# 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.

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?

• What's your definition of weak convergence? $F_n \to F$ pointwise for all contuinty points of $F$? – saz May 17 '15 at 7:44
• Yes, that is correct. – Keith May 17 '15 at 7:46
• Could you (for the sake of other interested readers) add the idea how to prove the uniform convergence on compact intervals to your question? – saz May 17 '15 at 9:27

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

$$\sup_{r \in \mathbb{R}} |F_n(r)-F(r)| \leq 3 \epsilon \qquad \text{for all n \geq N}.$$