I have a technical question regarding the measurability of functions of the form $R(A,y)$ where $R(\cdot,y)$ is a measure.
Let $(X,\Sigma)$ and $(Y,\mathfrak{A})$ be measurable spaces. Suppose we have a (regular) stochastic transition kernel $$ R: \Sigma \times Y \to \mathbb R \,.$$ Say stemming from a (regular) conditional probability w.r.t. a measurable mapping $\pi: (X,\Sigma) \to (Y,\mathfrak{A})$.
We can extend the $\sigma$-algebra $\Sigma$ to the completion w.r.t. $R(\cdot,y)$. Lets call this completion $\Sigma_y$.
Let $\nu$ be a probability measure on $(Y,\mathfrak A)$. Now suppose that $A,B \subseteq X$ are $\nu$-almost surely measurable w.r.t. $\Sigma_y$.
Now I can pose my question:
It can be that $y\mapsto R(A,y)$ as well as $y \mapsto R(B,y)$ are almost surely equal to $\Sigma$-measurable maps. (This is true by definition of a transition kernel for $A,B \in \Sigma$ but it can be true for many more sets.) Is it then also the case that $y\mapsto R(A\cap B, y)$ is also almost surely measurable? Does the same hold if we replace $\Sigma$ by some other $\sigma$-algebra?