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Let $(X,\mathcal{X},\mu)$ and $(Y,\mathcal{Y},\nu)$ be probability spaces, $\mathcal{X}_{0}\subset\mathcal{X}$ a (sub)sigma field and assume that $f=f(x,y)\in L^{1}_{\mu\otimes \nu}$ where $(X\times Y,\mathcal{X}\times \mathcal{Y},\mu\otimes\nu)$ is the product space.

By Fubini's theorem, the function $Y\to\mathbb{R}$ $$y\mapsto\int_{Y}|f(x,y)|d\mu(x) $$ is ($\nu$-a.e) well defined, $\mathcal{Y}-$measurable and $\nu-$integrable. In particular the function $F(x,y)$ given by $$F(x,y)=E[f(\cdot,y)|\mathcal{X}_{0}](x)$$ is well defined for $\nu-$a.e. $y$ where $E[\,\cdot\,|\mathcal{X}_{0}]$ denotes a version (depending on $y$) of the conditional expectation with respect to $\mathcal{X}_{0}$.

My question has several versions:

  1. (General version): is it possible in general to choose for (a.e) every $y$ a version $E[f(\cdot,y)|\mathcal{X}_{0}]$ in such a way that $F$ is $\mathcal{X}\times\mathcal{Y}-$measurable?
  2. (Particular version): 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 $E[g|\mathcal{X}_{0}]$. Does the choice of this version give $\mathcal{X}\times\mathcal{Y}$- measurability for $F$?
  3. (Another particular version) If not covered by the answers to 1. and 2. What happens when $Y$ is the interval $[0,1]$ with Lebesgue measure?
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1 Answer 1

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Here is a partial answer, valid in the case in which $\mathcal{X}_{0}$ is countably generated.

Let $\{X_{n}\}_{n\geq 1}$ be a (countable) family of $\mathcal{X}_{0}-$sets generating $\mathcal{X}_{0}$ that includes $\emptyset, X$ and is a $\pi-$system, and fix any version $E[f(\cdot,\cdot)|\mathcal{X}_{0}\times \mathcal{Y}]$ of the conditional expectation of $f$ given $\mathcal{X}_{0}\times\mathcal{Y}$.

Notice the following: given any $Y'\in\mathcal{Y}$ and any $n\geq 1$

$$\int_{Y'}\left(\int_{X_{n}}E[f(\cdot,\cdot)|\mathcal{X}_{0}\times\mathcal{Y}](x,y)d\mu(x)-\int_{X_{n}}f(x,y)d\mu(x)\right)d\nu(y)=0$$

by Fubini's theorem and the definition of conditional expectation. Therefore, there exists a set $Y_{n}$ with $\nu(Y_{n})=1$ with the following property: for all $y\in Y_{n}$. $$ \int_{X_{n}}E[f(\cdot,\cdot)|\mathcal{X}_{0}\times\mathcal{Y}](x,y)d\mu(x)=\int_{X_{n}}f(x,y)d\mu(x)\,\,\,\,\,\,\,\,\mbox{ (1)}$$

Let $Y_{0}':=\bigcap_{n\geq 1}Y_{n}$. Then (1) holds for all $y\in Y_{0}'$ and every $n\geq 1$.

Now consider the family $\mathcal{X}_{0}'$ of $\mathcal{X}_{0}$-sets $X'$ such that (1) holds for all $y\in Y_{0}'$ with $X'$ in place of $X_{n}$. It is not hard to see that this family is a $\lambda$-system (use the fact $X\in \mathcal{X}_{0}'$), and since it includes $\{X_{n}\}_{n\geq 0}$, $\mathcal{X}_{0}'=\mathcal{X}_{0}$ by Dynkin's $\pi-\lambda$ theorem.

Now let $Y_{0}'':=\{y\in Y: E[f|\mathcal{X}_{0}\times\mathcal{Y}](\cdot,y) \,\,\,\,\mbox{is ($\mathcal{X}_{0}$-measurable and) $\mu-$integrable}\}$ and let $Y_{0}=Y_{0}'\cap Y_{0}''$. We have just proved that for all $y\in Y_{0}$, $E[f|\mathcal{X}_{0}\times\mathcal{Y}](\cdot,y)$ is a version of $E[f(\cdot,y)|\mathcal{X}_{0}]$, which clearly implies the desired conclusion.

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  • $\begingroup$ I got this idea just a little time after posing the question, which I have been thinking of for a couple of days (I ignored even this case before). Now, is this a well-known/trivial fact? the question has been voted down and I wonder if that is the reason. On the other side, I have not found entries on books or in the web about this. I would appreciate any observation about the proof (does it have any minor/essential mistake? Can it be simplified?) $\endgroup$
    – David
    Apr 20, 2015 at 1:55
  • $\begingroup$ It is worth to comment the following: if the argument for the answer is correct then the function $F$ thus defined is not only $\mathcal{X}\times\mathcal{Y}$ measurable, but it is indeed measurable with repect to $\mathcal{X}_{0}\times\mathcal{Y}$. This generalizes the following elementary observation: if $\mathcal{X}_{0}=\{\emptyset, X\}$ then $F$ can be chosen constant in $x$, whereas for $\mathcal{X}_{0}=\mathcal{X}$, $F=f$. We ''gain'' $\mathcal{X}-$measurability as we decrease $\mathcal{X}_{0}$ from $\mathcal{X}$ to the trivial sigma-field. $\endgroup$
    – David
    Apr 20, 2015 at 2:19

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