# 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, 2015 at 7:44
• Yes, that is correct. May 17, 2015 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, 2015 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

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

• Can you also provide the proof for the convergence in compact intervals? Mar 31, 2016 at 0:34
• @Susan See this question: math.stackexchange.com/q/467976/36150
– saz
Mar 31, 2016 at 16:27
• By $r\in[-R,R]^{c}$ do you want to say that the convergence is out of that interval? And how do you pass from that to sup $r\in[-R,R]$ ?@saz Thank you and sorry if this question is so basic. Oct 31, 2021 at 16:41

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.