measure and integration theory

8/7/2019 Measure and Integration Theory

1/67

Measure and Integration Theory

Fabio Maccheroni 1

Department of Decision Sciences - U. Bocconi - Milano

First version: Agosto 2001,

Current version: September 2010.

1I thank Erio Castagnoli who read these notes over and over again and added punctuation;I also thank Laura Maspero who helped me to write them and took notes during the AdvancedAnalysis course held by Luigi Paganoni at "Universit degli Studi di Milano" during the academicyear 1991-1992, notes and course that inuenced these notes very much. Finally I thank GiuliaBrancaccio who drafted this rst English version of them in 2010. I take responsibility of all themistakes.


2/67

Contents

1 PREREQUISITES 2

2 CLASSES, CLASSES, CLASSES 2

3 SET FUNCTIONS 9

4 EXTENSION OF A MEASURE 12

5 COMPLETIONS 18

6 LEBESGUE-STIELTJES MEASURES ON R 20

7 MEASURABLE FUNCTIONS 23

8 INTEGRATION OF SIMPLE AND MEASURABLE FUNCTIONS WITHRESPECT TO A FINITE CHARGE 29

9 INTEGRATION OF BOUNDED MEASURABLE FUNCTIONS WITHRESPECT TO A FINITE CHARGE 32

10 INTEGRATION OF NONNEGATIVE FUNCTIONS WITH RESPECTTO A MEASURE 34

11 SUMMABLE FUNCTIONS 4012 PROPERTIES HOLDING ALMOST EVERYWHERE 43

13 RIEMANN (STIELTJES) Vs LEBESGUE (STIELTJES) 46

14 PRODUCT SPACES AND FUBINI TONELLI THEOREM 47

15 SIGNED MEASURES: HAHN AND JORDAN DECOMPOSITIONS 58

16 RADON-NIKODYM THEOREM 62

17 ESSENTIAL BIBLIOGRAPHY 66

1


3/67

1 PREREQUISITES

Set operations, topological and order properties of R , ("; ) arguments, properties of serieswith nonnegative terms.

2 CLASSES, CLASSES, CLASSES

Let X be a non empty set, and P (X ) the power set of X (sometimes denoted 2X ). We callclass a subset of P (X ).Definition. An algebra of subsets of X is a class A P (X ) s.t. (= such that)(a.i) X 2 A,

(a.ii) A 2 A )Ac 2 A,(a.iii) A; B 2 A )A [ B 2 A.

Some trivial algebras are: P (X ), f;; X g, f;;A;Ac; X gfor ; A X .Facts. Let Abe an algebra.

1.

; 2 A:

[In fact, ; = X c and X 2 A.]2. A1; A2;:::;AN 2 A )SN j =1 Aj 2 A:[By induction.]3. A; B 2 A )A \ B 2 A.

[In fact, A \ B = ( Ac [ B c)c.]4. A1; A2;:::;AN 2 A )TN j =1 Aj 2 A:[By induction.]5. A; B 2 A )A B 2 A.

[Exercise.]

6. A; B 2 A )A M B 2 A.[Exercise.]

2


4/67

Proposition. A P (X ) is an algebra if and only if (a.i) and (a.ii) hold, and A; B 2A )A \ B 2 A.Proof. It is enough to note that A [ B = ( Ac \ B c)

c.

Proposition. Let I = f(a; b] : 1 a < b < 1g [ f(b;1 ) : 1< b < 1g [ f;g [R ,thenE = (N [j =1 I j : N 2N and I j 2 I for j = 1 ; 2;:::;N )

is an algebra of subsets of R .

Proof. By assumption, R 2 E .If I 2 I , then I c is a nite disjoint union of elements of I . In fact, if I = ;, then I c = R ; if I = R , then I c = ;; if I = ( a; b] for a 2R , then I c = ( 1; a][ (b;1 ); if I = ( 1; b], thenI c = ( b;

1); nally if I = ( b;

1), then I c = (

1; b].

I; J 2 I ) I \ J 2 I . If I = ; or J = ;, and If I = R or J = R , the statement is straight-forward;If I = ( a; b] and J = ( c; d], x 2 I \ J i x > a; x b; x > c; x d i.e. x > max fa; cgandx min fb; dgi.e. x 2(max fa; cg; min fb; dg], that is I \ J = (max fa; cg; min fb; dg] 2 I .1If I = ( a; b] and J = ( c;1 ), x 2 I \ J i x > a; x b; x > c i.e. x > max fa; cgand x bi x 2(max fa; cg; b], that is I \ J = (max fa; cg; b] 2 I .If I = ( a; 1 ) and J = ( c;1 ), then I \ J = I or J 2 I .E; F 2 E )E \ F 2 E . In fact, E =

SN j =1 I j and F =

SM i=1 J i with I j ; J i 2 I for each j

and i, E \ F = SN j =1 I j \ S

M i=1 J i = S

N j =1

hI j \ S

M i=1 J i i= S

N j =1

SM i=1 I j \ J i =SN;M j =1 ;i =1 I j \ J i 2 E since I j \ J i 2 I for each i; j .By induction, E 1; E 2;:::;E N 2 E )TN j =1 E j 2 E .E 2 E )E c 2 E . In fact, E = SN j =1 I j with I j 2 I for each j and E c = TN j =1 I cj , but I cj 2 E for each j , therefore E c 2 E :

Definition. A semialgebra of subsets of X is a class S P (X ) s.t.(s.i) ;; X 2 S ,

(s.ii) A 2 S )Ac is a nite disjoint union of elements of S ,(s.iii) A; B 2 S )A \ B 2 S .

1With the convention ( ; ] = ; if .

3


5/67

With little abuse, we call the class I dened in the previous Proposition: class of real semi-intervals . I is a semialgebra (why?).Proposition. Let S be a semialgebra of subsets of X , then

A= (N

[j =1 Aj : N 2N and Aj 2 S for j = 1 ; 2;:::;N )is an algebra of subsets of X .

Proof. Exercise. Hint: look at the nal steps of the previous proof.

Definition. A -algebra (or tribu ) of subsets of X is a class F P (X ) s.t.( .i) X 2 F ,

( .ii) A 2 F )Ac 2 F ,( .iii) fAn gn 2 N F )S1n =1 An 2 F .

Some trivial -algebras are: P (X ), f;; X g, f;;A;Ac; X gfor ; A X . Each algebraof subsets of a nite set is a -algebra. E is not a -algebra (why?).Facts. Let F be a -algebra.

1. F is an algebra.[In fact, A; B 2 F )A [ B = A [ B [ B [ B:::: 2 F .]2. fAn gn 2 N F )T1n =1 An 2 F .[In fact, T1n =1 An = (S1n =1 Acn )c.]

Proposition. F P (X ) is a -algebra i ( .i), ( .ii) and fAn gn 2 N F )T1n =1 An 2 F hold.Proof. Exercise.

Let fAn gn 2 N P (X ); if An An +1 for all n 2N , the sequence fAn gn 2 N is increasing ,if An An +1 for all n 2 N , the sequence fAn gn 2 N is decreasing . If fAn gis increasing ordecreasing, it is said to be monotone . If fAn gn 2 N is increasing and A = S1n =1 An , we writeAn "A; analogously, if fAn gn 2 N is decreasing and A = T1n =1 An , we write An #A.Definition. A monotone class of subsets of X is a non empty class M P (X ) s.t.

4


6/67

(i) fAn gn 2 N Mand An "A ) A 2 M,(ii) fAn gn 2 N Mand An #A ) A 2 M.

Clearly a -algebra is a monotone class. The reverse is not true (why?), but:

Proposition. If a monotone class is an algebra, then it is a -algebra.

Proof. Let fAn gn 2 N M, then, for each N 2 N, BN = SN n =1 An 2 M(since Mis analgebra), and BN "S1n =1 An ; but Mis a monotone class, and therefore S1n =1 An 2 M.Definition. Let Cbe a non empty class of subsets of X . We call -algebra (resp. algebra ,resp. monotone class ) generated by Cthe smallest -algebra (resp. algebra, resp.monotone class) T containing C. Moreover, Cis said to be a system of generators of T .

In other words, the -algebra (resp. algebra, resp. monotone class) generated by Cis a-algebra (resp. algebra, resp. monotone class) T s.t.1. T C,2. if F is a -algebra (resp. algebra, resp. monotone class) s.t. F C, then F T .

Notice that, if T exists, then it is unique: If it existed T 0 6= T with the same properties, sinceT is a -algebra containing C, then T T 0; vice-versa, since T 0 is a -algebra containingC, then T 0 T . Therefore T 0 = T ! (= paradox). Such a -algebra is denoted by

(

C) (resp.

A(

C), resp.

M(

C)). The next proposition shows that (

C) (resp.

A(

C), resp.

M(C)) always exists.Proposition. Let Cbe a non empty class of subsets of X , the -algebra (resp. algebra,resp. monotone class) generated by Cis the intersection of all -algebras (resp. algebras,resp. monotone classes) containing C.Proof. We prove the -algebra case: the algebra and monotone class cases are similar(verify it). Let fF g2 be the set of all -algebras F such that F C. P (X ) Candit is a -algebra, therefore fF g2 6= ;. Let T =

T2 F . Since, for every , F is a

-algebra, then X

2 F for every and X

2 T . If A

2 T , then A

2 F for every , and

therefore Ac 2 F for every (F is a -algebra), then Ac 2 T . If An 2 T for every n 2N ,then An 2 F for every n 2N and for every , therefore S1n =1 An 2 F for every (F isa -algebra), nallyS1n =1 An 2 T . Therefore T is a -algebra. If C 2 C, then C 2 F forevery , and C 2 T . Thus, T is a -algebra and contains Cand, by denition, if F is a

-algebra s.t. F C, then F T .Facts. Let ; 6= C K P (X ).

5


7/67

1. A(C) (C).[Exercise.]

2. M(C) (C).[Exercise.]

3. (C) (K), A(C) A(K) and M(C) M(K) :[Exercise.]

Proposition. Let S be a semialgebra of subsets of X , then the following equalities hold:

A(S ) = (N [j =1 Aj : N 2N and Aj 2 S for j = 1 ; 2;:::;N )=

fnite disjoint unions of elements of

Sg:

Proof. Let A= nSN j =1 Aj : N 2N and Aj 2 S for j = 1 ; 2;:::;N oand A0 = fnite disjointunions of elements of Sg. We have already shown that Ais an algebra and since A S ,then A A(S ). Moreover, being A(S ) an algebra that contains S , it also contains allthe nite unions of elements of S , i.e. A(S ) A. Clearly A(S ) A0. It remains tobe shown that A0 is an algebra. Of course X 2 A0. If E; F 2 A0, E = SN j =1 Aj andF =

SM i=1 B i with A1; A2;:::;AN 2 S pairwise disjoint sets and B1; B2;:::;BM 2 S pairwise

disjoint sets, then E \ F = SN

j =1 Aj \ SM

i=1 B i = SN;M

j =1 ;i =1 Aj \ B i . Let (j; i ) ; (k; h) 2f1; 2;:::;N g f1; 2;:::;M gand (j; i ) 6= ( k; h), then either j 6= k or i 6= h, in the rst caseAj \ Ak = ;, in the second B i \ Bh = ;, in both cases (Aj \ B i ) \ (Ak \ Bh ) = ;. ThusE \ F = SN;M j =1 ;i =1 Aj \ B i 2 A0. By induction, E 1; E 2;:::;E N 2 A0 ) TN j =1 E j 2 A0. LetE 2 A0, then E = SN j =1 Aj with A1; A2;:::;AN 2 S pairwise disjoint sets and E c = TN j =1 Acj ,but, being S a semialgebra, for each j the Acj is a disjoint union of elements of S , i.e. Acj 2 A0,and TN j =1 Acj 2 A0.Monotone Classes Lemma. Let Abe an algebra, then

M(A) = (A) :

Proof. Clearly A M(A) (A). It remains to show that M(A) is an algebra, in thatcase, since it is also a monotone class, M(A) is a -algebra, and, containing A, it alsocontains (A).For each B X , set DB = fA 2 P (X ) : A\ B c; B \ Ac; A[ B 2 M(A)g. If fAn gn 2 N DB

6


8/67

and An #A, then An \ B c #A\ B c, B \ Acn "B \ Ac, and An [B #A[ B (verify it); moreoverfAn \ B cg;fB \ Acn g;fAn [ Bg M(A). It follows that A \ B c; B \ Ac; A [ B 2 M(A),and hence A 2 DB . Analogously, if fAn gn 2 N DB and An "A, then A 2 DB (verify it).Thus DB is a monotone class for each B X .If B 2 A, A DB (in fact, for each A 2 A, A \ B

c

; B \ Ac

; A[ B 2 A M(A)), but DB isa monotone class and therefore M(A) DB . This implies that, if B 2 Aand C 2 M(A),it is true that C \ B c; B \ C c; C [ B 2 M(A); that is, if B 2 Aand C 2 M(A), thenB \ C c; C \ B c; B [ C 2 M(A); that is, if B 2 Aand C 2 M(A), then B 2 DC . Thus, if C 2 M(A), A DC , but DC is a monotone class and M(A) DC .

Finally, if C; D 2 M(A), then D 2 DC , and D \ C c; C \ D c; D [ C 2 M(A); moreoverX 2 A M(A) and if C 2 M(A), then C c = X \ C c 2 M(A).Definition. A -class of subsets of X is a class H P (X ) s.t.( .i) X 2 H,

( .ii) A; B 2 H, A B ) B A 2 H,( .iii) An 2 Hand An "A ) A 2 H.

Observe that a -algebra is a -class. If Gis a non empty class of subsets of X , it existsa smallest -class containing G(why?), that we denote by (G).

Definition. A -class of subsets of X is a class ; 6= D P (X ) s.t.( .i) A; B 2 D )A \ B 2 D.

Observe that a semialgebra is also a -class.

Facts. Let Hbe a -class of subsets of X .1. If A 2 H, then Ac 2 H.

[Exercise.]2. If A; B 2 Hand A \ B = ;, then A [ B 2 H.

[In fact, A B c, and (A [ B)c = B c \ Ac = B c A.]3. His a monotone class.

[Exercise. Hint: if An 2 Hand An #A, then Acn 2 Hand Acn "Ac.]

7


9/67

4. If His a -class, then His a -algebra.[Exercise. Hint: (A [ B)

c = Ac \ B c.]

Dynkin s lemma. If

Dis a -class of subsets of X , then (

D) = (

D).

Proof. It is enough to show that (D) is a -class. Let

G1 = fA X : A \ D 2 (D) for each D 2 Dg.

G1 is a -class (why?) and contains D, then G1 (D). It follows that for each A 2 (D)and each D 2 D, A \ D 2 (D). Let

G2 = fA X : A \ E 2 (D) for each E 2 (D)g.

G2is a -class (why?) and contains

D, then

G2(

D). It follows that, if A

2(

D),

A \ E 2 (D) for each E 2 (D).If Y is a non empty subset of X and Cis a non empty class of subsets of X , set C \Y =

fC \ Y : C 2 Cg.Facts. Let ; 6= Y X and ; 6= C P (X ).

1. If Cis a semialgebra (resp. algebra, resp. -algebra) of subsets of X , then C \Y is asemialgebra (resp. algebra, resp. -algebra) of subsets of Y .

[In fact, ;; X 2 C ) ;\Y = ;; X \ Y = Y 2 C\Y . If E 2 C\Y , there exists C 2 Cs.t.E = C \ Y , but Cis a semialgebra: therefore there exist C 1; C 2;:::;C N 2 Cpairwisedisjoint, s.t. C c = SN j =1 C j ; it follows that Y E = Y \ E c = Y \ (C \ Y )c =Y \ (C c [ Y c) = Y \ C c = Y \ SN j =1 C j = SN j =1 (C j \ Y ). Then, if E 2 C \Y ,Y E (the complement of E in Y ) is a nite disjoint union of elements of C \Y .Finally, if E; F 2Y , there exist C; D 2 Cs.t. E = C \ Y and F = D \ Y , thereforeE \ F = C \ Y \ D \ Y = ( C \ D) \ Y , but C \ D 2 Cand E \ F 2 C \Y . Provethe similar results when Cis an algebra or a -algebra.]

2. (C) \ Y = (C \Y ) ;A(C) \ Y = A(C \Y ).[Exercise.]

Let I be the semialgebra of semi-intervals of R . The -algebra (I ) is called the Borel -algebra of R and it is denoted by Bor B(R ). The following classes generate B(prove it):

E = A(I ).

8


10/67

Closed sets in R .

Open sets in R .

Open intervals in R .

Closed intervals in R .

f( 1; a] : a 2Rg.

f( 1; a) : a 2Rg.

f(b;1 ) : b 2Rg.

f[b;1 ) : b 2Rg.

Let ; 6= I R ; the -algebra (I ) \ I = (I \ I ) is said Borel -algebra of I and itis denoted by B(I ). B(I ) is the -algebra generated by open sets in I , or by closed sets inI ,...(prove it).

3 SET FUNCTIONS

If A; B X and A\ B = ;, we write At B instead of A[B, analogously if fAn gn 2 N P (X )and Ai \ Aj = ; for i 6= j , we write

F1n =1 An instead of

S1n =1 An .

Let C P (X ) and A X . A nite partition (resp. countable ) of A in Cis a setfA1; A2;::;AN g(resp. fAn gn 2 N ) of pairwise disjoint elements of Cs.t. A = SN j =1 Aj (resp.A = S1n =1 Aj ):Definition. Let Cbe a class of subsets of X containing ; and let : C ! [0;1 ] be afunction, is called:

set function , if (;) = 0 .monotone , if A B

) (A) (B),

additive , if

(A) =N

Xj =1 (Aj ) .for each A 2 Cand each nite partition fA1; A2; : : ;AN gof A in C,

9


11/67

-additive , if (A) = Xj 2 J (Aj )

for each A 2 Cand each countable partition fAj gj 2 J of A in C.If Cis a semialgebra, an algebra or a -algebra, an additive set function is called charge ( orpositive charge) , while a -additive set function is called measure ( or positive meas-ure) .A measure (charge) is nite if (X ) < 1 , it is -nite if it exists a sequence fAn gn 2 NCs.t. X = S1n =1 An and (An ) < 1 for each n 2 N , nally a measure (charge) is aprobability measure (charge) if (X ) = 1 .Facts. Let Cbe a class of subsets of X containing ; and let : C ![0;1 ] be a function.

1. A -additive set function is additive. In particular, a measure is also a charge. (Doesthe converse hold?)

2. If Cis an algebra, then is a charge i (c.i) (;) = 0 ,

(c.ii) A; B 2 Cand A \ B = ; ) (A [ B) = (A) + (B).[Exercise.]

1. If Cis a semialgebra, then is a measure i (m.i) (;) = 0 ,

(m.ii) fAn gn 2 N C, S1n =1 An 2 C, Ai \ Aj = ; if i 6= j ) (S1n =1 An ) = P1n =1 (An ).[Exercise.]

1. If Cis a -algebra, is a measure on C, and F 2 C, then the function F : C ![0;1 ]dened as F (A) = (A \ F ) for each A 2 Cis a measure on C. If moreover (F ) < 1 ,then F is a nite measure.[Exercise.]

Proposition. Let Abe an algebra and : A ![0;1 ] be a charge.1. is monotone.

2. If A; B 2 A, A B, and (A) < 1 , then (B A) = (B) (A).10


12/67

3. SN j =1 Aj PN j =1 (Aj ) for each A1; A2;:::;AN 2 A.Proof. Let A B. Then B = ( B A) t A and (B) = (B A) + (A). It follows that

(B) (A) (and 1 holds). If also (A) < 1 , then (subtracting (A) from both sides) (B A) = (B) (A) (and 2 holds).

Let N = 2 . Then (A1 [ A2) = (A1 t (A2 A1)) = (A1) + (A2 A1) (A1) + (A2) by monotonicity. Assume the property holds for N M , we show it holds for N =

M + 1 . SM +1j =1 Aj = SM j =1 Aj [ AM +1 , then SM +1j =1 Aj SM j =1 Aj + (AM +1 )PM j =1 (Aj ) + (AM +1 ) = PM +1j =1 (Aj ).Proposition. Let Abe an algebra and : A ![0;1 ] be a measure.

1. fAn gn 2 N A, An "A, and A 2 A ) (A) = lim n (An ).2. fAn gn 2 N A, An #A, A 2 A, and (A1) < 1 ) (A) = lim n (An ).3. fAn gn 2 N Aand S1n =1 An 2 A ) (S1n =1 An ) P1n =1 (An ).The 3 rd property is called -subadditivity.

Proof. 1.) Let B1 = A1 and Bn = An An 1 = An Sn 1j =1 Aj for each n 2. Noticethat,for each n 2N, Bn 2 A,B i \ B j = ; if i 6= j ,

F1n =1 Bn = S1n =1 An = A,for each n 2N, An = Fnj =1 B j .

(Verify these statements.) Then

(A) =1

Xn =1

(Bn ) = limn

n

Xj =1

(B j ) = limn

n

Gj =1

B j!= limn (An ) :2.) If An #A, then A1 An "A1 A (verify it), then limn (A1) (An ) = (A1) (A),or limn (An ) = (A).3.) Let B1 = A1 and Bn = An Sn 1j =1 Aj for each n 2. Notice that,

for each n 2N, Bn 2 Aand Bn An ,B i \ B j = ; if i 6= j ,

11


13/67

F1n =1 Bn = S1n =1 An ,(Verify these statements.) Therefore (A) = P1n =1 (Bn ) P1n =1 (An ) .Proposition. Let

Abe an algebra and :

A ![0;

1] a charge. is a measure i at least

one of the following conditions is satised:

(c.iii) fAn gn 2 N A, An "A, and A 2 A ) (A) = lim n (An );(c.iii) is -subadditive.

Proof. If is a measure requirements (c.iii) and (c.iii) are satised. Let fAn gn 2 N AandA = F1n =1 An 2 A; for each N 2 N set BN = FN n =1 An 2 A. Suppose (c.iii) holds, sinceBn "A, it follows that

(A) = limN

(BN ) = limN

N

Xn =1 (An ) =1

Xn =1 (An ) :Suppose (c.iii) holds, monotonicity implies, as FN n =1 An A, that FN n =1 An (A) foreach N 2N , then

1

Xn =1 (An ) = limN N

Xn =1 (An ) = limN N

Gn =1 An! (A) :Finally -subadditivity implies that (A) = (F

1n =1 An ) P

1n =1 (An ).

4 EXTENSION OF A MEASURE

Let C Dbe two non empty classes of subsets of X , and consider the functions : C ![0;1 ]and : D ![0;1 ]; is an extension of if jC = .Proposition. Let S be a semialgebra and a charge on S . There exists a unique charge~ that extends to A(S ). Moreover if is a measure, then ~ is itself a measure.Proof. Each A 2 A(S ) can be written as A = F

N j =1 Aj with Aj 2 S for every j = 1 :::N ,

then

~ (A) =N

Xj =1 (Aj ) .The proof goes through several steps.Step 0. ~ : A(S ) ! R is well-dened.

12


14/67

Let A = FN j =1 Aj with Aj 2 S and A = FM k=1 Bk with Bk 2 S , thenN

Xj =1 (Aj ) =N

Xj =1 Aj \M

Gk=1 Bk!=N

Xj =1 M

Gk=1 (Aj \ Bk )!==

N

Xj =1M

Xk=1 (Aj \ Bk ) =M

Xk=1N

Xj =1 (Aj \ Bk ) ==

M

Xk=1 N

Gj =1 (Aj \ Bk )!=M

Xk=1 Bk \N

Gj =1 Aj!==

M

Xk=1 (Bk ) .Step 1. A; B 2 A(S ) and A \ B = ; ) ~ (A [ B) = ~ (A) + ~ (B).A = F

N j =1 Aj with Aj 2 S and B = F

M k=1 Bk with Bk 2 S , then

A [ B =N

Gj =1 Aj!tM

Gk=1 Bk!;it follows by the denition of ~ that

~ (A [ B) =N

Xj =1 (Aj ) +M

Xk=1 (Bk ) = ~ (A) + ~ (B) .Then ~ is a charge on A(S ) . Clearly ~ extends .Step 2. If is a charge on A(S ) and jS = , then = ~ . (That is: ~ is the unique exten-

sion of to A(S ) :)Let A 2 A(S ), then A = FN j =1 Aj with Aj 2 S for each j = 1 :::N , and

(A) =N

Xj =1 (Aj ) =N

Xj =1 (Aj ) = ~ (A) .Step 3. Assume is -additive, then ~ is -additive itself.Consider A

2 A(

S ) and

fA

n g A(

S ) s.t. A =

F1

n =1A

n; let

fC

1; C

2;:::;C

Lgbe a par-

tition of A in S and B j n ; B j n +1 ;:::;Bj n +1 1 (with j 1 = 1 ) a partition of An in S . When

P1n =1 ~ (An ) = 1 , then ~ (A) ~ FN n =1 An = PN n =1 ~ (An ) for each N 2N , and, passing

13


15/67

to the limit, ~ (A) = 1 . When instead P1n =1 ~ (An ) converges:1Xn =1 ~ (An ) =

1

Xn =1j n +1 1

Xj = j n (B j ) =1

Xj =1 (B j ) ==

1

Xj =1 (A \ B j ) =1

Xj =1L

Xk=1 (C k \ B j ) ==

L

Xk =11

Xj =1 (C k \ B j ) =L

Xk =1 (C k ) = ~ (A) ,that is ~ is -additive.

Let Abe an algebra of subsets of X and a measure on A. We aim at extendingto a measure on the -algebra (A) generated by Aand studying its properties. For eachA 2 P (X ) set

(A) = inf (1Xn =1 (Bn ) : fBn g Aand A1

[n =1 Bn);is called the outer measure induced by (beware! It might fail to be a measure on P (X )).

Clearly, : P (X ) ! [0;1 ] is a set function (verify it). Notice that, if (A) < 1 , then foreach " > 0 there exists fBn g As.t. A S1n =1 Bn and P1n =1 (Bn ) < (A) + " (why?).Proposition.

1. is monotone.

2. fAn gn 2 N P (X ) ) (S1n =1 An ) P1n =1 (An ).3. jA = .

Proof. 1.) Let A B X , for each fBn g As.t. B S1n =1 Bn , then A S1n =1 Bn , andhence (A) P1n =1 (Bn ), then(A) inf (

1

Xn =1 (Bn ) : fBn g Aand B

1

[n =1 Bn

)= (B) :2.) Set A = S1n =1 An . If P1n =1 (An ) = 1 , the result is trivial. Then suppose (An )P1j =1 (Aj ) < 1 for each n 2N. Let " > 0. For each n 2N there exists fBn;m g As.t.An S1m =1 Bn;m and P1m =1 (Bn;m ) < (An ) + "2n . Sn;m Bn;m = S1n =1 (S1m =1 Bn;m ) A(why?). It follows that (A) Pn;m (Bn;m ) = P1n =1 P1m =1 (Bn;m ) P1n =1 (An ) + "2n =P1n =1 (An ) + ". Since " is arbitrary, (A) P1n =1 (An ).14


16/67

3.) For each A 2 A, (A) (A) (consider fBn g= fA; ;; ;; ;; ;;::::g). If fBn g Aand A S1n =1 Bn , then A = S1n =1 (Bn \ A) for Bn \ A 2 A, then (A) P1n =1 (Bn \ A)P1n =1 (Bn ) and (A) (A).

In particular, for each A; E X , it holds that (E ) (E

\A) + (E

\Ac).

Definition. A subset A of X is -measurable if

(E ) = (E \ A) + (E \ Ac)for each E X .

Let C(A; ) be the class of all -measurable sets.Caratheodory s Theorem. C= C(A; ) is a -algebra containing Aand ~ = jC is ameasure.Proof. The proof is developped through several steps.Step 0. X 2 C.Let E X ,

(E ) = (E \ X ) + (E \ X c):Step 1. If A; B 2 C, then A [ B 2 C.Let E X , by denition

(E ) = (E

\B) + (E

\B c) (1)

Substituting E with E \ A(E \ A) = (E \ A \ B) + (E \ A \ B c) (2)

Substituting in (1) E with E \ Ac

(E \ Ac) = (E \ Ac \ B) + (E \ Ac \ B c) (3)Adding (2) and (3):

(E ) = (E \ A \ B) + (E \ A \ B c) + (E \ Ac \ B) + (E \ Ac \ B c) (4)Substituting E with E \ (A [ B) in (4)

(E \ (A [ B)) = (E \ (A [ B) \ A \ B) + (E \ (A [ B) \ A \ B c)+ (E \ (A [ B) \ Ac \ B) + (E \ (A [ B) \ Ac \ B c)= (E \ A \ B) + (E \ A \ B c) + (E \ Ac \ B) (5):

15


17/67

Substituting (4) in (5),

(E ) = (E \ (A [ B)) + (E \ (A [ B)c).Step 2. If A 2 C, then Ac 2 C.Let E X , by denition

(E ) = (E \ A) + (E \ Ac) = (E \ Ac) + (E \ (Ac)c):

Thus Cis an algebra .Step 3. Cis a -algebra.Let E X , and consider A; B 2 Cs.t. A \ B = ;. It follows from (5) that

(E \ (A [ B)) = (E \ A) + (E \ B);in particular (taking E = X ) is a charge on

C. By induction, if B1; B2;:::;BN

2 Care

pairwise disjoint:

(E \N

[j =1 B j ) =N

Xj =1 (E \ B j ) (6):Let A = S1j =1 Aj , with Aj 2 C. We can write A = S1j =1 B j , where B1 = A1 and B j =Aj Sj 1i=1 Ai for each n 2, moreover B i \ B j = ; if i 6= j . To show that A 2 Citsenough to prove that

(E ) E \1

[j =1

B j

!!+ E \

1

[j =1

B j

!c

!(7):

Since Cis an algebra, SN j =1 B j 2 Cfor each N 2N, thus(E ) E \

N

[j =1 B j!!+ E \N

[j =1 B j!c! (8):

Since E \ SN j =1 B jc

E \ S1j =1 B jc

= E \ Ac, monotonicity implies:

(E ) E

\

N

[j =1

B j

!!+ (E

\Ac) =

=N

Xj =1 (E \ B j ) + (E \ Ac) (9):Passing to the limit:

(E )1

Xj =1 (E \ B j ) + (E \ Ac) (10):16


18/67

But (E \ A) = E \ S1j =1 B j = S1j =1 (E \ B j ) P1j =1 (E \ B j ), this,together with (10), implies:(E ) (E \ A) + (E \ Ac) (10):

Thus A 2 C.Step 4. jC is a measure. In fact it is -subadditive.Step 5. C A.Let A 2 Aand E X: It remains to show that

(E ) (E \ A) + (E \ Ac) (11):If E 2 A, then fE \ A; E \ Acgis a partition of E 2 Aand (E ) = (E \ A) + (E \ Ac),but and coincide on Aand therefore (11) holds; as it does if (E ) = 1 . If E X ,and (E ) 0 s.t. f + a is greater than zero,therefore there exists a sequence f' n gof simple and measurable functions s.t. ' n " f + a,then ' n a "f (why?).Let f 0 and bounded, then there exists n 2 N s.t. f (x) < n for each x 2 X . Foreach k n and each x 2 X , f (x) < k , therefore there exists i 2 0; 2;:::;k 2k 1s.t. i2k f (x) 0 s.t. f + a is greater than zero.Then there exists a sequence f' n gof simple and measurable functions s.t. ' n

unif:

! f + a,and therefore ' n a

unif:

! f (why?).Finally for e generic measurable function f , f = f + f , and there exist two sequences

f' n g,f n gof simple and measurable functions s.t. ' n ! f + and n ! f , then ' n n !f + f = f (why?). We have already proved that the limit of measurable functions is itself measurable.

28


30/67

8 INTEGRATION OF SIMPLE AND MEASURABLEFUNCTIONS WITH RESPECT TO A FINITE CHARGE

Let (X; F ) be a measurable space and a nite charge (s.t. (X ) < 1 ) on F , then letB0 (X; F ) denote the set of all simple and measurable functions X .Definition. Let ' 2B0 (X; F ), and ' =

N

Pi=1 i A i be its standard representation. the realnumberZ X

'd =N

Xi=1 i (Ai )is called integral of ' with respect to .

Facts.

1. For each A 2 F and c 2R it is

Z X

c A d = c (A) :

[Obvious.]

2. For each ' 2B0 (X; F ) and c 2R it is

Z X (' + c) d =

Z X 'd + c (X ) :

[Exercise.]

Lemma. If B1; B2;:::;BM 2 F , there exist C 1; C 2;:::;C K 2 F pairwise disjoint s.t. for eachl = 1 ; 2;:::;K , there exists j s.t. C l B j and B j = Fl:C l B j C l for each j = 1 ; 2;:::;M .Proof. For M = 1 it is obvious. Assume the equality holds for M elements of F . ConsiderB1; B2;:::;BM ; BM +1 2 F , by induction, there exist C 1; C 2;:::;C K 2 F pairwise disjoint s.t.for each l = 1 ; 2;:::;K there exist j s.t. C l B j and B j = Sl:C l B j C l for each j = 1 ; 2;:::;M .Consider the 2K +1 sets C 1\ BM +1 ; C 1\ B C M +1 ; C 2\ BM +1 ; C 2\ B C M +1 ; :::; C K \ BM +1 ; C K \B C M +1 ; BM +1 n

M

Sj =1 B j 2 F , and rename them D1; D2;:::;D2K +1 : these are pairwise disjoint and

29


31/67

each of them is contained in B j for some j . If j < M +1 each B j = Sh :D h B j Dh and, moreover:BM +1 = D2K +1 [ BM +1 \

M

[j =1

B j!== D2K +1 [0@BM +1 \

M

[j =1 0@[h :D h B j Dh1A1A== D2K +1 [0@

M

[j =1 0@[h :D h B j BM +1 \ Dh1A1A== D2K +1 [0@

M

[j =1 0@[h odd and D h B j BM +1 \ Dh1A1A:Then BM +1 = Sh :D h B M +1 Dh .Proposition. Let ' 2 B0 (X; F ), and let ' =

M

Pj =1 j B j , with B j 2 F for each j =1; 2;:::;M , be a generic representation of ' . ThenZ X

'd =M

Xj =1 j (B j ) .Proof. Let ' =

N

Pi=1 i A i be the standard representation of ' .Let B1; B2;:::;BM be pairwise disjoint sets. Assume, without loss of generality, that j 6= 0for each j = 1 ; 2;:::;M . Then

M

Sj =1 B j = Si : i 6=0 Ai and i (Ai \ B j ) = j (Ai \ B j ), for eachj = 1 ; 2;:::;M and each i s.t. i 6= 0 . ThenZ X

'd =N

Xi=1 i (Ai ) = Xi : i 6=0 iM

[j =1 Ai \ B j!== Xi : i 6=0 i

M

Xj =1 (Ai \ B j ) =M

Xj =1 Xi : i 6=0 i (Ai \ B j ) ==

M

Xj =1 Xi : i 6=0 j (Ai \ B j ) =M

Xj =1 j Xi : i 6=0 (Ai \ B j ) ==

M

Xj =1 j (B j ) .30


32/67

If B1; B2;:::;BM are not pairwise disjoint, then there exist C 1; C 2;:::;C K 2 F pairwise disjoints.t. for each l = 1 ; 2;:::;K , there exists j s.t. C l B j and B j = Fl:C l B j C l for each j =1; 2;:::;M (it follows from the lemma we have just shown).Let

jl = (1 C l B j

0 else:

B j =K

Pl=1 jl C l and (B j ) =K

Pl=1 jl (C l).Consider the following representation of '' =

M

Xj =1 j B j =M

Xj =1 jK

Xl=1 jl C l =M

Xj =1K

Xl=1 j jl C l ==

K

Xl=1M

Xj =1 j

jl C l =

K

Xl=1M

Xj =1 j

jl

!C l ;

since C 1; C 2;:::;C K are pairwise disjoint,

Z X

'd =K

Xl=1M

Xj =1 j jl! (C l) =M

Xj =1K

Xl=1 j jl (C l) ==

M

Xj =1 jK

Xl=1 jl (C l)!=M

Xj =1 j (B j ) .Proposition. Let ' and 2B0 (X; F ), and ; 2R . Then

Z X

( ' + ) d = Z X

'd + Z X

d :

Proof. Let ' =N

Pi=1 i A i , =M

Pj =1 j B j :

Z X ( ' + ) d = Z X N

Xi=1 i A i +M

Xj =1 j B j :!d ==

N

Xi=1 i (Ai ) +M

Xj =1 j (B j ) == Z

X

'd + Z X

d :

31


33/67

Proposition. Let ' and 2B0 (X; F ), if ' , then

Z X 'd

Z X d :

Proof. Clearly ' 0, letN

Pi=1 i A i be its standard representation, then i 0 for eachi = 1 ; 2;:::;N andZ X

'd Z X

d = Z X

(' ) d =N

Xi=1 i (Ai ) 0:

9 INTEGRATION OF BOUNDED MEASURABLEFUNCTIONS WITH RESPECT TO A FINITE CHARGE

Let (X; F ) be a measurable space and a nite charge (s.t. (X ) < 1 ) on F . Let B (X; F )denote the set of all measurable and bounded functions on X .

Lemma. f n : X ! R . f nunif:

! f i there exists fdn g R + s.t. dn ! 0 and f n dn f f n + dn for each n 2N . In this case, f n dn

unif:

! f and f n + dnunif:

! f .Proof. If f n

unif:

! f , then supx 2 X jf (x) f n (x)j ! 0; set dn = sup x 2 X jf (x) f n (x)j, then:dn jf (x) f n (x)j f (x) f n (x) for each x 2X , and hence f (x) f n (x) + dn ,dn jf (x) f n (x)j f n (x) f (x) for each x 2X , and hence f n (x) dn f (x).

Thus, f n (x) dn f (x) f n (x) + dn .Vice-versa, if f n (x) dn f (x) f n (x) + dn for each x 2 X and each n 2 N, then

dn f (x) f n (x) dn for each x

2X and each n

2N , that is

jf (x) f n (x)

j dn for

each x 2X and each n 2N , thus supx 2 X jf (x) f n (x)j dn for each n 2N.Finally, set gn = f n dn , and d0n = 2 dn , then:

gn d0n = f n 3dn f n dnf f n + dn = gn + d0n ;

therefore f n dnunif:

! f , similarly, f n dnunif:

! f .32


34/67

Proposition (E). Let f be a measurable and bounded function, then:

sup8 0, then c' (x) < f (x) and there exists n 2N s.t.f n (x) > c' (x), that is x 2E n . It follows X = S

1

n =1 E n and E n "X . Then ( E n 1 foreach n)Z X

f n d Z X

f n E n d Z X

c' E n d = cZ X

' E n d = cZ E n

'd :

Passing to the limit ( ' is a measure):

cZ X

'd :

37


39/67

The inequality holds for each 0 < c < 1 and ' 2B +0 (X; F ) s.t. ' f . Then if you choosec = 1 1n and pass again to the limit,

Z X

'd :

for each ' 2B +0 (X; F ) s.t. ' f . Finally R X fd :Notice that, if f n #f , the Theorem doesnt hold (Consider as an example R with the

Lebesgue measure and the sequences f n (x) = 1n or f n (x) = [n; 1 )).

Proposition. Let f ;g;u n : X ! [0;1 ] be measurable functions, and take ; 0. Then:1.

R X

( f + g) d =

R X

fd +

R X

fd :

2. R X 1

Pn =1 un d =1

Pn =1R X un d .3. The function dened as f (E ) = R E fd for each E 2 F is a measure.4. If (E ) = 0 , then f (E ) = R E fd = 0 .5. If R X fd = 0 , then fx 2X : f (x) 6= 0 g= 0 .

Proof. 1.) Let f' n g;f n gbe sequences of simple measurable and nonnegative functions s.t.' n "f , n "g, then ' n + n "f + g (why?), and

Z X

(f + g) d = limn Z

X

(' n + n ) d =

= limn 0@Z X ' n d + Z X n d 1A=

= limn

Z X ' n d + lim

n

Z X n d =

= Z X

fd + Z X

gd .

Similarly it is possible to prove that: R X fd = R X fd .2.) We can write u =

1

Pn =1 un i N

Pn =1 un "u, and therefore u : X ! [0;1 ] is measurable (why?)38


40/67

and

Z X

ud = limn Z

X

N

Xn =1 un!d == lim

n

N

Xn =1Z X un d ==

1

Xn =1Z X un d .3.) Exercise. Hint: If Bn "B, then f B n "f B .4.) Exercise. Hint: f E 1 E and R X 1 E d = lim n R X n E d .5.) Let E = fx 2X : f (x) 6= 0 g= fx 2X : f (x) > 0g, E n = x 2X : f (x) > 1n , E n "E . But for each n 2

N;0 = Z

X

fd Z X

f E n d Z X

1n E n

=1n

(E n ) ,

and therefore (E ) = lim n (E n ) = 0 .

Fatou s Lemma. Consider f n : X ! [0;1 ] measurable functions, then

Z X

limn f n d limnZ X

f n d .

Proof. First of all observe that f = lim n f n : X ! [0;1 ] is measurable (why?). For eachx 2 X , limn f n (x) = sup k2 N (inf n k f n (x)), then for each k 2 N set gk (x) = inf n k f n (x).gk "f and f k gk for each k 2N . By the Monotone Convergence Theorem

Z X

limn f n d = limn Z X

gn d

= lim nZ X

gn d limnZ X

f n d .

It might be that the previous hold as a strict inequality. Consider, as an example,A; B 2 F s.t. A \ B = ;, (A) ; (B) > 0, and set

f n = ( A if n is evenB if n is odd :39


41/67

It is R X limn f n d = 0 and limnR X f n d = min( (A) ; (B)) (why?). The same doesnthold for limn ( as an example, with f n as above it is R X limn f n d = (A) + (B) >max( (A) ; (B)) = lim nR X

f n d ; on the other hand, for f n (x) = 1n ,

R X

limn f n d = 0 0,and for = 0 , = 1, then put themtogether.]

3. If f; g

2 Land f g, then

R X

fd

R X

gd .

[Exercise. Hint: f g i f g 0.]

4. If f 2 L, then R X fd R X jf jd .[Exercise. Hint: f; f jf j.]5. If f is summable on E 2 F and E D 2 F , then f is summable on D.

[Exercise.]

41


43/67

6. If f is summable on A; B 2 F with A \ B = ;, then f is summable on A t B and

Z A t B

fd = Z A

fd + Z B

fd :

[Exercise.]

7. If f 2 Land R A fd = 0 for each A 2 F , then (fx 2X : f (x) 6= 0g) = 0 .[Exercise. Hint: ff > 0g= Sn f > 1n .]Dominated Convergence Theorem. Let g;f ; f n : X ! R (or [ 1;1 ]) be measurablefunctions. If f n ! f , jf n j g and g is summable, then f n ; f are summable and

Z X

jf f n jd ! 0:

In particular, R X fd R X f n d R X jf f n jd ! 0, that is R X f n d ! R X fd .Proof. If jf n j ! jf j, then 0 jf n j; jf j g, and therefore f n ; jf n j 2 L.Set gn = 2 g jf n f j, gn 0; in fact jf n f j jf n j+ jf j 2g.gn ! 2g. By Fatous Lemma,

Z X

(2g) d = Z X

limn gn d limnZ X

gn d =

= lim nZ X 2g jf n f jd = lim n 0@Z X 2gd Z X jf

n f jd 1AZ X

2gd + lim n 0@Z X jf n f jd 1A= Z X 2gd limn 0@Z X jf n f jd 1A.Thus

0 limn 0@Z X jf n f jd 1Ai.e. limn 0@Z X jf n f jd 1A0,nally0 limn 0@Z X jf n f jd 1Alimn 0@Z X jf n f jd 1A0,and

limn 0@Z X jf n f jd 1A= 0 :

42


44/67

12 PROPERTIES HOLDING ALMOST EVERYWHERE

Let (X; F ) be a measurable space and be a complete measure on F .Definition. A property P holds -almost everywhere (briey -a.e. or a.e. ) if

fx 2X : on x P doesnt hold g 2 F and it has zero measure.Since is complete it is enough to show that there exists N 2 F s.t. fx 2X : on x P doesnt hold

N and (N ) = 0 .

As an example, take f n ; f ; g : X ! R (or [ 1;1 ]), h : Y ! R (or [ 1;1 ]) withY X .

f = g a.e. i fx 2X : f (x) 6= g (x)g 2 F and(

fx

2X : f (x)

6= g (x)

g) = 0 .

f g a.e. i fx 2X : f (x) < g (x)g 2 F and(fx 2X : f (x) < g (x)g) = 0 .

If X = R , f is continuous a.e. i fx 2X : f is discontinuous in xg 2 F and(fx 2X : f is discontinuous in xg) = 0 .

f n converges to f a.e. ( f na:e:

! f ) i fx 2X : f n (x) 6! f (x)g 2 F and(fx 2X : f n (x) 6! f (x)g) = 0 .

h is dened a.e. i X Y 2 F (in particular Y 2 F ) and (X Y ) = 0 .

Proposition. Consider f; g : X ! R (or [ 1;1 ]), with f = g a.e.1. If f is measurable, then g is measurable.

2. If f and g are measurable and nonnegative, then R X fd = R X gd .3. If f 2 L, then g 2 Land

R X

fd =

R X

gd .

Proof. Set N = fx 2X : f (x) 6= g (x)g.1.) Then fx 2X : g (x) ag= fx 2X : g (x) ag\(N [ N c) = ( fx 2X : g (x) ag \N ) [(fx 2X : g (x) ag \N c) = ( fx 2X : g (x) ag \N ) [ (fx 2X : f (x) ag \N c),but fx 2X : f (x) ag \N c 2 F (why?) and fx 2X : g (x) ag \N N is measurableand with zero measure.

43


45/67

2.) R X fd = R X N fd + R N fd = R X f N c d and R X gd = R X N gd + R N gd = R X g N c d , butf N c = g N c .3.) Set M = fx 2X : jf (x)j 6= jg (x)jg N , then jf j= jgj a.e., if f 2 L, then R X jgjd =R X j

f

jd g ,A = [ N [ M [ (Sn M n ) [ E [ (Sn E n )].

(A) = 0 . Let g; f; f n be the functions that coincide with g;f ; f n on Ac and are zero on A.g; f ; f n satisfy the hypothesis of the Dominated Convergence Theorem, therefore

Z X

f f n d ! 0;

but

R X

f f n d =

R X j

f f n

jd for each n

2N (why?).

Consider nite measure and bounded functions, then the following "counterpart" of Pro-position (E) holds

Proposition (E). Let f : X ! R be a bounded function s.t.

sup8


47/67

13 RIEMANN (STIELTJES) Vs LEBESGUE (STIELTJES)

Let X = [ a; b]. (E) delivers the following:

Corollary. Let be the Lebesgue measure on [a; b], and

F be the Lebesgue -algebra. If

the function f : [a; b] ! R is Riemann integrable, then f is summable and, if we denote byRR [a;b ] f (x) dx the Riemann integral of f , it is

Z [a;b ]

fd = RZ [a;b ]

f (x) dx.

Proof. Let RR [a;b ]f (x) dx and RR [a;b ]f (x) dx be the upper and lower Riemann integral of f . Noticethat

RZ [a;b ]

f (x) dx sup8


48/67

integral of f , it is

Z [a;b ]

fd = f (a) F (a) F a + RZ [a;b ]

f (x) dF (x) .

Proof. Let RR [a;b ]f (x) dF (x) and RR [a;b ]f (x) dF (x) the upper and lower Riemann-Stieltjes integ-ral of f . Notice that

f (a) F (a) F a + RZ [a;b ]

f (x) dF (x) sup8


49/67

and set A B = F1n =1 An Bn 2 F G. Then let fAn Bn g 6= ; for each n 2N .Fix x 2A. Set N x = fn 2N : x 2An g= fn 2N : A n (x) = 1g. For each y 2B, (x; y) 2A B, and there exists a unique n s.t. (x; y) 2An Bn . This implies that for each y 2Bthere exists an unique n s.t. x 2An and y 2Bn . It follows that for each y 2B there existsan unique n 2N x s.t. y 2Bn . Therefore B Sn 2 N x B

n , vice-versa, if there exists n 2N x s.t.y 2Bn , then x 2An and y 2Bn , i.e. (x; y) 2An Bn thus y 2B and Sn 2 N x Bn B. If thereexisted n; m 2N x s.t. y 2Bn \ Bm , then x 2An and y 2Bn , so (x; y) 2An Bn , but alsox 2Am and y 2Bm , so (x; y) 2Am Bm , which is absurd if n 6= m. Thus, B = Fn 2 N x Bn and

(B) A (x) = (B) = Xn 2 N x (Bn ) = Xn 2 N (Bn ) A n (x) .If we let x =2A, then A n (x) = 0 for each n 2 N (if it were x 2An for some n, choosingy 2Bn it would be (x; y) 2An Bn and x 2A, ! ), and

(B) A (x) = Xn 2 N (Bn ) A n (x) .Thus (by the Dominated Convergence Theorem), (B) A = Pn 2 N (Bn ) A n implies

(B) (A) = Z X (B) A d = Xn 2 NZ X (Bn ) A n d = Xn 2 N (Bn ) (An ) :If instead

fAn Bn

g=

;for some n

2N , then either A B =

;, and the result is

straightforward, or A B = Fn :A n B n 6= ; An Bn , and it is enough to apply the previousresult to derive the statement.The extension results guarantee that can be extended to a measure on

(F G), such a -algebra is denoted by F G.Definition. F Gis called the product -algebra of F and Gwhile is calledproduct measure of and .

If and are -nite, then is -nite (why?) and it is the unique extension of to F G.

From now on we will consider -nite and .

For each Q X Y , x 2X and y 2Y , setQx = fy 2Y : (x; y) 2QgandQy = fx 2X : (x; y) 2Qg.

48


50/67

It is straightforward to see that maps Q 7! Qx and Q 7! Qy preserve all the usual operationsbetween sets.

Proposition. If Q 2 F Gthen Qx 2 Gand Qy 2 F for each x 2X and each y 2Y .Proof. Its enough to prove that the class

T = fQ 2 F G: Qx 2 Gand Qy 2 F 8(x; y) 2X Y gcoincides with F G. Clearly T F G:If Q = A B 2 F G, then A 2 F and B 2 G, and Q 2 F G. Moreover:

if x 2A, Qx = fy 2Y : (x; y) 2A Bg= B 2 G,if x =2A, Qx = fy 2Y : (x; y) 2A Bg= ; 2 G,

thus Qx

2 Gfor each x

2X , a similar argument can be used for Qy for y

2Y , then

F G T .It remain to show that T is a -algebra.X Y 2 F G T .Let fQn g T and Q = S1n =1 Qn . Then Q 2 F G. Moreover: for each x 2X ,Qx = fy 2Y : (x; y) 2Qg=

= fy 2Y : 9n s.t. (x; y) 2Qn g== fy 2Y : 9n s.t. y 2(Qn )xg=

=

1

[n =1 (Qn )x 2 G;

similarly, for each y 2Y , Qy = S1n =1 (Qn )y 2 F . Then, Q 2 T .Let Q 2 T ; then Qc 2 F G. Moreover: for each x 2X ,(Qc)x = fy 2Y : (x; y) 2Qcg=

= fy 2Y : (x; y) =2Qg== ( Qx )c 2 G;

analogously, for each y 2Y , (Qc)y = ( Qy)c 2 F . Thus Qc 2 T .Attention: in general there exist subsets Q X Y s.t. Qx 2 Gand Qy 2 F for each

(x; y) 2X Y , but not belonging to F G.Let f : X Y ! R (or [ 1;1 ]). For each x 2X and y 2Y , set

f x (y) = f (x; y) 8y 2Y andf y (x) = f (x; y) 8x 2X .

49


51/67

f x is called x-section of f , analogously f y is called y-section of f . It is easy to see that mapsf 7! f x and f 7! f y preserve all the usual operations.Proposition. Let f : X Y ! R (or [ 1;1 ]) be measurable with respect to F G.Then, for each x

2X and y

2Y , f x is measurable with respect to

Gand f y is measurable

with respect to F .Proof. If A R (or [ 1;1 ]),

f 1x (A) = fy 2Y : f x (y) 2Ag== fy 2Y : f (x; y) 2Ag== y 2Y : (x; y) 2f 1 (A) == f 1 (A) x :

Choosing A = [a;

1) and letting a vary in R , from f 1 ([a;

1))

2 F Git follows

f 1x ([a; 1 )) = [ f 1 ([a; 1 ))]x 2 G, thus f x is measurable with respect to G. Analogously, foreach y 2Y , f y is measurable with respect to F .Proposition. Let Q 2 F G, and let p : X ! [0;1 ], q : Y ! [0;1 ] be dened as

p (x) = (Qx ) for each x 2X , andq (y) = (Qy ) for each y 2Y .

Then, p and q are measurable (with respect to F and G) and it is

Z X pd = ( ) (Q) = Z Y qd .Notice that p (x) = R Y ( Q )x d and q (y) = R X ( Q )y d .Proof. It is enough to show that the class T of subsets of Q 2 F Gs.t. p and q are

measurable and s.t. it is Z X pd = ( ) (Q) = Z Y qd ,coincides with F G. The proof is long: we develop it through several steps.Step 0. F G T .If Q = A B 2 F G, then A 2 F and B 2 G, and Q 2 F G. Moreover,

p (x) = (Qx ) = ( (B) if x 2A0 if x =2A = (B) A , andq (y) = (Qy) = ( (A) if y 2B0 if y =2B = (A) B .

50


52/67

Thus, p and q are measurable and

Z X pd = (A) (B) = Z Y qd ,therefore Q

2 T .

Step 1. If fQn g T , and Qn "Q, then Q 2 T .(Obviously Q 2 F G.) Let pn (x) = ((Qn )x ) and q n (y) = ((Qn )y). For each n 2N , pnand q n are measurable and

Z X pn d = ( ) (Qn ) = Z Y q n d .We show that pn "p. In fact, for each xedx 2X , for each n 2N

(Qn )x = fy 2Y : (x; y) 2Qn g fy 2Y : (x; y) 2Qn +1 g= ( Qn +1 )x ;and (Qn )x "S

1n =1 (Qn )x , but Q = S

1n =1 Qn implies Qx = S

1n =1 (Qn )x , that is (Qn )x "Qx .

Then, pn (x) = ((Qn )x ) " (Qx ) = p. Analogously, q n " q . It follows that p and q aremeasurable and

Z X pd = limn Z X pn d = limn ( ) (Qn ) = limn Z Y q n d = Z Y qd .Step 2. Let and be nite. If Q; R 2 T , Q R, then Q R 2 T .

We use weaker assumption: let Q; R 2 T be s.t. Q R, and Q A B 2 F Gwith(A) ; (B) M

which is absurd, according to the denition of M .

Thus = 0 , and (E ) = R E fd for each E 2 F . Finally if g 2 L(X; F ; ) and (E ) =R E gd for each E 2 F , then R E (f g) d = 0 for each E 2 F and fx 2X : f (x) g (x) 6= 0g=0 (why?).We conclude observing that if there exists f 2 L(X; F ; ) such that fx 2X : f (x) < 0g=

0 and (E ) = Z

E

fd

for each E 2 F , then the signed measure is positive (why?).Corollary. Let (X; F ) be a measurable space and let and be -nite measures onF with absolutely continuous with respect to . Then there exists a measurable functionf : X ! [0;1 ) such that

(E ) = Z E

fd

for each E 2 F . Moreover,if g 2X ! [0;1 ] is measurable,

Z X gd = Z X (g f ) d ;if g 2 L(X; F ; ), g f 2 L(X; F ; ) and

Z X

gd = Z X

(g f ) d :

Proof. It is an useful exercise. Hint: consider the countable partition fAn gof X in F s.t.(An ) + (An ) < 1 and set f (x) =

d A nd A n

(x) if x 2An .

65


67/67

17 ESSENTIAL BIBLIOGRAPHY

Berberian, S.K. (1965), Measure and Integration , The Macmillan Company, New York.

Chow, Y.S., e H. Teicher (1988), Probability Theory , Springer-Verlag, New York.

Kolmogorov, A.N., e S.V. Fomin (1975), Introductory Real Analysis , Dover, New York.(Translation of Elementy Teorii Funktsij i Funktsionalnogo Analiza , Nauka, Mosca,1968.)

Royden, H.L. (1988), Real Analysis , Prentice Hall, Englewood Clis.

Rudin, W. (1976), Principles of Mathematical Analysis , McGraw-Hill, New York.

Rudin, W. (1987), Real and Complex Analysis , McGraw-Hill, New York.

measure and integration theory

Documents