This is a question from Erhan Cinlar's "Probability and Stochastics" Chapter 1 Section 6 Question 6.30:
Let $\nu$ be a $\sigma$-finite measure on $(F,\mathscr{F})$, and let $k$ be a positive function in $\mathscr{E}\otimes\mathscr{F}$. Define $K$ by 6.2, that is, in differential notation, $K(x,dy)=\nu(dy)k(x,y)$ (integral notation: $\forall x\in E, \forall B\in\mathscr{F}, K(x,B)=\int_B\nu(dy)k(x,y)$. Show that $K$ is a transition kernel. Then, $k$ is called the transition density function $K$ with respect to $\nu$.
I was trying to show that $\forall B\in\mathscr{F}, g_B(x)=K(x,B)=\int_B \nu(dy)k(x,y)$ is $\mathscr{E}$-measurable.
I thought i could use Radon-Nikodym theorem to get a positive measurable function $p$ and show that $g_B=p$ by uniqueness from the theorem.
I define $\forall x\in E, \forall B\in\mathscr{F}, \mu_x(B) = K(x,B)$.
From there i was able to show that $\forall x\in E, \mu_x<<\nu$ ($\mu_x$ is absolutely continuous with respect to $\mu$).
After applying Radon-Nikodym theorem, i realised that that measurable function $p$ such that $\forall \mathscr{F}\text{-measurable }f,$ $\int_F\mu_x(dy)f(y)=\int_F\nu(dy)p(y)f(y)$, $p$ is defined on $(F,\mathscr{F})$ and not $(E,\mathscr{E})$, so $p$ cannot be equals to $g_B$.
Can I have a hint on how to solve this problem? Am I in the right direction?