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Let $(X,\mu_x)$ and $(Y,\mu_y)$ be two measure spaces endowed with $\sigma$-additive compete measures $\mu_x$ and $\mu_y$, respectively. Let $\mu:=\mu_x\otimes\mu_y$ be the Lebesgue extension of measure $\mu_x\times\mu_y$ defined by $(\mu_x\times\mu_y)(A\times B)=\mu_x(A)\mu_y(B)$ for any two measurable sets $A\subset X$, $B\subset Y$.

I wonder what we can say about the the $\mu$-measurability of a function $f:X\times Y\to\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$, if we know the $\mu_x$-measurability of $f(-,y):X\to\mathbb{K}$ for all $y\in Y$ and the $\mu_y$-measurability of $f(x,-):Y\to\mathbb{K}$ for all $x\in X$, and vice versa. Does any implication between the two exist?

The issue has arisen in my mind because I found the statement in problem 6 here that if $\int_X (\int_{A_x}|f(x,y)| d\mu_y)d\mu_x$, where $A_y=\{y\in Y:(x,y)\in A\}$, exists then $\int_A fd\mu$ does. I think it is implicitly intended that $f:A\to\mathbb{K}$ is measurable, but I think it would be interesting to explore the possibility of reciprocal implications. Thank you very much for any answer!

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    $\begingroup$ See the first "Example and Warning" here $\endgroup$ – Ilya Nov 25 '14 at 21:36
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    $\begingroup$ See this question on Mathoverflow: mathoverflow.net/questions/160329/…. Sierpinski constructed (in a paper listed in the answer to that question) that there is a non-measurable set $A \subset \Bbb{R}^2$ such that $A \cap L$ has at most two elements for every straight line $L$. In particular, for $f := \chi_A$ (indicator function of $A$), we have that $f(x,\cdot) = \chi_{N_x}$ and $f(\cdot,y) = \chi_{M_y}$ for sets $N_x, M_y \subset \Bbb{R}$ with at most two elements, so that the "slice maps" are measurable. But $f$ itself is not. $\endgroup$ – PhoemueX Jan 5 '15 at 21:59

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