If for every continuous distribution function $F$, $E[F(X_n)] \to E[F(X)]$ then $X_n \Rightarrow X$ $(X_n)_n$ and $X$ are random variables.
Show that $(X_n)_n$ converges in distribution to $X$ if and only if $$\lim_nE[F(X_n)]=E[F(X)]$$ for every continuous distribution function $F.$
$\implies$ is trivial, Any ideas for the converse?
 A: We know that $X_n\Rightarrow X$ iff $F_{X_n}(x) \to F_X(x)$ in all continuity points of $F_X$, so it is enough to prove the former convergence. This is done in quite a standard way; actually, the standard proof that the weak convergence implies such pointwise convergence uses monotone continuous functions.
Specifically, for any fixed $x<y$ define $$
F_{x,y} (z) = \begin{cases} 0,& z<x,\\
\frac{z-x}{y-x},& x\le z\le y\\
1, &z>y. 
\end{cases}
$$
Then, $\mathrm{P}(X_n>x)\ge \mathrm{E}[F_{x,y}(X_n)]$, so noting that $F_{x,y}$ is a continuous cdf and letting $n\to\infty$ we obtain
$$
\liminf_{n\to\infty} \mathrm{P}(X_n>x)\ge \mathrm{E}[F_{x,y}(X)] \ge \mathrm{P}(X>y).
$$
Now letting $y\to x+$ gives
$$
\liminf_{n\to\infty} \mathrm{P}(X_n>x)\ge \mathrm{P}(X>x).
$$
Similarly, for any $t<x$, from $\mathrm{P}(X_n>x)\le \mathrm{E}[F_{t,x}(X_n)]$, we get that
$$
\limsup_{n\to\infty} \mathrm{P}(X_n>x)\le \mathrm{P}(X\ge x).
$$
Consequently,
$$
\lim_{n\to\infty} \mathrm{P}(X_n>x)= \mathrm{P}(X>x)
$$
whenever $\mathrm{P}(X = x)=0$, and this is easily seen to be equivalent to the desired statement.
