Measure and Integration Full

8/10/2019 Measure and Integration Full

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MEASURE and INTEGRATION

Problems with Solutions

Anh Quang Le, Ph.D.

October 8, 2013


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NOTATIONS

A(X): The -algebra of subsets ofX.(X, A(X), ) : The measure space on X.B(X): The -algebra of Borel sets in a topological space X.ML : The -algebra of Lebesgue measurable sets in R.(R, ML, L): The Lebesgue measure space on R.L: The Lebesgue measure onR.L: The Lebesgue outer measure on R.1E orE: The characteristic function of the set E.

www.MATHVN.com - Anh Quang Le, PhD

www.MathVn.com - Math Vietnam


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Contents

Contents 1

1 Measure on a-Algebra of Sets 5

2 Lebesgue Measure on R 21

3 Measurable Functions 33

4 Convergence a.e. and Convergence in Measure 45

5 Integration of Bounded Functions on Sets of Finite Measure 53

6 Integration of Nonnegative Functions 63

7 Integration of Measurable Functions 75

8 Signed Measures and Radon-Nikodym Theorem 97

9 Differentiation and Integration 109

10 Lp Spaces 121

11 Integration on Product Measure Space 141

12 Some More Real Analysis Problems 151

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4 CONTENTS




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Chapter 1

Measure on a -Algebra of Sets

1. Limits of sequences of sets

Definition 1 Let(An)nN be a sequence of subsets of a setX.(a) We say that(An) is increasing if AnAn+1 for all nN, and decreasing ifAnAn+1 forallnN.(b) For an increasing sequence(An), we define

limn

An:=n=1

An.

For a decreasing sequence(An), we define

limn

An:=n=1

An.

Definition 2 For any sequence(An) of subsets of a set X, we define

liminfn

An:=nN

kn

Ak

limsupn

An:=nN

kn

Ak.

Proposition 1 Let(An

) be a sequence of subsets of a setX. Then

(i) lim inf n

An={xX : xAn for all but finitely many nN}.(ii) lim sup

nAn={xX: xAn for infinitely many n N}.

(iii) lim inf n

Anlimsupn

An.

2. -algebra of sets

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6 CHAPTER 1. MEASURE ON A -ALGEBRA OF SETS

Definition 3 (-algebra)Let X be an arbitrary set. A collectionA of subsets of X is called an algebra if it satisfies the

following conditions:

1. X A.2. A A Ac A.3. A, B A A B A.

An algebraA of a setXis called a -algebra if it satisfies the additional condition:4. An A,nN

nN AnnN.

Definition 4 (Borel-algebra)Let (X, O) be a topological space. We call the Borel -algebraB(X) the smallest -algebra of XcontainingO.

It is evident that open sets and closed sets in Xare Borel sets.

3. Measure on a -algebra

Definition 5 (Measure)LetA be a -algebra of subsets of X. A set function defined onA is called a measure if itsatisfies the following conditions:

1. (E)[0, ] for everyE A.2. () = 0.

3. (En)nN A, disjoint nNEn

=

nN(En).

Notice that ifE Asuch that (E) = 0, then Eis called a null set. If any subset E0 of a null setEis also a null set, then the measure space (X, A, ) is called complete.

Proposition 2 (Properties of a measure)A measure on a-algebraA of subsets ofX has the following properties:(1) Finite additivity: (E1, E2,...,En) A, disjoint= (

nk=1 Ek) =

nk=1 (Ek).

(2) Monotonicity: E1, E2 A, E1E2 =(E1)m(E2).(3) E1, E2 A, E1E2, (E1)


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4. Outer measures

Definition 7 (Outer measure)Let Xbe a set. A set function defined on the -algebraP(X) of all subsets of X is called anouter measure onXif it satisfies the following conditions:

(i) (E)[0, ] for everyE P(X).(ii) () = 0.

(iii) E, F P(X), EF (E)(F).(iv) countable subadditivity:

(En)nN P(X), nN

En

nN

(En).

Definition 8 (Caratheodory condition)We say thatE P(X) is-measurable if it satisfies the Caratheodory condition:

(A) = (A E) + (A Ec) for every A P(X).

We writeM() for the collection of all -measurableE P(X). ThenM() is a -algebra.

Proposition 3 (Properties of)(a) IfE1, E2 M(), thenE1 E2 M().(b) is additive onM(), that is,

E1, E2 M(), E1 E2 = =(E1 E2) = (E1) + (E2).




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Problem 1

LetA

be a collection of subsets of a setX with the following properties:

1. X A.2. A, B A A \ B A.

Show thatA is an algebra.

Solution

(i)X A.(ii)A A A

c

=X\ A A (by 2).(iii)A, B A A B=A \ Bc A since B c A (by (ii)).SinceAc, Bc A, (A B)c =Ac Bc A. Thus, A B A.

Problem 2

(a) Show that if(An)nN is an increasing sequence of algebras of subsets of a setX, then

nN An is an algebra of subsets ofX.

(b) Show by example that even ifAn in (a) is a-algebra for everynN, theunion still may not be a-algebra.

Solution

(a) LetA=nN An. We show thatAis an algebra.(i) Since X An,n N, so X A.

(ii) Let A A. Then A An for some n. And so Ac An ( sinceAn is analgebra). Thus, Ac A.

(iii) Suppose A, B A. We shall show A B A.Since

{An

} is increasing, i.e.,

A1

A2

... and A, B

nN An, there issome n0 N such thatA, B A0. Thus,A B A0. Hence, A B A.(b) LetX= N,An= the family of all subsets of{1, 2,...,n} and their complements.Clearly,An is a-algebra andA1 A2.... However,

nN An is the family of all

finite and co-finite subsets ofN, which is not a -algebra.




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Problem 3

LetXbe an arbitrary infinite set. We say that a subsetAofX is co-finite if itscomplementAc is a finite subset ofX. LetA consists of all the finite and theco-finite subsets of a setX.(a) Show thatA is an algebra of subsets ofX.(b) Show thatA is a-algebra if and only ifX is a finite set.

Solution

(a)

(i) X Asince X is co-finite.(ii) LetA A. IfA is finite then Ac is co-finite, so Ac A. IfA co-finite thenAc

is finite, so Ac A. In both cases,A A Ac A.

(iii) LetA, B A. We shall show A B A.IfAandB are finite, then A B is finite, so A B A. Otherwise, assumethatAis co-finite, then A B is co-finite, so A B A. In both cases,

A, B A A B A.

(b) IfXis finite thenA=P(X), which is a -algebra.To show the reserve, i.e., ifA is a -algebra then X is finite, we assume that Xis infinite. So we can find an infinite sequence (a1, a2,...) of distinct elements ofXsuch that X\ {a1, a2,...} is infinite. LetAn ={an}. Then An Afor any nN,while

nN An is neither finite nor co-finite. So

nN An / A. Thus,A is not a

-algebra: a contradiction!

Note:For an arbitrary collectionC of subsets of a set X, we write (C) for the smallest-algebra of subsets ofXcontaining

Cand call it the -algebra generated by

C.

Problem 4

LetC be an arbitrary collection of subsets of a set X. Show that for a givenA (C), there exists a countable sub-collectionCA ofC depending onA suchthatA(CA). (We say that every member of(C) is countable generated).




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Solution

Denote byB the family of all subsets A of X for which there exists a countablesub-collectionCA ofC such that A (CA). We claim thatB is a -algebra andthatC B.The second claim is clear, since A({A}) for any A C. To prove the first one,we have to verify thatBsatisfies the definition of a -algebra.

(i) Clearly,X B.(ii) If A B then A (CA) for some countable familyCA (C). Then

Ac (CA), so Ac B.

(iii) Suppose

{An

}n

N

B. ThenAn

(

CAn) for some countable family

CAn

C.

LetE=nN CAn thenEis countable andE C andAn(E) for alln N.By definition of-algebra,

nN An(E), and so

nN An B.

Thus,Bis a -algebra of subsets ofXandE B. Hence,(E) B.

By definition ofB, this implies that for every A (C) there exists a countableE C such that A(E).

Problem 5Leta set function defined on a-algebraAof subsets ofX. Show that it isadditive and countably subadditive onA, then it is countably additive onA.

Solution

We first show that the additivity ofimplies its monotonicity. Indeed, let A, B Awith AB. Then

B = A (B\ A) and A (B\ A) = .Sinceis additive, we get

(B) =(A) +(B\ A)(A).Now let (En) be a disjoint sequence inA. For everyN N, by the monotonicityand the additivity of, we have

nN

En

Nn=1

En

=

Nn=1

(En).




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Since this holds for every N N, so we have

(i) nN

En nN

(En).

On the other hand, by the countable subadditivity of, we have

(ii)

nN

En

nN

(En).

From (i) and (ii), it follows that

nN

En= nN

(En).

This proves the countable additivity of .

Problem 6

LetXbe an infinite set andAbe the algebra consisting of the finite and co-finitesubsets of X (cf. Prob.3). Define a set function onA by setting for everyA A:

(A) = 0 ifA is finite1 ifA is co-finite.(a) Show that is additive.(b) Show that whenX is countably infinite, is not additive.(c) Show that whenX is countably infinite, thenXis the limit of an increasingsequence{An: n N} inA with(An) = 0 for every n N, but(X) = 1.(d) Show that whenX is uncountably, the is countably additive.

Solution

(a) Suppose A, B

Aand A

B=

(i.e., A

Bc andB

Ac).

If A is co-finite then B is finite (since B Ac). So AB is co-finite. We have(A B) = 1, (A) = 1 and (B) = 0. Hence, (A B) =(A) +(B).IfB is co-finite then A is finite (since ABc). So A B is co-finite, and we havethe same result. Thus, is additive.(b) Suppose X is countably infinite. We can then put X under this form: X ={x1, x2,...}, xi=xj if i=j . LetAn={xn}. Then the family{An}nN is disjointand (An) = 0 for every nN. So

nN (An) = 0. On the other hand, we have




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nN An= X, and(X) = 1. Thus,

nN

An= nN

(An).

Hence, is not additive.(c) Suppose Xis countably infinite, and X ={x1, x2,...}, xi=xj if i= j as in(b). LetBn ={x1, x2,...,xn}. Then (Bn) = 0 for every nN, and the sequence(Bn)nN is increasing. Moreover,

limn

Bn =nN

Bn= X and (X) = 1.

(d) Suppose X is uncountably. Consider the family of disjoint sets

{Cn

}n

N in

A.

SupposeC=nN Cn A. We first claim: At most one of theCns can be co-finite.Indeed, assume there are two elements Cn andCm of the family are co-finite. SinceCmCcn, soCm must be finite: a contradiction.Suppose Cn0 is the co-finite set. Then since CCn0,Cis also co-finite. Therefore,

(C) =

nN

Cn

= 1.

On the other hand, we have

(Cn0) = 1 and (Cn) = 0 for n

=n0.

Thus,

nN

Cn

=nN

(Cn).

If allCn are finite then

nN Cn is finite, so we have

0 =

nN

Cn

=nN

(Cn).

Problem 7

Let (X, A, ) be a measure space. Show that for any A, B A, we have theequality:

(A B) +(A B) =(A) +(B).




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Solution

If(A) =or (B) =, then the equality is clear. Suppose (A) and (B) arefinite. We have

A B= (A \ B) (A B) (B\ A),A= (A \ B) (A B)B= (B\ A) (A B).

Notice that in these decompositions, sets are disjoint. So we have

(A B) =(A \ B) +(A B) +(B\ A),(1.1)(A) +(B) = 2(A B) +(A \ B) +(B\ A).(1.2)

From (1.1) and (1.2) we obtain

(A B) (A) (B) =(A B).The equality is proved.

Problem 8

The symmetry difference ofA, B P(X) is defined by

A B= (A \ B) (B\ A).

(a) Prove that

A,B,C P(X), A B(A C) (C B).

(b) Let(X, A, ) be a measure space. Show that

A,B,C A, (A B)(A C) +(C B).

Solution

(a) Let xA B. Suppose xA \ B. IfxCthen xC\ B so xC B. Ifx /C, then xA \ C, so xA C. In both cases, we have

xA Bx(A C) (C B).The case xB\ A is dealt with the same way.(b) Use subadditivity of and (a).




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Problem 9

LetXbe an infinite set and the counting measure on the-algebraA

=P

(X).Show that there exists a decreasing sequence(En)nN inAsuch that

limn

En= with limn

(En)= 0.

Solution

SinceXis a infinite set, we can find an countably infinite set{x1, x2,...} Xwithxi=xj ifi=j . LetEn={xn, xn+1,...}. Then (En)nN is a decreasing sequence in

Awith

limn

En= and limn

(En) = 0.

Problem 10 (Monotone sequence of measurable sets)Let(X, A, ) be a measure space, and(En) be a monotone sequence inA.(a) If(En) is increasing, show that

limn

(En) =

limn

En

.

(b) If(En) is decreasing, show that

limn

(En) =

limn

En

,

provided that there is a setA A satisfying(A)


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Thus

limn

En= = lim

n(En).

Consider the next case where (En)


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Since

nN En(E1)(A)


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Problem 11 (Fatous lemma for )Let(X,

A, ) be a measure space, and(En) be a sequence in

A.

(a) Show that

lim infn

Enlim inf

n(En).

(b) If there exists A A withEn A and (A)


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Now

limn

kn Ek= lim supn kn Ek ,since the limit of a sequence, if it exists, is equal to the limit superior of the sequence.Then by

kn EkEn we have

kn

Ek

(En).

Thus

limsupn

kn

Ek

lim sup

n(En).

It follows that

lim supn

Enlimsup

n(En).

Problem 12

Let be an outer measure on a setX. Show that the following two conditionsare equivalent:(i) is additive onP(X).(ii) Every element ofP(X) is-measurable, that is,M() =P(X).

Solution

Suppose is additive onP(X). LetE P(X). Then for anyA P(X),A= (A E) (A Ec) a n d (A E) (A Ec) = .

By the additivity of onP(X), we have(A) =(A E) +(A Ec).

This show that E satisfies the Caratheodory condition. Hence E M(). SoP(X) M(). But by definition,M() P(X). Thus

M() =P(X). Conversely, supposeM() =P(X). Since is additive onM() by Proposi-tion 3, so is additive onP(X).




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Problem 13

Let be an outer measure on a setX.(a) Show that the restriction of on the -algebraM() is a measure onM().(b) Show that if is additive onP(X), then it is countably additive onP(X).

Solution

(a) By definition, is countably subadditive onP(X). Its restriction onM()is countably subadditive onM(). By Proposition 3b, is additive onM().Therefore, by Problem 5, is countably additive onM(). Thus, is a measureonM(). But is the restriction of onM(), so we can say that is ameasure onM().(b) If is additive onP(X), then by Problem 11,M() =P(X). So is ameasure onP(X) (Problem 5). In particular, is countably additive onP(X).




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Chapter 2

Lebesgue Measure on R

1. Lebesgue outer measure onR

Definition 9 (Outer measure)Lebesgue outer measure onR is a set functionL:P(R)[0, ] defined by

L(A) = inf

k=1

(Ik) : Ak=1

Ik, Ik is open interval in R

.

Proposition 4 (Properties ofL)

1. L(A) = 0 ifA is at most countable.

2. Monotonicity: ABL(A)L(B).3. Translation invariant: L(A + x) =

L(A),x R.

4. Countable subadditivity: L(n=1 An)

n=1

L(An).

5. Null set: L(A) = 0L(A B) = L(B) and L(B\ A) = L(B)for allB P(R).

6. For any intervalI R, L(I) = (I).7. Regularity:

E

P(R), >

0,

O open set in R :

O

E and

L(

E)

L(

O)

L(

E) +

.

2. Measurable sets and Lebesgue measure on R

Definition 10 (Caratheodory condition)A setE Ris said to be Lebesgue measurable (orL-measurable, or measurable) if, for allA R,we have

L(A) = L(A E) + L(A Ec).

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22 CHAPTER 2. LEBESGUE MEASURE ONR

SinceL is subadditive, the sufficient condition for Caratheodory condition is

L(A)

L(A

E) + L(A

Ec).

The family of all measurable sets is denoted byML. We can see thatML is a -algebra. Therestriction ofL onML is denoted by L and is called Lebesgue measure.

Proposition 5 (Properties ofL)

1. (R, ML, L) is a complete measure space.2. (R, ML, L) is-finite measure space.3. BR ML, that is, every Borel set is measurable.

4. L(O)> 0 for every nonempty open set inR.5. (R, ML, L) is translation invariant.6. (R, ML, L) is positively homogeneous, that is,

L(E) =||L(E),R, E ML.

Note onF andG sets:Let (X, T) be a topological space. A subset EofXis called a F-set if it is the union of countably many closed sets. A subset EofXis called a G-set if it is the intersection of countably many open sets.

IfEis aG-set thenE

c is aF-set andvice versa. EveryG-set is Borel set, so is every F-set.

Problem 14

IfE is a null set in(R, ML, L), prove thatEc is dense inR.

Solution

For every open interval I inR

, L

(I) > 0 (property of Lebesgue measure). IfL(E) = 0, then by the monotonicity ofL,Ecannot contain any open interval asa subset. This implies that

Ec I= for any open interval I inR. Thus Ec is dense inR.




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Problem 15

Prove that for everyER, there exists aG-setG

R such that

GE and L(G) =L(E).

Solution

We use the regularity property ofL (Property 7).For = 1n , n N, there exists an open set On Rsuch that

On

E and L(E)

L(On)

L(E) +1

n.

LetG =

nN On. Then G is a G-set and GE. Since GOn for every nN,we have

L(E)L(G)L(On)L(E) +1

n.

This holds for every n N, so we haveL(E)L(G)L(E).

Therefore(G) =(E).

Problem 16

LetE R. Prove that the following statements are equivalent:(i)E is (Lebesgue) measurable.(ii) For every >0, there exists an open setOEwithL(O \ E).(iii) There exists aG-setGE withL(G \ E) = 0.

Solution

(i)(ii) Suppose that Eis measurable. Then >0, open set O : OE and L(E)L(O)L(E) +. (1)

SinceEis measurable, withOas a testing set in the Caratheodory condition satisfiedbyE, we have

L(O) =L(O E) +L(O Ec) =L(E) +L(O \ E). (2)




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IfL(E)0, there is an open set On such that

OnEn and L(On \ En) 13

.

2|n|.

LetO =nZ)On, thenO is open and OE, andO \ E =

nZ

On

\

nZEn

=

nZ

On

nZEn

c

=nZ

On

nZ

En

c=

nZ

On \

nZ

En

nZ

(On \ En).

Then we have

L(O \ E) L

nZ(On \ En)

nZ

L(On \ E)

nZ

1

3.

2|n| =

1

3+ 2

nN

1

3.

2n

= 1

3+

2

3= .

This shows that (ii) satisfies.

(ii)(iii) Assume that Esatisfies (ii). Then for = 1n , n N, there is an opensetOn such that

OnEn and L(On \ En) 1n

,n N.

LetG =

nN On. Then G is a G-set containingE. Now

GO =L(G \ E)L(On \ E) 1

n,n N.




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ThusL(G \ E) = 0. This shows thatEsatisfies (iii).

(iii)

(i) Assume thatEsatisfies (iii). Then there exists a G-setG such that

GE and L(G \ E) = 0.Now L(G\E) = 0 implies that G\E is (Lebesgue) measurable. SinceE G,we can write E = G\(G\E). Then the fact that G and G\Eare (Lebesgue)measurable implies thatE is (Lebesgue) measurable.

Problem 17(Similar problem)LetE R. Prove that the following statements are equivalent:(i) Eis (Lebesgue) measurable.(ii) For every >0, there exists an closed set CE withL(E\ C).

(iii) There exists aF-set

F

Ewith

L(

E\

F) = 0.

Problem 18

LetQ be the set of all rational numbers inR. For any >0, construct an opensetO R such that

OQ and L(O).

SolutionSinceQ is countable, we can write Q ={r1, r2,...}. For any > 0, let

In=

rn 2n+1

, rn+

2n+1

, n N.

ThenIn is open and O =

n=1 In is also open. We have, for everyn N, rnIn.ThereforeOQ.Moreover,

L(O) = L

n=1

In

n=1L(In)

=

n=1

2

2n+1

=

n=1

1

2n =.




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Problem 19

LetQ be the set of all rational numbers inR.(a) Show thatQ is a null set in(R, BR, L).(b) Show thatQ is aF-set.(c) Show that there exists aG-setG such thatG Q andL(G) = 0.(d) Show that the set of all irrational numbers inR is aG-set.

Solution

(a) SinceQ is countable, we can writeQ ={r1, r2,...}. Each{rn}, n Nis closed,so{rn} BR. SinceBR is a-algebra,

Q =

n=1{

rn} B

R.

SinceL({rn}) = 0, we have

L(Q) =

n=1

L({rn}) = 0.

Thus,Q is a null set in (R, BR, L).(b) Since{rn} is closed and Q =

n=1{rn}, Qis a F-set.

(c) By (a), L(Q) = 0. This implies that, for every nN, there exists an open setGn such that

Gn Q and L(Gn)< 1n

.

IfG =

n=1 Gn then Gis a G-set andG Q. Furthermore,

L(G)L(Gn)< 1n

, n N.This implies that L(G) = 0.

(d) By (b), Q is a F-set, so R\Q, the set of all irrational numbers in R, is aG-set.

Problem 20Let E ML with L(E) > 0. Prove that for every (0, 1), there exists a

finite open intervalIsuch that

L(I)L(E I)L(I).




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Solution

Consider first the case where 0< L(E)0, 0< = aL(E)


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Problem 21

Letfbe a real-valued function on(a, b) such thatf exists and satisfies

|f(x)| M for all x(a, b) and for some M0.

Show that for everyE(a, b) we have

L(f(E))M L(E).

Solution

IfM= 0 then f(x) = 0,x(a, b). Hence, f(x) =y0,x(a, b). Thus, for anyE(a, b) we haveL(f(E)) = 0.

The inequality holds. Suppose M > 0. For all x, y (a, b), by the Mean ValueTheorem, we have

|f(x) f(y)| = |x y||f(c)|, for some c(a, b) M|x y|. ()

By definition of the outer measure, for any E(a, b) we have

L(E) = inf

n=1

(bn an),

where{In= (an, bn), n N} is a covering class ofE. By (*) we have

n=1

|f(bn) f(an)| M

n=1

|bn an|

Minf

n=1

|bn an|

ML(E).

Infimum takes over all covering classes ofE. Thus,

L(f(E)) = inf

n=1

|f(bn) f(an)| M L(E).




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Problem 22

(a) Let ER. Show that

F =

{, E , E c,R

} is the-algebra of subsets ofR

generated by{E}(b) IfS andT are collections of subsets ofR, then

(S T) =(S) (T).

Is the statement true? Why?

Solution

(a)It is easy to check thatF is a -algebra.Note first that

{E

} F. Hence

({E}) F. (i)On the other hand, since ({E}) is a -algebra, so ,R ({E}). Also, sinceE({E}), so Ec ({E}). Hence

F ({E}). (ii)From (i) and (ii) it follows that

F=({E}).(b) No. Here is why.Take

S=

{(, 1]

}and

T =

{(1, 2]

}. Then, by part (a),

(S) ={, (0, 1], (0, 1]c,R} and (T) ={, (1, 2], (1, 2]c,R}.Therefore

(S) (T) ={, (0, 1], (0, 1]c, (1, 2], (1, 2]c,R}.We have

(0, 1] (1, 2] = (0, 2] /(S) (T).Hence(S) (T) is not a -algebra. But, by definition, (S T) is a -algebra.And hence it cannot be equal to (S) (T).

Problem 23

ConsiderF={E R : eitherE is countable orEc is countable}.(a) Show thatF is a-algebra andF is a proper sub--algebra of the-algebraBR.(b) Show thatF is the-algebra generated by{x}: x R.(c) Find a measure:F [0, ] such that the only-null set is.




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Solution

(a) We check conditions of a -algebra:

It is clear that is countable, so F. Suppose E F. ThenE R and E is countable or Ec is countable. This isequivalent toEc Rand Ec is countable or Eis countable. Thus,

E F Ec F. Suppose E1, E2,... F. Either allEns are countable, so

n=1 En is countable.

Hence

n=1 En F. Or there exists some En0 F which is not countable. Bydefinition,Ecn0 must be countable. Now

n=1 Enc

=

n=1 Ecn

En0.

This implies that (

n=1 En)c

is countable. Thus

n=1

En F.

Finally,F is a -algebra. Recall thatBR is the -algebra generated by the family of open sets in R. It is alsogenerated by the family of closed sets inR. Now suppose E F. IfE is countablethen we can write

E={x1, x2,...}= n=1

{xn}.

Each{xn}is a closed set inR, so belongs toBR. HenceE BR. Therefore,F BR.

F is a proper subset ofBR. Indeed, [0, 1] BR and [0, 1] / F. (b) LetS= {x}: x R. Clearly,S F, and so

(S) F.

Now take E F and E= . IfEis countable then we can write

E=

n=1

{xn}S

(S).

HenceF (S).




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31

Thus(S) =F.

(c) Define the set function :F [0, ] by

(E) =

|E| ifEis finite otherwise.

We can check that is a measure. IfE= then(E)> 0 for every E F.

Problem 24

ForE ML withL(E)


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Problem 25

Let E be a Lebesgue measurable subset ofR with L(E) = 1. Show that thereexists a Lebesgue measurable setAE such thatL(A) = 12 .

Solution

Define the function f : R [0, 1] by

f(x) =L

E (, x], x R.By Problem 23, we have

|f(x)

f(y)

| |x

y

|,

x, y

R.

Hencef is (uniformly) continuous onR. Since

limx

f(x) = 0 and limx

f(x) = 1,

by the Mean Value Theorem, we have

x0 Rsuch thatf(x0) =12

.

SetA = E (, x0]. Then we have

AE and L(A) =12

.




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Chapter 3

Measurable Functions

Remark:From now on, measurable means Lebesgue measurable. Alsomeasuremeans Lebesgue measure,and we write instead ofL for Lebesgue measure.

1. Definition, basic properties

Proposition 6 (Equivalent conditions)Let f be an extended real-valued function whose domain D is measurable. Then the followingstatements are equivalent:

1. For each real numbera, the set{xD : f(x)> a} is measurable.

2. For each real numbera, the set{xD : f(x)a} is measurable.3. For each real numbera, the set{xD : f(x)< a} is measurable.4. For each real numbera, the set{xD : f(x)a} is measurable.

Definition 11 (Measurable function)An extended real-valued function f is said to be measurable if its domain is measurable and if itsatisfies one of the four statements of Proposition 6.

Proposition 7 (Operations)Let f, g be two measurable real-valued functions defined on the same domain and c a constant.Then the functionsf+ c,cf,f+ g, andf g are also measurable.

Note:A functionfis said to beBorel measurableif for each Rthe set{x: f(x)> }is a Borel set.Every Borel measurable function is Lebesgue measurable.

2. Equality almost everywhere

A property is said to holdalmost everywhere(abbreviated a.e.) if the set of points where it failsto hold is a set of measure zero. We say that f = g a.e. iff and g have the same domain and {xD : f(x)= g (x)}= 0.Also we say that the sequence (fn) converges to fa.e. if the set{x: fn(x) f(x)} is a null set.

33




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34 CHAPTER 3. MEASURABLE FUNCTIONS

Proposition 8 (Measurable functions)If a functionf is measurable andf=g a.e., theng is measurable.

3. Sequence of measurable functions

Proposition 9 (Monotone sequence)Let (fn) be a monotone sequence of extended real-valued measurable functions on the same mea-surable domainD. Thenlimn fn exists onD and is measurable.

Proposition 10 Let(fn) be a sequence of extended real-valued measurable functions on the samemeasurable domainD. Thenmax{f1,...,fn}, min{f1,...,fn}, limsupn fn, liminfn fn, supnN, infnNare all measurable.

Proposition 11 Iffis continuous a.e. on a measurable setD, thenfis measurable.

Problem 26

LetD be a dense set inR. Letfbe an extended real-valued function onR suchthat

{x: f(x)>

}is measurable for each

D. Show thatf is measurable.

Solution

Letbe an arbitrary real number. For eachnN, there exists nD such that < n< +

1n

by the density ofD. Now

{x: f(x)> }=

n=1

x: f(x)+ 1

n

=

n=1

{x: f(x)> n}.

Sincen=1

{x: f(x)> n

}is measurable (as countable union of measurable sets),

{x: f(x)> } is measurable. Thus, f is measurable.

Problem 27

Let f be an extended real-valued measurable function onR. Prove that{x :f(x) =} is measurable for any R.




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35

Solution

For R, we have

{x: f(x) =}={x: f(x)} measurable

\ {x: f(x)< } measurable

.

Thus{x: f(x) =} is measurable. For =, we have

{x: f(x) =}= R \ {x: f(x)}= R \nN

{x: f(x) n} measurable

.

Thus{x: f(x) =} is measurable.

Problem 28

(a). LetD andEbe measurable sets andfa function with domainD E. Showthatfis measurable if and only if its restriction to D andEare measurable.(b). Letfbe a function with measurable domainD. Show thatf is measurableif and only if the functiong defined by

g(x) =

f(x) forxD0 forx /D

is measurable.

Solution

(a) Suppose thatfis measurable. SinceD andEare measurable subsets ofD E,the restrictions f|D and f|Eare measurable.Conversely, suppose f|D andf|E are measurable. For any R, we have

{x: f(x)> }={xD : f|D(x)> } {xE: f|E(x)> }.Each set on the right hand side is measurable, so{x : f(x) > } is measurable.Thus, f is measurable.




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(b) Suppose thatfis measurable. If0, then{x: g(x)> }={x: f(x)> },which is measurable. If }={x: f(x)> } Dc, whichis measurable. Hence, g is measurable.Conversely, suppose that g is measurable. Since f=g|D and D is measurable, f ismeasurable.

Problem 29

Letfbe measurable andB a Borel set. Thenf1(B) is a measurable set.

Solution

Let

Cbe the collection of all sets Esuch thatf1(E) is measurable. We show that

C is a-algebra. Suppose E C. Sincef1(Ec) =

f1(E)

c,

which is measurable, so Ec C. Suppose (En) is a sequence of sets inC. Since

f1

n

En

=n

f1(En),

which is measurable, so

n En C. Thus,C is a -algebra.

Next, we show that all intervals (a, b), for any extended real numbers a, b with

a < b, belong toC. Since f is measurable,{x: f(x)> a} and{x: f(x)< b} aremeasurable. It follows that (a, ) and (, b) C. Furtheremore, we have

(a, b) = (, b) (a, ),so (a, b) C. Thus,C is a -algebra containing all open intervals, so it contains allBorel sets. Hence f1(B) is measurable.

Problem 30

Show that if f is measurable real-valued function and g a continuous function

defined onR, theng f is measurable.

Solution

For any R,

{x: (g f)(x)> }= (g f)1((, )) = f1

g1

(, ).




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37

By the continuity ofg, g1

(, ) is an open set, so a Borel set. By Problem 24,the last set is measurable. Thus, g f is measurable.

Problem 31

Letfbe an extended real-valued function defined on a measurable setDR.(a) Show that if{xD : f(x)< r} is measurable inR for everyrQ, thenf is measurable onD.(b) What subsets ofR other thanQ have this property?(c) Show that iffis measurable onD, then there exists a countable sub-collectionC ML, depending onf, such thatf is(C)-measurable onD.(Note: (C) is the -algebra generated byC.)

Solution

(a) To show that f is measurable on D, we show that{x D : f(x) < a} ismeasurable for every aR. Let I={rQ : r < a}. Then I is countable , andwe have

{xD : f(x)< a}=rI

{xD : f(x)< r}.

Since{xD : f(x)< r} is measurable,rI{xD : f(x)< r} is measurable.Thus,{xD : f(x)< a} is measurable.(b) Here is the answer to the question:

Claim 1 : If E R is dense inR, then E has the property in (a), that is, if{xD : f(x)< r} is measurable for everyrE thenf is measurable onD.Proof.Given any a R, the interval (a 1, a) intersects E since E is dense. Pick somer1E (a 1, a). Now the interval (r1, a) intersects E for the same reason. Picksomer2E (r1, a). Repeating this process, we obtain an increasing sequence (rn)inEwhich converges to a.By assumption,{xD : f(x)< rn} is measurable, so we have

{xD : f(x)< a}= nN

{xD : f(x)< rn} is measurable.

Thus, fis measurable on D.

Claim 2 : IfE R is not dense inR, thenEdoes not have the property in (a).Proof.Since E is not dense in R, there exists an interval [a, b] E. Let F be a non




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measurable set inR. We define a function fas follows:

f(x) = a ifxFc

b ifxF .ForrE, by definition ofF, we observe that

Ifr < a then f1([, r)) =. Ifr > b thenf1([, r)) = R. Ifr = a+b2 then f1([, r)) =Fc.

SinceFis non measurable, Fc is also non measurable. Through the above observa-tion, we see that

xD : f(x)< a+b2

non measurable.

Thus, fis not measurable.

Conclusion : Only subsets ofR which are dense inR have the property in (a).

(c) LetC={Cr}rQ where Cr ={xD : f(x)< r} for every r Q. Clearly,C isa countable family of subsets ofR. Sincef is measurable, Cr is measurable. Hence,C ML. SinceML is a -algebra, by definition, we must have (C) ML. Leta R. Then

{xD : f(x)< a}= r


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39

Solution

(a) For anya R, letE={xD : f(x)< a}.

Ifa >1 then E= R, so E BR (Borel measurable). If 0< a1 then E= Q, so E BR (Borel measurable). Ifa0 then E= , so E BR (Borel measurable).

Thus, f is Borel measurable.

(b) Considerg1 defined on Q byg1(x) = x, then g|Q = g1. Considerg2 defined onR\Q byg(x) =x, theng|R\Q= g2. Notice that R, R\Q BR(Borel measurable).For anya R, we have

{xD : f1(x)< a}= [, a) Q BR (Borel measurable),and

{xD : f2(x)< a}= [, a) (R \Q) BR (Borel measurable).Thus, g is Borel measurable.

(c) Use the same way as in (b).

Problem 33

Letfbe a real-valued increasing function onR. Show thatfis Borel measurable,

and hence Lebesgue measurable also onR.

Solution

For anya R, letE={xD : f(x)a}. Let= infE. Sincef is increasing, if Im(f)(, a) then E= . if Im(f) (, a) then Eis either (, ) or [, ).

Since, (, ), [, ) are Borel sets, so f is Borel measurable.

Problem 34

If(fn) is a sequence of measurable functions onD R, then show that

{xD : limn

fn(x) exists} is measurable.




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Solution

Recall that if fns are measurable, then lim supn fn, liminfn fn and g(x) =

lim supn fn liminfn fn are also measurable, and if h is measurable then{xD : h(x) =} is measurable (Problem 22).Now we have

E={xD : limn

fn(x) exists}={xD : g(x) = 0}.Thus, E is measurable.

Problem 35

(a) If g : R R is continuous and f : R R is measurable then gf ismeasurable.

(b) Iff is measurable then|f| is measurable. Does the converse hold?

Solution

(a) For anya R, thenE={x: (g f)(x)< a} = (g f)1(, a)

= f1

g1(, a) .Since g is continuous, g1(, a) is open. Then there is a family of open disjointintervals{In}nN such thatg1(, a) =

nNIn. Hence,

E=f1

nNIn

=nN

f1(In).

Sincef is measurable, f1(In) is measurable. Hence Eis measurable. This tells usthatg f is measurable.(b) Ifg(u) =|u| theng is continuous. We have

(g f)(x) =g(f(x)) =|f(x)|.By part (a), g f=|f| is measurable.The converse is not true.

LetEbe a non-measurable subset ofR. Consider the function:

f(x) =

1 ifxE1 ifx /E.

Then f1( 12 , ) = E, which is not measurable. Since ( 12 , ) is open, so f is notmeasurable, while|f|= 1 is measurable.




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41

Problem 36

Let (fn : nN) and f be an extended real-valued measurable functions on a

measurable setD R such that limn fn = f onD. Then for every Rprove that:

(i) {xD : f(x)> } lim infn

{xD : fn(x)}(ii) {xD : f(x)< } liminf

n{xD : fn(x)}.

Solution

Recall that, for any sequence (En)nN of measurable sets,(lim inf

nEn)lim inf

n(En), ()

liminfn

En=nN

kn

Ek = limn

kn

Ek.

Now for every R, letEk ={xD : fk(x)} for each k N. Thenliminf

nEn = lim

n

kn

Ek

= limn

kn{x

D : fk(x)

}= {xD : f(x)> } since fk(x)f(x) on D.

Using (*) we get

{xD : f(x)> } liminfn

{xD : fn}.

For the second inequality, we use the similar argument.LetFk ={xD : fk(x)}for eachk N. Then

liminfn

En = limn kn

Fk

= limn

kn

{xD : fk(x)}

= {xD : f(x)< } since fk(x)f(x) on D.Using (*) we get

{xD : f(x)< } lim infn

{xD : fn}.




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Simple functions

Definition 12 (Simple function)A function: X R is simple if it takes only a finite number of different values.

Definition 13 (Canonical representation )Let be a simple function onX. Let{a1,...,an} the set of distinct valued assumedby onD. LetDi ={xX : (x) =ai} fori= 1,...,n. Then the expression

=n

i=1

aiDi

is called the canonical representation of.

It is evident thatDi Dj = fori=j andn

i=1 Di= X.

Problem 37

(a). Show that

AB =A BAB =A+B A B

Ac = 1 A.

(b). Show that the sum and product of two simple functions are simple functions.

Solution

(a). We have

AB(x) = 1 xA and xB A(x) = 1 =B(x).Thus,

AB =A B.We have

AB(x) = 1xA B.




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43

IfxA B then A(x) +B(x) A(x) B(x) = 1 + 1 1 = 1.If x / AB , then x A\B or x B\ A. Then A(x) + B(x) = 1 andA BA(x) +B(x) = 0.Also,

AB(x) = 0x /A B.Then

A(x) =B(x) =A(x) B(x) = 0.Thus,

AB =A+B A B.IfAc(x) = 1, then x /A, soA(x) = 0.IfAc(x) = 0, then xA, soA(x) = 1. Thus,

Ac = 1 A. (b). Let be a simple function having values a1,...,an. Then

=n

i=1

aiAi where Ai={x: (x) =ai}.

Similarly, if is a simple function having values b1,...,bm. Then

=m

j=1bjBj where Bj ={x: (x) =bj}.

DefineCij :=Ai Bj . Then

AiX=m

j=1

Bj and so Ai = Ai m

j=1

Bj =m

j=1

Cij.

Similarly, we have

Bj =n

i=1

Cij.

Since the Cij s are disjoint, this means that (see part (a))

Ai =m

j=1

Cij and Bj =n

i=1

Cij .

Thus

=n

i=1

mj=1

aiCij and =n

i=1

mj=1

bjCij .




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Hence

+=n

i=1m

j=1 (ai+bj)Cij and =n

i=1m

j=1 aibjCij .They are simple function.

Problem 38

Let: R R be a simple function defined byn

i=1

aiAi where Ai={x R : (x) =ai}.

Prove that is measurable if and only if all theAis are measurable.

Solution

Assume thatAi is measurable for alli= 1,...,n. Then for any c R, we have

{x: (x)> c}=ai>c

Ai.

Since every Ai is measurable,

ai>cAi is measurable. Thus{x : (x) > c} is

measurable. By definition, is measurable.

Conversely, suppose is measurable. We can suppose a1 < a2 < ... < an. Givenj {1, 2,...,n}, choose c1 and c2 such thataj1 < c1 < aj < c2 < aj+1. (Ifj = 1 orj=n, part of this requirement is empty.) Then

Aj =

ai>c1

Ai

\

ai>c2

Ai

= {x: (x)> c1} measurable

\ {x: (x)> c2} measurable

.

Thus, Aj is measurable for all j {1, 2,...,n}.




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Chapter 4

Convergence a.e. and Convergence

in Measure

1. Convergence almost everywhere

Definition 14 Let(fn) be a sequence extended real-valued measurable functions on a measurablesetD R.1. We say that limn fn exists a.e. on D if there exists a null set N such that N D andlimn fn(x) exists for everyxD \ N.2. We say that(fn) converges a.e. onD if limn fn(x) exists and limn fn(x)R for everyxD \ N.

Proposition 12 (Uniqueness)Let(fn) be a sequence extended real-valued measurable functions on a measurable set DR. Letg1 andg2 be two extended real-valued measurable functions onD. Then

limn

fn= g1 a.e. onD and limn

fn= g2 a.e. onD

=g1 = g2 a.e. onD.

Theorem 1 (Borel-Cantelli Lemma)For any sequence(An) of measurable subsets inR, we have

nN(An)0 there exists a measurable setED such that (E)< and(fn) converges uniformly to fonD \ E.

Theorem 2 (Egoroff)LetD be a measurable set with(D)


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46 CHAPTER 4. CONVERGENCE A.E. AND CONVERGENCE IN MEASURE

2. Convergence in measure

Definition 16 Let(fn) be a sequence extended real-valued measurable functions on a measurablesetDR. We say that(fn) converges in measureonD if there exists a real-valued measurablefunctionf onD such that for every >0 we have

limn

{D: |fn f| }:= limn

{xD : |fn(x) f(x)| }= 0.

That is,

>0, >0,N(, ) N : {D: |fn f| }< for nN(, ).

We write fnf on D for this convergence.

Proposition 13 (Uniqueness)Let(fn) be a sequence extended real-valued measurable functions on a measurable setDR. Letf andg be two real-valued measurable functions on D. Then

[fnf on D andfn g onD] =f=g a.e. onD.

Proposition 14 (Equivalent conditions)

(1) [fnf onD] >0,N()N : {D: |fn f| }< fornN().

(2) [fnf onD] m N,N(m) :

D: |fn f| 1

m

0 there exists N N such that {X : |fnf| < } fornN.

Solution

LetZbe a null set such that f is finite on X\ Z. Since fnf a.e. on X,fnfa.e. onX\ Z. For every >0 we have1

(lim supn

{X\ Z: |fn f| }) = 0

lim sup

n

{X

\Z:

|fn

f

|

}= 0

limn

{X\ Z: |fn f| }= 0

The last condition is equivalent to

limn

{X\ Z: |fn f|< }= (X\ Z) =(X) >0, N N : (X) {X\ Z: |fn f|< } for all nN.

Let us take = (X) >0. Then we haveN N : {X\ Z: |fn f|< } for all nN.

Since{X: |fn f|< } {X\ Z : |fn f|< }, so we have >0, N N : nN({X: |fn f|< }).

1See Problem 11b. We have

limsupn

Enlimsup

n(En).




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Problem 41

(a) Show that the condition

limn

{xD : |fn(x) f(x)|>0}= 0

implies thatfnf onD.

(b) Show that the converse is not true.(c) Show that the condition in (a) implies that for a.e. xD we havefn(x) =f(x) for infinitely manyn N.

Solution

(a) Given any >0, for every n N, letEn={xD : |fn(x) f(x)|> }; Fn={xD : |fn(x) f(x)|> 0}.

Then we have for all n NxEn |fn(x) f(x)|>

|fn(x) f(x)|> 0 xFn.

Consequently, En Fn and (En)(Fn) for all nN. By hypothesis, we havethat limn (Fn) = 0. This implies that limn (En) = 0. Thus, fn

f.

(b) The converse of (a) is false.Consider functions:

fn(x) = 1

n, x[0, 1] n N.

f(x) = 0, x[0, 1].Then fnf (pointwise) on [0, 1]. By Lebesgue Theorem fn fon [0, 1]. But foreveryn N

|fn(x) f(x)|= 1n

>0,x[0, 1].In other words,

{xD : |fn(x) f(x)|> 0}= [0, 1].Thus,

limn

{xD : |fn(x) f(x)|> 0}= 1= 0.(c) Recall that (Problem 11a)

(lim infn

En)liminfn

(En). ()




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49

LetEn ={xD : fn(x)=f(x)}andE= lim infn En. By (a),lim inf

n(En) = lim

n(En) = 0.

Therefore, by (), (E) = 0. By definition, we have

E=nN

kn

Ek.

Hence,x /E wheneverxEcn for infinitely many ns, that is fn(x) =f(x) a.e. inD for infinitely many ns.

Problem 42

Supposefn(x)fn+1(x) for alln NandxD \ Zwith(Z) = 0. Iffn fonD , then prove thatfnf a.e. onD.

Solution

LetB = D \ Z. Sincefn f onD, fn f onB . Then, By Riesz theorem, thereexists a sub-sequence (fnk) of (fn) such that fnkf a.e. onB .LetC={xB : fnk f}. Then(C) = 0 and fnkf onB\ C.Fromfn(x)fn+1(x) for alln N, and sincenkk, we getfkfnk for allk N.Therefore

|fk f| |fnk f|.This implies thatfkf onB\ C. Since B\ C=D \ (Z C) and(Z C) = 0,it follows that fnf a.e. on D .

Problem 43

Show that iffnf onD andgn g onD thenfn+gn f+g onD.

Solution

Sincefn

f and gn

g on D, for every >0,lim

n{D: |fn f|

2}= 0(4.1)

limn

{D: |gn g| 2}= 0.(4.2)

Now|(fn+gn) (f+g)| |fn f| + |gn g|.




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By the triangle inequality above, if|(fn+ gn) (f+g)| is true, then at leastone of the two following inequalities must be true:

|fn f| 2

or |gn g| 2

.

Hence

{D: |(fn+gn) (f+g)| }

D:|fn f| 2

D:|gn g| 2

.

Therefore,

{D: |(fn+gn) (f+g)| }

D:|fn f| 2

+

D:|gn g|

2

.

From (4.1) and (4.2) we obtain

limn

{D: |(fn+gn) (f+ g)| }= 0.

That is, by definition, fn+gnf+g onD.

Problem 44

Show that iffnf onD andgn g onD and(D) 0 and >0, we want {|fngn f g| }< forn large enough.Notice that

() |fngn f g| |fngn f gn| + |f gn f g| |fn f||gn| + |f||gn g|.

For anyN

N, let

EN={D: |f|> N} {D: |g|> N}.

It is clear that ENEN+1 for every N N. Since (D)


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51

It follows that, we can take Nlarge enough to get, for every >0,

()

2N N+ 1}

D: |gn g| 2N

EN

(since|gn| |gn g| + |g|). Now if we have

|fn f| 2(N+ 1)

;|gn|> N+ 1;|gn g| 2N

, and |f|> N,

then (*) implies

{D: |fngn f g| } D: |fn f| 2(N+ 1)

EN

D: |gn g|

2N

{D: |gn|> N+ 1}.

By assumption, given >0, >0, forn > N, we have

D: |fn f| 2(N+ 1)

0 such that

{xD : |fn(x) f(x)|> 0} 0 as n .




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We can choose n1 < n2< ...such that

{

x

D : |

fnk(x)

f(x)|> 0

} r for some r >0 and

kN.

Now since fnfa.u on D,

ED such that (E)< r2

and fnfuniformly on D \ E.

Let C ={x D : |fnk(x)f(x)| > 0} k N. Then (C) r. Sincefnf uniformly on D \ E,

N : nN |fn(x) f(x)| 0,xD \ E.

Thus,C(D \ E)c =E.

Hence,

0< r(C)(E)< r2

.

This is a contradiction.

(c) Since fn f a.u. on D, for every n N, there exists En D such that(En) M2.

Show thatg is measurable onD.

Solution

Leta R. We need to show that the set E={xD : g(x)> a} is measurable.There are three cases to consider:

1. IfaM2 thenE= which is measurable.

2. Ifa < M1 then E= D which is measurable.

3. IfM1a < M2 thenE={xD : f(x)> a} which is measurable.

Thus, in all three cases Eis measurable, so g is measurable.

Problem 47

Given a measure space(X, A, ). Letfbe a bounded real-valuedA-measurablefunction on D A with (D) 0.(a) Show that if

Df d= M (D), thenf=M a.e. onD.

(b) Show that iff < M a.e. onD and if(D)>0, then

Dfd < M(D).

Solution

(a) For every nN, let En ={x D : f(x) < M 1n}. Then, sincef M onD \ En, we have

D

f d =

En

f d+

D\En

f d

M1n

(En) +M (D \ En).




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56CHAPTER 5. INTEGRATION OF BOUNDED FUNCTIONS ON SETS OF FINITE MEASU

SinceEnD, we have(D

\En) =(D)

(En).

Therefore, D

f d

M1n

(En) +M(D) M(En)

= M(D) 1n

(En).

By assumption

Df d= M (D), it follows that

0 1n

(En)0,n N,

which implies (En) = 0,n N.Now let E =n=1 En then E ={x D : f(x) < M}. We want to show that(E) = 0. We have

0(E)

n=1

(En) = 0.

Thus, (E) = 0. Since|f| M, the last result implies f=M a.e. onD.(b) First we note that|f| M on D implies that Df dM (D). Assume that

Df d= M (D). By part (a) we have f=M a.e. onD. This contradicts the fact

thatf < M a.e. onD. Thus

Dfd < M(D).

Problem 48

Consider a sequence of functions(fn)nN defined on [0, 1] by

fn(x) = nx

1 +n2x2 for x[0, 1].

(a) Show that(fn) is uniformly bounded on[0, 1] and evaluate

limn

[0,1]

nx

1 +n2x2 d.

(b) Show that(fn) does not converge uniformly on[0, 1].

Solution

(a) For all nN, for all x[0, 1], we have 1 +n2x2 2nx0 and 1 +n2x2 >0,hence

0fn(x) = nx1 +n2x2

12

.




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57

Thus, (fn) is uniformly bounded on [0, 1].Since eachfn is continuous on [0, 1], fis Riemann integrable on [0, 1]. In this case,

Lebesgue integral and Riemann integral on [0, 1] coincide:[0,1]

nx

1 +n2x2 d =

10

nx

1 +n2x2dx

= 1

2n

1+n21

1

tdt (with t= 1 +n2x2)

= 1

2nln(1 +n2) =

ln(1 +n2)

2n .

Using LHospital rule we get limxln(1+x2)

2x = 0. Hence,

limn

[0,1]

nx1 +n2x2

d= 0.

(b) For each x[0, 1],lim

nnx

1 +n2x2 = 0.

Hence, fn f 0 pointwise on [0, 1]. To show fn does not converge to f 0uniformly on [0, 1], we find a sequence (xn) in [0, 1] such thatxn0 andfn(xn) f(0) = 0 as n . Indeed, take xn= 1n . Then fn(x) = 12 .Thus,

limn fn(xn) =1

2=f(0) = 0.

Problem 49

Let(fn)nN andfbe extended real-valued measurable functions onD ML with(D)0,there exists anN N such that fornN ,

EnD : (En)< 2

and |fn f|< 2(D)

on D \ En.




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FornNwe have

(

) D |fn f|1 + |fn f| d= En|fn f|

1 + |fn f| d+ D\En |fn f|1 + |fn f| d.

Note that for all n N, we have 0 |fnf|1+|fnf|1 on En and

0 |fn f|1 + |fn f| =|fn f|

1

1 + |fn f| |fn f|

2(D) on D \ En.

So fornN, we can write (*) as

0

D

|fn f|1 + |fn f| d

En

1d+

D\En

2(D) d

= (En) +

2(D)

(D

\En)

(En) + 2

0, for n N, letEn={xD : |fn f| }. We have

|fn f| |fn f|1 + |fn f|

1 +

( since the function (x) = x1+x

, x >0 is increasing).

It follows that

0

En

1 + d

En

|fn f|1 + |fn f|d

D

|fn f|1 + |fn f|d.

Hence,

0 1 +

(En)

D

|fn f|1 + |fn f|d.

Since limn

D|fnf|

1+|fnf|d= 0, limn (En) = 0. Thus, fnf on D.

Problem 50Let (X, A, ) be a finite measure space. Let be the set of all extended real-valuedA-measurable function on Xwhere we identify functions that are equala.e. onX. Let

(f, g) =

X

|f g|1 + |f g| d for f, g.




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59

(a) Show that is a metric on.(b) Show that is complete w.r.t. the metric.

Solution

(a) Note that (X) is finite and 0 |fg|1+|fg| 0. Forn, mN, define Am,n ={X : |fn fm| }. For every > 0, thereexists an N Nsuch that

() n, mN (fn, fm)< 1 +

.




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While we have that

(fn, fm) = X |fn

fm

|1 + |fn fm| d Am,n |fn

fm

|1 + |fn fm| d

1 + (Am,n).

Forn, mN, from (*) we get

1 + >

1 + (Am,n).

This implies that (Am,n)< . Thus, (fn) is Cauchy in measure. We know that if(fn) is Cauchy in measure then (fn) converges in measure to some f.

Next we prove that (fn, f) 0. Sincefn

f, for any >0 there exists E Aand an N Nsuch that(E)0, there exists >0 such that

E

fd <

for every measurableED with(E)< .

Problem 52

Let f1 and f2 be nonnegative extended real-valued measurable functions on ameasurable setDR. Supposef1 f2 andf1 is integrable onD. Prove thatf2 f1 is defined a.e. onD and

D

(f2 f1)d=

D

f2d

D

f1d.

SolutionSincef1 is integrable onD,f1 is real-valued a.e. on D. Thus there exists a null setN D such that 0f1(x)


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65

Since

Df1d 0}={xX: f(x)= 0}. We shall show that(A) = 0.LetAn={xX: f(x) 1n}for everyn N. ThenA =

nN An. Now onAn we

have

f 1n

An

f d 1n

(An)

(An)n An f d= 0 (by assumption) (An) = 0 for every n N.Thus, 0 (A)nN (An) = 0. Hence, (A) = 0. This tells us thatf = 0a.e.

Problem 54

Let(fn: n N) be a sequence of non-negative real-valued measurable functionsonR such thatfnf a.e. onR.Suppose lim

n R fnd = R fd


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66 CHAPTER 6. INTEGRATION OF NONNEGATIVE FUNCTIONS

Sincegn= fn fnE0, n N and fnfa.e., we have, by Fatous lemma,

R limn gnd lim infn R gndR

(f f E)d lim infn

R

(fn fnE)dR

f d

E

fd limn

R

fnd lim supn

E

fnd.

From the last inequation and assumption we get

(6.1)

E

fdlimsupn

E

fnd.

Lethn= fn

fnE

0. Using the similar calculation, we obtain

(6.2)

E

f dlim infn

E

fnd.

From (6.1) and (6.2) we have

limn

E

fnd=

E

fd.

Problem 55

Given a measure space (X,

A, ). Let (fn) and f be extended real-valued

A-

measurable functions on D A and assume that f is real-valued a.e. on D.Suppose there exists a sequence of positive numbers(n) such that

1.

nN n


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67

Using assumptions we get

D

n=1

|fn f|p

d = limN

Nn=1

D |fn f|

p

d

=

n=1

D

|fn f|pd

n=1

n


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Problem 57

Let (X,A

, ) be a measure space and let f be an extended real-valuedA

-measurable function onX such that

X|f|pd X

|f|pd=

n=0

En

|f|pd

n=0np(En).

Since

n=0 np(En)0, there exists N N such that for nN

we have n=N

np(En)< .

Note thatnp Np sincep >0. So we have

Np

n=N(En)< .

But

n=NEn={X: |f| N}. So with the above N, we have

Np

n=N

En

= Np{X: |f| N}< .

Thus,lim

p{X: |f| }= 0.

Problem 58

Let (X, A, ) be a-finite measure space. Letfbe an extended real-valuedA-measurable function onX. Show that for everyp(0, ) we have

X

|f|pd=

[0,)pp1{X : |f|> }L(d). ()




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69

Solution

We may suppose f0 (otherwise we set g =|f| 0).1. Iff=E, E A, then

X

fpd=

X

(E)pd= (E).

[0,)pp1{X : E> }L(d) =

10

pp1(E)d= (E).

Thus, the equality () holds.2. Iff=

ni=1 aiEi (simple function), withai0, Ei A, i= 1,...,n., then the

equality () holds because of the linearity of the integral.

3. Iff 0 measurable, then there is a sequence (n) of non-negative measurablesimple functions such thatnf. By the Monotone Convergence Theorem we haveX

fpd = limn

X

pnd

= limn

[0,)

pp1{X: n> }L(d)

=

[0,)

pp1{X: f > }L(d).

Notes:

1. A={X: E> }={xX : E(x)> }. If 0


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Solution

From the expression Dn ={xD : f(x)n} with f A-measurable, we deducethatDn Aand

D:= D0D1D2...DnDn+1...Moreover, all the sets Dn \ Dn+1={D: nf < n+ 1, n N} are disjoint and

D=nN

(Dn \ Dn+1).

It follows that

n(Dn \ Dn+1)

Dn\Dn+1f d(n+ 1)(Dn \ Dn+1)

n=0

n(Dn \ Dn+1) nN(Dn\Dn+1)

f d

n=0

(n+ 1)(Dn \ Dn+1)

n=0

n[(Dn) (Dn+1)]

D

f d

n=0

(n+ 1)[(Dn) (Dn+1)]. (i)

Some more calculations:

n=0

n[(Dn) (Dn+1)] = 1[(D1) (D2)] + 2[(D2) (D3)] +...

=

n=1 (Dn),and

n=0

(n+ 1)[(Dn) (Dn+1)] = 1[(D0) (D1)] + 2[(D1) (D2)] +...

= (D) +

n=1

(Dn).

With these, we rewrite (i) as follows

n=1

(Dn) D f d(D) +

n=1(Dn).

Since(D)


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71

Problem 60

Given a measure space (X,A

, ) with (X) 2n}2n}for eachn = 0, 1, 2,...Then it is clear that

E0E1...EnEn+1...En \ En+1={X : 2n < f2n+1} and are disjointX\ E0 ={X: 0f1}

X= (X\ E0)

n=0

(En \ En+1).

Now we have

X f d = X\E0 f d+ n=0(En\En+1) f d=

X\E0

f d+

n=0

En\En+1

fd.

This implies that

(6.3)

n=0

En\En+1

f d=

X

f d

X\E0fd.

On the other hand, for n = 0, 1, 2,..., we have

2n(En \ En+1)

En\En+1f d2n+1(En \ En+1).

Therefore,

n=0

2n(En \ En+1)

n=0

En\En+1

f d

n=0

2n+1(En \ En+1).




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From (6.3) we obtain

n=0

2n(En \ En+1) + X\E0 f d

X

f d

n=02n+1(En \ En+1) +

X\E0 fd.

Since

0

X\E0fd(X\ E0)(X)


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73

Since(X)


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Chapter 7

Integration of Measurable

Functions

Given a measure space (X, A, ). Let fbe a measurable function on a set D A. We define thepositive and negative parts off by

f+ := max{f, 0} and f := max{f, 0}.Then we have

f=f+ f and |f|= f+ + f.

Definition 20 Letfbe an extended real-valued measurable function onD. The functionfis said

to be integrable on D iff+ andf are both integrable onD. In this case we defineD

f d=

D

f+d D

fd.

Proposition 20 (Properties)

1. f is integrable onD if and only if|f| is integrable onD.2. Iff is integrable onD thencf is integrable onD, and we have

D

cfd= cD

f d, wherec is a constant inR.

3. Iff andg are integrable onD thenf+ g are integrable onD, and we haveD(f+ g)d=Df d + Dgd.4. fg Df d Dgd.5. Iff is integrable onD then|f|


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76 CHAPTER 7. INTEGRATION OF MEASURABLE FUNCTIONS

Theorem 8 (generalized monotone convergence theorem)Let(fn) be a sequence of integrable extended real-valued functions onD.1. If(fn) is increasing and there is a extended real-valued measurable functiong such thatfn

g

for everyn N, thenlimn

fnd=

D

gd.

2. If(fn) is decreasing and there is a extended real-valued measurable functiong such thatfngfor everyn N, then

limn

fnd=

D

gd.

Theorem 9 (Lebesgue dominated convergence theorem theorem - D.C.T)Let (fn) be a sequence of integrable extended real-valued functions on D and g be an integrablenonnegative extended real-valued function on D such that|fn| g on D for every n N. Iflimn fn= f exists a.e. onD, thenf is integrable onD and

limn

D

fnd=D

fd and limn

D

|fn f|d= 0.

Problem 62

Prove this statement:Letf be extended real-valued measurable function on a measurable setD. Iffis integrable onD, then the set{D: f= 0} is a-finite set.

Solution

For every n NsetDn=

xD : |f(x)| 1

n

.

Then we have

{xD : f(x)= 0}={xD : |f(x)|>0}=nN

Dn.

Now for each n Nwe have1

n(Dn) Dn |f|d D |f|d


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77

Problem 63

Letfbe extended real-valued measurable function on a measurable setD. If(En)is an increasing sequence of measurable sets such thatlimn En= D, then

D

f d = limn

En

fd.

Solution

Since (En) is an increasing sequence with limit D, so by definition, we have

D=

n=1

En.

LetD1 = E1 and Dn= En \ En+1, n2.

Then{D1, D2,...} is a disjoint collection of measurable sets, and we haven

i=1

Di = En and

n=1

Dn=

n=1

En= D.

Hence D

fd =

n=1

Dn

f d= limn

ni=1

Di

f d

= limn

ni=1

Di

f d = limn

En

fd.

Problem 64

Let(X,A

, )be a measure space. Letf andg be extended real-valued measurablefunctions onX. Suppose thatf andg are integrable onXandEf d= Egdfor everyE A. Show thatf=g a.e. onX.

Solution

Case 1: f and g are two real-valued integrable functions on X.Assume that the statement f=g a.e. onX is false. Then at least one of the two




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sets E={X : f < g} andF ={X : f > g} has a positive measure. Consider thecase(E)>0. Now since both f andg are real-valued, we have

E=kN

Ek where Ek =E=

X: g f 1k

.

Then 0< (E)kN (Ek). Thus there exists k0N such that (Ek0)> 0, sothat

Ek0

(g f)d 1k0

(Ek0)> 0.

Therefore

Ek0 gd Ek0 f d+ 1

k0(Ek0)> Ek0 fd.

This is a contradiction. Thus(E) = 0. Similarly,(F) = 0. This shows thatf=ga.e. on X.

Case 2: General case, where f and g are two extended real-valued integrablefunctions on X. The integrability off and g implies that f and g are real-valueda.e. onX. Thus there exists a null set NXsuch that f andg are real-valued onX\ N. Set

f=

f on X\ N,0 on N.

and g=

g on X\ N,0 on N.

Then fand g are real-valued on X, and so on every E Awe haveE

f d=

E

f d=

E

gd =

E

gd.

By the first part of the proof, we have f = g a.e. on X. Since f =f a.e. on Xand g= g a.e. on X, we deduce that

f=g a.e. on X.

Problem 65

Let (X, A, ) be a -finite measure space and let f, g be extended real-valuedmeasurable functions onX. Show that if

E

f d=

Egd for everyE Athen

f=g a.e. onX. (Note that the integrability off andg is not assumed.)




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2. (E(2))> 0. Let

E(2)l =

{Xn:

< f

l; g=

}.

ThenE(2) =

lN

E(2)l .

By assumption and the subadditivity ofwe have

0< (E(2))lN

(E(2)l ).

This implies that there is some l0 N such that(E

(2)l0

)>0.

LetE= E(2)l0 . Then E

gd =>

Efd.

This contradicts the assumption that

Ef d=

E

gd for every E A.3. (E(3))> 0. Let

E(2)m ={Xn: = f;mg 0.

LetE = E(3)m0. ThenE

gd m(E)>= E

f d:

This contradicts the assumption.

Thus, (E) = 0. Similarly, we get (F) = 0. That is f=g a.e. on X. .




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81

Problem 66

Given a measure space(X,A

, ). Letfbe extended real-valued measurable andintegrable function onX.1. Show that for any >0 there exists >0 such that ifA A with(A)< then

A

f d

< .2. Let (En) be a sequence inA such that limn (En) = 0. Show thatlimn

En

f d= 0.

Solution

1. For everyn N, set

fn(x) =

f(x) iff(x)nn otherwise.

Then the sequence (fn) is increasing. Each fn is bounded and fn f pointwise.By the Monotone Convergence Theorem,

>0,N Nsuch that

X

fNd

X

f d

X

|f|d=

n=0

En\En+1

|f|d

n=0

n(En \ En+1).

Some more calculations for the last summation:

n=0n(En \ En+1) =

n=0n[(En) (En+1)]

= (E1) (E2) + 2[(E2) (E3)] + 3[(E3) (E4)] +...

=

n=1

(En)


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In particular, forxIn= (xn , xn+), we have

|f(xn) f(x)||f(xn)|

2

2=

2.

Since xn+1 xn > 1 and 0 < < 12 , In In+1 =. Moreover,

n=1 In [0, ).By assumption, fis integrable on [0, ), so we have

> [0,) f d

n=1 In fd >

n=1 In

2

d=

.

This is a contradiction. Thus,

limx

f(x) = 0.

Problem 70

Let(X, A, )be a measure space and let(fn)nN, andf, gbe extended real-valued

A-measurable and integrable functions onD

A. Suppose that

1. limn fn= f a.e. on D.

2. limn

Dfnd=

D

f d.

3. eitherfng onD for alln N orfng onD for alln N.

Show that, for everyE A andED, we have

limnE

fnd=

Efd.

Solution

(a) First we solve the problem in the case the condition 3. is replaced by fn0 onD for all n N.Lethn =fn fnE for every E Aand ED. Then hn0 andA-measurable




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85

and integrable on D. Applying Fatous lemma to hn and using assumptions, we get

D f d E f d= D(f f E)d lim infn D(fn fnE)d

= limn

D

fnd limsupn

D

fnEd

=

D

f d lim supn

E

fnd.

Sincef is integrable onD,

Dfd


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Problem 71(An extension of the Dominated Convergence Theorem)Let(X,

A, ) be a measure space and let(fn)n

N, (gn)n

N, andf, g be extended

real-valuedA-measurable functions onD A. Suppose that

1. limn fn= f and limn gn= g a.e. on D.

2. (gn) andg are all integrable onD and limn

Dgnd=

D

gd.

3.|fn| gn onD for everyn N.

Prove thatf is integrable onD and limn

Dfnd=

Df d.

SolutionConsider the sequence (gn fn). Since|fn| gn, and (fn) and (gn) are sequencesof measurable functions, the sequence (gn fn) consists of non-negative measurablefunctions. Using the Fatous lemma we have

D

liminfn

(gn fn)dlim infn

D

(gn fn)dD

limn

(gn fn)d limn

D

gnd lim supn

D

fnd

D gd D f d D gd lim supn D fndD

f dlimsupn

D

fnd. () (since

D

gd


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87

Problem 72

Given a measure space (X,A

, ). Let (fn)nN and f be extended real-valued

A-measurable and integrabe functions onD A. Suppose that

limn

fn= f a.e. on D.

(a) Show that if limn

D|fn|d =

D

|f|d, then limn

Dfnd =

Df d.(b) Show that the converse of (a) is false by constructing a counter example.

Solution

(a) We will use Problem 71 for

gn= 2(|fn| + |gn|) and hn=|fn f| + |fn| |f|, n N.We have

hn0 a.e. on D,gn4|f| a.e. on D,|hn|= hn2|fn| gn,

limn

D

gnd= 2 limn

D

|fn|d+ 2

D

|f|d=

D

4|f|d.

So all conditions of Problem 71 are satisfied. Therefore,

limn

D

hnd=

D

hd= 0 (h= 0).

limn

D

|fn f|d+ limn

D

|fn|d

D

|f|d= 0.

Since limn

D|fn|d

D

|f|d= 0 by assumption, we have

limn

D

|fn f|d= 0.

This implies that

limn

D

fnd D

f d = 0.

Hence, limn

Dfnd=

Df d.

(b) We will give an example showing that it is not true that

limn

D

fnd=

D

f d limn

D

|fn|d=

D

|f|d.




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fn(x) =

n if 0x < 1n

0 if 1nx1 1nn if 1 1n < x1.

And so

|fn|(x) =

n if 0x < 1n

or 1 1n

< x10 if 1

nx1 1

n.

Then we have

fn00 and

[0,1]

fnd= 00 =

[0,1]

0d

while [0,1]

|fn|d= 22= 0.

Problem 73

Given a measure space(X, A, ).(a) Show that an extended real-valued integrable function is finite a.e. onX.(b) If (fn)nN is a sequence of measurable functions defined on X such that

nN X|fn|d 0. Since f is integrable

> X |f|d E |f|d=.

This is a contradiction. Thus, (E) = 0.

(b) First we note thatN

n=1 |fn| is measurable since fn is measurable for n N.Hence,

limN

Nn=1

|fn|=

n=1

|fn|




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89

is measurable. Recall that (for nonnegative measurable functions)

X

n=1

|fn|d=

n=1

X

|fn|d.

By assumption,

n=1

X

|fn|d


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Problem 74

Letfbe a real-valued Lebesgue measurable function on [0,

) such that

1. f is Lebesgue integrable on every finite subinterval of[0, ).2. limx f(x) =c R.

Show that

lima

1

a

[0,a]

f dL = c.

SolutionBy assumption 2. we can write

() >0, N : x > N |f(x) c|< .Now, for a > Nwe have1a

[0,a]

f dL c =

1a

[0,a]

(f c)dL

1a [0,a]

|f c|dL

= 1

a

[0,N]

|f c|dL+

[N,a]

|f c|dL

.

By (*) we havex[N, a] |f(x) c|< .

Therefore,

()

1

a

[0,a]

f dL c 1

a

[0,N]

|f c|dL+(a N)a

.

It is evident thatlim

a(a N)

a = .

By assumption 1.,|fc|is integrable on [0, N], so [0,N]

|fc|dL is finite and doesnot depend on a. Hence

lima

1

a

[0,N]

|f c|dL = 0.




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91

Thus, we can rewrite (**) as

lima 1a [0,a] f dL c .Since >0 is arbitrary, this implies that

lima

1a

[0,a]

f dL c = 0.

Problem 75

Let f be a non-negative real-valued Lebesgue measurable onR. Show that ifn=1 f(x+n) is Lebesgue integrable onR, thenf= 0 a.e. onR.Solution

Recall these two facts:

1. Iffn0 is measurable on D then

D(

n=1 fn) d=

n=1

D

fnd.

2. Iffis defined and measurable on R thenR

f(x+h)d=R

f(x)d.

From these two facts we haveR

n=1

f(x+n)

dL =

n=1

R

f(x+n)dL

=

n=1

R

f(x)dL.

Since

n=1 f(x+n) is Lebesgue integrable onR,R

n=1

f(x+n)

dL


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Problem 76

Show that the Lebesgue Dominated Convergence Theorem holds ifa.e. convergence is replaced by convergence in measure.

Solution

We state the theorem:Given a measure space(X, A, ). Let(fn : nN) be a sequence of extended real-valuedA-measurable functions onD Asuch that|fn| g onD for everynN

for some integrable non-negative extended real-valuedA-measurable function g onD. Iffn

f onD, thenf is integrable onD and

limnD fnd= D fd.Proof:

Let (fnk) be any subsequence of (fn). Then fnk f since fn f. By Riesz

theorem, there exists a subsequence (fnkl ) of (fnk) such that fnkl f a.e. onD.And we have also|fnkl | g onD. By the Lebesgue D.C.T. we have

()

D

f d= liml

D

fnkld.

Letan= Dfndanda = Df d. Then (*) can be written asliml

ankl =a.

Hence we can say that any subsequence (ank) of (an) has a subsequence (ankl )converging to a. Thus, the original sequence, namely (an), converges to the samelimit (See Problem 51): limn an= a. That is,

limn

D

fnd=

D

fd.

Problem 77Given a measure space (X, A, ). Let (fn)nN and f be extended real-valuedmeasurable and integrable functions onD A.Suppose thatlimn

D

|fn f|d= 0. Show that(a) fn

f onD.(b)limn

D

|fn|d=

D|f|d.




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93

Solution

(a) Given any >0, for each n N, letEn={D: |fn f| }. ThenD

|fn f|d

En

|fn f|d(En).

Since limn

D|fn f|d= 0, limn (En) = 0. That isfn f onD.

(b) Since fn andfare integrableD

(|fn| |f|)d=

D

|fn|d

D

|f|d

D

|fn f|d.

By this and the assumption, we get

limn

D

|fn|d

D

|f|d

limn

D

|fn f|d= 0.

This implies

limn

D

|fn|d=

D

|f|d.

Problem 78

Given a measure space (X,

A, ). Let (fn)nN and f be extended real-valued

measurable and integrable functions onD A. Assume thatfnf a.e. onDandlimn

D

|fn|d=

D|f|d. Show that

limn

D

|fn f|d= 0.

Solution

For eachn N, let hn =|fn| + |f| |fn f|. Then hn0 for all n N.Sincefn

f a.e. onD, hn

2

|f

|a.e on D. By Fatous lemma,

2

D

|f|d liminfn

D

(|fn| + |f|)d lim supn

D

|fn f|d

= limn

D

|fn|d+ limn

D

|f|d lim supn

D

|fn f|d

= 2

D

|f|d limsupn

D

|fn f|d.




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95

SinceCc(R) is dense inL1(R), we can find a continuous functionvanishing outside[M, M] such that

MM

|f |dx 0 is arbitrary, we have

limh0

f(x+h) f(x)1= limh0

R

|f(x+h) f(x)|dx= 0.




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Chapter 8

Signed Measures and

Radon-Nikodym Theorem

1. Signed measure

Definition 21 (Signed measure)A signed measure on a measurable space(X, A) is a function:A [, ] such that:(1) () = 0.(2) assumes at most one of the values.(3) is countably additive. That is, if{En}nN A is disjoint, then

nNE

n= nN(E

n).

Definition 22 (Positive, negative, null sets)Let(X, A, ) be a signed measure space. A setE A is said to be positive (negative, null) for thesigned measure if

F A, F E=(F)0 (0, = 0).

Proposition 21 (Continuity)Let(X, A, ) be a signed measure space.1. If(En)nN A is an increasing sequence then

limn

(En) = limn

nN

En

=

limn

En

.

2. If(En)nN A is an decreasing sequence and(E1)


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98 CHAPTER 8. SIGNED MEASURES AND RADON-NIKODYM THEOREM

Proposition 22 (Some more properties)Let(X, A, ) be a signed measure space.1. Every measurable subset of a positive (negative, null) set is a positive (negative, null) set.

2. IfE is a positive set andF is a negative set, thenE F is a null set.3. Union of positive (negative, null) sets is a positive (negative, null) set.

Theorem 10 (Hahn decomposition theorem)Let(X, A, ) be a signed measure space. Then there is a positive set A and a negative setB suchthat

A B = and A B= X.Moreover, ifA andB are another pair, thenA A andB B are null sets.{A, B} is called a Hahn decomposition of(X, A, ).

Definition 23 (Singularity)Two signed measure1 and2 on a measurable space(X, A) are said to be mutually singular andwe write12 if there exist two setE, F Asuch thatE F= , E F=X, E is a null set

for1 andF is a null set for2.

Definition 24 (Jordan decomposition)Given a signed measure space(X, A, ). If there exist two positive measures and, at least oneof which is finite, on the measurable(X, A) such that

and = ,then{, } is called a Jordan decomposition of.

Theorem 11 (Jordan decomposition of signed measures)Given a signed measure space(X, A, ). A Jordan decomposition for(X, A, ) exists and unique,that is, there exist a unique pair{, } of positive measures on (X, A), at least one of which is

finite, such that and = .Moreover, with any arbitrary Hahn decomposition{A, B} of(X, A, ), if we define two set functions andby setting

(E) = (E A) and (E) =(E B) for E A,then{, } is a Jordan decomposition for(X, A, ).

2. Lebesgue decomposition, Radon-Nikodym Theorm

Definition 25 (Radon-Nikodym derivative)Let be a positive measure and be a signed measure on a measurable space(X, A). If there existsan extended real-valuedA-measurable functionf onXsuch that

(E) =

E

f d for every E A,

thenf is called a Radon-Nikodym derivative of with respect to, and we write dd for it.




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99

Proposition 23 (Uniqueness)Let be a -finite positive measure and be a signed measure on a measurable space (X, A). Iftwo extended real-valued

A-measurable functionsf andg are Radon-Nikodym derivatives of with

respect to, thenf=g -a.e. onX.

Definition 26 (Absolute continuity)Let be a positive measure and be a signed measure on a measurable space(X, A). We say that is absolutely continuous with respect to and write if

E A), (E) = 0 =(E) = 0.

Definition 27 (Lebesgue decomposition)Let be a positive measure and be a signed measure on a measurable space(X, A). If there existtwo signed measuresa ands on(X, A) such that

a, s and = a+ s,then we call{a, s} a Lebesgue decomposition of with respect to . We call a and s theabsolutely continuous part and the singular part of with respect to .

Theorem 12 (Existence of Lebesgue decomposition)Let be a -finite positive measure and be a -finite signed measure on a measurable space(X, A). Then there exist two signed measuresa ands on(X, A) such that

a, s, = a+ s and a is defined by a(E) =E

fd,E A,

wheref is an extended real-valued measurable function onX.

Theorem 13 (Radon-Nikodym theorem)Let be a -finite positive measure and be a -finite signed measure on a measurable space(X, A). If , then the Radon-Nikodym derivative of with respect to exists, that is, thereexists an extended real-valued measurable function onX such that

(E) =

E

fd,E A.

Problem 80

Given a signed measure space (X, A, ). Suppose that{, } is a Jordan de-composition of , and E and F are two measurable subsets of X such thatE F =, E F = X, E is a null set for andF is a null set for.Showthat{E, F} is a Hahn decomposition for(X, A, ).




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100 CHAPTER 8. SIGNED MEASURES AND RADON-NIKODYM THEOREM

Solution

We show that Eis a positive set for and Fis a negative set for . Since{, }isa Jordan decomposition of, we have

(E) =(E) (E),E A.LetE0 A, E0E. Since Eis a null set for , E0 is also a null set for . Thus(E0) = 0. Consequently, (E0) = (E0)0. This shows thatE is a positive setfor.Similarly, let F0 A, F0E. Since F is a null set for , F0 is also a null set for. Thus(F0) = 0. Consequently, (F0) =(F0) 0. This shows thatF is anegative set for .We conclude that{E, F} is a Hahn decomposition for (X, A, ).

Problem 81

Consider a measure space([0, 2],

Measure and Integration Full

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