Weak convergence in probability implies uniform convergence in distribution functions 
Exercise 1: Let $\mu_n$, $\mu$ be probability measures on $\left(\mathbb{R}, \mathcal{B}\left(\mathbb{R}\right)\right)$ with distribution functions $F_n$, $F$. Show: If $\left(\mu_n\right)$ converges weakly to $\mu$ and $F$ is continuous, then $\left(F_n\right)$ converges uniformly on $\mathbb{R}$ to $F$.

This is a problem that I am totally stuck at. I know the fact that $F_n$ converges pointwise to $F$ in this question. Also, I looked through Google and found out that I have to show first that $F_{n}(t_{n})$ converges to $F(t)$ if $t_{n}$ converges to $t$. But, no matter how I tried, I keep failing to prove the fact.
Also, I have no idea how to use the fact to get the uniform convergence. So I'm just stymied.
Could anyone please help me with this?
Add : I managed to show that $F_n$ converges uniformly to $F$ on any compact intervals. However, the generalization to the whole real line is still not solved...Could anyone at least help me with the generalization to the real line?
 A: Here's another proof for the compact case which I find a bit more intuitive. We are trying to show that $F_n \rightarrow F$ uniformly on an arbitrary closed interval $[a,b]$.
Fix $\epsilon > 0$. Let $d = F(b)-F(a)$, and take $k$ large enough so that $\frac{d}{k} \leq \frac{\epsilon}{5}$. By the continuity of $F$, we can apply the intermediate value theorem to show there exist real numbers $a := x_0 < x_1 < ... < x_k := b$ such that $F(x_i) = F(a) + i\frac{d}{k}$ for each $i \in \{0,1,...,k\}$.
Since $F_n \rightarrow F$ pointwise, for each $i$ in $\{0,1,...,k\}$ there exists $N_i$ such that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N_i$. Taking $N = \max(N_0,N_1,...,N_k)$, we conclude that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N$, $i \in \{0,1,...,k\}$. This gives
$F_n(x_{i+1})-F_n(x_i) \leq |F_n(x_{i+1})-F(x_{i+1})| + |F(x_{i+1})-F(x_i)| + |F(x_i)-F_n(x_i)| \leq \frac{\epsilon}{5} + \frac{d}{k} + \frac{\epsilon}{5} \leq \frac{3\epsilon}{5}$
For any $x \in [a,b)$, we have $x_i \leq x < x_{i+1}$ for some $i \in \{0,1,...,k\}$. Then for all $n \geq N$, we have
$$
\begin{align}
|F_n(x)-F(x)| & \leq |F_n(x)-F_n(x_i)| + |F_n(x_i) - F(x_i)| + |F(x_i)-F(x)| \\\\
& \leq |F_n(x_{i+1})-F_n(x_i)| + |F_n(x_i)-F(x_i)| + |F(x_{i+1})-F(x)| \\\\
& \leq \frac{3\epsilon}{5} + \frac{\epsilon}{5} + \frac{d}{k} \\\\
& \leq \epsilon \\\\
\end{align}
$$
where the second inequality uses monotonicity of $F$ and $F_n$.
Of course, $|F_n(b)-F(b)| \leq \frac{\epsilon}{5} \leq \epsilon$ since $b = x_k$.
Thus, we've shown $|F_n(x)-F(x)| \leq \frac{5\epsilon}{3}$ for all $x \in [a,b]$, concluding the proof.
A: Fix $\epsilon>0$. Since $F$ is a distribution function, there exists $R >0$ such that $F(r) \leq \epsilon$ for all $r \leq -R$ and $F(r) \geq 1-\epsilon$ for all $r \geq R$. As $F_n \to F$ pointwise, we can choose $N \in \mathbb{N}$ such that
$$|F_n(-R) - F(-R)| \leq \epsilon \qquad \text{and} \qquad |F_n(R)-F(R)| \leq \epsilon$$
for all $n \geq N$. Hence, by the monotonicity of $F_n$ and $F$,
$$\begin{align*} |F_n(r)-F(r)| \leq |F_n(r)|+|F(r)| &\leq F_n(-R)+F(-R) \\ &= (F_n(-R)-F(-R)) + 2 F(-R) \\ &\leq 3\epsilon \tag{1} \end{align*}$$
for all $r \leq -R$. Similarly, it follows from
$$1 \geq F_n(r) \geq F_n(R) = (F_n(R)-F(R))+F(R) \geq 1-2\epsilon, \qquad r \geq R,$$
that
$$|F_n(r)-F(r)| \leq |F_n(r)-(1-\epsilon))|+ |F(r)-(1-\epsilon)| \leq 2 \epsilon \tag{2}$$
for all $r \geq R$. Combining $(1)$ and $(2)$ yields
$$\sup_{r \in [-R,R]^c} |F_n(r)-F(r)|  \leq 3 \epsilon$$
for all $n \geq N$. Since you have already shown that $F_n$ converges to $F$ uniformly on compact intervals, there exists $N' \in \mathbb{N}$ such that
$$\sup_{r \in [-R,R]} |F_n(r)-F(r)| \leq \epsilon$$
for all $n \geq N'$. Setting $\tilde{N} := \max\{N,N'\}$, we get
$$\sup_{r \in \mathbb{R}} |F_n(r)-F(r)| \leq 3 \epsilon \qquad \text{for all $n \geq N$}.$$
