Assume $(\Omega, \mathcal{F}, \Pr)$ is a measure space and $X, Y$ are two continuous random variable defined on it. $\mu$ is Lebesgue measure on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$.
I have known that the Randon-Nikodym derivative of $(\Pr X)^{-1}$ w.r.t Lebesgue measure $\frac{d(\Pr X)^{-1}}{d\mu}$ is $X$'s density function because of the uniqueness of Randon-Nikodym derivative.
What confused me is whether the conditional density function $\frac{p(x, y)}{p(x)}$ is Randon-Nikodym derivative of the distribution of $Y \vert X= x$ and how to illustrate if this statement is right?
I've already made some attempts to handle this problem and find following relating exercise in some textbook:
Let $(X,Y)$ be $\mathbb{R}^2$-valued random variable of density function $p(x,y)$, thus there exists a measurable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $$ E[X\vert \sigma(Y)] = f(Y), $$ as $E[X\vert Y]$ is $\sigma(Y)$-measurable. Please try to show $f(y) = \frac{\int_{\mathbb{R}}xp(x,y)dx}{\int_{\mathbb{R}}p(x,y)dx}$.
I have tried to illustrate above $f(Y)$ satisfies the equation shown as follow: $$ \int_{A}Xd\Pr = \int_{A} f(Y)d\Pr, \quad \forall A \in \sigma(Y), $$ which can follow the conclusion that we need because of the uniqueness of conditional expectation, but unfortunately, I encountered some troubles in the process of proving it.
Please help me to above the statement or tell me another way to illustrate $\frac{p(x, y)}{p(x)}$ is Randon-Nikodym derivative of the distribution of $Y \vert X= x$.