# Measurability according to a sub-sigma algebra

Let $(X,\mathcal{A},p)$ and $(Y,\mathcal{B},q)$ be two probability spaces. One can assume that the first is the unit interval with its Borel sigma algebra and the Lesbegue measure. Let $\mathcal{C} \subset \mathcal{B}$ be a sub-sigma-algebra. Consider the product space $(X \times Y, \mathcal{A} \times \mathcal{B}, p \times q)$ (here, $p\times q$ is the product probability measure). Let $E \in \mathcal{A} \times \mathcal{B}$ satisfy the property that all its $x$-sections are in $\mathcal{C}$: for every $x \in X$ one has $E_x := \{ y \in Y \colon (x,y) \in E \} \in \mathcal{C}$.

Question: is it true that the function $y \mapsto p(E^y)$ is $\mathcal{C}$-measurable, where $E^y := \{ x \in X \colon (x,y) \in E \}$ is the $y$-section of $E$.

The answer is positive if $X$ is a finite (or countable) set, but recall that it is the unit interval.

One possible direction for a proof is to prove that $E \in \mathcal{A} \times \mathcal{C}$, but I think that this is incorrect in general. For example, assume that $(Y,\mathcal{B},q)$ is the unit interval with its Borel sigma algebra, $\mathcal{C}$ is the sigma algebra of all countable and co-countable sets, and $E$ is the diagonal.

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Theorem 30 here probability.net/WEBfubini.pdf can help. –  Davide Giraudo Jul 18 '12 at 14:03
Thanks, Davide. The theorem you refer to does not talk about sub-sigma algebras, and therefore it is not clear to me how it can be used to prove that the function $y \mapsto p(E^y)$ is $\mathcal{C}$-measurable. It does prove that this function is $\mathcal{B}$-measurable, but this is not what we need to show :( –  Eilon Jul 18 '12 at 14:07
Eilon, you might get more help at mathoverflow.net , which has a focus on research mathematics. –  Michael Greinecker Jul 22 '12 at 16:10
Thanks for the suggestion, Michael. –  Eilon Jul 23 '12 at 6:19
@Eilon Have you asked the question at mathoverflow? –  Davide Giraudo Aug 6 '12 at 9:40