Product spaces and Fubini s theorem. Presentation for M721 Dec 7 th 2011 Chai Molina

Transcription

1 Product spaces and Fubini s theorem Presentation for M721 Dec 7 th 2011 Chai Molina

2 Motivation When can we exchange a double integral with iterated integrals? When are the two iterated integrals equal? In R n : 2 x e dx x y x y dx dy 2 2 dy dx x y 0 0 x y dx dy 0 0 x x y y 2 2 2

3 Setting and contents Generally, it is unclear what a double integral is, let alone when it can be replaced by iterated integrals. X, A, Y, B- measurable spaces Define measureable sets in (σ-algebra) Relating XY -meas. to - and Y meas. Product measure Fubini s theorem X X Y

4 Product spaces and measurability (X, A), (Y, B) measurable spaces Def ns: Measurable rectangle: A B with A A B B A B is the smallest σ-algebra containing all meas. rects. (why does this exist?) This defines meas. functions on X Y Elementary sets 1 i n Monotone class M PZ on a set Z: E = A B A A, B B i i i i E M, E E n E M n n n1 i n E M, E E n E M n n1 n n n A B

5 Minimality of A B Th m: A B is the smallest monotone class on X Y containing E. Strategy: Let M be the smallest monotone class containing E. A B is a monotone class so E M A B showing M is a σ-algebra is sufficient.

6 Minimality of A B Proof (1) If A i A B i B R i =A i B i E (i=1,2) (A 1 B 1 ) c = (A 1c B 1 ) (A 1 B 1c ) (A 1c B 1c ) E (A 1 B 1 ) (A 2 B 2 ) = (A 1 A 2 ) (B 1 B 2 ) E R 1 \R 2 = R 1 R c 2 E R 1 R 2 = R 1 (R 2 \R 1 ) E A 1 E is closed under finite unions, complements, intersections, differences B 1

7 Minimality of A B Proof (2) P X Y define: M P ={Q P(X Y) P\Q, Q\P,P Q M} Q M P iff P M Q M P are monotone classes (since M is one). P,Q E: (1) Q M P E M P M M P Q M, P E: Q M P P M Q E M Q M M Q So: Q,P M, we have P\Q, Q\P,P Q M (closure under finite unions, complements, intersections, differences)

8 Minimality of A B Proof (3) M is a σ-algebra X Y E M P M P c =X Y\P M P i M i N n N, Q n =P 1 P n M and Q n Q n+1 so P= i N P i = n N Q n M (monotonicity) E M A B M Thus: A B=M

9 Definition: Slices Given E X Y, f:x Y S=C or [0, ] x-slices of E&f: given x X, define E x :={y (x,y) E}, f x :Y S by f x (y)=f(x,y) y-slices of E&f: given y Y, define E y :={x (x,y) E} f y :Y S by f y (x)=f(x,y). Also called x- (y-) sections Ex Y f x, y Y x E X Y x X

10 Measurability of slices of A B-meas. sets Th m: E A B E x B and E y A x X, y Y. Proof: (same for E y ) Ω:={E A B E x B x X} is a σ-algebra: X Y Ω E Ω (E c ) x =(E x ) c B E c Ω E i Ω i N ( E i ) x = (E i ) x B E i Ω Ω contains all measurable rectangles: B B, A A (A B) x =B B A B Ω. A B Ω A B Ω =A B

11 Measurability of slices of A B-meas. functions Recall: f:x Y S=C or [0, ] is meas. Iff V S open, G=f -1 (V) A B Th m: f-a B-measurable x X, and y Y, f x, f y are B- and A measurable (resp.) Proof: V S open, G=f -1 (V) A B =f x -1 (V)=G x B f x is B-meas. Similarly for f y. χ Ex =(χ E ) x, χ E y=(χ E ) y are B- and A meas. (resp.) E A B

12 Product measure...? Let (X, A, μ) and (Y, B, ν) be meas. spaces, and E A B For a sensible definition of a measure λ on A B induced by μ&ν, we would like λ(e) = X ν(e x )dμ(x) Or is it λ(e) = Y μ(e y )dν(y)? If all is well and good in the world, X ν(e x )dμ(x)= Y μ(e y )dν(y) Or: X { Y (χ E ) x (y)dν(y)}dμ(x) = Y { X (χ E ) y (x)dμ(x)}dν(y)

13 Fubini s theorem for char. functs. (FCF) Def n: A measure space (X, A, μ) is called σ- finite iff X= n N X n with X n p.d. and μ(x n )< Th m: Let (X, A, μ) and (Y, B, ν) be σ-finite meas. spaces and E A B. Then x X, y Y: φ(x):=ν(e x )= Y (χ E ) x (y)dν is A- meas. ψ(y):=μ(e y )= X (χ E ) y (x)dμ is B- meas. X φ(x)dμ = Y ψ(y)dν.

14 Comments on FCF X { Y (χ E ) x (y) dν(y)}dμ(x) = Y { X (χ E ) y (x)dμ(x)}dν(y) φ(x) and ψ(y) exist ((χ E ) y and (χ E ) x are meas.) When are φ(x) and ψ(y) measurable? If they are, the iterated integrals can still differ. Example: X=Y=R. μ=lebesge measure on L, ν=counting measure on P(R), E={(x,x) x [0,1]}. E A B. X ν(e x )dμ = X 1 χ [0,1] (x)dμ=1, but Y μ(e y )dν= Y 0dν=0, P, y 1,1 x, L,

15 Proof of FCF - strategy Let Ω = set of all E A B for which the conclusions of the FCF hold. Let M be the set of all E A B for which E (X i Y j ) Ω j,i N. We will show M is a monotone class containing E, and so M=A B. Then any E M will be decomposed into E (X i Y j ) s (which Ω). Ω will be closed under ctbl. unions so Ω=A B. X i Y j

16 Proof of FCF: E Ω B B, A A R=A B Ω: φ(x)=ν(r x )=ν(b)χ A (x) X φ(x)dμ=μ(a)ν(b) ψ(y)=μ(r y )=μ(a)χ B (y) Y ψ(y)dν=μ(a)ν(b) Closure to finite p. d. unions follows from R Q= χ R Q =χ R +χ Q, + linearity of the integral 1 R Q 0 R X Y Q Thus FCF works for elementary sets:e Ω

17 Proof of FCF: -monotonicity of Ω Let Q i Ω i N s.t. Q i Q i+1 and Q:= i N Q i φ i (x):= ν((q i ) x )and ψ i (y):=μ((q i ) y ) are meas. φ(x):= ν(q x )and ψ(y):=μ(q y ) 0 φ i (x) φ i+1 (x) φ(x) as i so apply MCT φ meas. and X φ i (x)dμ X φ(x)dμ as i ψ meas. and Y ψ i (y)dν Y ψ(y)dν as i By assumption X φ i (x)dμ = Y ψ i (y)dν so X φ(x)dμ = Y ψ(y)dν

18 Proof of FCF: finite-meas. -monotonicity of Ω Let Q i Ω i N s.t. Q i+1 Q i A B with μ(a), ν(b)<. Set Q:= n N Q n φ i (x), ψ i (y) (meas.) and φ(x), ψ(y) as before 0 φ i+1 (x) φ i (x) ν(b) χ A (x); φ i (x) φ(x) as n From DCT : φ meas. and X φ i (x)dμ X φ(x)dμ as n ψ meas. and Y ψ i (y)dν Y ψ(y)dν as n By assumption X φ i (x)dμ = Y ψ i (y)dν so X φ(x)dμ = Y ψ(y)dν

19 Proof of FCF: M=A B M is a monotone class: Q n M, Q n Q n+1, Q:= n N Q i Fix i,j N. Set E n =Q n (X i Y j ) Ω (def. of M ). E n Q (X i Y j ) and from -monotonicity of Ω, Q (X i Y j ) Ω. Similar for Q n+1 Q n. E M: B n B, A n A ( n N (A n B n )) (X i Y j )= n N ((A n B n ) (X i Y j )) Ω E M A B M A B A B=M

20 Proof of FCF: Finale closure of of Ω to p.d. ctbl. follows from monotonicity + closure to finite unions. E M E (X i Y j ) Ω i,j N E= i,j N (E (X i Y j )) Ω

21 Product measure Let (X, A, μ) and (Y, B, ν) be σ-finite meas. spaces and E A B. Note that ν(e x )= Y χ E (x,y)dν(y) so the following makes sense Def n: μ ν:a B [0, ] μ ν(e)= X dμ Y χ E (x,y)dν (= Y dν X χ E (x,y)dμ) μ ν is called the product of the measures μ ν is a measure: MCT applied to sums of char functs. ctbl. additivity. (Nondegeneracy is trivial) Notes: (1) by definition μ ν(e)= X Y χ E (x,y)dμ ν (2) μ ν is also σ-finite

22 Fubini s Theorem (part I) Let (X, A, μ) and (Y, B, ν) be σ-finite meas. spaces and f:x Y S=C, R or [0, ] A B - meas. I. If S= [0, ] then φ(x):= Y f x (y)dν is A- meas. ψ(y):= X f y (x)dμ is B- meas. X φ(x)dμ = X Y f(x,y)d(μ ν) = Y ψ(y)dν Can write X φ(x)dμ = X dμ(x) Y f(x,y)dν(y) (iterated integral)

23 Fubini s Theorem (parts II &III) II. If S=C, and φ*(x):= Y f x (y)dν then X Y f(x,y) d(μ ν) = X φ*(x)dμ X φ*(x)dμ< f L 1 (μ ν) III. If f L 1 (μ ν) then f x L 1 (ν) for a.e. x X and f y L 1 (μ) for a.e. y Y φ(x):= Y f x (y)dν (a.e.) φ L 1 (μ) ψ(y):= X f y (x)dμ (a.e.) ψ L 1 (ν) X φ(x)dμ = X Y f(x,y)d(μ ν) = Y ψ(y)dν

24 Fubini s Theorem -comments (II+III): if at least X dμ(x) Y f(x,y) dν(y) < : f L 1 (μ ν)) X φ(x)dμ = X Y f(x,y)d(μ ν) = Y ψ(y)dν (finite) (I) is called Tonelli s th m no < Most of the work was for FCF; now it s just a lot of MCT

25 Fubini s theorem Proof (I) FCF + linearity of (I) for simple functions. f:x Y [0, ] meas. {s n } n N of simple functs. such that s n (x,y) s n+1 (x,y) f(x,y) s n (x,y) f(x,y) as n Letting φ n (x):= Y (s n ) x (y)dν, MCT gives: φ n (x) φ(x) as n and φ(x) meas. X φ n (x)dμ = X φ(x)dμ X φ n (x)dμ= X Y s n (x,y)d(μ ν) X Y f(x,y)d(μ ν) as n Similarly for ψ gives the result.

26 Fubini s theorem Proof (II+III) (I) for f (II) For (III): suffices to show for S=R (I) for f +, f -, along with f ± f implies: (*) X φ ± (x)dμ = X Y f ± (x,y)d(μ ν)< so φ ± (x) L 1 (μ) f x = (f + ) x- (f - ) x gives f x L 1 (ν) x s.t. φ ± (x)< (= a.e.) φ ± (x)< a.e. so φ(x)=φ + (x)-φ - (x) a.e. φ(x) L 1 (μ) Subtract (*) (a.e.): X φ(x)dμ= X Y f(x,y)d(μ ν) < Similar for ψ.

27 Counterexample: f L 1 (μ ν) When f:x Y R is A B meas. but L 1 (μ ν), (even if other hypotheses hold) iterated integrals may differ (0 f): X=Y=[0,1) with Leb. meas. Set I(n):=[2 -n, 2 -(n-1) ) f(x,y):= 1 n [2 n+1 χ I(n+1) (x)- 2 n χ I(n) (x)] 2 n χ I(n) (y) x I(1)=[1/2, 1) f x (y)= -4χ I(1) (y) f x (y)dy=-2 x I(m+1)=[2 -(m+1), 2 -m ) for 0<n f x (y)=2 2m+1 χ I(m) (y)-2 2(m+1) χ I(m+1) (y) f x (y)dy=0 dx( f x (y)dy ) = [1/2,1) (-2)dy+0=-1

28 Counterexample: f L 1 (μ ν) (cont d) y I(n) f y (x)= [2 n+1 χ I(n+1) (x)- 2 n χ I(n) (x)] 2 n f y (x)dx= [2 n+1 2 -(n+1) -2 n 2 -n ] 2 n =0 dy( f y (x)dx )=0 Can easily replace 2 n χ I(n) with cont. 0 g n, supp(g n ) I(n)with unit to get f cont. Note: Simpler example in the beginning

29 Counterexample: f not μ ν meas. f:x Y [0,K] (bounded) not A B meas. other hypotheses hold but slices are meas. and iterated integrals exist Iterated integrals may differ: X=Y=[0,1) with Lebesgue meas. Assume the continuum hypothesis. a bijection Γ:[0,1] X\{ω 1 } (HW2-Q4) (X, ) is the well-ordered set with max(x)=1 st unctbl. ordinal ω 1.

30 Counterexample: f not μ ν meas. (cont d) Let E={(x,y) [0,1] 2 Γ(x) Γ(y)} α X\{ω 1 }, P α is ctbl E y =0, E x =1 [0,1] { [0,1] (χ E ) x (y)dy }dx=1 [0,1] { [0,1] (χ E ) y (x)dx }dy=0

31 Conclusion Defined A B Product measure for σ-finite measure spaces Fubini s theorem now tells that for meas. functs. in the product of σ-finite spaces: if either 0 f or when one of the iterated integrals of f is < we can turn convert double integrals to iterated interals the order of integration doesn t matter