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

Question

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 $E[f|\mathcal{X}_{0}\times\mathcal{Y}]$?).

Two comments:

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!

Answer 1

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.