

f d = limn

nd = lim

n

nd.

(iii) Show that f R[a, b] and f d =

ba

f(x)dx.

(4.46) Letf : [0, 1][0, ) be Riemann integrable on [, 1] for all >0.

Show thatfL1[0, 1] iff lim01

f(x)dxexists, and in that case f(x)d(x) = lim

0

1

f(x)dx.

(4.47) Letf(x) = 1/xp if 0 < x 1, andf(0) = 0.Find necessary and suf-

ficient condition onpsuch thatfL1[0, 1].Compute 1

0 f(x)d(x)in that case.(Hint: Use exercise 4.46.)

(4.48) (Mean value property):Let f : [a, b] R be a continuousfunction and let E [a, b], E L, be such that (E) > 0. Showthat there exists a real number such that

Ef(x)(x) = (E).

(4.49) Let fL1(R), and let g : R R be a measurable function suchthat g(x) for a.e. x(). Show that f g L1(R) and thereexists [, ] such that

|f|gd =

|f|d.

(4.50) LetfL1(R, L, ) and let a R be fixed. Define

F(x) :=

[a,x] f(t) d(t) for x a,[x,a] f(t)d(t) for x a.

Show that F is continuous.


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58 4. Integration

(Hint: Without loss of generality takef 0 and show that F is

continuous from the left and right. In fact, Fis actually uniformlycontinuous.)

(4.51) Let f L1(R, L, ) and let c R be a point of continuity of f.Show that

limn

n

[c, c + 1/n]

f(x) d(x) = f(c).

(4.52) (Arzelas theorem):Let{fn}n1 be a sequence of Riemann integrable functions on[a, b]such that for some M > 0, |fn(x)| < M x [a, b] and n =1, 2, . . . . Let fn(x) f(x) x [a, b] and let f be Riemann

integrable on [a, b]. Then

limn

ba

fn(x)dx=

ba

f(x)dx.

(4.53) Let f : [a, b] R be any constant function. Show that f L1[a, b].

(4.54) Let f : [a, b] R be any bounded measurable function. Showthat fL1[a, b].

(4.55) Let f : [a, b] R be any continuous function. Show that f L1[a, b].

(4.56) Let f L1(R) be such that

Kf d = 0 for every compact setK R. Show that f(x) = 0 for a.e. x().

(4.57) Let {fn}n1 be a decreasing sequence of nonnegative functions inC(a, b) and let f1 L1(a, b). If

n=1(1)

n1fn C(a, b), showthat b

a

n=1

(1)n1fn(x)

dx =

n=1

(1)n1

ba

fn(x)dx

.

(Hint: For every n,

nk=1(1)

kfk fL1(a, b)).

(4.58) Give examples to show that analogues of the monotone convergence

theorem and the dominated convergence theorem do not hold forthe Riemann integral.

(4.59) LetfL1(R) and, t [0, ),

g(t) = sup

|f(x + y) f(x)| d(x) t y t

.

Show that g is continuous at t = 0.(Hint: Use the simple function technique).


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4.6. L1[a, b] as the completion ofR[a, b]


Recall that,R[a, b] is not a complete metric space under the L1-metric

d(f, g) :=

ba

|f(x) g(x)|dx,

for f, g R[a, b]. Since every metric space has a completion, the questionarises: what is the completion ofR[a, b] under this metric? You might haveseen the abstract construction of the completion of a metric space. We shall

show that for R[a, b] this completion is nothing but L1[a, b]. Thus L1[a, b]is the concrete realization of the completion ofR[a, b]. We shall first showthatL1[a, b] is a complete metric space, and then show that R[a, b] is denseinL1[a, b].

4.6.1. Definition:For f , g L1[a, b], we say f is equivalent to g , and write f g, if the set{x [a, b]| f(x)=g(x)} has Lebesgue measure zero.

It is easy to see that the relation is an equivalence relation onL1[a, b].We denote the set of equivalence classes again by L1[a, b]. In other words,we identify two functions f, g with each other if f g. Thus, for g, f

L1[a, b], f=g ifff(x) =g(x) for a.e. x(), as functions.

4.6.2. Definition:For fL1[a, b], we define

f1:=

|f(x)|d(x).

Clearlyf1 is well-defined, and it is easy to check that the function f f1, fL1[a, b], has the following properties:

(i) f1 0 fL1[a, b].

(ii) f1

= 0 ifff = 0.

(iii) af1= |a| f1 a R and fL1[a, b].

(iv) f+ g1 f1+ g1 f, g L1[a, b].

The function 1 is called a norm on L1[a, b]. (Geometrically it is thedistance of the vector f L1[a, b] from the vector 0 L1[a, b].) Forf, g L1[a, b], if we define

d(f, g) := f g1,


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60 4. Integration

thend is a metric onL1[a, b], called the L1-metric . The most important

property of this metric is given by the next theorem.

4.6.3. Theorem (Riesz-Fischer):L1[a, b] is a complete metric space in theL1-metric.

We show next that C[a, b], the space of continuous function on [a, b],(and hence R[a, b]) is dense in L1[a, b].

4.6.4. Theorem:The spaceC[a, b] is a dense subset ofL1[a, b].

Since C[a, b] R[a, b], theorems 4.6.3 and 4.6.4 together prove the fol-lowing theorem:

4.6.5. Theorem:L1[a, b] is the completion ofR[a, b].

4.6.6. Notes:

(i) In the proof of theorem 4.6.3, one does not use require anywhere thefact that the functions are defined on an interval. One can define 1for functions defined on (E, L E, ), where E L is arbitrary, and theproof of theorem 5.6.1 will show that L1(E) :=L1(E, L E, ) is also a

complete metric space under the metric

f g1:=

E

|f(x) g(x)|d(x).

A closer look will show that the metric d makes sense on L1(X, S, ),where (X, S, ) is any complete measure space and we identify functionswhich agree for a.e. (). Further, theorem 4.6.3 also remains true forL1(X, S, ).

(ii) Parts of the proof of theorem 4.6.4 also exhibit the following facts whichare of independent interest. Let f L1[a, b] and > 0 be given. Thenthere exists a simple function such that f1 < and a step function

hsuch thatf h1< .Thus simple functions in L1[a, b] are dense in it.

Exercises:

(4.60) LetfL1(R) and let >0 be given. Prove the following:(i) There exists a positive integer n such that

f [n,n]f1< .


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(ii) There exists a continuous function g on R such that g is zero

outside some finite interval andf [n,n] g1< .(iii) For f : R R, let

supp (f) := closure {x R |f(x)= 0}.

The set supp(f) is called thesupportoffand is the smallestclosed s

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