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Tonelli-Fubini Theorem.

Let $(\mathbb{X},\mathscr{X},\mu)$ and $(\mathbb{Y},\mathscr{Y},\nu)$ be probability spaces and let $\mathscr{Z}$ be the $\sigma$-field product i.e. the $\sigma$-field generated by $\{A\times B : A\in\mathscr{X}, B\in\mathscr{Y}\}$. Let $f:\mathbb{X}\times\mathbb{Y}\rightarrow [0,+\infty]$ be $\mathscr{Z}$-mensurable, and let $\eta=\mu\times\nu$ be the product probability on $\mathscr{Z}$. Then it is true that

a) $f(x,\cdot)$ is $\mathscr{Y}$-mensurable $\forall x\in\mathbb{X}$, where $f(x,\cdot):\mathbb{Y}\rightarrow[0,+\infty]$ is defined by $y\mapsto f(x,y)$. Analogously, $f(\cdot,y)$ is $\mathscr{X}$-mensurable $\forall y\in\mathbb{Y}$, where $f(\cdot,y):\mathbb{X}\rightarrow[0,+\infty]$ is defined by $x\mapsto f(x,y)$.

b) $\phi$ is $\mathscr{Y}$-mensurable where $\phi:\mathbb{Y}\rightarrow[0,+\infty]$ is defined by $y\mapsto \int_{\mathbb{X}}f(\cdot,y) d\mu$. Analogously, $\psi$ is $\mathscr{X}$-mensurable, where $\psi:\mathbb{X}\rightarrow[0,+\infty]$ is defined by $x\mapsto \int_{\mathbb{Y}} f(x,\cdot) d\nu$.

c)$ \int_{\mathbb{X}\times\mathbb{Y}}f d(\nu\times\mu)=\int_{\mathbb{X}}\psi d\mu=\int_{\mathbb{Y}}\phi d\nu$

My question. If $\mathscr{B}\subset\mathscr{Z}$ is the sub-$\sigma$-field generated by $\{A\times\mathbb{Y} : A\in\mathscr{X}\}$ and $f:\mathbb{X}\times\mathbb{Y}\rightarrow [0,+\infty]$ is $\mathscr{B}$-measurable, then:

1) $f(\cdot,y_1)=f(\cdot,y_2)$ for all $y_1,y_2\in\mathbb{Y}$?

2) $\phi(y_1)=\phi(y_2)$ for all $y_1,y_2\in\mathbb{Y}$?

  • 0
    Two minor observations: Isn't $\{A \times \mathbb Y\}$ already a $\sigma$-field? Is it unclear that (1) implies (2)?2011-12-19
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    Maybe my head is foggy, but isn't (1) clear? If $\{A \times \mathbb Y\}$ is already a $\sigma$-field/algebra, then for $x \in \mathbb X$ we have that $f^{-1}(f(x, y_1))$ is $A \times \mathbb Y$ for some $A \in \mathscr X$, $x \in A$. And clearly $(x, y_2) \in A \times \mathbb Y$.2011-12-19
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    On the first point yes it's true. $\{A\times\mathbb{Y} : A\in\mathscr{X}\}$ is clearly a sub-sigma-field but that does not undertakes the issue. On the second point yes it's true that (1) implies (2). But (1) may be false and yet (2) be true. So I posted (1) and (2).2011-12-19
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    I could not understand the objection of his last post. Could I clarify? Thank you.2011-12-19
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    Well, it seems like I've given a proof of (1). And it seemed pretty easy. It appears that you know much more about measure theory than I do, so I'm somewhat concerned that I've said something incorrect. Does the proof that for all $x \in \mathbb X$ we have $f(x, y_1) = f(x, y_2)$ above make sense?2011-12-20
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    @DylanMoreland: To me, you are totally right! Why don't you post your answer?2011-12-22
  • 0
    @אליהו צלע: Whose objection? To whose last post?2011-12-22
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    I thought it was an objection about the need to question 2) as the questão1) implies the question 2).2011-12-22
  • 0
    It is a pleasure to meet you here André.2011-12-22

1 Answers 1

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@DylanMoreland: Why didn't you post your answer? @theOP: What is the motivation for your question?

$\mathcal{B}=\{A\times Y:A\in\mathcal{X}\}$ is already a $\sigma$-field, in fact, it is the product of $\mathcal{X}$ with the trivial $\sigma$-field $\{\emptyset,Y\}$ on $Y$. Therefore we have $f^{-1}(D)\in\mathcal{B}$ for any Borel set $D\subseteq[0,\infty]$, i.e. $f^{-1}(\{D\})=\emptyset$ or $f^{-1}(\{D\})=A\times Y$ for some $A\in\mathcal{B}$. From this we easily get (1): Assume (1) is not true, then there exist some $x\in X$ and $y_1,y_2\in Y$ such that $z_1=f(x,y_1)\not= f(x,y_2)=z_2$. Therefore there exists $A\in\mathcal{X}$ s.t. $f^{-1}(\{z_1\})=A\otimes Y$ (since $(x,y_1)\in f^{-1}(\{z_1\})$, it can not be empty). But then we also have $(x,y_2)\in A\times Y$, which contradicts our assumption $f(x,y_1)\not=f(x,y_2)$.

(2) follows from (1).