Product Spaces

7/26/2019 Product Spaces

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MEASURE AND PROBABILITY

by

Ricardo Huamn Aguilar

Class notes for the the course Medida y Probabilidad in the Maestraen Matemticas Aplicadas at PUCP, March-July 2016.

All your comments are very welcome at [email protected]

PUCP


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Contents

1 Measure and Integration in Product Spaces 11.1 Exterior measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Product measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 The Fubini-Tonelli Theorem . . . . . . . . . . . . . . . . . . . . 12

1.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1


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Chapter 1

Measure and Integration inProduct Spaces

Rectangles, product-algebra, product measure space, productprobability space, Tonellis theorem, Fubinis theorem, independence


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1.1 Exterior measure

1.1 Definition. An exterior measure on a nonempty set is a function : () [0, ] such that

1. () = 0;

2. If A B, then (A) (B);

3. (j=1

Aj)

j(A

j).

1.2 Definition. A set E is said to be -measurable if

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

1.3 Theorem. Let be an exterior measure on and let be thecollection of all measurable subsets of . Then is a -algebra and

(, ,) is a complete measure space, where := |.


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Proof. Denote := (,). Then . (i) If E , then Ec, since

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

If E1, E2 , then

(A) = (A E1) +(A Ec

1)

= (A E1 E2) +(A E1 E

c2

) +(A Ec1

E2) +(A Ec

1Ec

2)

= (A (E1 E2)) +(A (E1 \E2)) +(A (E2 \E1)) +(A (E1 E2)c)

(A (E1 E2)) +(A (E1 E2)

c)

So E1 E2 .

(ii) If Fj for j , we prove that jFj . Define E1 = F1 and

Ej:=Fj \ (j1k=1Fk), j= 2, 3, . Then Ej are pairwise disjoint and njEj= nj=1Fjand Ej= jFj.


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Hence to prove (ii) it suffices to show that if Ej for j and all Ejare pairwise disjoint, then Ej . Let Bn:=

njEj and B := j=1Ej. Then

Bn

. For any A ,

(A Bn) = (A Bn En) +

(A Bn Ecn

) = (A En) +(A Bn1).

By induction, we have (A Bn) =n

j=1(A Ej). So

(A) = (A Bn) +(A Bc

n)

n

j=1

(A Ej) +(A Bc),

and letting n , we obtain

(A) j=1

(A Ej) +(A Bc) (j(A Ej)) +

(A Bc)

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

Thus, B= Ej and

(Ej) =

j

(Ej).


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Hence, is a -algebra and :=| is a measure on . If (E) =0,

then E (,), since

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

Therefore, (, (,),) is a complete measure space.

1.4 Proposition. Let () and : [0, ] be a function such

that

(1) and () = 0;

(2) = j=1

Aj for some Aj , j .

For E , define

(E):=inf

j=1(Aj) | Aj j , E

j=1Aj

. (1.1)

Then is an exterior measure.


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Proof. (i) Since 0 () () =0, we have () =0. (ii) If A B, then

any countable cover of B by elements of is also a cover of A. Hence

(A) (B).

(iii) For any given > 0, by the definition of(Aj), there exists a countable

cover kEj,k Aj such that all Ej,k andk

(Ej,k) (Aj) +

2j.

Hence jAj j kEj,k and

(jAj)

j

k

(Ej,k)

j

((Aj) +

2j) = +

j

(Aj).

Taking 0, we conclude that (jAj) j (Aj) and is an exteriormeasure.


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1.2 Product measure

1.5 Theorem. Let (, ,) and (S, ,) be measure spaces. Denote

:= {A B |A ,B } and (A B):= (A)(B).

For each subset E S, define

(E):=inf

j=1

(Aj)(B

j) | E

j=1A

jB

j, A

j , B

j .

(1) Then is an exterior measure on S.

(2) Let S denote the collection of all -measurable subsets of S.

Then S is a -algebra.

(3) If we define := |S,

then ( S, S, ) is a complete measure space.


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1.6 Lemma. If E and F , then

EF S and (EF):= (EF) = (E)(F).

Proof. Let A S. For any > 0, there exists Ej , Fj such that

A j(Ej Fj) and

j

(Ej)(Fj) (A) + .

Note that

A (EF) [j(Ej Fj)] (EF) = j[(Ej E) (Fj F)]

and

A (EF)c

j

(Ej E) (Fj F

c) (Ej Ec) Fj

.


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Consequently,

(A (EF)) +(A (EF)c) j

(Ej E)(Fj F)

+

j

(Ej E)(Fj Fc) +

j

(Ej Ec)(Fj)

=j

(Ej E)(Fj) +j

(Ej Ec)(Fj) =

j

(Ej)(Fj) (A) + .

Taking 0, we see that

(A (EF)) +(A (EF)c) (A).

So, EF is -measurable, that is, EF S.

Lets show the other claim. By EF EF, it is trivial that (EF):=(EF) (E)(F).


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Conversely, if EF j(Ej Fj), Ej , Fj , we have

1E()1F(s) j

1Ej()1Fj (s), ,s S.

Integrating both sides of the above inequality, first with respect to and

then with respect to , and applying twice the MCT, we have

(E)(F) =

S

1E()1F(s)d(s)d()

Sj

1Ej()1Fj (s)d(s)d()

=

j

S

1Ej()1Fj (s)d(s)d() =

j

1Ej()

S

1Fj(s)d(s)d()

j

1Ej()(Fj)d() =

j

1Ej(Fj)d() =

j

(Ej)(Fj).

This proves (E)(F) (EF) =: (EF).


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1.7 Definition. Let be the -algebra generated by the sets

:= {A B |A , B }.

The measure space

( S, , )

is called the product space of (, ,) and (S, ,).

It is also denoted as ( S, , ).

1.1 Exercise. The measure is the unique measure on such

that

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


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1.3 The Fubini-Tonelli Theorem

1.8 Definition. The -section E and s-section Es

of a subset E Sare

E:= {s S | (,s) E} and Es:= { | (,s) E}.

For a function f : S , the -section f and s-section fs are

f

(s) := f(,s), fs() := f(,s).

1.9 Proposition. (i) If E , then E for all and Es

for all s S.

(ii) If f is -measurable, then f is -measurable for all and

fs

is -measurable for all s S.


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Proof. Let be the collection

:= {E | E , ;Es , s S}.

Obviously, A B for all A and B . SincejEj

= j(Ej), (Ec)= (E)

c,

and likewise for s-sections, is a -algebra. Therefore, . This

proves (i).

(ii) Follows from (i) since for any B

(f)1(B) = (f1(B)), (f

s)1(B) = (f1(B))s.


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1.10 Theorem. Suppose that (, ,) and (S, ,) are -finite spaces.

If E , then the function

(E) is -measurable,

and the function

s (Es) is -measurable;

and

(E

) =

(E

)d

(

) = S

(E

s)d

(s

).

Proof. First, we assume that () < and (S) < . Let be the set of

all E for which the conclusions of the theorem are true. IfE= AB,

then (E) = 1A()(B) and (Es) =(A)1B(s). So clearly, E=A B for

all A and B . Hence contains the -system . We note that

is a -system (see exercise below). By the Theorem, = .


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If and are -finite, we can write S= j(j Sj) of rectangles with

(j)(Sj) < , j and Sj S. IfE , we have (see exercise below)

(E (j Sj)) =

1j()(E Sj)d() =

S

1Sj(s)(Es j)d(s). (1.2)

Noting that 1j ()(E Sj) 1()(E), the conclusion now follows from

the MCT.

1.2 Exercise. (i) Prove that defined in the above theorem is a -

system. (ii) Prove (1.2).


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1.11 Theorem. Suppose that (, ,) and (S, ,) are -finite spaces.

Define

g():=

S

fd and h(s):=

fsd.

[Tonelli] If f : S [0, ] is -measurable function, then g and h

are nonnegative and measurable; and

S f(,s)d (,s) =

S f(,s)d(s)

d()

=

S

f(,s)d()

d(s). (1.3)

[Fubini] If f L1( ), then

(1) f L1(S, ,) for -a.e. , fs L1(, ,) for -a.e. s S,

(2) the a.e.-defined functions g L1(, ,), h L1(S, ,), a n d (1.3)

holds.


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Proof. By Theorem 1.10, Tonellis theorem holds when f is a character-

istic function. By linearity of integration, it holds for nonnegative simple

functions. If f is a nonnegative measurable function, there is a sequence

(n)n of nonnegative simple functions such that n(,s) f(,s) for all

(,s) S. The MCT implies first gn() :=

S(n)d

S

fd and

hn(s):=

(n)

sd

fsd, and again

S

f dd =

g d =limn

gnd =limn

S

n(,s)d(s)()

=limn

S

nd( ) =

S

f d( ).

and similarly,

S

f dd =

S

hd =limn

S

hnd =limn

S

n(,s)d(s)()


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=limn

S

nd( ) =

S

f d( ).

This proves Tonellis theorem. Moreover, if

Sf d( )< , this also

shows that g< a.e. and h< a.e., that is f L1(S, ,) for a.e.

and fs L1(, ,) for a.e. s S.

If f L1( S, , ), the conclusion of Fubinis theorem follows by

applying Tonellis theorem to the positive and negative parts of f.

1.3 Exercise. In general, ( S, , ) is not complete.

Notation. The product space is also denoted as ( S, , ).


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1.4 Applications

Throughout this section, the probability space (, ,P) is given.1.12 Theorem. Let X1, X2, , Xn be random variables where Xk has

distribution P Xk. Let P X be the distribution of the random vector X =

(X1, ,Xn). The random variables X1, X2, , Xn are independent if and

only if

P X = P X1 P Xn.

Proof. For the sake of notation, we consider n=2. We want to show that

(2, (2),PX) = ( , ,PX1 PX2). By Exercise 1.5 below (2) =

. We need to proof that PX =PX1 PX2. Given the uniqueness of the

product measure, it suffices to show that it holds only on the rectangles

= {A B |A, B }. Indeed, by independence,

PX(A B) = P

(X1,X2) A B

= P(X1 A)P(X2 B)

=PX1(A)PX2(B) = PX1 PX2(A B).


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To prove the converse, we note that PX(A B) = P(X1 A,X2 B) and

PX1(A)PX2(B) = P(X1 A)P(X2 B).

1.13 Theorem. Let X and Y be independent random variables with

distributions P X and P Y, respectively. Let h : 2 be measurable. If

either h 0, or E|h(X, Y)| < , then

Eh(X, Y) =

h(x,y)d P X(x) d PY(y). ()

In particular, if h(x,y) = f(x)g(y) where f,g : measurable with either

f,g 0 or E|f(X)|, E|g(Y)| < , then

Ef(X)g(Y) = Ef(X)Eg(Y). ()

Proof. Consider h 0 first. By the Change of Variable Formula and

Tonellis theorem,

Eh(X, Y) =

2

h(x,y)PX P Y(d x, d y) =

h(x,y)PX(d x)P Y(d y).


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To prove the second claim, consider first f 0 and g 0. By the above

equality,

Ef(X)g(Y) =

f(x)g(y)PX(d x)P Y(d y) = Ef(X)Eg(Y).

For the other case, we apply the above argument to |f(X)| and |g(Y)|,

E|f(X)g(Y)| = E|f(X)|E|g(Y)| < .

Now Fubini can be applied to f(X)g(Y) to obtain the desired result.

1.14 Theorem. Let X1, X2, , Xn be independent random variables. If

either Xi 0 for all i, or E|Xi| < for all i, then

Eni=1

Xi = ni=1

EXi.

Proof. Follows directly from the previous theorem.


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1.5 Exercises

1.4 Exercise. Let (, ,P) be a probability space. SupposeX : [0, )is a random variable. Then

E [X] =

[0,)

P(X>t)(d t),

where is the Lebesgue measure.

1.5 Exercise. Prove that () () = (2).

1.6 Exercise. If X and Y are independent, then

FX+Y

(z):=P(X+Yz) =

FX

(z y)P Y(d y), z .

1.7 Exercise. Prove that the converse of Theorem1.14 does not hold.

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