Convergence in law implies uniform convergence of cdf's

Question

Let $F_n, \ F$ be distribution functions with respect to some variables $X_n,\ X$ (in a not necessarily common probability space). Suppose that $F$ is continuous and $F_n \overset{d}{\rightarrow}F$ (i.e in law). Prove that $(F_n)$ converges uniformly to $F$, i.e. $$\displaystyle \lim_{n\rightarrow +\infty}\sup_{x\in \mathbb{R}}|F_n(x)-F(x)|=0$$

Comments. On a proof by contadiction, I 'd suppose that for some $\varepsilon>0$ for all $n$ there is $x_n$ such that $|F_n(x_n)-F(x_n)|\geq \varepsilon.$ But a classic analytic approach throught Bolzano - Weierstrass theorem can not be applied here since there is no information on the boundness of $(x_n).$

Thanks a lot in advance for the help!

Answer 1

Hint: (as stated in Parzen, 1960). Convergence in distribution is defined for points $x$ of continuity of $F$, so here since $F$ is continuous, that is for every $x\in \mathbb R$. Now, to any $ε>0$, choose points $$-\infty=x_0<x_1<\dots<x_K=+\infty$$ so that $F(x_j)-F(x_{j-1})<ε, \text{ for } \ j=1,2,\dots,K$ (recall that $F$ in non-decreasing). Verify that $$\sup_{x\in \Bbb R}|F_{n}(x)-F(x)|\le \max_{j=0,1,\dots,K}|F_n(x_j)-F(x_j)|+ε$$ If $F$ is not continuous, this statement fails. Bernoulli distributions provide already counterexamples, in this case.