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 $\nu_n$.


1 Answer 1


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 $\nu(C)=0$.

  • $\begingroup$ 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... $\endgroup$
    – Dzoooks
    Commented Mar 18, 2021 at 10:29
  • 1
    $\begingroup$ @Dzoooks Yes, that argument works. $\endgroup$ Commented Mar 18, 2021 at 11:35

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