Let $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ be two measurable spaces, let $\mu$ be a measure on $(X,\Sigma_X)$, and let $(K_x)_{x \in X}$ be a transition kernel from $(X,\Sigma_X)$ to $(Y,\Sigma_Y)$, that is, each $K_x$ is a measure on $(Y,\Sigma_Y)$ and for each $B \in \Sigma_Y$, the map $$ X \ni x \mapsto K_x(B) \in [0,\infty] $$ is $\Sigma_X$-measurable. Let us denote the product of $\mu$ and $(K_x)_{x \in X}$, which then is a measure on $\Sigma_X \otimes \Sigma_Y$, by $\mu \times K$.
For each $x \in X$, we denow the space Banach space $L^2(Y,\Sigma_Y, K_x)$ with it's Borel-$\sigma$-algebra $\mathcal{B}(L^2(Y,\Sigma_Y,K_x))$ induced by the corresponding norm. We now form the disjoint union $$ Z := \bigsqcup_{x \in X} L^2(Y,\Sigma_Y,K_x) := \big\{(h,x) : x \in X, h \in L^2(Y,\Sigma_Y,K_x)\big\}. $$ For an arbitrary subset $E \subseteq Z$, we then write $E_x := \{h: (h,x) \in E\}$ for $x \in X$. We then endow the disjoint union $Z$ with the $\sigma$-algebra $$ \Sigma_Z := \{E \subseteq Z : E_x \in \mathcal{B}(L^2(Y,\Sigma_Y,K_x)) \text{ for each $x \in X$}. \} $$ The question is now the following:
Suppose $f \in L^2(X \times Y, \mu \times K)$. Is the map $$ X \ni x \mapsto (f(x,\cdot),x) \in \bigsqcup_{x \in X} L^2(Y,\Sigma_Y,K_x) = Z $$ measurable with respect to $\Sigma_X$ and $\Sigma_Z$ in general?
I know that in case $(K_x)_{x \in X}$ reduces to a single measure $K$ on $(Y,\Sigma_Y)$, this has been asked before in here and here, but the case for a transition kernel seemed to be lacking as far as i know.
Follow-up question, in case the statement is not true:
Are there assumptions we can put on $(X,\Sigma_X,\mu)$ or $(Y,\Sigma_Y,(K_x)_{x \in X})$ such that it becomes measurable? Is there a different reasonable $\sigma$-algebra we could construct on $Z$ or even on each $L^2(Y,\Sigma_Y,K_x)$ such that this map becomes measurable?
