Fubini in practice. Tonelli. Sums and integrals. Fubini

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Willis Ford
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1 Tonelli Fubini in practice Tonelli s Theorem: If (X, E, µ and (Y, K, ν are two σ-finite measure spaces and f M + (X Y, E K then ( fdµ ν ( f(x, ydν(y dµ(x A product of Lebesgue measures is a Lebesgue measure: m k m... m }{{} k times f(x, ydµ(x dν(y. and we can put parentheses as we like (associative law of. In particular m p m q m p+q, and we can use this to show that all hyperplanes in R k Note that to use Fubini we need to first verify that f is integrable w.r.t. µ ν and then compute the integral using either of the successive integration orders. Verification of integrability follows from Tonelli s theorem. First verify that f is integrable by Tonelli ( f dµ ν f(x, y dν(y dµ(x... < then compute the value of the integral by Fubini ( fdµ ν f(x, ydν(y dµ(x. are m k -nullsets.. p./32 This is Fubinelli s theorem.. p.3/32 Fubini Sums and integrals Fubini s Theorem: If (X, E, µ and (Y, K, ν are two σ-finite measure spaces and if f M(X Y, E K is integrable w.r.t. µ ν then fdµ ν A ( ( f(x, ydν(y dµ(x B f(x, ydµ(x dν(y. Here A E with µ(a c and B K with ν(b c are given by and A {x X f(x, y dν(y < } B {y Y f(x, y dµ(x < }. If f, f 2,... : X R are function we can define the function f : X N R by f(x, n f n (x. Then f is E P(N-B-measurable if and only if all the functions f n are E-B-measurable (Exercise 4.8. Recall that integration w.r.t. the counting measure τ on N is the same as infinite sums. Thus f n (x f(x, ndτ(n whenever the sum and the integral make sense.. p.2/32. p.4/32

2 Interchanging sums and integrals Interchanging the order of integration Let (X, E be a mesurable space. Theorem: If f, f 2,... : X R are measurable functions and µ is a measure on (X, E then if f n dµ < each f n is integrable, the sum f n(x is finite for all x A where A {x X f n (x < } and µ(a c. Moreover, f n(x is µ-a.e. (on A equal to an integrable function and f n (xdµ(x f n (xdµ(x. A. p.5/32 Using Fubini on gives We find from this that though f(x, y sinxe xy for x (, K, y (, K K sinx x dx + y dy sinx x dx π 2 sinx x dx for K. p.7/32 Interchanging sums and integrals Image measures Proof: Fubinelli with the product measure µ τ. Catch, requires a priori µ to be σ-finite! Workaround: The purpose of σ-finiteness is to assure uniqueness of the product measure as well as existence in terms of the technical measurability lemma 8.6. If we can check that by other means, Tonelli and Fubini applies! For all A E P(N we have the disjoint decomposition A X {n} A A n {n}, A n {x X (x, n A} thus µ τ(a µ(an, which defines this particular product measure and shows that it is uniquely specified by it s values on product sets. Let (X, E, µ be a measurable space. Let (Y, K be a measurable, and let t : X Y be E-K-measurable. Definition: The image measure t(µ is the measure on (Y, K, which is given by t(µ(b µ ( t (B for all B K X t B Y Alternative proof: Use dominated convergence as in the proof of Theorem 7... p.6/32. p.8/32

3 Image measures Let (X, E, µ be a measurable space. Let (Y, K be a measurable, and let t : X Y be E-K-measurable. Familiar example Let m (, denote the Lebesgue measure restricted to the positive halfline; formally m (, (A m(a (, Definition: The image measure t(µ is the measure on (Y, K, which is given by t(µ(b µ ( t (B for all B K X t Y for all A B. Define t : R R by log(x for x > t(x for x t (B B Then t(m (, ((, x] m (, (t ((, x] m(t ((, x] (, m((, e x ] e x.. p.9/32. p./32 The image measure is a measure Lemma: The image measure t(µ is a measure on (Y, K. Proof: Obviously t(µ(a µ(t (A [, ] and we see that t(µ( µ ( t ( µ ( Total mass of the image measure Lemma: It holds that t(µ(y µ(x. Proof: Observe that t(µ(y µ ( t (Y µ(x If B, B 2,... are disjoint K-sets then t (B i t (B j t (B i B j t ( and therefore ( ( ( ( t(µ B n µ t B n µ t (B n µ ( t (B n t(µ(b n. p./32 Corollary: If µ is a probability measure then t(µ is a probability measure. Probability theory is essentially a theory about image measures given one measure and a map t, what is t(µ? The abstract simplicity cheats the eye. Finding t(µ, characterizing t(µ or just computing certain characteristics of t(µ can by arbitrarily complicated for concrete t and µ.. p.2/32

4 Let µ be the uniform distribution on X {, 2, 3, 4, 5, 6} {, 2, 3, 4, 5, 6}. Every point (i, j X has probability. Interpretation: µ has something to do with throwing two dice. The map: t : X {,, 2, 3, 4, 5} is given by t(i, j i j Image measure: Interpretation: t represents the difference between the two dice. t(µ({} µ ( {(,, (2, 2, (3, 3, (4, 4, (5, 5, (6, 6} 6. p.3/32. p.5/ Image measure: t(µ({} µ ( {(,, (2, 2, (3, 3, (4, 4, (5, 5, (6, 6} 6 Image measure: t(µ({} µ ( {(,, (2, 2, (3, 3, (4, 4, (5, 5, (6, 6} 6. p.4/32. p.6/32

5 Succesive transformations Likewise, t(µ({} Lemma: Let t : X Y and s : Y Z be measurable. Let µ be a measure on X. Then Proof: X t A t (B s ( t(µ (s t(µ s t Y s B s (C s (t(µ(c t(µ ( s (C µ ( t ( s (C µ ( (s t (C (s t(µ(c C Z. p.7/32. p.9/32 : Marginalization If λ is a measure on (X Y, E K the image measures ˆX(λ og Ŷ (λ are the marginal measures of λ s. Let X X X 2, and consider ˆX : X X 3 X, ˆX : X (X 2 X 3 X and ˆX : X X 2 X. Then ˆX ˆX ˆX and by the theorem on successive transformation And so on and so forth. k t(µ({k} p.8/32 ˆX (λ ˆX ( ˆX(λ. Moral: For marginalization on multiple product spaces it does not matter if we do several succesive marginalizations or one combined. p.2/32 marginalization.

6 : Marginalization Integral transformation for M + : Consider m 2 m m on (R 2, B 2. Then if m(a > ˆX(m 2 (A m 2 (A R m(am(r if m(a Theorem: Let t : X Y be measurable. Let µ be a measure on X. Then g d t(µ g t dµ for all g M + (Y, K. : If λ is a probability measure on (X Y, E K then the marginals are probability measures and λ is a product measure if and only if it is a product of it s marginals; Indicator functions: Let B K. Then B d t(µ t(µ(b µ ( t (B So the formula is correct in this case. t (B dµ B t dµ λ ˆX(λ Ŷ (λ.. p.2/32. p.23/32 Integral transformation for M + Integral transformation for M + Theorem: Let t : X Y be measurable. Let µ be a measure on X. Then g d t(µ g t dµ Theorem: Let t : X Y be measurable. Let µ be a measure on X. Then g d t(µ g t dµ for all g M + (Y, K. for all g M + (Y, K. Proof: Strategy: Show the formula for Simple functions: Consider indicator functions 2 simple functions 3 M + -functions Point 3 is shown from 2 via monotone convergence. Then n g d t(µ c i i g Bi d t(µ n c i Bi i n c i Bi t dµ i n i c i Bi t dµ. p.22/32 So the formula is correct in this case.. p.24/32

7 Integral transformation for M + Theorem: Let t : X Y be measurable. Let µ be a measure on X. Then for all g M + (Y, K. g d t(µ g t dµ All functions: Let g M + and choose S + -functions s n such that s n ր g. Then s n t ր g t and by the theorem on monotone convergence g d t(µ lim n s n d t(µ lim n s n t dµ g t dµ The function x (, (xe αx is M + and we find that e αx dµ(x (, (t(xe αt(x dm(x e α log(x dm(x x α dm(x α for α > for α In conclusion, the formula holds for all g M +.. p.25/32. p.27/32 Consider the measure µ with µ((, x] e x for all x R. We know that µ t(m (, The function x e αx is M + and we find that e αx dµ(x e αt(x dm(x for all α R e α log(x dm(x x α dm(x Integral transformation for L Theorem: Let t : X Y be measurable. Let µ be a measure on X. A function g M(Y, K is integrable w.r.t. t(µ if and only if in which case g t d µ <, g d t(µ g t dµ. Proof: We know that g is integrable if and only if g dt(µ <. But this integral is computed from the previous theorem. In case we have integrability, the actual integral is computed as: g d t(µ g + d t(µ (g t + dµ g d t(µ (g t dµ g t dµ g + t dµ g t dµ. p.26/32. p.28/32

8 Translations in R k Translation and integrals Definition: The translation with w R k is the map τ w : R k R k defined by τ w (x x + w for all x R k, If f M + (R k or if f L(R k, m k then in general f(x + wdµ(x f τ w (xdµ(x f(xdτ w (µ(x Definition: A measure µ on (R k, B k is translation invariant if τ w (µ µ for all w R k and for the Lebesgue measure f(x + w dx f(x dx for all w R k.. p.29/32. p.3/32 The Lebesgue measure Linear transformations on R k Theorem: A measure µ on (R k, B k, which is finite on bounded sets is translation invariant if and only if A linear map s : R k R k is given in terms of a k k matrix A, s(x Ax. for some constant c. µ cm k The map is an isomorphism if A is invertible, that is, if deta, in which case the inverse map is given by A. Theorem: If s is a linear transformation given by an invertible matrix A then s(m k deta m k. Remark: Note that for an orthonormal matrix Q (QQ T Q T Q I we have detq ±, which shows that m k is invariant under orthonormal transformations.. p.3/32. p.32/32

The Γ-function. Continuity. Systematic computations with improper integrals

The Γ-function. Continuity. Systematic computations with improper integrals Continuity Let f be a function on U with U open in R k and (, E, µ) a measure space. Define ϕ(y) = f (x, y)dµ(x). Theorem: If all the sections y f (x, y) are continuous in y 0 U, and if there is an µ-integrable