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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?

Aryeh Kontorovich
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    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. – Nate Eldredge Jan 05 '19 at 20:24
  • In the title you say "sections" but in the first paragraph you say "projections"? – Nate Eldredge Jan 05 '19 at 20:25
  • Changed to "projections", thanks. – Aryeh Kontorovich Jan 05 '19 at 20:30
  • Really? Just having the projections measurable (i.e. $\pi_X(A) = { x : \exists y((x,y) \in A)}$) is a really weak condition and can't possibly guarantee product measurability of $A$. For instance, take your favorite non-measurable $A$ and union it with the diagonal. Or do you mean something else? – Nate Eldredge Jan 05 '19 at 20:33
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    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. – user44191 Jan 05 '19 at 20:37
  • @NateEldredge ack -- edited back to "sections". What about the conditions on X and Y that I proposed? – Aryeh Kontorovich Jan 05 '19 at 21:05
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    Well, my first comment shows that $\mathcal{X} = 2^X$ isn't sufficient, at least not in ZFC. – Nate Eldredge Jan 05 '19 at 21:25
  • Another example ... For $X = Y = \mathbb R$, let $\phi$ be a non-Borel-measurable bijection of $\mathbb R$ (say, permute a Hamel basis and $\phi$ could even be additive $\phi(x+y) = \phi(x)+\phi(y)$), and let $A$ be the graph of $\phi$. Then all sections in both directions are singletons, hence measurable. But $A$ is not Borel measurable in $\mathbb R \times \mathbb R$. – Gerald Edgar Jul 24 '22 at 11:24

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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.

DCK
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