A sequence of random variables $\{X_n\}$ in $[0,1]$ converges to a constant $c\in (0,1)$ almost surely. The sequence is also uniformly integrable.
Let $p$ be a probability density function on $[0,1]$, which is continuous and smooth in $(0,1)$.
Then $p(X_n)\to p(c)$ a.s.
Is it true that $\mathbb{E}[p(X_n)]\to p(c)$?
If $p$ is bounded the result is trivial, however $p$ might explode at $0$ or $1$. It would be enough to show that the sequence $p(X_n)$ is uniformly integrable but I didn't manage to do that. In particular I am interested in proving this result for a sequence of Beta random variables: $X_n=Beta(nx,n(1-x))$ which converges almost surely to $x\in(0,1)$. It seems to me that this problem shouldn't be very hard to solve, but I'm stuck.
No, it's not. Take $$p(x) =C \frac{1}{\sqrt{x}} 1_{(0,1)}(x)$$ for a normalizing constant $C>0$ such that $p$ is a probability density function. On $\Omega=(0,1)$ (with Lebesgue measure) consider $$X_n(\omega) := \begin{cases} \frac{1}{n^4}, & \omega \in (0,\frac{1}{n}), \\ \frac{1}{2}, & \omega \in [\frac{1}{n},1), \end{cases}$$ then $X_n \to \frac{1}{2}$ almost surely. On the other hand,
$$\mathbb{E}p(X_n) = C n^2 \frac{1}{n} + C \sqrt{2} \left(1-\frac{1}{n} \right)$$
and so
$$\lim_{n \to \infty} \mathbb{E}p(X_n) = \infty \neq C \sqrt{2} = p(\frac{1}{2})$$