Let $X$ be a compact metric space. Take a sequence $\{\mu_n\}_{n=1}^\infty$ of Borel probability measures on $X$. Assume that this sequence converges (weak-$\ast$) to a Borel probability measure $\mu$ on $X$.

Let $A$ be a Borel subset of $A$ such that $\mu_n(A)=0$ for all $n\geq 1$. Is it necessarily true that $\mu(A)=0$?


No: take $X=[0,1]$, $\mu_n=\delta_{1/n}$, $\mu=\delta_0$, and $A=\{0\}$.


If $\mu_n\to\mu$ weak* there's not much that can be said about convergence of $\mu_n(A)$. If I recall correctly, assuming of course we're talking about regular Borel measures:

  1. If $A$ is compact then $\mu(A)\ge\limsup\mu_n(A)$.

  2. If $A$ is open then $\mu(A)\le\liminf\mu_n(A)$,

and I think that's about the whole story.

Thanks to @Mindlack for pointing out something I should have included here:

  1. Hence, if $\mu(\partial A)=0$ then $\mu(A)=\lim\mu(A_n)$.

Proof: $A\cup\partial A=\overline A$, so, noting that $\mu_n(\overline A)\ge\mu_n(A)$, (1) shows that $$\mu(A)=\mu(\overline A)\ge\limsup\mu_n(\overline A)\ge\limsup\mu_n(A).$$

Similarly $A\setminus\partial A=A^0$, the interior of $A$, so (2) shows $$\mu(A)=\mu(A^0)\le\liminf\mu_n(A^0)\le\liminf\mu_n(A).$$And for any sequence $t_n$, if $\limsup t_n\le\liminf t_n$ then $(t_n)$ is convergent, with limit $\limsup t_n=\liminf t_n$.

  • 1
    $\begingroup$ And as a consequence, if $\mu(\partial A)=0$, then $\mu_n(A) \rightarrow \mu(A)$. $\endgroup$
    – Mindlack
    Aug 19 '19 at 17:49
  • $\begingroup$ @Mindlack Ah, right - thanks. $\endgroup$ Aug 19 '19 at 19:53

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