Uniform convergence and weak convergence Assume $F_{n},F$ are distribution functions of r.v.$X_{n}$ and $X$, $F_{n}$ weakly converge to $F$. If $F$ is pointwise continuous in the interval $[a,b]\subset\mathbb{R}$, show that
$$\sup_{x\in[a,b]}|F_{n}(x)-F(x)|\rightarrow 0,n\rightarrow \infty.$$Thanks.
 A: The key is to prove that $F_n(x_n)\to F(x)$ for each sequence $(x_n,n\geqslant 1)$ converging to $x$. Indeed, once it is done, suppose that $F_n$ does not converge uniformly to $F$ on $[a,b]$. This means that for some positive $\delta$ and some sequence of integers $n_k\uparrow \infty$, $\sup_{x\in [a,b]}|F_{n_k}(x)-F(x)|>\delta$. Take $x_{n_k}$ such that 
$|F_{n_k}(x_{n_k})-F(x)|>\delta$.
Extract a convergent subsequence of $(x_{n_k})$ to get a contradiction. 
Passing to subsequences, one can deal with two cases:


*

*$x_n\uparrow x$;

*$x_n\downarrow x$.


We will thereat the first one. The idea is the following: write
$$F_n(x_n)-F(x_n)=\mu(X_n\leqslant x)-\mu(x_n\lt X_n\leqslant x)-\mu(X\leqslant x)+\mu(x_n\lt X\leqslant x).$$
Notice that for each $n_0$ and $n\geqslant n_0$,
$$\mu(x_n\lt X_n\leqslant x)\leqslant \mu(x_{n_0}\lt X_n\leqslant x),$$
hence 
$$\limsup_{n\to \infty}\mu(x_n\lt X_n\leqslant x)\leqslant\mu(x_{n_0}\lt X\leqslant x),$$
and since $F$ is continuous and $n_0$ arbitrary, we get $\lim_{n\to \infty}\mu(x_n\lt X_n\leqslant x)=0$. 

Notice that we didn't use compactness: actually, we can prove that if $F$ is continuous on $\mathbb R$, there is uniform convergence of the sequence of cumulative distribution functions on the whole real line. In the argument, once we get the $\delta$, we take $R$ such that $\sup_n1-F_n(R)\lt\delta$ and $F(-R)\lt\delta$ in order to work with a compact interval.
