# Distances between measures

Suppose that $$\mu_n$$ and $$\nu_n$$ are two probability measures with $$\lim_{n \to \infty} \sup_{a \in \mathbb{R}} | \mu_n(-\infty,a)-\nu_n(-\infty,a)|=0.$$

Then is it true that $$\mu_n(B)- \nu_n(B) \to 0$$ for all Borel sets $$B$$?

The usual approach to this sort of thing is to define $$\mathcal{B}:=\{B \subset \mathbb{R}: \mu_n(B)-\nu_n(B) \to 0 \},$$ and then prove that $$\mathcal{B}$$ is a $$\sigma$$-algebra that contains all $$(-\infty,a)$$, or something generating the Borel sets.

$$\mathcal{B}$$ does contain all $$(-\infty,a)$$ and is closed under taking complements because $$\mu_n$$ and $$\nu_n$$ are probability measures, but I am not sure how to show closure under countable and unions and intersections without assuming some uniformity accross all $$B$$. (Maybe I just need to add that hypothesis to $$\mathcal{B}$$?) One could also rephrase my question in terms of the cdf's for $$\mu_n$$ and $$\nu_n$$.

This is not true. By CLT there exist discrete distributions $$\mu_n$$ converging to standard normal distribution. The fact that standard normal distribution $$\nu$$ is continuous implies that the hypotheis is satisfied with $$\nu_n=\nu$$ for all $$n$$. But there is a countable set $$C$$ such that $$\mu_n(C)=1$$ for all $$n$$ whereas $$\nu(C)=0$$.
• Got it, thank you. Follow up question: what if both $\mu_n$ and $\nu_n$ are discrete? Perhaps we make the same argument, showing there are discrete $\mu_n$ and $\nu_n$ both converging in dist. to normal distribution, but defined on different discrete sets... Commented Mar 18, 2021 at 10:29