Showing that $\lim\limits_{n \to \infty} F_{\mu_n}(x) = F_\mu(x)$

Suppose $$(\mu_n)_{n \in \mathbb{N}}$$ is a sequence of probability measures (on $$\mathcal{B}(\mathbb{R}$$)) that I wish to show converges weakly to $$\mu$$. One way of doing is of course to show that $$\lim\limits_{n \to \infty} F_{\mu_n}(x) = F_\mu(x)$$, where $$F$$ denotes the distribution function) for every continuity point $$x \in \mathbb{R}$$.

Say I have already proven that this is true for $$x$$ with either $$F_\mu(x) \geq F_{\mu_n}(x)$$ for all $$n \in \mathbb{N}$$ or $$F_\mu(x) \leq F_{\mu_n}(x)$$ for all $$n \in \mathbb{N}$$. Can I use this to show this for an arbitray continuity point $$x$$? I was thinking of taking an arbitrary subsequence and showing that this sequence has a convergent subsequence - but I am not sure how to utilize what I have shown.

• The answer seems to be no. For example, fix $m\in\mathbb{N}$ an conducer $\mu_n=\delta_{\frac{1}{m}+\frac{1}{n}}$ and $\mu=\delta_0$. We have that $\mu_n\Longrightarrow\nu:=\delta_{\frac{1}{m}}$, and $\lim_n F_{\mu_n}(x)=F_\nu(x)\leq F_\mu(x)$ Jun 6 '21 at 11:03

It might not be sufficient. Take a standard normal random variable $$Z$$, and for $$n\geq 1$$ define a sequence of random variables $$Z_n:=Z+\frac{(-1)^n}{n}$$, Clearly $$Z_n \to Z$$ in distribution. Now pick an $$x\in \mathbb{R}$$, note that for even $$n$$, $$F_{Z_n}(x)=\mathbb{P}(Z+\frac{1}{n}\leq x)$$ is strictly smaller than $$F_{Z}(x)=\mathbb{P}(Z\leq x)$$, and for odd $$n$$,$$F_{Z_n}(x)=\mathbb{P}(Z-\frac{1}{n}\leq x)$$ is strictly larger than $$F_{Z}(x)=\mathbb{P}(Z\leq x)$$. Therefore there is no $$x$$ satisfying your assumptions. Hope this clarifies your doubts.