# Is the following "section-wise" defined function measurable in the product space?

I asked this question in mathstackexchange a couple of days ago. Almost right after posing it a partial (affirmative) answer came to my mind in the following form

Proposition: Assume that $(X,\mathcal{X},\mu)$ and $(Y,\mathcal{Y},\nu)$ are probability spaces and that $f\in L^{1}_{\mu\times\nu}$. Then for every countably generated $\sigma-$field $\mathcal{X}_{0}\subset \mathcal{X}$ and any version $E[f|\mathcal{X}_{0}\times\mathcal{Y}]$ of the conditional expectation of $f$ with respect to $\mathcal{X_{0}}\times\mathcal{Y}$, there exists $Y'\subset Y$ with $\nu(Y')=1$ such that for all $y\in Y'$ $E[f|\mathcal{X}_{0}\times\mathcal{Y}](\cdot,y)$ is a version of $E[f(\cdot,y)|\mathcal{X}_{0}]$.

While I thought that this solved my particular problem (which has to do with the proof of a quenched functional limit theorem for Fourier Transforms in which I am working), I realized later that this proposition is not enough for my needs. Rather I need to know whether the following -the second version of the question in stackexchange- is true:

Question: Assume that $E[\,\cdot\,|\mathcal{X}_{0}]$ admits a regular version. This is: there exists a family of measures $\{\mu_{x}\}_{x\in X}$ such that for every $g\in L^{1}_{\mu}$ $$x\mapsto \int_{X}g(z)d\mu_{x}(z)$$ defines a version of $E[g|\mathcal{X}_{0}]$. Is the function $$(x,y)\mapsto \int_{X}f(z,y)d\mu_{x}(z)\,\,\,\,\,\,\mbox{(1)}$$ (which is $\mu-$a.e well defined for $\nu-$a.e $y$ by the integrability of $x\mapsto f(x,y)$ for $\nu$-almost every $y$) extensible to a $\mathcal{X}\times \mathcal{Y}$ measurable map, or even to a $\mathcal{X}_{0}\times \mathcal{Y}$ measurable map? (in other words for this last case, can we think of (1) as a version of $E[f|\mathcal{X}_{0}\times\mathcal{Y}]$?).

1. I am posing the question here because it did not go very far in Stackexchange. Indeed it was voted down a couple of times. If you think that the question does not deserve an answer I would appreciate any reference or comment explaining why (is it trivial?, please say why in such case. Is the answer well known? please give a reference then...).

2. (For your curiosity) I need a (positive) answer to this question in order to ''integrate'' certain quenched results to obtain a quite general (expected) limit theorem for the Fourier transforms

$$S_{n}(\theta,\omega)=\sum_{k=0}^{n-1}X_{k}(\omega)e^{ik\theta}$$ of an ergodic process $(X_{k})_{k}$ in $L^{2}$, but I'm stuck at a certain step related to this. Before giving up (or looking for an alternative statement to my ''theorem'') I'd like to se if anyone can help.

Thanks for your attention!

The actual answer is yes. And the function obtained is $\mathcal{X}_{0}\times\mathcal{Y}$ measurable.

To see why start by considering the following reductions: given a function $f$, denote by $\bar{f}$ the corresponding function obtained by the procedure just explained.

$$\bar{f}(x,y)=\int_{X}f(z,y)d\mu_{x}(z)$$

If $f=\sum_{n}f_{n}$ is a finite sum of integrable functions, then $$\bar{f}=\sum_{n}\bar{f}_{n}$$ thus if each $\bar{f}_{n}$ is $\mathcal{X}_{0}\times \mathcal{Y}$ measurable, so is $\bar{f}$.

Now $f=f\,\chi_{f\geq 0}-(-f\,\chi_{f<0})$ so it suffices to assume that f is nonnegative.

It is well known that every nonnegative function $f$ can be approximated by simple functions $f_{n}$ with $f_{n}$ increasing to $f$. Then, by the monotone convergence theorem. $$\bar{f}=\lim_{n}\bar{f}_{n}.$$

Thus it suffices to see that for each simple function $f$, $\bar{f}$ is $\mathcal{X}_{0}\times\mathcal{Y}-$measurable and therefore it suffices (again by linearity) to prove the result if $f=\chi_{A}$ for any $A\in \mathcal{X}\times\mathcal{Y}$.

To do so start by noticing that if $A=B\times C$ then $$\bar{\chi}_{A}(x,y)=\mu_{x}(B)\chi_{C}(y)$$

which is clearly $\mathcal{X}_{0}\times\mathcal{Y}-$measurable. Now consider the family $\mathcal{F}$ of sets $E\in \mathcal{X}\times\mathcal{Y}$ for which $\bar{\chi}_{E}$ is $\mathcal{X}_{0}\times \mathcal{Y}-$measurable. It is easy to see that this family is a $\lambda-$system, and since it includes every finite union of (disjoint) rectangles in $\mathcal{X}\times\mathcal{Y}$, $\mathcal{F}= \mathcal{X}\times\mathcal{Y}$ by Dynkin's $\pi-\lambda$ theorem. This gives the desired conclusion.