weak convergence: Show that $\mathbf{P}(X_n = q) \to 0$ as $n \to \infty$ when the cdf $F$ of the limit $X$ is continuous at $q$

Question

The question might seem easy to some of you, but I'm still pretty new to probabilty theory (and also not very deep in mathematics) and appreciate any help. Here is the full statement I'm trying to prove:

Let $(X_n)_{n \in \mathbf{N}}$ be a sequence of random variables with cdfs $F_n$ and let $X$ be a random variable with cdf $F$. Show that if $X_n \overset{d}\to X$ (i.e. if $F_n \overset{w}\to F$) it holds that

$$\mathbf{P}(X_n = q) \overset{n \, \to \, \infty}\to 0 $$

for any continuity point $q$ of $F$.

My attempt to prove it reads as follows:

Let $q$ be a point of continuity of $F$. Then there exists an $\epsilon > 0$ such that $F$ is continuous at $(q - \epsilon)$. We have

\begin{align*} \mathbf{P}(X_n = q) &= \lim_{\epsilon \to 0} \, \mathbf{P}(X_n \in (q - \epsilon, q]) \\ &= \lim_{\epsilon \to 0} \, F_n(q) - F_n(q - \epsilon) \\ &\overset{n \, \to \, \infty}\to \lim_{\epsilon \to 0} \, F(q) - F(q - \epsilon) \\ &= 0 \end{align*}

The problem that I encounter is that I apply a limit within another limit and I feel that this needs some justification. Since I know nothing about the theory of exchanging limits, I'm wondering whether my approach is reasonable at all?

If you know an alternative easy solution, I'd be glad to see it. If not I'd appreciate your help in justifying the exchage of limits.

Answer 1

Your idea is good: in order to avoid the use of a double limit, start from $$ \mathbf{P}(X_n = q)\leqslant \mathbf{P}(X_n \in (q - \epsilon, q]) $$ which is valid for each positive $\epsilon$ (but we will also assume that $q-\epsilon$ is a continuity point of $F$). Then your argument showed that $$ \limsup_{n\to +\infty}\mathbf{P}(X_n = q)\leqslant F(q)-F(q-\varepsilon) $$ and since $F$ is continuous at $q$, and there is a sequence of continuity points of $F$ which converges to $q$, the right hand side of the previous inequality can be made arbitrarily small.

Answer 2

For $\epsilon>0$ find some $\delta>0$ such that $F(q)-F(x-\delta)<\epsilon$ and such that $F$ is continuous at $q-\delta$ is.

This is possible because $q$ is a continuity point of $F$ and every interval contains continuity points of $F$.

Then: $$\limsup P(X_n=q)\leq\limsup (F_n(q)-F_n(q-\delta))=F(q)-F(q-\delta)<\epsilon$$

This for every $\epsilon$ justifying the conclusion that:$$\limsup P(X_n=q)=0$$ or equivalently:$$\lim_{n\to\infty}P(X_n=q)=0$$