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One step in the proof of Tonelli-Fubini Theorem

$Q$ is a probability measure on $(F,\mathcal{F})$.

Suppose $C\in\mathcal{E}\bigotimes\mathcal{F}$ and $C(x)=\{y:(x,y)\in C\}$

Let $\mathcal{H}=\left\{ C\in\mathcal{E}\bigotimes\mathcal{F}:\: x\rightarrow Q[C(x)]\textrm{ is }\mathcal{E}\textrm{-measurable}\right\} $ .

Show that $\mathcal{H}$ is closed under increasing limit and differences.

I can prove it is closed under the increasing limit because $C_{n}(x)\uparrow C(x)\Rightarrow Q[C_{n}(x)]\uparrow Q[C(x)]$.

However, I don't know how to show the difference because

(1) The difference of two measurable functions might not be measurable.

(2) $Q[C_{1}(x)\backslash C_{2}(x)]\neq Q[C_{1}(x)]-Q[C_{2}(x)]$ .

Thank you!

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The difference of two measurable functions might not be measurable... Really? – Did Mar 4 '12 at 9:37
By closed under differences, does the author mean that if $A\subset B$ and $A,B\in\mathcal H$ then $B\setminus A\in\mathcal H$? – Davide Giraudo Mar 4 '12 at 10:47
@Didier Piau. Oh I take it back. Since their sum is measurable, their difference should be measurable as well. – user16859 Mar 4 '12 at 16:36
@Davide Giraudo Oh, I just checked the textbook. That is what they meant! Thanks. – user16859 Mar 4 '12 at 16:40
up vote 2 down vote accepted

Let $S$ be a set and $\mathcal F\subset 2^S$. $\mathcal F$ is closed by differences means that if $A$ and $B$ are in $\mathcal F$ and $A\subset B$ then $B\setminus A\in\mathcal F$.

We apply this to $S=E\times F$ and $\mathcal F=\mathcal H$. We take $C_1, C_2\in\mathcal H$ such that $C_1\subset C_2$. Then $Q(C_2(x)\setminus C_1(x))=Q(C_2(x))-Q(C_1(x))$. Since the maps $x\mapsto Q(C_2(x))$ and $x\mapsto -Q(C_1(x))$ are $\mathcal E$-measurable, so is their sum, which is $x\mapsto Q(C_2(x)\setminus C_1(x))$.

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So if $C$ is measurable, the slice $C(x)$ is measurable? – Rudy the Reindeer Mar 4 '12 at 21:44
Yes, let $\mathcal A:=\bigcap_{x\in E}\{C\subset E \times F, C(x)\in \mathcal F\}$. Then $\mathcal A$ is a $\sigma$-algebra which contains all the sets of the form $E'\times F'$ with $E'\in\mathcal E'$ and $F'\in\mathcal F'$ – Davide Giraudo Mar 4 '12 at 21:51
Thank you. But I still don't fully understand: We defined $\mathcal{H}$ in terms of the functions $f_C : x \mapsto Q[{C(x)}]$, namely we take it to be all the sets $C$ that make $f_C$ measurable. Since $Q$ is a probability measure, $f_C$ is a function into $[0,1]$. Now for some reason that I don't yet see, $f_C^{-1}$ maps (Lebesgue?) measurable sets in $[0,1]$ to sets in $\mathcal{E}$. Is $\mathcal{A} = \mathcal{H}$? – Rudy the Reindeer Mar 5 '12 at 11:49

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