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I have a question about the joint measurability of a function defined as below:
Let $(S, \Sigma)$ be a measurable space, and let $X$ and $Y$ be topological spaces with Borel $\sigma-$algebras. Suppose that $X$ is countable, and for each $x\in X$, $f^x=f(\cdot, x):S\rightarrow Y$ is measurable. Can one say $f(\cdot, \cdot): S\times X\rightarrow$ is jointly measurable?
Denote by $\mathcal{B}(Y)$ the Borel-$\sigma$-algebra on $Y$. In order to show measurability of $f:S \times X \to Y$ we have to show $\{f \in B\} \in \Sigma \otimes \mathcal{B}(X)$ for all $B \in \mathcal{B}(Y)$.
For fixed $B \in \mathcal{B}(Y)$ we have
$$\{f \in B\} = \bigcup_{x \in X} \underbrace{\{f(\cdot,x) \in B\}}_{\in \Sigma} \times \underbrace{\{x\}}_{\in \mathcal{B}(X)}.$$
As $X$ is countable, this shows that $\{f \in B\} \in \Sigma \otimes \mathcal{B}(X)$ as a countable union of $\Sigma \otimes \mathcal{B}(X)$-measurable sets.
Is your definition of measurability that the preimage of a Borel set in $Y$ is a measurable set in $S \times X$? And does $S \times X$ have the product measure? And does $X$ have the discrete topology, so that singleton sets are open and hence Borel measurable in $X$?
If so, I think the answer to your question is yes. Given a Borel subset $V \subset Y$, its preimage is $\cup_{x \in X} \left( (f^x)^{-1}(V) \times \{ x\}\right)$, which by hypothesis, is a countable union of measurable sets in $S \times X$, hence is measurable.
