Several conjectures on measurable functions defined on a product space

Question

Let $X$ be equipped with a $\sigma$-field $\mathcal{A}$, and $Y$ be equipped with a $\sigma$-field $\mathcal{B}$, and $f$ be a function from $X\times Y$ to $\mathbb{R}$.

It can be shown that if $f(x,y)$ is $\mathcal{A}\otimes\mathcal{B}\backslash\mathcal{B}(\mathbb{R})$-measurable, then the map $x\mapsto f(x,y)$ is $\mathcal{A}$-measurable for each fixed $y\in Y$ and $y\mapsto f(x,y)$ is $\mathcal{B}$-measurable for each fixed $x\in X$, using an analog of $\pi$-$\lambda$ theorem for classes of functions.

My first question is whether its converse is true. That is, is it possible to show that if both $x\mapsto f(x,y)$ and $y\mapsto f(x,y)$ are measurable then $f(x,y)$ is product measurable?

My second question is whether the proposition is still true for more general measurable functions. $f(x,y)$ need not be a real valued function. Let it be a function from $X\times Y$ to a topological space, say $Z$, equipped with a $\sigma$-field $\mathcal{C}$. Then does the $\mathcal{A}\otimes\mathcal{B}\backslash\mathcal{C}$-measurability of $f(x,y)$ imply the $\mathcal{A}\backslash\mathcal{C}$-measurability of $x\mapsto f(x,y)$ and $\mathcal{B}\backslash\mathcal{C}$-measurability of $y\mapsto f(x,y)$? And converse?

Answer 1

Not true. Let $B$ be a subset of $\mathbb R$ which is not a Borel set and $f(x,y)=1$ if $x=y$ and $x \in B$, $0$ otherwise. You can easily check that $x\to f(x,y)$ is measurable for each $y$ and $y\to f(x,y)$ is measurable for each $x$ but $f$ is not Borel measurable [ Let me know if you need some details].

Answer 2

The answer to your second question is yes. For fixed $y \in Y$, define $i_y : X \to X \times Y$ by $i_y(x) = (x,y)$. Now verify, as an exercise, that $i_y$ is $\mathcal{A} \backslash \mathcal{A} \otimes \mathcal{B}$-measurable. Finally, note that the map $x \mapsto f(x,y)$ is simply $f \circ i_y$, and the composition of measurable maps is measurable.