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Let $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ be measurable spaces and let $\mathcal{Z}$ be the product $\sigma$-algebra on $X\times Y$. Suppose $A\subset X\times Y$ is such that its $X$- and $Y$-sections are measurable w.r.t. their respective $\sigma$-algebras.

Question: Under what natural assumptions on $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ can we conclude that $A\in\mathcal{Z}$?

For example, does $\mathcal{X}=2^X$ suffice? What if we add the condition that $Y=\{0,1\}^X$, and $\mathcal{Y}$ is the cylindrical $\sigma$-algebra?

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    $\begingroup$ As I mentioned here, $\mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R})$ is consistently not equal to $\mathcal{P}(\mathbb{R}^2)$. If so, take $A \subset \mathbb{R}^2$ with $A \notin \mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R})$, then trivially all its sections are measurable but it is not in the product $\sigma$-algebra. $\endgroup$
    – Nate Eldredge
    Commented Jan 5, 2019 at 20:24
  • $\begingroup$ In the title you say "sections" but in the first paragraph you say "projections"? $\endgroup$
    – Nate Eldredge
    Commented Jan 5, 2019 at 20:25
  • $\begingroup$ Changed to "projections", thanks. $\endgroup$
    – Aryeh Kontorovich
    Commented Jan 5, 2019 at 20:30
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    $\begingroup$ A stronger condition that may keep the spirit of the original condition: for any measurable set $S$, the projection of $A \cap S \times Y$ to $Y$ must be measurable. $\endgroup$
    – user44191
    Commented Jan 5, 2019 at 20:37
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    $\begingroup$ Well, my first comment shows that $\mathcal{X} = 2^X$ isn't sufficient, at least not in ZFC. $\endgroup$
    – Nate Eldredge
    Commented Jan 5, 2019 at 21:25

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Let $(X,\Sigma)$ be a measurable space and let $Y,Z$ be two separable metric spaces.

If $f:X\times Y \rightarrow Z$ meets that $f_x$ is continuous and $f^y$ is measurable for all $(x,y)\in X\times Y$ then $f$ is measurable.

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    $\begingroup$ Sorry for the comment on an old post, but could I request a reference for this result? $\endgroup$
    – Nate River
    Commented May 14 at 10:43
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    $\begingroup$ @NateRiver For your reference, please see this comment. $\endgroup$
    – Akira
    Commented May 15 at 9:25

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