Suppose $(Y, {\cal T}, \nu)$ is a measure space and $(X, {\cal S}, \mu_y)$ is a measure space for each fixed $y \in Y$; $\mu_y$ depends on $y$. What can we say about the ${\cal T}-{\cal B}$ measurability of the function $g(y) = \int f d \mu_y$ if we know that $f$ is ${\cal S}-{\cal B}$ measurable?
For any $B \in {\cal B}$, $g^{-1} (B) = \{ y \in Y : g(y) \in B \} = \{ y \in Y : \int f d \mu_y \in B \}$... not sure where to go.
I am asking this question because I am having trouble proving that the product of transition kernel is measurable: https://en.wikipedia.org/wiki/Transition_kernel.
If $\phi_1$ is a transition kernel from $(E, {\cal E})$ to $(X, {\cal X})$ and $\phi_2$ is a transition kerenel from $(X, {\cal X})$ to $(Y, {\cal Y})$, then the product of $\phi_1$ and $\phi_2$ defined as $$\phi_1 \phi_2 (e, A) \equiv \int \phi_1(e, d x) \phi_2(x, A) \equiv \int \phi_2(x, A) d \phi_1(e, \cdot ) \text{ for } e \in E, A \in {\cal Y}$$.
I want to show that for a fixed $A \in {\cal Y}$, $g_A (e) = \int \phi_2(x, A) d \phi_1(e, \cdot )$ is ${\cal E}$ measurable.