Let $X$ and $Y$ be two random variables, let $p(\cdot|y)$ be the probability density function of $X|Y=y$ and let $h(X,Y)$ be a function of $X$ and $Y$.
Question: Suppose that $p(x|y)$ is differentiable in $y$, that $h(\cdot,\cdot)$ is continuous and $|h(x,y)|\le B$ for all $x,y\in\mathbb{R}^2$. Do we have the following? If $y_n\to y$ then \begin{align*} \lim_{n\to\infty}\mathbb{E}(h(X,Y)\mid Y = y_n) = \mathbb{E}(h(X,Y)\mid Y = y) \end{align*}
My thoughts: I would like to write
\begin{align*} \lim_{n\to\infty}\mathbb{E}(h(X,Y)|Y = y_n) &= \lim_{n\to\infty}\int h(x,y_n)p(x|y_n) dx \\ &= \int \lim_{n\to\infty} h(x,y_n)p(x|y_n) dx \\ &= \int h(x,y)p(x|y) dx \\ &= \mathbb{E}(h(X,Y)|Y = y). \end{align*} However to move the limit in the integral I wanted to use the dominated convergence theorem, which doesn't clearly apply because $p(\cdot|y_n)$ might not be necessarily bounded over $(y_n)_{n\in\mathbb{N}}$ by an integrable function.
Thank you in advance!
I think I might have found the answer to my own question, would love it if someone could confirm. The proof is as follows: \begin{align} |\mathbb{E}(h(X,Y)&|Y = y_n) - \mathbb{E}(h(X,Y)|Y = y)| \\ &\le \int |h(x,y_n)p(x|y_n) - h(x,y)p(x|y)|dx \\ &\le \int |h(x,y_n)|\times|p(x|y_n) - p(x|y)| dx \\ &+ \int |h(x,y_n) - h(x,y)|\times p(x|y) dx \end{align} All functions within the integrals are continuous in $y$. The first integral is bounded by $$ \int |h(x,y_n)|\times|p(x|y_n) - p(x|y)| dx \le B\int |p(x|y_n) - p(x|y)| dx, $$ which converges to zero by Scheffe's lemma applied to $p(x|y_n)\to p(x|y)$. The function inside the second integral is bounded by $B\times p(x\mid y)$, so that the integral converges to zero by the dominated convergence theorem.