Let $(\Omega_1, \mathcal{G}_1), (\Omega_2, \mathcal{G}_2), (\Omega, \mathcal{G})$ be measurable spaces and let
$$(\Omega_1 \times \Omega_2, \mathcal{G}_1 \otimes \mathcal{G_2}) $$
the canonical product measurable space of $(\Omega_1, \mathcal{G}_1)$ and $(\Omega_2, \mathcal{G}_2)$.
Now let
$$f : \Omega_1 \times \Omega_2 \rightarrow \mathbb{\Omega} $$
be some function.
Is the following statement true?
$$ f \text{ is measurable} \iff $$ $$ \forall \tilde{\omega_1} \in \Omega_1: \text{} \omega_2 \mapsto f(\tilde{\omega_1},\omega_2) \text{ is measurable and } \forall \tilde{\omega_2} \in \Omega_2: \text{ } \omega_1 \mapsto f(\omega_1,\tilde{\omega}_2) \text{ is measurable} $$
I think $ "\Rightarrow "$ can be shown using canonical injections, but I am not sure how to show the other direction.
Best Answer
The question is about how for $F \subset \Omega_1 \times \Omega_2$ two sentences "$F$ is measurable" $(1)$ and "every intersection of $F$ with "lines" $(\omega_1, \cdot)$ and $(\cdot, \omega_2)$ is measurable" $(2)$ are connected.
$(1) \Rightarrow (2)$. Any measurable subset of $\Omega_1 \times \Omega_2$ can be constructed by countable many steps of taking unions and intersections of sets of form $A_i \times B_i$ where $A_i$ is $\mathcal{G_1}$ measurable and $B_i$ is $\mathcal{G_2}$ measurable. Take some $\omega_1 \in \Omega_1$. Repeat steps used to build $F$ on $\Omega_2$, replacing $B_i$ with $\varnothing$ if $\omega_1 \notin A_i$ - resulting set is $\mathcal{G_2}$ measurable, and is exactly $F \cap (\omega_1, \cdot)$. Similarly for $F \cap (\cdot, \omega_2)$.
$(2) \not \Rightarrow (1)$: there exists a non measurable subset of $\mathbb R^2$ such that it's intersection with any line consists of at most two points (see, for example, "Counterexamples in analysis", chapter 10, example 21). [it's probably overkill here, but I couldn't find any simple example specifically for your case]