Measure and Integration Theory

Fabio Maccheroni1

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 Advanced

Analysis course held by Luigi Paganoni at "Università degli Studi di Milano" during the academic

year 1991-1992, notes and course that in�uenced these notes very much. Finally I thank Giulia

Brancaccio who drafted this �rst English version of them in 2010. I take responsibility of all the

mistakes.

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 INTEGRATIONOF SIMPLEANDMEASURABLE FUNCTIONSWITHRESPECT 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 40

12 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

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 call

class 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;; Xg, f;; A;Ac; Xg for ; � A � X.

Facts. Let A be an algebra.

1. ; 2 A:[In fact, ; = Xc and X 2 A.]

2. A1; A2; :::; AN 2 A )SNj=1Aj 2 A:

[By induction.]

3. A;B 2 A ) A \B 2 A.[In fact, A \B = (Ac [Bc)c.]

4. A1; A2; :::; AN 2 A )TNj=1Aj 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

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 \Bc)c. �

Proposition. Let I = f(a; b] : �1 � a < b <1g [ f(b;1) : �1 < b <1g [ f;g [ R,then

E =(

N[j=1

Ij : N 2 N and Ij 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 Ic is a �nite disjoint union of elements of I. In fact, if I = ;, then Ic = R; ifI = R, then Ic = ;; if I = (a; b] for a 2 R, then Ic = (�1; a] [ (b;1); if I = (�1; b], thenIc = (b;1); �nally if I = (b;1), then Ic = (�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; cg andx � min fb; dg i.e. x 2 (max fa; cg ;min fb; dg], that is I \ J = (max fa; cg ;min fb; dg] 2 I.1

If I = (a; b] and J = (c;1), x 2 I \ J i¤ x > a; x � b; x > c i.e. x > max fa; cg and x � b

i¤ 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 =

SNj=1 Ij and F =

SMi=1 Ji with Ij; Ji 2 I for each j

and i, E \ F =�SN

j=1 Ij

�\�SM

i=1 Ji

�=SNj=1

hIj \

�SMi=1 Ji

�i=SNj=1

�SMi=1 Ij \ Ji

�=SN;M

j=1;i=1 Ij \ Ji 2 E since Ij \ Ji 2 I for each i; j.By induction, E1; E2; :::; EN 2 E )

TNj=1Ej 2 E .

E 2 E ) Ec 2 E . In fact, E =SNj=1 Ij with Ij 2 I for each j and Ec =

TNj=1 I

cj , but I

cj 2 E

for each j, therefore Ec 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

With little abuse, we call the class I de�ned in the previous Proposition: class of realsemi-intervals. I is a semialgebra (why?).

Proposition. Let S be a semialgebra of subsets of X, then

A =(

N[j=1

Aj : N 2 N 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) fAngn2N � F )S1n=1An 2 F .

Some trivial �-algebras are: P (X), f;; Xg, f;; A;Ac; Xg for ; � A � X. Each algebra

of 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. fAngn2N � F )T1n=1An 2 F .

[In fact,T1n=1An = (

S1n=1A

cn)c.]

Proposition. F � P (X) is a �-algebra i¤ (�.i), (�.ii) and fAngn2N � F )T1n=1An 2 F

hold.

Proof. Exercise. �

Let fAngn2N � P (X); if An � An+1 for all n 2 N, the sequence fAngn2N is increasing,if An � An+1 for all n 2 N, the sequence fAngn2N is decreasing. If fAng is increasing ordecreasing, it is said to be monotone. If fAngn2N is increasing and A =

S1n=1An, we write

An " A; analogously, if fAngn2N is decreasing and A =T1n=1An, we write An # A.

Definition. A monotone class of subsets of X is a non empty classM� P (X) s.t.

4

(i) fAngn2N �M and An " A) A 2M,

(ii) fAngn2N �M and An # A) A 2M.

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 fAngn2N � M, then, for each N 2 N, BN =SNn=1An 2 M (since M is an

algebra), and BN "S1n=1An; butM is a monotone class, and therefore

S1n=1An 2M. �

Definition. Let C be a non empty class of subsets ofX. We call �-algebra (resp. algebra,resp. monotone class) generated by C the smallest �-algebra (resp. algebra, resp.

monotone class) T containing C. Moreover, C is said to be a system of generators of T .

In other words, the �-algebra (resp. algebra, resp. monotone class) generated by C is 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 C be a non empty class of subsets of X, the �-algebra (resp. algebra,resp. monotone class) generated by C is 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�g�2� be the set of all �-algebras F� such that F� � C. P (X) � C andit is a �-algebra, therefore fF�g�2� 6= ;. Let T =

T�2�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 �, andtherefore Ac 2 F� for every � (F� is a �-algebra), then Ac 2 T . If An 2 T for every n 2 N,then An 2 F� for every n 2 N and for every �, therefore

S1n=1An 2 F� for every � (F� is

a �-algebra), �nallyS1n=1An 2 T . Therefore T is a �-algebra. If C 2 C, then C 2 F� for

every �, and C 2 T . Thus, T is a �-algebra and contains C and, by de�nition, if F� is a�-algebra s.t. F� � C, then F� � T . �

Facts. Let ; 6= C � K � P (X).

5

1. A (C) � � (C).[Exercise.]

2. M (C) � � (C).[Exercise.]

3. � (C) � � (K), A (C) � A (K) andM (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 2 N and Aj 2 S for j = 1; 2; :::; N)

= f�nite disjoint unions of elements of Sg :

Proof. Let A =nSN

j=1Aj : N 2 N and Aj 2 S for j = 1; 2; :::; Noand A0 = f�nite disjoint

unions of elements of Sg. We have already shown that A is 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 =

SNj=1Aj and

F =SMi=1Bi with A1; A2; :::; AN 2 S pairwise disjoint sets and B1; B2; :::; BM 2 S pairwise

disjoint sets, then E \ F =�SN

j=1Aj

�\�SM

i=1Bi

�=SN;Mj=1;i=1Aj \ Bi. Let (j; i) ; (k; h) 2

f1; 2; :::; Ng � f1; 2; :::;Mg and (j; i) 6= (k; h), then either j 6= k or i 6= h, in the �rst case

Aj \ Ak = ;, in the second Bi \ Bh = ;, in both cases (Aj \Bi) \ (Ak \Bh) = ;. ThusE \ F =

SN;Mj=1;i=1Aj \ Bi 2 A0. By induction, E1; E2; :::; EN 2 A0 )

TNj=1Ej 2 A0. Let

E 2 A0, then E =SNj=1Aj with A1; A2; :::; AN 2 S pairwise disjoint sets and Ec =

TNj=1A

cj,

but, being S a semialgebra, for each j the Acj is a disjoint union of elements of S, i.e. Acj 2 A0,and

TNj=1A

cj 2 A0. �

Monotone Classes Lemma. Let A be an algebra, then

M (A) = � (A) :

Proof. Clearly A � M (A) � � (A). It remains to show thatM (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\Bc; B\Ac; A[B 2M (A)g. If fAngn2N � DB

6

and An # A, then An\Bc # A\Bc, B\Acn " B\Ac, and An[B # A[B (verify it); moreoverfAn \ Bcg; fB \ Acng; fAn [ Bg � M (A). It follows that A \ Bc; B \ Ac; A [ B 2 M (A),and hence A 2 DB. Analogously, if fAngn2N � 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\Bc; B \Ac; A[B 2 A �M (A)), but DB isa monotone class and thereforeM (A) � DB. This implies that, if B 2 A and C 2 M (A),it is true that C \ Bc; B \ Cc; C [ B 2 M (A); that is, if B 2 A and C 2 M (A), thenB \ Cc; C \ Bc; B [ C 2 M (A); that is, if B 2 A and C 2 M (A), then B 2 DC . Thus, ifC 2M (A), A � DC , but DC is a monotone class andM (A) � DC .Finally, if C;D 2M (A), then D 2 DC , and D \Cc; C \Dc; D [ C 2M (A); moreover

X 2 A �M (A) and if C 2M (A), then Cc = X \ Cc 2M (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 H and An " A) A 2 H.

Observe that a �-algebra is a �-class. If G is 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 H be a �-class of subsets of X.

1. If A 2 H, then Ac 2 H.[Exercise.]

2. If A;B 2 H and A \B = ;, then A [B 2 H.[In fact, A � Bc, and (A [B)c = Bc \ Ac = Bc � A.]

3. H is a monotone class.[Exercise. Hint: if An 2 H and An # A, then Acn 2 H and Acn " Ac.]

7

4. If H is a �-class, then H is a �-algebra.[Exercise. Hint: (A [B)c = Ac \Bc.]

Dynkin�s lemma. If D is 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 .

G2 is 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 C is 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 C is a semialgebra (resp. algebra, resp. �-algebra) of subsets of X, then C \ Y is a

semialgebra (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 C s.t.E = C \ Y , but C is a semialgebra: therefore there exist C1; C2; :::; CN 2 C pairwisedisjoint, s.t. Cc =

SNj=1Cj; it follows that Y � E = Y \ Ec = Y \ (C \ Y )c =

Y \ (Cc [ Y c) = Y \ Cc = Y \SNj=1Cj =

SNj=1 (Cj \ 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 2 Y , there exist C;D 2 C s.t. E = C \ Y and F = D \ Y , thereforeE \ F = C \ Y \D \ Y = (C \D) \ Y , but C \D 2 C and E \ F 2 C \ Y . Provethe similar results when C is 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 B or B (R). The following classes generate B (prove it):

� E = A (I).

8

� Closed sets in R.

� Open sets in R.

� Open intervals in R.

� Closed intervals in R.

� f(�1; a] : a 2 Rg.

� f(�1; a) : a 2 Rg.

� f(b;1) : b 2 Rg.

� f[b;1) : b 2 Rg.

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 AtB instead of A[B, analogously if fAngn2N � P (X)and Ai \ Aj = ; for i 6= j, we write

F1n=1An instead of

S1n=1An.

Let C � P (X) and A � X. A �nite partition (resp. countable) of A in C is a setfA1; A2; ::; ANg (resp. fAngn2N) of pairwise disjoint elements of C s.t. A =

SNj=1Aj (resp.

A =S1n=1Aj):

Definition. Let C be 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) =NXj=1

� (Aj) .

for each A 2 C and each �nite partition fA1; A2; ::; ANg of A in C,

9

� �-additive, if� (A) =

Xj2J

� (Aj)

for each A 2 C and each countable partition fAjgj2J of A in C.

If C is 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 fAngn2N �C s.t. X =

S1n=1An and � (An) < 1 for each n 2 N, �nally a measure (charge) is a

probability measure (charge) if � (X) = 1.

Facts. Let C be 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. (Does

the converse hold?)

2. If C is an algebra, then � is a charge i¤

(c.i) � (;) = 0,

(c.ii) A;B 2 C and A \B = ; ) � (A [B) = � (A) + � (B).

[Exercise.]

1. If C is a semialgebra, then � is a measure i¤

(m.i) � (;) = 0,

(m.ii) fAngn2N � C,S1n=1An 2 C, Ai \ Aj = ; if i 6= j ) � (

S1n=1An) =

P1n=1 � (An).

[Exercise.]

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

[Exercise.]

Proposition. Let A be 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

3. ��SN

j=1Aj

��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 =

�SMj=1Aj

�[ AM+1, then �

�SM+1j=1 Aj

�� �

�SMj=1Aj

�+ � (AM+1) �PM

j=1 � (Aj) + � (AM+1) =PM+1

j=1 � (Aj). �

Proposition. Let A be an algebra and � : A ! [0;1] be a measure.

1. fAngn2N � A, An " A, and A 2 A ) � (A) = limn � (An).

2. fAngn2N � A, An # A, A 2 A, and � (A1) <1) � (A) = limn � (An).

3. fAngn2N � A andS1n=1An 2 A ) � (

S1n=1An) �

P1n=1 � (An).

The 3rd property is called �-subadditivity.

Proof. 1.) Let B1 = A1 and Bn = An � An�1 = An ��Sn�1

j=1 Aj

�for each n � 2. Notice

that,

� for each n 2 N, Bn 2 A,

� Bi \Bj = ; if i 6= j,

�F1n=1Bn =

S1n=1An = A,

� for each n 2 N, An =Fnj=1Bj.

(Verify these statements.) Then

� (A) =1Xn=1

� (Bn) = limn

nXj=1

� (Bj) = limn�

nGj=1

Bj

!= lim

n� (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�1

j=1 Aj

�for each n � 2. Notice that,

� for each n 2 N, Bn 2 A and Bn � An,

� Bi \Bj = ; if i 6= j,

11

�F1n=1Bn =

S1n=1An,

(Verify these statements.) Therefore � (A) =P1

n=1 � (Bn) �P1

n=1 � (An) . �

Proposition. Let A be an algebra and � : A ! [0;1] a charge. � is a measure i¤ at leastone of the following conditions is satis�ed:

(c.iii) fAngn2N � A, An " A, and A 2 A ) � (A) = limn � (An);

(c.iii�) � is �-subadditive.

Proof. If � is a measure requirements (c.iii) and (c.iii�) are satis�ed. Let fAngn2N � A andA =

F1n=1An 2 A; for each N 2 N set BN =

FNn=1An 2 A. Suppose (c.iii) holds, since

Bn " A, it follows that

� (A) = limN� (BN) = lim

N

NXn=1

� (An) =1Xn=1

� (An) :

Suppose (c.iii�) holds, monotonicity implies, asFNn=1An � A, that �

�FNn=1An

�� � (A) for

each N 2 N, then

1Xn=1

� (An) = limN

NXn=1

� (An) = limN�

NGn=1

An

!� � (A) :

Finally �-subadditivity implies that � (A) = � (F1n=1An) �

P1n=1 � (An). �

4 EXTENSION OF A MEASURE

Let C � D be two non empty classes of subsets ofX, 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 =FNj=1Aj with Aj 2 S for every j = 1:::N ,

then

~� (A) =

NXj=1

� (Aj) .

The proof goes through several steps.

Step 0. ~� : A (S)! R is well-de�ned.

12

Let A =FNj=1Aj with Aj 2 S and A =

FMk=1Bk with Bk 2 S, then

NXj=1

� (Aj) =

NXj=1

�

Aj \

MGk=1

Bk

!=

NXj=1

�

MGk=1

(Aj \Bk)!=

=

NXj=1

MXk=1

� (Aj \Bk) =MXk=1

NXj=1

� (Aj \Bk) =

=MXk=1

�

NGj=1

(Aj \Bk)!=

MXk=1

�

Bk \

NGj=1

Aj

!=

=

MXk=1

� (Bk) .

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

FNj=1Aj with Aj 2 S and B =

FMk=1Bk with Bk 2 S, then

A [B =

NGj=1

Aj

!t

MGk=1

Bk

!;

it follows by the de�nition of ~� that

~� (A [B) =NXj=1

� (Aj) +MXk=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 =

FNj=1Aj with Aj 2 S for each j = 1:::N , and

�� (A) =NXj=1

�� (Aj) =NXj=1

� (Aj) = ~� (A) .

Step 3. Assume � is �-additive, then ~� is �-additive itself.

Consider A 2 A (S) and fAng � A (S) s.t. A =F1n=1An; let fC1; C2; :::; CLg be a par-

tition of A in S and�Bjn ; Bjn+1; :::; Bjn+1�1

(with j1 = 1) a partition of An in S. WhenP1

n=1 ~� (An) = 1, then ~� (A) � ~��FN

n=1An

�=PN

n=1 ~� (An) for each N 2 N, and, passing

13

to the limit, ~� (A) =1. When insteadP1

n=1 ~� (An) converges:

1Xn=1

~� (An) =

1Xn=1

jn+1�1Xj=jn

� (Bj) =1Xj=1

� (Bj) =

=1Xj=1

� (A \Bj) =1Xj=1

LXk=1

� (Ck \Bj) =

=

LXk=1

1Xj=1

� (Ck \Bj) =LXk=1

� (Ck) = ~� (A) ,

that is ~� is �-additive. �

Let A be an algebra of subsets of X and � a measure on A. We aim at extending �

to a measure on the �-algebra � (A) generated by A and studying its properties. For eachA 2 P (X) set

�� (A) = inf

( 1Xn=1

� (Bn) : fBng � A and A �1[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 fBng � A s.t. A �

S1n=1Bn and

P1n=1 � (Bn) < �� (A) + " (why?).

Proposition.

1. �� is monotone.

2. fAngn2N � P (X)) �� (S1n=1An) �

P1n=1 �

� (An).

3. ��jA = �.

Proof. 1.) Let A � B � X, for each fBng � A s.t. B �S1n=1Bn, then A �

S1n=1Bn, and

hence �� (A) �P1

n=1 � (Bn), then

�� (A) � inf( 1Xn=1

� (Bn) : fBng � A and B �1[n=1

Bn

)= �� (B) :

2.) Set A =S1n=1An. If

P1n=1 �

� (An) = 1, the result is trivial. Then suppose �� (An) �P1j=1 �

� (Aj) < 1 for each n 2 N. Let " > 0. For each n 2 N there exists fBn;mg � A s.t.An �

S1m=1Bn;m and

P1m=1 � (Bn;m) < �� (An) +

"2n.Sn;mBn;m =

S1n=1 (

S1m=1Bn;m) � A

(why?). It follows that �� (A) �P

n;m � (Bn;m) =P1

n=1

P1m=1 � (Bn;m) �

P1n=1

��� (An) +

"2n

�=P1

n=1 �� (An) + ". Since " is arbitrary, �� (A) �

P1n=1 �

� (An).

14

3.) For each A 2 A, �� (A) � � (A) (consider fBng = fA; ;; ;; ;; ;; ::::g). If fBng � AandA �

S1n=1Bn, thenA =

S1n=1 (Bn \ A) forBn\A 2 A, then � (A) �

P1n=1 � (Bn \ A) �P1

n=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 A and ~� = ��jC is a

measure.

Proof. The proof is developped through several steps.

Step 0. X 2 C.Let E � X,

��(E) = ��(E \X) + ��(E \Xc):

Step 1. If A;B 2 C, then A [B 2 C.Let E � X, by de�nition

��(E) = ��(E \B) + ��(E \Bc) (1)

Substituting E with E \ A

��(E \ A) = ��(E \ A \B) + ��(E \ A \Bc) (2)

Substituting in (1) E with E \ Ac

��(E \ Ac) = ��(E \ Ac \B) + ��(E \ Ac \Bc) (3)

Adding (2) and (3):

��(E) = ��(E \ A \B) + ��(E \ A \Bc) + ��(E \ Ac \B) + ��(E \ Ac \Bc) (4)

Substituting E with E \ (A [B) in (4)

��(E \ (A [B)) = ��(E \ (A [B) \ A \B) + ��(E \ (A [B) \ A \Bc)

+ ��(E \ (A [B) \ Ac \B) + ��(E \ (A [B) \ Ac \Bc)

= ��(E \ A \B) + ��(E \ A \Bc) + ��(E \ Ac \B) (5):

15

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 de�nition

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

Thus C is an algebra .Step 3. C is a �-algebra.Let E � X, and consider A;B 2 C s.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 C arepairwise disjoint:

��(E \N[j=1

Bj) =NXj=1

��(E \Bj) (6):

Let A =S1j=1Aj, with Aj 2 C. We can write A =

S1j=1Bj, where B1 = A1 and Bj =

Aj ��Sj�1

i=1 Ai

�for each n � 2, moreover Bi \ Bj = ; if i 6= j. To show that A 2 C it�s

enough to prove that

��(E) � ��

E \

1[j=1

Bj

!!+ ��

E \

1[j=1

Bj

!c!(7):

Since C is an algebra,SNj=1Bj 2 C for each N 2 N, thus

��(E) � ��

E \

N[j=1

Bj

!!+ ��

E \

N[j=1

Bj

!c!(8):

Since E \�SN

j=1Bj

�c� E \

�S1j=1Bj

�c= E \ Ac, monotonicity implies:

��(E) � ��

E \

N[j=1

Bj

!!+ �� (E \ Ac) =

=

NXj=1

��(E \Bj) + �� (E \ Ac) (9):

Passing to the limit:

��(E) �1Xj=1

��(E \Bj) + �� (E \ Ac) (10):

16

But ��(E \ A) = ���E \

�S1j=1Bj

��= ��

�S1j=1 (E \Bj)

��P1

j=1 �� (E \Bj), 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 A and E � X: It remains to show that

��(E) � ��(E \ A) + ��(E \ Ac) (11):

If E 2 A, then fE \ A;E \ Acg is a partition of E 2 A and �(E) = �(E \A) + �(E \Ac),but � and �� coincide on A and therefore (11) holds; as it does if ��(E) = 1. If E � X,

and ��(E) <1, for each " > 0 there exists fEjg � A s.t. E �S1j=1Ej and

P1j=1 � (Ej) <

�� (E) + ". Since Ej = (Ej \ A) [ (Ej \ Ac), it follows that

�(Ej) = �(Ej \ A) + �(Ej \ Ac);

moreover

E \ A �1[j=1

(Ej \ A) and E \ Ac �1[j=1

(Ej \ Ac) .

Monotonicity and �-subadditivity imply

�� (E \ A) � ��

1[j=1

(Ej \ A)!�

1Xj=1

� (Ej \ A) and

�� (E \ Ac) � ��

1[j=1

(Ej \ Ac)!�

1Xj=1

� (Ej \ Ac) .

Thus

�� (E \ A) + �� (E \ Ac) �1Xj=1

(� (Ej \ A) + � (Ej \ Ac)) =

=1Xj=1

� (Ej) < �� (E) + ":

Since " is arbitrary, the statement follows. �

A direct consequence of the previous Theorem is the following:

Corollary. Let S be a semialgebra of subsets of X and � a measure on S; then thereexists a �-algebra T � S and a measure ~� on T s.t. ~�jS = �.

17

In particular, a measure on a semialgebra S can be extended to a measure on � (S). Nexttheorems concern the uniqueness of this extension.

Theorem. Let S be a semialgebra of subsets of X and �; � measures on � (S). If �jS and�jS are �-�nite (as measures on S) and �jS = �jS , then � = �.

Proof. Since �jA and �jA extend �jS and �jS to A = A (S) � � (S), �jA = �jA. Let � and �

be �nite and set

M = fA 2 � (A) : � (A) = � (A)g � � (A) :

M � A. An 2 M and An " A implies � (A) = limn � (An) = limn � (An) = � (A)

and hence A 2 M. Similarly An 2 M and An # A implies (since � and � are �nite)

� (A) = limn � (An) = limn � (An) = � (A) and hence A 2 M. Then M is a monotone

class containing A, and thereforeM�M (A). According to the Monotone Classes Lemma,M� � (A).If �jS and �jS are �-�nite, then �jA and �jA are �-�nite and there exists fAngn2N � A

s.t. � (An) < 1 for each n 2 N andS1n=1An = X. Replacing An with

Snj=1Aj, we can

assume that An " X. Let A 2 � (A), then An \ A " A, hence � (A) = limn � (An \ A), and� (A) = limn � (An \ A). It remains to show that � (An \ A) = � (An \ A) for each n 2 N.In this regard notice that, for each n 2 N, the measures de�ned as �n (B) = � (An \B) and�n (B) = � (An \B) for each B 2 � (A) coincide on A (if B 2 A, An \ B 2 A) and are�nite, therefore, according to what we have just proved, they coincide on � (A). �

Corollary. Let S be a semialgebra of subsets of X and � a �-�nite measure on S; thenthere exists an unique measure �� that extends � to � (S). Moreover, denoting by �� theouter measure induced by the unique extension of � to A (S), it is �� = ��j�(S).

Corollary. Let A be an algebra of subsets of X and � a �-�nite measure on A; then thereexists an unique measure �� that extends � to � (A). Moreover, denoting by �� the outermeasure induced by A, it is �� = ��j�(A).

5 COMPLETIONS

Definition. A measure � on a �-algebra F is complete if A 2 F , � (A) = 0 and N � A

imply N 2 F (and � (N) = 0).

Proposition. Let A be an algebra and � a measure on A. The measure ~� on C (A; �) iscomplete.

18

Proof. Let A 2 C (A; �), �� (A) = 0 and N � A; then, by monotonicity, for each E � X:

0 � �� (E \N) � �� (E \ A) � �� (A) = 0:

Again by monotonicity:

�� (E) � 0 + �� (E \N c) = �� (E \N) + �� (E \N c) .

Thus N 2 C (A; �). �

Denote by C (A; �) the class of �-measurable sets.

Theorem. Let F be a �-algebra, � be a measure on F and N be the class de�ned as

fN 2 P (X) : 9B 2 F s.t. N � B and � (B) = 0g .

Then

F 0 = fA4N : A 2 F ; N 2 Ng

is a �-algebra containing F and the set function

�0 (A4N) = � (A)

is a complete measure on F 0 that extends �.

Proof. See Berberian p.29-30. �

�0 is called completion of �, and F 0 is called the �-algebra of �0-measurable sets (orcompletion of F with respect to �).

Theorem. Let

� A be an algebra of subsets of X,and � be a �-�nite measure on A,

� �� be the unique extension of � to � (A),

� ��0 be the completion of �� and � (A)0 be the �-algebra of ��0-measurable sets.

� ~� be the restriction of the outer measure �� (induced by �) to the �-algebra C (A; �)��-measurable sets.

Then

� (A)0 = C (A; �) and ��0 = ~�.

Proof. See Berberian p.30-31. �

19

6 LEBESGUE-STIELTJES MEASURES ON R

Consider the semialgebra

I = f(a; b] : �1 � a < b <1g [ f(b;1) : �1 < b <1g [ f;g [ R

of real semi-intervals, and let F : R! R be an increasing and right continuous function,

then set

� F (�1) = limt!�1 F (t),

� F (1) = limt!1 F (t),

� (b;1] = (b;1) for each b 2 R,

� (b; b] = ; for each b 2 R.

With this notation

I = f(a; b] : �1 � a � b � 1g

then set

�F ((a; b]) = F (b)� F (a) :

As an example using id (t) = t for each t 2 R, it is �id ((a; b]) = b� a, which is the length of(a; b].

Lemma. Let F : R! R be an increasing and right continuous function; consider (a; b] and(an; bn] 2 I s.t. (a; b] �

S1n=1 (an; bn], then

F (b)� F (a) �1Xn=1

(F (bn)� F (an)) :

Proof. It is based on Heine-Borel theorem: see, as an example, Royden p.301-302. �

Proposition. Let F : R! R be an increasing and right continuous function, the function�F : I ! [0;1] de�ned as

�F ((a; b]) = F (b)� F (a)

for each �1 � a � b � 1, is a �-�nite measure on I. Moreover, its (unique) extension��F to B (= � (I)) is �nite on each bounded set in B. If G : R! R is increasing and rightcontinuous, then �F = �G i¤ there exists c 2 R s.t. F = G+ c.

Proof. Clearly �F (;) = 0 and �F ((a; b]) 2 [0;1]. Let (a; b] =FNj=1 (aj; bj].

If (aj; bj] 6= ; for each j = 1; 2; :::; N it is possible to reorder the intervals in such a way that

20

a = a1 < b1 � a2 < b2 � :::::bN�1 � aN < bN = b, but if bj < aj+1 for some j � N � 1, (a; b]wouldn�t be connected, which is absurd. It follows that bj = aj+1 for every j � N � 1; setaN+1 = bN obtaining:

NXj=1

�F ((aj; bj]) =

NXj=1

� ((aj; aj+1]) =

NXj=1

(F (aj+1)� F (aj)) =

=N+1Xj=2

F (aj)�NXj=1

F (aj) = F (aN+1)� F (a1) = F (b)� F (a) =

= �F ((a; b]) .

If (aj; bj] = ; for some j = 1; 2; :::; N , then

(a; b] =G

j:(aj ;bj ] 6=;

(aj; bj] [G

j:(aj ;bj ]=;

(aj; bj] =G

j:(aj ;bj ] 6=;

(aj; bj]

and as a consequence

�F ((a; b]) =X

j:(aj ;bj ] 6=;

�F ((aj; bj]) =

=NXj=1

�F ((aj; bj]) .

Thus �F is additive and hence it is a charge.

Let (a; b] =F1j=1 (aj; bj], for each N 2 N,

�F ((a; b]) �NXj=1

�F ((aj; bj])

and, passing to the limit,

�F ((a; b]) �1Xj=1

�F ((aj; bj]) .

The reverse inequality is granted by the previous Lemma. Thus �F is a measure on the

semialgebra I and, as R =S1n=1 (�n; n], �F is �-�nite. As a consequence there exists an

unique measure ��F that extends �F to B. Clearly, if B 2 B and B is bounded, then there

exist a; b 2 R s.t. B � (a; b] and ��F (B) � �F ((a; b]) = F (b)� F (a) <1.Notice that, if t � 0, then:

F (t) = �F ((0; t]) + F (0) :

If t < 0, then:

F (t) = F (0)� �F ((t; 0]) .

21

Let �F = �G. If t � 0, then:

F (t) = �F ((0; t]) + F (0) = �G ((0; t]) + F (0) =

= �G ((0; t]) +G (0) + F (0)�G (0) == G (t) + F (0)�G (0) .

If t < 0, then:

F (t) = F (0)� �F ((t; 0]) = F (0)�G (0) +G (0)� �G ((t; 0]) == F (0)�G (0) +G (t) .

Then F (t) = G (t) + F (0) � G (0) for each t 2 R. Vice-versa, if F (t) = G (t) + c for each

t 2 R, then F (�1) = G (�1) + c, and

�F ((a; b]) = F (b)� F (a) = G (b)�G (a) = �G ((a; b])

for each �1 � a � b � 1. �

Definition. A measure � on B is called Lebesgue-Stieltjes (Baire) measure if it is�nite on each bounded interval in R.

Let F : R! R be an increasing and right continuous fnction and let ��F be the extensionof �F to B. We use �F instead of ��F , and we call it Lebesgue-Stieltjes measure induced byF . The Lebesgue-Stieltjes measure induced by id is called Borel measure, its completion

Lebesgue measure and the �-algebra of �0id-measurable sets is called Lebesgue �-algebra.

Attention: not every subset of R belong to the Lebesgue �-algebra.

Facts. Let F : R! R be an increasing and right continuous function and �F be the inducedLebesgue-Stieltjes measure. We write F (b�) = limt!b� F (t) for each b 2 R, then

1. �F (fbg) = F (b)� F (b�) :[Exercise; in this regards, why fbg 2 B?]

2. �F (fbg) = 0 i¤ F is continuous in b:

3. �F ([a; b]) = F (b)� F (a�) :[Exercise.]

4. �F ((a; b)) = F (b�)� F (a) :[Exercise.]

22

Proposition. Let � be a Lebesgue-Stieltjes measure on B, then there exists a functionF� : R! R increasing and right continuous, s.t. � = �(F�).

Proof. If t � 0, setF� (t) = � ((0; t]) ;

if t < 0,

F� (t) = �� ((t; 0]) .

Prove (as an exercise) that F� is increasing and right continuous. �

F is called Cumulative Distribution Function of � and it is unique up to an additive

constant (why?).

Instead of considering the whole real line R we can consider an interval I, and the �-algebra B (I), the the Lebesgue-Stieltjes measure induced by F on I, is precisely (�F )B(I).

7 MEASURABLE FUNCTIONS

Let X be a set and F a �-algebra of subsets of X. The pair (X;F) is called measurablespace.

Definition. Consider a measurable space (X;F) and a function f : X ! R (or [�1;1]).f is called measurable if fx 2 X : f(x) > ag 2 F for each a 2 R.

Notice that fx 2 X : f(x) > ag = f�1 ((a;1)): it can be simply denoted by ff > ag.

Proposition. Let (X;F) be a measurable space, and consider the function f : X ! R (or[�1;1]). The following are equivalent:

1. f is measurable.

2. fx 2 X : f(x) � ag 2 F for each a 2 R.

3. fx 2 X : f(x) < ag 2 F for each a 2 R.

4. fx 2 X : f(x) � ag 2 F for each a 2 R.

It follows that

5. fx 2 X : f(x) = ag 2 F for each a 2 R (or [�1;1]):

23

Proof. Exercise. Use (after verifying) the following identities

fx 2 X : f(x) � ag =1\n=1

�x 2 X : f(x) > a� 1

n

�;

fx 2 X : f(x) < ag = X � fx 2 X : f(x) � ag ;

fx 2 X : f(x) � ag =1\n=1

�x 2 X : f(x) < a+

1

n

�;

fx 2 X : f(x) > ag = X � fx 2 X : f(x) � ag ;

fx 2 X : f(x) = ag = fx 2 X : f(x) � ag � fx 2 X : f(x) > ag if a 2 R,

fx 2 X : f(x) =1g =1\n=1

fx 2 X : f(x) > ng ;

fx 2 X : f(x) = �1g =1\n=1

fx 2 X : f(x) < �ng :

�

Facts. Let (X;F) be a measurable space, consider f : X ! R, ; 6= H � P (R), andf�1 (H) = ff�1 (H) : H 2 Hg.

1. The class T = fA � R t.c. f�1 (A) 2 Fg is a �-algebra.[Exercise.]

2. If H is a �-algebra, f�1 (H) is �-algebra.[Exercise.]

3. � (f�1 (H)) = f�1 (� (H)).[In fact, according to 2., f�1 (� (H)) is a �-algebra, moreover if H 2 H, H 2 � (H),and f�1 (H) 2 f�1 (� (H)); thus f�1 (H) � f�1 (� (H)) and therefore � (f�1 (H)) �f�1 (� (H)). Vice-versa, we want to show that each �-algebra C containing f�1 (H) alsocontains f�1 (� (H)). Set G = fA � R : f�1 (A) 2 Cg, according to1. G is a �-algebra,since f�1 (H) � C, H � G, and � (H) � G, it follows that f�1 (� (H)) � C.]

Proposition. Let (X;F) be a measurable space, and consider f : X ! R (or [�1;1]).The following are equivalent:

1. f is measurable.

2. f�1 (B) � F .

24

3. f�1 (G) � F for each G � P (R) s.t. B = � (G).

4. f�1 (G) � F for each G � P (R) s.t. B � � (G).

Proof. 2.)3.) If B = � (G), G � B, and therefore f�1 (G) � f�1 (B) � F .3.)4.) Obvious.4.)2.) f�1 (G) � F , therefore � (f�1 (G)) � F , but � (f�1 (G)) = f�1 (� (G)) � f�1 (B),that is f�1 (B) � F .2.)1.) It is obvious, in fact (a;1) 2 B for each a 2 R, and f�1 ((a;1)) 2 F for each a 2 R.1.)4.) Thanks to the measurability of f; f�1

�f(a;1)ga2R

�� F . It is enough to show that

��f(a;1)ga2R

�= B. �

�f(a;1)ga2R

�contains (a;1), (�1; b] = (b;1)c, R, ;, (a; b] =

(�1; b] � (�1; a], for each �1 < a < b < 1. Thus I � ��f(a;1)ga2R

�, and B �

��f(a;1)ga2R

�. �

Proposition. Let (X;F) be a measurable space and fn : X ! R (or [�1;1]) be meas-urable functions for each n 2 N. Then the functions de�ned for each x 2 X as supn2N fn(x),

infn2N fn(x), limnfn(x), limnfn(x), and (when it exists) limn fn(x), are measurable.

Proof. Set g (x) = supn2N fn (x), then for each a 2 R

fx 2 X : g(x) > ag =1[n=1

fx 2 X : fn(x) > ag

(why?), and hence g is measurable (why?).

Set g = inf n2Nfn, then for each a 2 R

fx 2 X : g(x) < ag =1[n=1

fx 2 X : fn(x) < ag

(why?), and hence g is measurable (why?).

For each x 2 X, set g (x) = limnfn(x) = infk2N�supn�k fn(x)

�, for each k 2 N, set gk (x) =

supn�k fn(x) for each x 2 X, gk is measurable and therefore g is so.For each x 2 X, set g (x) = limnfn(x) = supk2N (infn�k fn(x)), for each k 2 N, set gk (x) =infn�k fn(x) for each x 2 X, gk is measurable and therefore g is so.For each x 2 X, g (x) = limn fn(x) exists, then g (x) = limnfn(x) which is measurable. (If

you want to consider only real valued functions is necessary to also require the existence of

supn2N fn(x), infn2N fn(x), limnfn(x), limnfn(x) for each x 2 X.) �

Remember that f+ = max ff; 0g, f� = �min ff; 0g, and the following equalities hold:jf j = f+ + f�, f = f+ � f�.

25

Proposition. Let (X;F) be a measurable space; consider f; g : X ! R (or [�1;1]setting 0 = 1 � 1 = �1 +1) measurable functions and c 2 R. Then the functions:h (x) = c for each x 2 X, f + c, cf , f + g, fg, max ff; gg, min ff; gg, f+, f�, and jf j, aremeasurable.

Proof. For each a 2 R,

fx 2 X : h(x) > ag = fx 2 X : c > ag = X or ;

then h is measurable. For each a 2 R,

fx 2 X : f(x) + c > ag = fx 2 X : f(x) > a� cg

for each a 2 R, then f + c is measurable. For each a 2 R,

fx 2 X : cf(x) > ag =nx 2 X : f(x) >

a

c

oif c > 0,

= fx 2 X : 0 > ag if c = 0,

=nx 2 X : f(x) <

a

c

oif c > 0,

and therefore cf is measurable.

Let a 2 R, if f (x) + g (x) < a, then f (x) < a � g (x), and there exists q 2 Q s.t. f (x) <

q < a � g (x), then f (x) < q and g (x) < a � q, and hence x 2 fx 2 X : f (x) < qg \fx 2 X : g (x) < a� qg. Thus

fx 2 X : f (x) + g (x) < ag =[q2Q

fx 2 X : f (x) < qg \ fx 2 X : g (x) < a� qg 2 F

and f + g is measurable.

For each a 2 R,�x 2 X : [f (x)]2 > a

= X if a < 0

=�x 2 X : f (x) >

pa[�x 2 X : f (x) < �

paif a > 0

and f 2 is measurable, similarly fg = 12

�(f + g)2 � f 2 � g2

�is measurable. (This discussion,

with some re�nement, can be applied to functions taking values on the extended real line.)

Measurability for max ff; gg, min ff; gg, f+, f�, and jf j can be proved following the stepsof the previous proposition (do it). �

Definition. A function ' : X ! R is called simple if it takes a �nite number of values.

26

Let A � X, denote by �A the characteristic function of A de�ned as

�A (x) =

(1 if x 2 A0 if x 2 Ac

.

Facts. Let (X;F) be a measurable space

1. The sum of simple functions is itself simple.

[Exercise.]

2. A function ' : X ! R is simple i¤ it can be written as

' =

NXj=1

�i�Ai

with �1; �2; :::; �N 2 R and A1; A2; :::; AN � X. In general this representation is not

unique, however if we impose �1 > �2 > ::: > �N and fA1; A2; :::; ANg partition of X,the representation is unique and we call it standard representation of '.

[Exercise. Hint: consider the values �1; �2; :::; �N that ' takes, rename them so that

�1 > �2 > ::: > �N , then set Ai = '�1 (�i).]

3. A simple function is measurable i¤ A1; A2; :::; AN 2 F .[Exercise.]

Consider gn; g : X ! R (or [�1;1]), then

� gn converges pointwise to g (gn ! g) if gn (x)! g (x) for each x 2 X;

� gn is monotone increasing (gn ") if gn (x) � gn+1 (x) for each x 2 X and each n 2 N,if also gn ! g, we write gn " g.

� gn converges uniformly to g (gnunif:! g) if supx2X jg (x)� gn (x)j ! 0 (this de�nition

applies only for real valued functions).

Proposition. Let (X;F) be a measurable space. f : X ! R (or [�1;1]) is measurablei¤ there exists a sequence f'ng of simple and measurable functions s.t. 'n (x) ! f (x) for

each x 2 X. Moreover, if f has a lower bound, it is possible to choose an increasing sequencef'ng; if f is bounded it is possible to choose a sequence f'ng s.t. it converges uniformly tof .

27

Proof. Let f � 0. For each n 2 N and x 2 X, set

'n(x) =

8>>><>>>:n if x 2 f�1 ([n;1]) i.e. f (x) � n

j2nif x 2 f�1

��j2n; j+12n

��i.e. j

2n� f (x) < j+1

2n

for j = 0; 2; :::; n2n � 1

:

Clearly 'n is simple and measurable. If f (x) =1, then 'n(x) = n!1.If f (x) <1, then there exists n 2 N s.t. f (x) < n and therefore for each k � n, f (x) < k;

thus there exists i 2�0; 2; :::; k2k � 1

s.t. i

2k� f (x) < i+1

2kand 'k (x) = i

2k, therefore

jf (x)� 'k (x)j < 12k, and 'k (x)! f (x).

We now prove that 'n is increasing. Fix an arbitrary x 2 X and let m < n.

� If n � f (x), then 'm (x) = m < n = 'n (x).

� If m � f (x) < n, then 'm (x) = m and 'n (x) =j2n. Suppose it were 'm (x) > 'n (x),

that ism > j2n; then we would have j

2n< m � f (x) < j+1

2n, and therefore j

2n< m < j+1

2n,

and j < m2n < j + 1 ! .

� If 0 � f (x) < m, then 'm (x) = i2mand 'n (x) =

j2n, in this case i

2m� f (x) < i

2m+ 12m

and j2n� f (x) < j

2n+ 1

2n. Then 2n�mi � 2nf (x) < 2n�m (i+ 1) and j � 2nf (x) <

j +1. Suppose it were 'm (x) > 'n (x), i.e. i2m

> j2n, then 2n�mi > j � j +1, it would

imply that 2nf (x) � 2n�mi � j + 1 > 2nf (x) ! .

Concluding 'm (x) � 'n (x) for each x 2 X and 'n " f .If f has a lower bound (not necessarily � 0) there exists a > 0 s.t. f +a is greater than zero,therefore there exists a sequence f'ng of 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. For

each k � �n and each x 2 X, f (x) < k, therefore there exists i 2�0; 2; :::; k � 2k � 1

s.t. i

2k� f (x) < i+1

2kand 'k (x) =

i2k. Thus, for each k � �n and each x 2 X it is

jf (x)� 'k (x)j < 12k, then for each k � �n supx2X jf (x)� 'k (x)j < 1

2kand 'n converges

pointwise to f .

If f is bounded (not necessarily � 0), then there exists a > 0 s.t. f + a is greater than zero.Then there exists a sequence f'ng of simple and measurable functions s.t. 'n

unif:! f + a,

and therefore 'n � aunif:! f (why?).

Finally for e generic measurable function f , f = f+ � f�, and there exist two sequences

f'ng,f ng of 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

8 INTEGRATIONOF SIMPLEANDMEASURABLE

FUNCTIONSWITHRESPECTTOAFINITECHARGE

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 ' 2 B0 (X;F), and ' =NPi=1

�i�Ai be its standard representation. the real

number ZX

'd� =

NXi=1

�i� (Ai)

is called integral of ' with respect to �.

Facts.

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

c�Ad� = c� (A) :

[Obvious.]

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

('+ c) d� =

ZX

'd�+ c� (X) :

[Exercise.]

Lemma. If B1; B2; :::; BM 2 F , there exist C1; C2; :::; CK 2 F pairwise disjoint s.t. for eachl = 1; 2; :::; K, there exists j s.t. Cl � Bj and Bj =

Fl:Cl�Bj

Cl 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 C1; C2; :::; CK 2 F pairwise disjoint s.t.for each l = 1; 2; :::; K there exist j s.t. Cl � Bj and Bj =

Sl:Cl�Bj

Cl for each j = 1; 2; :::;M .

Consider the 2K+1 sets C1\BM+1; C1\BCM+1; C2\BM+1; C2\BC

M+1; :::; CK\BM+1; CK\

BCM+1; BM+1n

MSj=1

Bj 2 F , and rename themD1; D2; :::; D2K+1: these are pairwise disjoint and

29

each of them is contained in Bj for some j. If j < M+1 each Bj =S

h:Dh�BjDh and, moreover:

BM+1 = D2K+1 [ BM+1 \

M[j=1

Bj

!=

= D2K+1 [

0@BM+1 \M[j=1

0@ [h:Dh�Bj

Dh

1A1A =

= D2K+1 [

0@ M[j=1

0@ [h:Dh�Bj

BM+1 \Dh

1A1A =

= D2K+1 [

0@ M[j=1

0@ [h odd and Dh�Bj

BM+1 \Dh

1A1A :

Then BM+1 =S

h:Dh�BM+1

Dh. �

Proposition. Let ' 2 B0 (X;F), and let ' =MPj=1

�j�Bj , with Bj 2 F for each j =

1; 2; :::;M , be a generic representation of '. ThenZX

'd� =MXj=1

�j� (Bj) .

Proof. Let ' =NPi=1

�i�Ai be the standard representation of '.

Let B1; B2; :::; BM be pairwise disjoint sets. Assume, without loss of generality, that �j 6= 0

for each j = 1; 2; :::;M . ThenMSj=1

Bj =S

i:�i 6=0Ai and �i� (Ai \Bj) = �j� (Ai \Bj), for each

j = 1; 2; :::;M and each i s.t. �i 6= 0. ThenZX

'd� =

NXi=1

�i� (Ai) =Xi:�i 6=0

�i�

M[j=1

Ai \Bj

!=

=Xi:�i 6=0

�i

MXj=1

� (Ai \Bj) =MXj=1

Xi:�i 6=0

�i� (Ai \Bj) =

=MXj=1

Xi:�i 6=0

�j� (Ai \Bj) =MXj=1

�jXi:�i 6=0

� (Ai \Bj) =

=MXj=1

�j� (Bj) .

30

IfB1; B2; :::; BM are not pairwise disjoint, then there exist C1; C2; :::; CK 2 F pairwise disjoints.t. for each l = 1; 2; :::; K, there exists j s.t. Cl � Bj and Bj =

Fl:Cl�Bj

Cl for each j =

1; 2; :::;M (it follows from the lemma we have just shown).

Let

jl =

(1 Cl � Bj

0 else:

�Bj =KPl=1

jl �Cl and � (Bj) =KPl=1

jl � (Cl).

Consider the following representation of '

' =MXj=1

�j�Bj =MXj=1

�j

KXl=1

jl �Cl =

MXj=1

KXl=1

�j jl �Cl =

=KXl=1

MXj=1

�j jl �Cl =

KXl=1

MXj=1

�j jl

!�Cl ;

since C1; C2; :::; CK are pairwise disjoint,ZX

'd� =KXl=1

MXj=1

�j jl

!� (Cl) =

MXj=1

KXl=1

�j jl � (Cl) =

=MXj=1

�j

KXl=1

jl � (Cl)

!=

MXj=1

�j� (Bj) .

�

Proposition. Let ' and 2 B0 (X;F), and �; � 2 R. ThenZX

(�'+ � ) d� = �

ZX

'd�+ �

ZX

d�:

Proof. Let ' =NPi=1

�i�Ai, =MPj=1

�j�Bj :

ZX

(�'+ � ) d� =

ZX

NXi=1

��i�Ai +MXj=1

��j�Bj :

!d� =

=NXi=1

��i� (Ai) +MXj=1

��j� (Bj) =

= �

ZX

'd�+ �

ZX

d�:

31

�

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

'd� �ZX

d�:

Proof. Clearly ' � � 0, letNPi=1

�i�Ai be its standard representation, then �i � 0 for eachi = 1; 2; :::; N and Z

X

'd��ZX

d� =

ZX

('� ) d� =NXi=1

�i� (Ai) � 0:

�

9 INTEGRATION OF BOUNDED MEASURABLE

FUNCTIONS WITHRESPECTTOAFINITECHARGE

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. fn : X ! R. fnunif:! f i¤ there exists fdng � R+ s.t. dn ! 0 and fn � dn � f �

fn + dn for each n 2 N. In this case, fn � dnunif:! f and fn + dn

unif:! f .

Proof. If fnunif:! f , then supx2X jf (x)� fn (x)j ! 0; set dn = supx2X jf (x)� fn (x)j, then:

� dn � jf (x)� fn (x)j � f (x)� fn (x) for each x 2 X, and hence f (x) � fn (x) + dn,

� dn � jf (x)� fn (x)j � fn (x)� f (x) for each x 2 X, and hence fn (x)� dn � f (x).

Thus, fn (x)� dn � f (x) � fn (x) + dn.

Vice-versa, if fn (x) � dn � f (x) � fn (x) + dn for each x 2 X and each n 2 N, then�dn � f (x) � fn (x) � dn for each x 2 X and each n 2 N, that is jf (x)� fn (x)j � dn for

each x 2 X and each n 2 N, thus supx2X jf (x)� fn (x)j � dn for each n 2 N.Finally, set gn = fn � dn, and d0n = 2dn, then:

gn � d0n = fn � 3dn � fn � dn �� f � fn + dn = gn + d0n;

therefore fn � dnunif:! f , similarly, fn � dn

unif:! f . �

32

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

sup

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=; = inf

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=; .If f'ng is a sequence of simple and measurable functions s.t. 'n

unif:! f , then the common

value for the two terms is limn

RX

'nd�.

Proof. Set:

I� (f) = sup

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=;and

I� (f) = inf

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=; :

Obviously �1 < I� (f) � I� (f) < 1. Let f'ng be a sequence of simple and measurablefunctions s.t. 'n

unif:! f , then there exists fdng � R+ s.t. dn ! 0 and 'n�dn � f � 'n+dn,

then for each n 2 N;ZX

('n � dn) d� � I� (f) � I� (f) �ZX

('n + dn) d�

and therefore:

� �dn� (X) � I� (f)�RX

'nd� � �dn� (X) and

� �dn� (X) � I� (f)�RX

'nd� � �dn� (X),

thus I� (f) = limn

RX

'nd� = I� (f) : �

Definition. Let f 2 B (X;F). The real number

S

ZX

fd� = sup

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=;is called Stieltjes integral of f with respect to �.

If f is simple considering the sequence 'n = f for each n it�s easy to obtain

S

ZX

fd� = limn

ZX

'nd� =

ZX

fd�.

Consistently with that, for f 2 B (X;F) we useRX

fd� instead of SRX

fd�.

Proposition. Let f; fn,g 2 B (X;F), and �; � 2 R. Then:

33

1.RX

(�f + �g) d� = �RX

fd�+ �RX

gd�:

2. f � g )RX

fd� �RX

gd�:

3. fnunif:! f )

RX

fnd�unif:!

RX

fd�.

Proof. Exercise. Hints: 1.) Notice that if f'ng ; f ng are sequences of simple and measurablefunctions s.t. 'n

unif:! f and nunif:! g, then �'n

unif:! �f for each � 2 R (verify it), and n + n

unif:! f + g (verify it). 2.) Notice that ' 2 B0 (X;F) and ' � f implies ' � g. 3.)

Use the Lemma we have shown at the beginning of the section. �

Proposition. Let I : B (X;F)! R s.t for each f; g 2 B (X;F) and each �; � 2 R:

1. I (�f + �g) = �I (f) + �I (g) :

2. f � g ) I (f) � I (g) :

Then there exists one and only one �nite charge � on F s.t.

I (f) =

ZX

fd�

for each f 2 B (X;F).Moreover, if for each sequence ffng in B (X;F) of nonnegative functions s.t. fn " f 2

B (X;F) it holds that I (fn)! I (f), then � is a measure.

Proof. Exercise. Hint: set � (A) = I (�A) for each A 2 F , show that I (') =RX

'd� for

each ' 2 B0 (X;F), then consider a sequence f'ng of simple and measurable functions s.t.'n

unif:! f and use the Lemma of the beginning of the section and Proposition (E). �

10 INTEGRATIONOFNONNEGATIVE FUNCTIONS

WITH RESPECT TO A MEASURE

Let (X;F) be a measurable space and � a measure on F (we say (X;F ; �) is a measurespace). Let B+

0 (X;F) denote the set of all simple, measurable and nonnegative functionson X.

Definition. Let ' 2 B+0 (X;F), and let ' =

NPi=1

�i�Ai be its standard representation. The

extended real number ZX

'd� =

NXi=1

�i� (Ai)

34

is called integral of ' with respect to �.

Facts.

1. For each A 2 F and c � 0 it is ZX

c�Ad� = c� (A) :

[Obvious.]

2. For each ' 2 B+0 (X;F) and c � 0 it isZ

X

('+ c) d� =

ZX

'd�+ c� (X) :

[Exercise.]

Proposition. Let ' 2 B+0 (X;F), and let ' =

MPj=1

�j�Bj , with Bj 2 F for each j =

1; 2; :::;M be a generic representation for '. ThenZX

'd� =MXj=1

�j� (Bj) .

Proof. Repeat exactly that designed for the case of � �nite charge. �

Proposition Let '; 2 B+0 (X;F), �; � � 0. Then:

1.RX

(�f + �g) d� = �RX

fd�+ �RX

gd�:

2. ' � )RX

'd� �RX

d�:

Proof. Exercise. Hints: 1.) Repeat exactly that designed for the case with � �nite charge.

2.) Let ' =NPi=1

�i�Ai, =MPj=1

�j�Bj , and let C1; C2; :::; CK be a partition for fAi \Bjg; it

can be written ' =KPk=1

��k�Ck , =KPk=1

��k�Ck , moreover ' � implies ��k � ��k for each

k = 1; 2; :::; K, then.... �

Notice that if ' is simple and measurable, then '�E is so (�A�B = �A\B).

35

Definition. If ' 2 B+0 (X;F) and E 2 F , setZ

E

'd� =

ZX

'�Ed�:

This extended real number is called Lebesgue integral of ' with respect to � over E.

Proposition. Let '; 2 B+0 (X;F), �; � � 0, Then:

1. If ' =NPi=1

�i�Ai, thenRE

'd� =NPi=1

�i� (Ai \ E) for each E 2 F :

2.RE

(�'+ � ) d� = �RE

'd�+ �RE

d� for each E 2 F :

3. ' � )RE

'd� �RE

d� for each E 2 F :

4. The function de�ned as �' (E) =RE

'd� for each E 2 F is a measure.

Proof. Steps 1.,2.,3. are an exercise. Obviously �' (E) =RE

'd� 2 [0;1] for each E 2 F and

�' (;) =R;'d� =

RX

'�;d� =RX

0d� = 0: If A;B 2 F and A \B = ;, then

�' (A [B) =ZX

'�AtBd� =

ZX

' (�A + �B) d� =

=

ZX

'�Ad�+

ZX

'�Bd� = �' (A) + �' (B) .

Let ' =NPi=1

�i�Ai and fBngn2N � F , Bn " B. Notice that, for each i = 1; 2; :::; N ,

fAi \Bngn2N � F , Ai \Bn " Ai \B, and limn � (Ai \Bn) = � (Ai \B). Then:

limn�' (Bn) = lim

n

ZX

NXi=1

�i�Ai

!�Bnd� =

= limn

ZX

NXi=1

�i�Ai�Bnd� = limn

ZX

NXi=1

�i�Ai\Bnd� =

= limn

NXi=1

�i� (Ai \Bn) =NXi=1

�i limn� (Ai \Bn) =

=

NXi=1

�i� (Ai \B) = �' (B) :

�

36

Definition. Consider a measurable function f : X ! [0;1]. The extended real number

L

ZX

fd� = sup

8<:ZX

'd� : ' 2 B+0 (X;F) ; ' � f

9=;is called the Lebesgue integral of f with respect to �. If E 2 F , we set

RE

fd� =RX

f�Ed�:

If f is simple, clearlyRX

fd� � LRX

fd�. For the reverse inequality consider that for each

' 2 B+0 (X;F), ' � f implies

RX

'd� �RX

fd�, and therefore LRX

fd� �RX

fd�. If f is bounded

and � �nite, by de�nition SRX

fd� = sup

�RX

'd� : ' 2 B0 (X;F) ; ' � f

�� L

RX

fd�; on

the other hand, for each ' 2 B0 (X;F), ' � f , '+ 2 B+0 (X;F), ' � '+ � f , thenR

X

'd� �RX

'+d� � LRX

fd� and SRX

fd� = sup

�RX

'd� : ' 2 B0 (X;F) ; ' � f

�� L

RX

fd�.

Consistently with this, for f � 0 measurable function it is possible to writeRX

fd� instead of

LRX

fd�.

Proposition. Let f; g : X ! [0;1] be measurable functions. If f � g, thenRX

fd� �RX

gd�:

Proof. Notice that ' 2 B+0 (X;F) and ' � f implies ' � g. �

Monotone Convergence Theorem. Let fn : X ! [0;1] be measurable functions. Iffn " f , then

RX

fnd� "RX

fd�.

Proof. According to the previous Proposition,RX

fnd� is an increasing sequence, supposeRX

fnd� " �, since fn � f for each n 2 N, � �RX

fd�. It remains to prove that � �RX

fd�.

Take 0 < c < 1 and ' 2 B+0 (X;F) s.t. ' � f . Set

En = fx 2 X : fn (x) � c' (x)g

For each n 2 N, En = fx 2 X : fn (x)� c' (x) � 0g 2 F . En " in fact x 2 En ) fn (x) �c' (x) ) fn+1 (x) � fn (x) � c' (x) ) x 2 En+1. If f (x) = 0, then fn (x) = 0 for each

n and ' (x) = 0, and x 2 En; if f (x) > 0, then c' (x) < f (x) and there exists �n 2 N s.t.fn (x) > c' (x), that is x 2 E�n. It follows X =

S1n=1En and En " X. Then (�En � 1 for

each n) ZX

fnd� �ZX

fn�End� �ZX

c'�End� = c

ZX

'�End� = c

ZEn

'd�:

Passing to the limit (�' is a measure):

� � c

ZX

'd�:

37

The inequality holds for each 0 < c < 1 and ' 2 B+0 (X;F) s.t. ' � f . Then if you choose

c = 1� 1nand pass again to the limit,

� �ZX

'd�:

for each ' 2 B+0 (X;F) s.t. ' � f . Finally � �

RX

fd�: �

Notice that, if fn # f , the Theorem doesn�t hold (Consider as an example R with theLebesgue measure and the sequences fn (x) = 1

nor fn (x) = �[n;1)).

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

1.RX

(�f + �g) d� = �RX

fd�+ �RX

fd�:

2.RX

� 1Pn=1

un

�d� =

1Pn=1

RX

und�.

3. The function de�ned as �f (E) =RE

fd� for each E 2 F is a measure.

4. If � (E) = 0, then �f (E) =RE

fd� = 0.

5. IfRX

fd� = 0, then � fx 2 X : f (x) 6= 0g = 0.

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

X

(f + g) d� = limn

ZX

('n + n) d� =

= limn

0@ZX

'nd�+

ZX

nd�

1A =

= limn

ZX

'nd�+ limn

ZX

nd� =

=

ZX

fd�+

ZX

gd�.

Similarly it is possible to prove that:RX

�fd� = �RX

fd�.

2.) We can write u =1Pn=1

un i¤NPn=1

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

38

and ZX

ud� = limn

ZX

NXn=1

un

!d� =

= limn

NXn=1

ZX

und� =

=1Xn=1

ZX

und�.

3.) Exercise. Hint: If Bn " B, then f�Bn " f�B.4.) Exercise. Hint: f�E � 1�E and

RX

1�Ed� = limn

RX

n�Ed�.

5.) Let E = fx 2 X : f (x) 6= 0g = fx 2 X : f (x) > 0g, En =�x 2 X : f (x) > 1

n

, En "

E. But for each n 2 N;

0 =

ZX

fd� �ZX

f�End� �ZX

1

n�En =

1

n� (En) ,

and therefore � (E) = limn � (En) = 0. �

Fatou�s Lemma. Consider fn : X ! [0;1] measurable functions, thenZX

limnfnd� � limn

ZX

fnd�.

Proof. First of all observe that f = limnfn : X ! [0;1] is measurable (why?). For eachx 2 X, limnfn(x) = supk2N (infn�k fn(x)), then for each k 2 N set gk (x) = infn�k fn(x).

gk " f and fk � gk for each k 2 N. By the Monotone Convergence TheoremZX

limnfnd� = limn

ZX

gnd�

= limn

ZX

gnd� � limn

ZX

fnd�.

�

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

fn =

(�A if n is even

�B if n is odd:

39

It isRX

limnfnd� = 0 and limn

RX

fnd� = min (� (A) ; � (B)) (why?). The same doesn�t

hold for limn ( as an example, with fn as above it isRX

limnfnd� = � (A) + � (B) >

max (� (A) ; � (B)) = limn

RX

fnd�; on the other hand, for fn (x) = 1n,RX

limnfnd� = 0 <

1 = limn

RX

fnd�).

11 SUMMABLE FUNCTIONS

Let (X;F) be a measurable space and � be a measure on F . If we consider a genericmeasurable function f : X ! R (or [�1;1]), it is f = f+ � f� where f+; f� : X ! [0;1]are measurable functions, smart as a fox...

Definition. Let f : X ! R (or [�1;1]) be a measurable function. When it exists, theextended real number

L

ZX

fd� =

ZX

f+d��ZX

f�d�

is called the Lebesgue integral of f with respect to �. If E 2 F , it is LRE

fd� = LRX

f�Ed�.

WhenRX

f+d�;RX

f�d� <1, f is said to be summable. If E 2 F , f is said to be summable

on E if f�E is summable.

Notice that, if f is bounded and � �nite, it is

S

ZX

fd� = S

ZX

f+ � f�d� = S

ZX

f+d�� SZX

f�d� =

=

ZX

f+d��ZX

f�d� = L

ZX

fd�.

Consistently with this, for f measurable we can write (if it exists)RX

fd� instead of LRX

fd�.

Proposition. Let f : X ! R (or [�1;1]) be a measurable function. The function f issummable i¤ Z

X

jf j d� <1.

40

Proof. As jf j = f+ + f� thenZX

jf j d� <1 ,ZX

f+d�+

ZX

f�d� <1

,ZX

f+d� andZX

f�d� <1

�

The set of all summable functions is denoted by L (X;F ; �) or, shortly, L.

Facts.

1. If f; g 2 L and �; � 2 R, then �f + �g 2 L.[Exercise. Hint: j�f + �gj � j�j jf j+ j�j jgj.]

2. If f; g 2 L, thenRX

(�f + �g) d� = �RX

fd�+ �RX

gd�.

[Exercise. Hint: let h = f + g, h = h+ � h�, then h+ � h� = f+ � f� + g+ � g� i.e.h+ + f� + g� = f+ + g+ + h�, thusZ

X

�h+ + f� + g�

�d� =

ZX

�f+ + g+ + h�

�d�,

and thereforeZX

h+d�+

ZX

f�d�+

ZX

g�d� =

ZX

f+d�+

ZX

g+d�+

ZX

h�d�.

Reordering we can concludeRX

(f + g) d� =RX

fd� +RX

gd� (verify that each step

is legitimate thanks to properties of integrals for nonnegative functions). To showRX

(�f) d� = �RX

fd�, verify it for � > 0,and for � = 0, � = �1, then put them

together.]

3. If f; g 2 L and f � g, thenRX

fd� �RX

gd�.

[Exercise. Hint: f � g i¤ f � g � 0.]

4. If f 2 L, then����RX

fd�

���� � RX

jf j d�.

[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

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

fd� =

ZA

fd�+

ZB

fd�:

[Exercise.]

7. If f 2 L andRA

fd� = 0 for each A 2 F , then � (fx 2 X : f (x) 6= 0g) = 0.

[Exercise. Hint: ff > 0g =Sn

�f > 1

n

.]

Dominated Convergence Theorem. Let g; f; fn : X ! R (or [�1;1]) be measurablefunctions. If fn ! f , jfnj � g and g is summable, then fn; f are summable andZ

X

jf � fnj d�! 0:

In particular,

����RX

fd��RX

fnd�

���� � RX

jf � fnj d�! 0, that isRX

fnd�!RX

fd�.

Proof. If jfnj ! jf j, then 0 � jfnj ; jf j � g, and therefore fn; jfnj 2 L.Set gn = 2g � jfn � f j, gn � 0; in fact jfn � f j � jfnj+ jf j � 2g.gn ! 2g. By Fatou�s Lemma,Z

X

(2g) d� =

ZX

limngnd� � limn

ZX

gnd� =

= limn

ZX

2g � jfn � f j d� = limn

0@ZX

2gd��ZX

jfn � f j d�

1A ��ZX

2gd�+ limn

0@�ZX

jfn � f j d�

1A =

ZX

2gd�� limn

0@ZX

jfn � f j d�

1A .Thus

0 � �limn

0@ZX

jfn � f j d�

1A i.e. limn

0@ZX

jfn � f j d�

1A � 0,�nally

0 � limn

0@ZX

jfn � f j d�

1A � limn

0@ZX

jfn � f j d�

1A � 0,and

limn

0@ZX

jfn � f j d�

1A = 0:

�

42

12 PROPERTIES HOLDINGALMOST EVERYWHERE

Let (X;F) be a measurable space and � be a complete measure on F .

Definition. A property P holds �-almost everywhere (brie�y �-a.e. or a.e.) if

fx 2 X : on x P doesn�t holdg 2 F and it has zero measure.

Since � is complete it is enough to show that there existsN 2 F s.t. fx 2 X : on x P doesn�t holdg �N and � (N) = 0.

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

� f = g a.e. i¤ fx 2 X : f (x) 6= g (x)g 2 F and� (fx 2 X : f (x) 6= g (x)g) = 0.

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

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

� fn converges to f a.e. (fna:e:! f) i¤ fx 2 X : fn (x) 6! f (x)g 2 F and

� (fx 2 X : fn (x) 6! f (x)g) = 0.

� h is de�ned 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, thenRX

fd� =RX

gd�.

3. If f 2 L, then g 2 L andRX

fd� =RX

gd�.

Proof. Set N = fx 2 X : f (x) 6= g (x)g.1.) Then fx 2 X : g (x) � ag = fx 2 X : g (x) � ag\ (N [N c) = (fx 2 X : g (x) � ag \ N)[(fx 2 X : g (x) � ag \ N c) = (fx 2 X : g (x) � ag \ N) [ (fx 2 X : f (x) � ag \ N c),

but fx 2 X : f (x) � ag \ N c 2 F (why?) and fx 2 X : g (x) � ag \ N � N is measurable

and with zero measure.

43

2.)RX

fd� =R

X�Nfd� +

RN

fd� =RX

f�Ncd� andRX

gd� =R

X�Ngd� +

RN

gd� =RX

g�Ncd�, but

f�Nc = g�Nc .

3.) Set M = fx 2 X : jf (x)j 6= jg (x)jg � N , then jf j = jgj a.e., if f 2 L, thenRX

jgj d� =RX

jf j d� < 1 and g 2 L. Then setting M+ = fx 2 X : f+ (x) 6= g+ (x)g and M� =

fx 2 X : f� (x) 6= g� (x)g, it is M+;M� � N (why?), and thereforeRX

fd� =RX

gd� (why?).

�

As a consequence, if f is de�ned almost everywhere, then it has a measurable completion

i¤ all its completions are measurable (why?); moreover all its completions have the same

integral (when it exists, why?). Thus, measurability, summability, the value of the integral

and its properties are concept pertaining functions de�ned almost everywhere. Furthermore:

Facts. Let f; fn; g; f 0; f 0n; g0 : X ! R (or [�1;1]) s.t. f = f 0 a.e., fn = f 0n and g = g0 a.e.

1. f + g = f 0 + g0 a.e. (if they are at least de�ned a.e.).

[Exercise.]

2. �f = �g a.e. for each � 2 R.[Exercise.]

3. If fn ! f a.e., then f 0n ! f 0 a.e..

[Exercise.]

4. If f � 0, f is measurable andRXfd� <1, then f is �nite a.e..

[Set A = fx : f (x) =1g, f � n�A for each n 2 N, if � (A) > 0, thenRXfd� � n� (A)

for each n 2 N impliesRXfd� =1 ! .]

1. If f is summable, then f is �nite a.e..

[Exercise.]

Proposition we have proved so far continue to hold, as an example:

Dominated Convergence Theorem (a.e.). Consider g; f; fn : X ! R (or [�1;1])measurable functions which are de�ned a.e.. If fn ! f a.e., jfnj � g a.e. and g is summable,

then fn; f are summable and ZX

jf � fnj d�! 0:

Proof. Let:

44

� N be the set on which g is not de�ned,

� M (resp. Mn) be the set on which f (resp. fn) is not de�ned,

� E be the set in which fn 6! f ,

� En be the set in which jfnj > g,

� A = [N [M [ (SnMn) [ E [ (

SnEn)].

� (A) = 0. Let �g; �f; �fn be the functions that coincide with g; f; fn on Ac and are zero on A.

�g; �f; �fn satisfy the hypothesis of the Dominated Convergence Theorem, thereforeZX

�� �f � �fn�� d�! 0;

butRX

�� �f � �fn�� d� = R

X

jf � fnj d� for each n 2 N (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.

sup

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=; = inf

8<:ZX

'd� : ' 2 B0 (X;F) ; ' � f

9=; .Then f is measurable.

Proof. There exists a sequence fb'ng � B0 (;F) s.t. b'n � f for each n 2 N, andlimn!1

R

b'nd� = I (f) (why?). Let 'n = max fb'1; b'2; :::; b'ng for each n 2 N, then f'ng �B0 (;F), and b'n � 'n � 'n+1 � f for each n 2 N. Thus

RX

b'nd� � RX

'nd� �RX

'n+1d� �

I (f) for each n 2 N, and henceR

'nd� " I (f). Similarly, it is possible to build a sequence

f ng � B0 (;F) s.t. f � n+1 � n andRX

nd� # I (f). Set f = lim'n and f = lim n,

then

'n � f � f � f � n

for each n 2 N, thus

0 �ZX

�f � f

�d� �

ZX

( n � 'n) d� =ZX

nd� �ZX

'nd�

for each n 2 N, and f�f = 0 a.e. (why?), then f = f a.e.. Notice that�x 2 X : f (x) 6= f (x)

��

x 2 X : f (x) 6= f (x), and therefore f = f a.e. �

45

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. Ifthe 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� = R

Z[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 . Notice

that

R

Z[a;b]

f (x) dx � sup

8<:ZX

'd� : ' 2 B0 ([a; b] ;F) ; ' � f

9=; �� inf

8<:ZX

'd� : ' 2 B0 ([a; b] ;F) ; ' � f

9=; �� R

Z[a;b]

f (x) dx (why?),

but RR

[a;b]

f (x) dx = RR

[a;b]

f (x) dx. �

Moreover it holds the following result:

Proposition. Let � be the Lebesgue measure on [a; b], F be the Lebesgue �-algebra;

consider a bounded function f : [a; b]! R. Then, f : [a; b]! R is Riemann integrable i¤ itis continuous a.e..

Proof. See, as an example, Royden p.85. �

Similarly, if F : R! R is an increasing and right continuous function, �F is (the comple-tion of) the Lebesgue-Stieltjes measure induced by F on B ([a; b]), and FF is the �-algebraof �F -measurable sets.

Corollary. If the function f : [a; b] ! R is Riemann-Stieltjes integrable on [a; b] with

respect to F , then f is summable and, denoted by RR

[a;b]

f (x) dF (x) the Riemann-Stieltjes

46

integral of f , it is Z[a;b]

fd� = f (a)�F (a)� F

�a���+R

Z[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���+R

Z[a;b]

f (x) dF (x) � sup

8<:ZX

'd� : ' 2 B0 ([a; b] ;F) ; ' � f

9=; �� inf

8<:ZX

'd� : ' 2 B0 ([a; b] ;F) ; ' � f

9=; �� f (a)

�F (a)� F

�a���+R

Z[a;b]

f (x) dF (x) (why?),

but RR

[a;b]

f (x) dF (x) = RR

[a;b]

f (x) dF (x). �

14 PRODUCT SPACESANDFUBINI TONELLI THE-

OREM

Let (X;F ; �) and (Y;G; �) be two measure spaces.

Proposition. The class

F � G = fA�B : A 2 F ; B 2 Gg

is a semialgebra of subsets of X � Y . Moreover the function

�� � : F � G ! [0;1]A�B 7! � (A)� � (B)

is a measure on F � G.

Proof. Since X 2 F and Y 2 G, X � Y 2 F � G; similarly ; = ; � Y 2 F � G. IfA�B;C�D 2 F � G, then (A�B)\(C �D) = (A \ C)�(B \D) 2 F � G. Finally, if A�B 2 F � G, then (A�B)c = (A�Bc)t(Ac �B)t(Ac �Bc). Thus F � G is a semialgebra.It can be easily veri�ed that (�� �) (;) = 0. Consider the sequence fAn �Bngn2N � F � G

47

and set A�B =F1n=1An �Bn 2 F � G. Then let fAn �Bng 6= ; for each n 2 N.

Fix x 2 A. Set Nx = fn 2 N : x 2 Ang = fn 2 N : �An (x) = 1g. For each y 2 B, (x; y) 2A� B, and there exists a unique �n s.t. (x; y) 2 A�n � B�n. This implies that for each y 2 Bthere exists an unique �n s.t. x 2 A�n and y 2 B�n. It follows that for each y 2 B there exists

an unique �n 2 Nx s.t. y 2 B�n. Therefore B �S

n2NxBn, vice-versa, if there exists �n 2 Nx s.t.

y 2 B�n, then x 2 A�n and y 2 B�n, i.e. (x; y) 2 A�n�B�n thus y 2 B andS

n2NxBn � B. If there

existed �n; �m 2 Nx s.t. y 2 B�n \B �m, then x 2 A�n and y 2 B�n, so (x; y) 2 A�n �B�n, but alsox 2 A �m and y 2 B �m, so (x; y) 2 A �m�B �m, which is absurd if �n 6= �m. Thus, B =

Fn2Nx

Bn and

� (B)�A (x) = � (B) =Xn2Nx

� (Bn) =Xn2N

� (Bn)�An (x) .

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

� (B)�A (x) =Xn2N

� (Bn)�An (x) .

Thus (by the Dominated Convergence Theorem), � (B)�A =Pn2N

� (Bn)�An implies

� (B)� (A) =

ZX

� (B)�Ad� =Xn2N

ZX

� (Bn)�And� =Xn2N

� (Bn)� (An) :

If instead fAn �Bng = ; for some n 2 N, then either A � B = ;, and the result isstraightforward, or A � B =

Fn:An�Bn 6=;An � Bn, and it is enough to apply the previous

result 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 G is called the product �-algebra of F and G while � � is called

product 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 2 X and �y 2 Y , set

Q�x = fy 2 Y : (�x; y) 2 Qg andQ�y = fx 2 X : (x; �y) 2 Qg .

48

It is straightforward to see that maps Q 7! Q�x and Q 7! Q�y preserve all the usual operations

between sets.

Proposition. If Q 2 F G then Qx 2 G and Qy 2 F for each x 2 X and each y 2 Y .

Proof. It�s enough to prove that the class

T = fQ 2 F G : Qx 2 G and Qy 2 F 8 (x; y) 2 X � Y g

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

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

thus Qx 2 G for each x 2 X, a similar argument can be used for Qy for y 2 Y , then

F � G � T .It remain to show that T is a �-algebra.X � Y 2 F � G � T .Let fQng � T and Q =

S1n=1Qn. Then Q 2 F G. Moreover: for each x 2 X,

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

=1[n=1

(Qn)x 2 G;

similarly, for each y 2 Y , Qy =S1n=1 (Qn)

y 2 F . Then, Q 2 T .Let Q 2 T ; then Qc 2 F G. Moreover: for each x 2 X,

(Qc)x = fy 2 Y : (x; y) 2 Qcg == fy 2 Y : (x; y) =2 Qg == (Qx)

c 2 G;

analogously, for each y 2 Y , (Qc)y = (Qy)c 2 F . Thus Qc 2 T .

Attention: in general there exist subsets Q � X � Y s.t. Qx 2 G and Qy 2 F for each(x; y) 2 X � Y , but not belonging to F G.

Let f : X � Y ! R (or [�1;1]). For each �x 2 X and �y 2 Y , set

f�x (y) = f (�x; y) 8y 2 Y and

f �y (x) = f (x; �y) 8x 2 X.

49

f�x is called �x-section of f , analogously f �y is called �y-section of f . It is easy to see that maps

f 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 2 X and y 2 Y , fx is measurable with respect to G and f y is measurablewith respect to F .

Proof. If A � R (or [�1;1]),

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

=

=�f�1 (A)

�x:

Choosing A = [a;1) and letting a vary in R, from f�1 ([a;1)) 2 F G it followsf�1x ([a;1)) = [f�1 ([a;1))]x 2 G, thus fx is measurable with respect to G. Analogously, foreach y 2 Y , 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 de�ned as

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

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

pd� = (� �) (Q) =ZY

qd�.

Notice that p (x) =RY(�Q)x d� and q (y) =

RX(�Q)

y d�.

Proof. It is enough to show that the class T of subsets of Q 2 F G s.t. p and q are

measurable and s.t. it is ZX

pd� = (� �) (Q) =ZY

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 2 A0 if x =2 A

= � (B)�A, and

q (y) = � (Qy) =

(� (A) if y 2 B0 if y =2 B

= � (A)�B.

50

Thus, p and q are measurable andZX

pd� = � (A) � (B) =

ZY

qd�,

therefore Q 2 T .Step 1. If fQng � T , and Qn " Q, then Q 2 T .(Obviously Q 2 F G.) Let pn (x) = � ((Qn)x) and qn (y) = � ((Qn)

y). For each n 2 N, pnand qn are measurable and Z

X

pnd� = (� �) (Qn) =ZY

qnd�.

We show that pn " p. In fact, for each �xed x 2 X, for each n 2 N

(Qn)x = fy 2 Y : (x; y) 2 Qng � fy 2 Y : (x; y) 2 Qn+1g = (Qn+1)x ;

and (Qn)x "S1n=1 (Qn)x, but Q =

S1n=1Qn implies Qx =

S1n=1 (Qn)x, that is (Qn)x " Qx.

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

X

pd� = limn

ZX

pnd� = limn(� �) (Qn) = lim

n

ZY

qnd� =

ZY

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 � G with

� (A) ; � (B) <1:(Obviously Q�R 2 F G.) Fix x 2 X, then

(Q�R)x = fy 2 Y : (x; y) 2 Q�Rg == fy 2 Y : (x; y) 2 Q and (x; y) 2 Rcg == fy 2 Y : (x; y) 2 Qg \ fy 2 Y : (x; y) 2 Rcg == Qx \ (Rc)x = Qx \ (Rx)c = Qx �Rx

analogously, (Q�R)y = Qy �Ry for each y 2 Y . Let:

� pQ (x) = � (Qx) 8x 2 X, it is 0 � pQ (x) � � (B)�A 8x 2 X, thus pQ (x) 2 L (X;F ; �).

� pR (x) = � (Rx) 8x 2 X, it is 0 � pR (x) � � (B)�A 8x 2 X, thus pR (x) 2 L (X;F ; �).

� pQ�R (x) = � ((Q�R)x) 8x 2 X,

� qQ (y) = � (Qy) 8y 2 Y , it is 0 � qQ (y) � � (A)�B 8x 2 X, thus qQ (y) 2 L (Y;G; �).

� qR (y) = � (Ry) 8y 2 Y , it is 0 � qR (y) � � (A)�B 8x 2 X, thus qQ (y) 2 L (Y;G; �).

51

� qQ�R (x) = � ((Q�R)y) 8y 2 Y .

Notice that

pQ�R = pQ � pR and qQ�R = qQ � qR;

but pQ, pR, qQ and qR are summable, thus pQ � pR and qQ�R = qQ � qR are summable, andit is Z

X

pQd� = (� �) (Q) =ZY

qQd� andZX

pRd� = (� �) (R) =ZY

qRd�.

Subtracting in each term it follows Q�R 2 T .Step 3. Let � and � be �nite. Then T = F G.By steps 1 and 2 and since T � F � G 3 X � Y , then T is a �-class containing the �-classF � G, by Dynkin�s Lemma, T � � (F � G) = F G, the reverse inclusion is trivial.Step 4. T = F G.Let � and � be �-�nite. There exists a sequence fAngn2N � F s.t. � (An) < 1 for each

n 2 N and X =S1n=1An, analogously there exists a sequence fBngn2N � F s.t. � (Bn) <1

for each n 2 N and X =S1n=1Bn. We can assume that An " X and Bn " Y , therefore

An �Bn " X � Y . Set

T 0 = fQ 2 F G : Q \ (An �Bn) 2 T for each n 2 Ng .

If Q 2 F � G, then Q = A�B, and

Q \ (An �Bn) = (A \ An)� (B \Bn) 2 T

thus F � G � T 0.If fQmg � T 0, andQm " Q, thenQ 2 T 0. Fixing n 2 N, for eachm 2 N Qm\(An �Bn) 2 T ,but (for m ! 1) Qm \ (An �Bn) " Q \ (An �Bn) and by Step 1, Q \ (An �Bn) 2 T .Since n 2 N is arbitrary, Q 2 T 0.IfQ;R 2 T 0, Q � R, thenQ�R 2 T 0. Fixing n 2 N, Q\(An �Bn) 2 T , R\(An �Bn) 2 T ,R\ (An �Bn) � Q\ (An �Bn) � An�Bn, by Step 2, Q\ (An �Bn)�R\ (An �Bn) 2 T ,then (Q�R)\ (An �Bn) = Q\ (An �Bn)�R\ (An �Bn) 2 T . Since n 2 N is arbitrary,Q�R 2 T 0.Thus T 0 is a �-class containing F � G and, by Dynkin�s Lemma, T 0 � � (F � G) = FG,

the reverse inclusion is trivial, and T 0 = F G. It follows that, for each Q 2 F G,Q\ (An �Bn) 2 T for each n 2 N, but Q\ (An �Bn) " Q and therefore, by Step 1, Q 2 T .Then we have F G � T ; The reverse inclusion is trivial. �

52

If h : X ! R (or [�1;1]) we sometimes writeRXh (x) d� (x) instead of

RXhd�.

Tonelli�s Theorem. Let f : X � Y ! [0;1] be measurable with respect to F G. Then

a) the function y 7! f (x; y) is measurable with respect to G for each x 2 X,

b) the function x 7! f (x; y) is measurable with respect to F for each y 2 Y ,

c) the function x 7!RYf (x; y) d� (y) is measurable with respect to F ,

d) the function y 7!RXf (x; y) d� (x) is measurable with respect to G,

e) the following equality holds:ZX

�ZY

f (x; y) d� (y)

�d� (x) =

ZX�Y

f (x; y) d (� �) (x; y)

=

ZY

�ZX

f (x; y) d� (x)

�d� (y) .

Proof. We can rewrite statements a),b),c),d),e) using the sections of f :

a) the function fx : y 7! f (x; y) is measurable with respect to G for each x 2 X,

b) the function f y : x 7! f (x; y) is measurable with respect to F for each y 2 Y ,

c) the function x 7!RYfx (y) d� (y) is measurable with respect to F ,

d) the function y 7!RXf y (x) d� (x) is measurable with respect to G,

e) the following equality holds:ZX

�ZY

fx (y) d� (y)

�d� (x) =

ZX�Y

f (x; y) d (� �) (x; y)

=

ZY

�ZX

f y (x) d� (x)

�d� (y) .

Denote M the class of functions f : X � Y ! [0;1] s.t. a),b),c),d),e) are satis�ed. Wewant to show that the set of all measurable and nonnegative functions lies inM.

Step 0. If f = �Q, Q 2 F G, then f 2M.

(Since f is measurable, a) and b) hold) Denote p : X ! [0;1], q : Y ! [0;1], as

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

53

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

pd� = (� �) (Q) =ZY

qd�.

But

p (x) = � (Qx) =

ZY

�Qx (y) d� (y) for each x 2 X, and

q (y) = � (Qy) =

ZX

�Qy (x) d� (x) for each y 2 Y ,

then ZX

�ZY

�Qx (y) d� (y)

�d� (x) =

ZX�Y

�Q (x; y) d (� �) (x; y)

=

ZY

�ZX

�Qy (x) d� (x)

�d� (y) .

For each x 2 X and y 2 Y , (�Q)x (y) = �Q (x; y) and

�Qx (y) =

(1 if y 2 Qx i.e. (x; y) 2 Q0 if y =2 Qx i.e. (x; y) =2 Q

= �Q (x; y)

thus (�Q)x = �Qx, analogously �Qy = (�Q)y.

Therefore, for each Q 2 F G, the functions de�ned as

p (x) =

ZY

(�Q)x (y) d� (y) for each x 2 X, and

q (y) =

ZX

(�Q)y d� (x) for each y 2 Y .

are measurable (with respect to F and G), thus c) and d) holds and the functions arenonnegative, moreoverZ

X

�ZY

(�Q)x (y) d� (y)

�d� (x) =

ZX�Y

�Q (x; y) d (� �) (x; y)

=

ZY

�ZX

(�Q)y d� (x)

�d� (y) ,

and e) holds as well. It follows that f = �Q 2M.

Step 1. If f; g 2M and �; � � 0, then �f + �g 2M.

Notice that:

a) fx : y 7! f (x; y) is measurable with respect to G for each x 2 X and gx : y 7! f (x; y)

is measurable with respect to G for each x 2 X. Then, (�f + �g)x (y) = �fx + �gx is

measurable with respect to G for each x 2 X.

54

b) Analogously the function (�f + �g)y = �f y + �gy is measurable with respect to F foreach y 2 Y .

c) pf (x) =RYfxd� is measurable with respect to F and pg (x) =

RYgxd� is measurable

with respect to F , then is measurable with respect to F the function �pf +�pg de�nedas

(�pf + �pg) (x) = �

ZY

fxd� + �

ZY

gxd� =

=

ZY

(�fx + �gx) d� =

=

ZY

(�f + �g)x d�

or as x 7!RY(�f + �g)x d�.

d) Analogously the function y 7!RX(�f + �g)y d� is measurable with respect to G.

e) The following equalities hold:ZX

�ZY

fxd�

�d� (x) =

ZX�Y

fd (� �) =ZY

�ZX

f yd�

�d� (y) andZ

X

�ZY

gxd�

�d� (x) =

ZX�Y

gd (� �) =ZY

�ZX

gyd�

�d� (y)

it followsZX

�ZY

(�f + �g)x d�

�d� (x) =

ZX

(�pf + �pg) (x) d� (x) =

= �

ZX

pf (x) d� (x) + �

ZX

pg (x) d� (x) =

= �

ZX

�ZY

fxd�

�d� (x) + �

ZX

�ZY

gxd�

�d� (x) =

= �

ZX�Y

fd (� �) + �

ZX�Y

gd (� �) =

=

ZX�Y

(�f + �g) d (� �) =

=

ZY

�ZX

(�f + �g)y d�

�d� (y) .

Step 2. If fn 2M and fn " f , then f 2M.

a) Fixing and arbitrary x 2 X, (fn)x : y 7! fn (x; y) is measurable with respect to G foreach n 2 N, moreover (fn)x (y) = fn (x; y) " f (x; y) = fx (y) for each y 2 Y , and

(fn)x " fx. Thus fx is measurable for each x 2 X.

55

b) Analogously the function f y is measurable with respect to F for each y 2 Y .

c) pfn (x) =RY(fn)x d� is measurable with respect to F for each n 2 N. For each x 2 X,

(fn)x " fx; then, by the Monotone Convergence Theorem, pfn (x) =RY(fn)x d� "R

Yfxd� = pf (x), that is pfn " pf , and pf (x) =

RYfxd� is measurable with respect to

F .

d) Analogously the function y 7!RXf yd� is measurable with respect to G.

e) The following equalities hold:ZX

�ZY

fxd�

�d� (x) =

ZX

pfd� = limn

ZX

pfnd� =

= limn

ZX

�ZY

(fn)x d�

�d� (x) =

= limn

ZX�Y

(fn) d (� �) =

=

ZX�Y

fd (� �) =

=

ZY

�ZX

f yd�

�d� (y) .

Step 3. If f is measurable and nonnegative, then f 2M.

M contains all the characteristic functions of elements of F G (Step 0), then (Step 1) allthe simple functions which are nonnegative and measurable with respect to F G, (Step 2)all the functions which are nonnegative and measurable with respect to F G (why?). �

Corollary. Let f : X � Y ! R (or [�1;1]) be measurable with respect to F G. thefollowing are equivalent:

� f is summable,

�RX

�RYjf (x; y)j d� (y)

�d� (x) <1;

�RX

�RYjf (x; y)j d� (y)

�d� (x) <1:

Proof. Exercise. �

Fubini�s Theorem. Let � and � be complete measures, and let f : X � Y ! R (or

[�1;1]) be summable with respect to � �. Then

a) the function y 7! f (x; y) is summable with respect to � for a.e. every x 2 X,

56

b) the function x 7! f (x; y) is summable with respect to � for a.e. every y 2 Y ,

c) the function x 7!RYf (x; y) d� (y) is de�ned a.e. and is summable with respect to �,

d) the function y 7!RXf (x; y) d� (x) is de�ned a.e. and is summable with respect to �,

e) it holds the equality:ZX

�ZY

f (x; y) d� (y)

�d� (x) =

ZX�Y

f (x; y) d (� �) (x; y)

=

ZY

�ZX

f (x; y) d� (x)

�d� (y) .

Proof. We use the notation already introduced for the Proof of Tonelli�s theorem. We only

prove claims a), c), and the �rst line of the equality e) (the others can be obtained in a

similar fashion, using Y instead of X).

Notice that jf jx = jfxj, (f+)x = (fx)+ and (f�)x = (fx)

� for each x 2 X. f+ and f� aresummable and nonnegative since 0 � f+; f� � jf j. The functions de�ned for each x 2 X as

pf+ (x) =RYf+x d� and pf� (x) =

RYf�x d� are F measurable and nonnegative,moreover it isZ

X

pf+d� =

ZX

�ZY

f+x d�

�d� (x) =

ZX�Y

f+d (� �) <1 andZX

pf�d� =

ZX�Y

f�d (� �) <1.

Thus pf+ and pf� are summable and �nite a.e.. For each x s.t. pf+ (x) ; pf� (x) < 1, itisRYf+x d�;

RYf�x d� < 1 (in particular fx is summable for a.e. x 2 X), then

RYfxd� =R

Yf+x d� �

RYf�x d�, is well-de�ned and pf : x 7!

RYf (x; y) d� (y) is de�ned a.e.. Moreover

it isZX

jpf j d� =ZX

����ZY

fxd�

���� d� (x) � ZX

�ZY

jfxj d��d� (x) =

ZX�Yjf j d (� �) <1.

And x 7!RYf (x; y) d� (y) is summable.

Finally, note thatRYfxd� =

RYf+x d� �

RYf�x d� is the same as pf = pf+ � pf�, andZ

X

�ZY

f (x; y) d� (y)

�d� (x) =

ZX

pfd� =

=

ZX

pf+d��ZY

pf+d� =

=

ZX�Y

f+d (� �)�ZX�Y

f�d (� �) =

=

ZX�Y

fd (� �) :

57

�

If � and � are not complete, it is always possible to complete them. If � and � are

complete, in general � � is not complete but is always possible to complete it. Denoted by(� �)0 the completion of � �, consider functions f that are measurable with respect tothe �-algebra (F G)0 of subsets (� �)0-measurable, then Tonelli ad Fubini theorems stillhold with the following

a) the function y 7! f (x; y) is measurable with respect to G for a.e. every x 2 X,

b) the function x 7! f (x; y) is measurable with respect to F for a.e. every y 2 Y ,

c) the function x 7!RYf (x; y) d� (y) is de�ned (also is f � 0) a.e. and measurable with

respect to F ,

d) the function y 7!RXf (x; y) d� (x) is de�ned (also if f � 0) a.e. and measurable with

respect to G.

When instead of 2, we are facing N measurable spaces f(Xi;Fi; �i)gNi=1 it is straightfor-ward to extend de�nitions and theorems. (Useful exercise.)

In particular, if Xi = R, Fi = B, and � is the Borel measure, then the product �-algebraNi=1B is called the Borel �-algebra of RN and it is denoted by BN , the product measure is

called the Borel measure on RN , its completion is the Lebesgue measure of RN , and thecompletion of BN with respect to the Borel measure is called the Lebesgue �-algebra of RN .

15 SIGNEDMEASURES: HAHNAND JORDANDE-

COMPOSITIONS

Let (X;F) be a measurable space.

Proposition. Let � be a measure on F and f 2 L (X;F ; �). Consider the function

�f : F ! R de�ned as

�f (E) =

ZE

fd�

8E 2 F , then:

1. �f (;) = 0.

58

2. fAngn2N � F and Ai \ Aj = ; for each i 6= j implies �f (S1n=1An) =

P1n=1 �f (An).

3. � (E) = 0 implies �f (E) = 0.

Proof. 1.) Exercise.

2.) Exercise. Hint: set Bn =Fn1 Aj, then f�Bn ! f�F1

n=1 An, and the Dominated Conver-

gence Theorem can be applied.

3.) Exercise. Hint:RE

fd� =RE

f+d��RE

f�d�. �

1. Definition. A function � : F ! R is called a signed (�nite) measure if

(i) � (;) = 0,

(ii) fAngn2N � F and Ai \Aj = ; for each i 6= j implies � (S1n=1An) =

P1n=1 � (An).

Facts.

1. A signed measure is positive i¤ it is �nite.

[Obvious.]

2. If � and � are �nite measures, then �� � is a signed measure.[Exercise.]

3. If � is a signed measure, fAngn2N � F and An " A (or An # A), then � (A) =

limn � (An).

[Exercise. Hint: set A0 = ; and Bn = An � An�1, A =F1n=1Bn.]

Let � be a signed measure on F .

Definition. A 2 F is:

� positive (with respect to �), if � (E) � 0 for each E 2 F s.t. E � A,

� negative (with respect to �), if � (E) � 0 for each E 2 F s.t. E � A,

� null (with respect to �), if � (E) = 0 for each E 2 F s.t. E � A.

59

Lemma. If A is positive, then each E 2 F s.t. E � A is positive. If An is positive 8n 2 N,then

SnAn is positive.

Proof. The �rst part is obvious. Let B1 = A1 and Bn = An ��Sn�1

j=1 Aj

�for each n � 2.

For each n 2 N, Bn is positive andF1n=1Bn =

S1n=1An (why?). 8E 2 F s.t. E �

S1n=1An,

� (E) =P1

n=1 � (E \Bn) � 0 . �

Lemma. If E 2 F and � (E) > 0, there exists A 2 F s.t. A � E, A is positive and

� (A) � � (E).

Proof. Reductio ab absurdum. Suppose there existed E 2 F with � (E) > 0 and s.t. for

each positive subset A of E (; is positive) it is � (A) < � (E).

In particular, E is not positive, let n1 be the smallest positive integer s.t. there exists

F 3 E1 � E with

� (E1) � �1

n1

(show that n1 always exists). With the convention 10=1; for each B � E � E1

� (B) � � 1

n1 � 1

(why?).

Suppose that for �xed k there existed E1; E2; :::; Ek 2 F e n1; :::; nk 2 N s.t.:

� Ej � E ��Sj�1

i=1 Ei

�for each j = 1; :::; k,

� � (Ej) � � 1njfor each j = 1; :::; k,

� for each B � E ��Sj

i=1Ei

�,

� (B) � � 1

nj � 1for each j = 1; :::; k.

Then, ��E �

Fki=1Ei

�= � (E)�

Pki=1 � (Ei) > � (E), and E �

Fki=1Ei is not positive;

let nk+1 the smallest integer s.t. it exists F 3 Ek+1 � E �Fki=1Ei with

� (Ek+1) � �1

nk+1:

With the convention1

0=1; for each B �

�E �

Fki=1Ei

�� Ek+1 = E �

Fk+1i=1 Ei

� (B) � � 1

nk+1 � 1:

60

Let A = E �F1k=1Ek, then � (A) = � (E) �

P1k=1 � (Ek) � � (E) and A is not positive.

SinceP1

k=1 � (Ek) converges, � (Ek)! 0 as k !1. Thus � (Ek) � � 1nk� 0 implies 1

nk! 0

for k ! 1, nk ! 1 for k ! 1, and 1nk�1 ! 0. If B � A, then B � E �

�Skj=1Ek

�for

each k 2 N, therefore� (B) � � 1

nk � 1and, passing to the limit for k !1, � (B) � 0. It follows that A is positive ! . �

Analogous results hold for negative sets. It follows that

supA2Fj� (A)j <1:

(Exercise.)

Hahn�s Decomposition theorem. Let (X;F) be a measurable space and � a signedmeasure on F , then there exist A;B 2 F s.t. A is positive, B is negative, A \ B = ;,A [B = X

Proof. Let � = supA2F � (A). If � = 0, X is negative and we are done. Otherwise, consider

a sequence fAng 2 F s.t. � (An) ! �. For n big enough � (An) > 0 and, choosing Bn 2 Fwith Bn � An, Bn positive, � (Bn) � � (An); it follows that � (Bn)! �.

S1k=1Bk is positive

and � (Bn) � � (S1k=1Bk) � � for each n 2 N, passing to the limit � (

S1k=1Bk) = �. Set

A =S1k=1Bk. If C � Ac and � (C) > 0, then � (A t C) > � ! . Set B = Ac. �

Definition. Let � be a signed measure. If A;B 2 F , A is positive, B is negative, A\B = ;,A [B = X, then fA;Bg is called Hahn decomposition of �.

Facts.

1. If fA;Bg and fA0; B0g are two Hahn decomposition of �, then A \ B0 and B \ A0 arenull sets. Thus

�A = �A\A0 = �A0 and �B = �B\B0 = �B0.

Moreover �A and ��B are (positive) �nite measures and � = �A + �B = �A � (��B).[Exercise.]

2. � is a signed measure i¤ there exist two �nite measures � and � s.t. � = �� �.[Obvious.]

3. Let � and � be two �nite measures and fA;Bg be a partition of X s.t. � (B) = � (A) =

0 (we say � and � are mutually singular), then fA;Bg is an Hahn decomposition of� = �� �. Moreover � = �A and � = ��B.

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[Exercise.]

Definition. Let � be a signed measure and fA;Bg an Hahn decomposition of �, the(positive) �nite measures de�ned as

�+ = �A; �� = ��B; j�j = �+ + ��

are called upper variation, lower variation and total variation of �.

Observations up to now can be summed up in the following:

Jordan Decomposition Theorem. Let (X;F) be a measurable space and � a signedmeasure on F , then

� = �+ � ��

and (�+; ��) is the unique pair (�; �) of mutually singular measures s.t. � = �� �.

16 RADON-NIKODYM THEOREM

Let (X;F) be a measurable space.

Definition. Let � be a measure on F and � a measure or a signed measure on F , � isabsolutely continuous with respect to � if A 2 F and � (A) = 0 implies � (A) = 0. Itis written � � �.

Facts.

1. Let f : X ! R (or [�1;1]) be a measurable function. If f is nonnegative, thefunction �f : F ! [0;1] de�ned as

�f (E) =

ZE

fd�

8E 2 F , is a measure absolutely continuous with respect to �. If f is summable, thefunction �f : F ! R de�ned as

�f (E) =

ZE

fd�

8E 2 F , is a signed measure absolutely continuous with respect to �.[It simply follows by formalizing already obtained results.]

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2. If � is a signed measure and � � �, then �+; ��; j�j � �.

[Exercise. Hint: let fA;Bg be an Hahn decomposition of �, if E 2 F and � (E) = 0,then �+ (E) = � (E \ A), but � (E) = 0) � (E \ A) = 0.]

Lemma. If � and � are �nite (positive) measures on F , � 6= 0 and � � �, then there exist

n 2 N and E 2 F s.t. � (E) > 0 and E is positive with respect to � � 1n�.

Proof. Let X = An t Bn be an Hahn decomposition of � � 1n� for each n 2 N (with An

positive and Bn negative with respect to � � 1n�). Set A =

S1n=1An and B =

T1n=1Bn

(=T1n=1A

cn = Ac). For each n 2 N,

� (Bn) �1

n� (Bn) ;

then � (B) � � (Bn) � 1n� (Bn) � 1

n� (X), therefore � (B) = 0. Then � (A) > 0 (� (A) +

� (B) = � (X) > 0), but � (A) �P1

n=1 � (An), and there exists n s.t. � (An) > 0. An is

positive with respect to � � 1n� and � (An) > 0 implies � (An) > 0. Set E = An. �

Radon-Nikodym Theorem. Let (X;F ; �) be a measure space with � �nite measure

and let � be a signed measure absolutely continuous with respect to �. Then there exists

f 2 L (X;F ; �) such that� (E) =

ZE

fd�

for each E 2 F . Moreover,

� � is a �nite (positive) measure i¤ � fx 2 X : f (x) < 0g = 0,

� if g 2 L (X;F ; �) and � (E) =RE

gd� for each E 2 F , then � fx 2 X : f (x) 6= g (x)g =0.

Such f is sometimes denoted byd�

d�and is called the Radon-Nikodym derivative or

density of � with respect to �.

Proof. It is enough to prove the theorem for � �nite (positive) measure and � 6= 0 (why?).Set

K =

8<:f : X ! [0;1] measurable :ZA

fd� � � (A) 8A 2 F

9=; .K � 0 and 8f 2 K,

RX

fd� � � (X) < 1 i.e. f 2 L (X;F ; �). Let M = supf2KRX

fd�, there

exists fn 2 K s.t.RX

fnd�!M for n!1. For each n 2 N and each x 2 X set

gn (x) = max ff1 (x) ; f2 (x) ; :::; fn (x)g

63

and denote by N the set f1; 2; :::; ng. For each E 2 F set

E1 = fx 2 E : f1 (x) � fj (x) 8j 2 Ng = E \\j2Nff1 � fjg

E2 = (E � E1) \\j2Nff2 � fjg

:::

En =

E �

n�1[j=1

E1

!\\j2Nffn � fjg

fE1; E2; :::; Eng is a partition of E in F (if x 2 E and j is the smallest element of N s.t.

fj (x) = max ff1 (x) ; f2 (x) ; :::; fn (x)g, then x 2 Ej) and gn (x) = fj (x) if x 2 Ej. ThusZE

gnd� =nXj=1

ZEj

gnd� =nXj=1

ZEj

fjd� �nXj=1

� (Ej) = � (E) ,

and gn 2 K for each n 2 N. Moreover gn (x) " sup fn (x) for each x 2 X, set f = sup fn.Applying the Monotone Convergence Theorem, for each E 2 F ,Z

E

fd� = limn

ZE

gnd� � � (E)

and f 2 K, moreover ZX

fnd� �ZX

gnd� �M

impliesRX

fd� =M .

The set function de�ned as � (E) = � (E)�RE

fd�, for each E 2 F , is a �nite (positive)

measure and it is absolutely continuous with respect to �. If � 6= 0, then there exist " > 0and A 2 F such that � (A) > 0 and A is positive with respect to � � "�, that is

� (E \ A) � "� (E \ A)

for each E 2 F . Let h = f + "�A, for each E 2 F , it isZE

hd� =

ZE

fd�+ "� (E \ A)

�ZE

fd�+ � (E \ A)�ZE\A

fd�

=

ZE\Ac

fd�+ � (E \ A) � � (E)

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and h 2 K, but ZX

hd� =

ZX

fd�+ "� (A) > M

which is absurd, according to the de�nition of M .

Thus � = 0, and � (E) =RE

fd� for each E 2 F . Finally if g 2 L (X;F ; �) and � (E) =RE

gd� for eachE 2 F , thenRE

(f � g) d� = 0 for eachE 2 F and � fx 2 X : f (x)� g (x) 6= 0g =

0 (why?).

We conclude observing that if there exists f 2 L (X;F ; �) such that � fx 2 X : f (x) < 0g =0 and

� (E) =

ZE

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) =

ZE

fd�

for each E 2 F . Moreover,

� if g 2 X ! [0;1] is measurable,ZX

gd� =

ZX

(g � f) d�;

� if g 2 L (X;F ; �), g � f 2 L (X;F ; �) andZX

gd� =

ZX

(g � f) d�:

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

d�And�An

(x) if x 2 An. �

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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 Funktsional�nogo Analiza, Nauka, Mosca,

1968.)

� Royden, H.L. (1988), Real Analysis, Prentice Hall, Englewood Cli¤s.

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

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

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