I have a very general question about how to show that a subset $A$ of $X \times Y$ lies in a $\sigma$ algebra generated by $S \times T$ where $S, T$ are $\sigma$ algebras of $X, Y$ respectively.
Is it enough to show that $A$ restricted to $x$-component is $T$ measurable and $A$ restricted to $y$-component is $S$ measurable?
Thank you.
No, it is not enough to show that the sections of $A$ are measurable in the respective components.
For example let $X=Y=\mathbb{R}$ with $S=T=\mathcal{B}_{\mathbb{R}}$ (Borel sets)
Let $N \subset \mathbb{R}$ be any set which is not a Borel set (i.e, $N \notin \mathcal{B}_{\mathbb{R}}$).
Define $A=\{(x,x):x \in N \} \subset \mathbb{R}^2$. Then all of the sections of $A$ with respect to $x$ and $y$ are one-point sets, hence they are Borel sets in $\mathbb{R}$. However $A$ itself is not a Borel set in $\mathbb{R}^2$. (To prove this, note that if $A$ was a Borel set, then the preimage of $A$ under the measurable mapping $f:\mathbb{R} \to \mathbb{R}^2$ given by $x \mapsto (x,x)$ would be a Borel set. But this preimage is $N$, which is not a Borel set, so $A$ cannot be a Borel set).
For an even stronger example, assume that $N \subset [0, 2 \pi )$, and then consider the set $A':= \{ (\cos x, \sin x) : x \in N \} $. Then $A' \cap L$ is Borel for every line $L \subset \mathbb{R}^2$ (because $A' \cap L$ can cntain at most two points), but $A'$ is not Borel (otherwise the preimage of $A$ under $f: [0,2 \pi) \to \mathbb{R}^2$, which is $N$, would be Borel).