TITLE: If $f\left(x,\cdot\right)$ is measurable for every $x$ and $f\left(\cdot,y\right)$ is measurable for every $y$, is $f$ necessarily measurable?
QUESTION [5 upvotes]: If $f\left(x,\cdot\right)$ is measurable for every $x$ and $f\left(\cdot,y\right)$ is measurable for every $y$, is $f$ necessarily measurable?
More precisely, let $\left(\Omega_i,\mathcal{A}_i\right)$ be measurable spaces, $i\in\left\{1,2,3\right\}$. Let $f:\Omega_1\times\Omega_2\rightarrow\Omega_3$. For every $\omega_1\in\Omega_1$, $\omega_2\in\Omega_2$ define
$$
f_1^{\left(\omega_1\right)}:\Omega_2\rightarrow\Omega_3,\space\space f_1^{\left(\omega_1\right)}\left(\omega_2\right):=f\left(\omega_1,\omega_2\right)
$$
$$
f_2^{\left(\omega_2\right)}:\Omega_1\rightarrow\Omega_3,\space\space f_2^{\left(\omega_2\right)}\left(\omega_1\right):=f\left(\omega_2,\omega_1\right)
$$
It is known (e.g. Schilling, Theorem 13.10 iii) that if $f$ is $\mathcal{A}_1\otimes\mathcal{A}_2/\mathcal{A}_3$-measurable, then $f_1^{\left(\omega_1\right)}$ is $\mathcal{A}_2/\mathcal{A}_3$-measurable for every $\omega_1\in\Omega_1$ and $f_2^{\left(\omega_2\right)}$ is $\mathcal{A}_1/\mathcal{A}_3$-measurable for every $\omega_2\in\Omega_2$.
But does the converse hold as well?
In comparison, both directions hold in the following, related result (Schilling, Theorem 13.10 ii): $f:\Omega_3\rightarrow\Omega_1\times\Omega_2$ is $\mathcal{A}_3/\mathcal{A}_1\otimes\mathcal{A}_2$-measurable iff $\pi_i\circ f$ is $\mathcal{A}_3/\mathcal{A}_i$-measurable ($i\in\left\{1,2\right\}$), with
$$\pi_i:\Omega_1\times\Omega_2\rightarrow\Omega_i,\space\space \pi_i\left(\left(\omega_1,\omega_2\right)\right):=\omega_i$$
References
Schilling, René L. Measures, Integrals and Martingales. 2005

REPLY [2 votes]: Assuming CH... there is a set $E$ in the unit square $[0,1]\times[0,1]$ such that every vertical cross-section
$$
\{y : (a,y) \in E\}
$$
is countable, and every horizontal cross-section
$$
\{x:(x,b) \in E\}
$$
has countable complement.  Then by Fibini's theorem, the set is not Borel measurable.  Not even Lebesgue measurable.