TITLE: A.s. convergence of densities implies convergence in distribution?
QUESTION [9 upvotes]: Problem. Let $X,X_1,X_2,...$ be random variables with distribution functions $F, F_1, F_2,...$ and $\lambda_1$ densities $f, f_1,f_2,..$ respectively. Is it true that $$\lim_{n\to \infty} f_n = f \text{ a.s.} \implies X_n\to_dX?$$
  Is the converse true?

I think that "$\implies$" is true and the converse is false.
Suppose $f_n\to f$ a.s. By Polya's theorem, $X_n\to_dX$ is equivalent to $\sup_{x\in \Bbb R} |F_n(x) - F(x)| \to 0$, so let's try to show that.
$$\begin{eqnarray} \sup_{x\in \Bbb R} |F_n(x) - F(x)|&=&\sup_{x\in \Bbb R} \left| \int_{(-\infty,x)}f_n(t)dt - \int_{(-\infty,x)}f(t)dt\right|
\\&\leq& \sup_{x\in \Bbb R} \int_{(-\infty, x)}|f_n(t)-f(t)| dt
\\&=& \int_{\Bbb R}|f_n(t)-f(t)| dt
\end{eqnarray}$$
which converges to $0$ as $n\to \infty$ by Scheffé's Lemma.
Is that an acceptable proof for "$\implies$"?
I was not able to prove the converse, nor find a counter example to it, but I suspect it is false. Could you please share a hint?
Thank you.

REPLY [2 votes]: As for a converse result, in the following references:
https://projecteuclid.org/download/pdf_1/euclid.aos/1176346604
https://projecteuclid.org/download/pdf_1/euclid.aos/1176350065
pointwise convergence on densities is obtained from weak convergence of the corresponding probability measures, under suitable equicontinuity assumptions on the sequence of densities.