I have a candidate proof for this result, but the solution given in the solution manual for this exercise seems much more complicated than mine so I wonder if I did something wrong.
I argue:
Let $(F_n)_n$ be a sequence of distributions such that $F_n \implies F$. And suppose $F$ is continuous. We want to show that
$$\sup_x|F_n(x)-F(x)| \rightarrow 0$$
$\text{as }n \rightarrow \infty$.
Let $\varepsilon > 0$, let $M_\varepsilon$ be such that
$$1-F(M_\varepsilon)<\varepsilon\text{ and } F(-M_\varepsilon)<\varepsilon\tag{$\star$}$$
Using the fact that a sequence of cumulative distributions is $\textbf{tight}$ if and only if every subsequential (vague) limit is a cumulative distribution function for a probability measure, we may conclude that $(F_n)_n$ is tight. That is, for arbitrary $\varepsilon$ we may find $M_\varepsilon'$ such that
$$\limsup_{n \rightarrow \infty} F_n(-M_{\varepsilon}')+\big(1-F_n(M_\varepsilon')\big) < \varepsilon.\tag{$\star\star$}$$
Taking the maximum between $M_\varepsilon$ and $M_\varepsilon'$ we may suppose $M_\varepsilon=M_\varepsilon'$.
From $(\star)$ and $(\star\star)$ and the fact that the $F_n$ and $F$ are nondecreasing, it readily follows that there exists $N\in \mathbb{N}$ such that
$$\sup_{n\geq N} \sup_{|x|\geq M_\varepsilon}|F_n(x)-F(x)|<\varepsilon$$
additionally, given that convergence is uniform in the compact interval $[-M_\varepsilon,M_\varepsilon]$, there is an $N'$ such that
$$\sup_{n\geq N'} \sup_{|x|\leq M_\varepsilon}|F_n(x)-F(x)|<\varepsilon.$$
Supposing without loss of generality that $N=N'$, we get that
$$\sup_{n\geq N} \sup_{x\in \mathbb{R}}|F_n(x)-F(x)|<\varepsilon$$
which shows uniform convergence.
Am I making some mistake?
Thank you for any feedback.
Best Answer
Even if your functions are continuous it does not follow that the convergence is uniform on $[-M_{\epsilon},M_{\epsilon}]$.
Note that $F$ is uniformly continuous on this interval. Choose $\delta >0$ such that $|F(x)-F(y)| <\epsilon$ for $|x-y| <\delta$. Choose $N$ such that $\frac 2 N M_{\epsilon} <\delta$ and divide the interval $[-M_{\epsilon},M_{\epsilon}]$ in to $N$ equal parts to get a partition $(x_i)$. Now note that if $$x_{i-1} \leq x \leq x_i$$ then $$F_n(x)-F(x) \leq F_n(x_i)-F(x_i)+ (F(x_i)-F(x))< 2 \epsilon $$ if $n$ is sufficiently large since $F_n(x_i) \to F(x_i)$ for each $i$. Use a similar inequality for $F(x)-F_n(x)$ to finish the proof.