# Weak-* convergence and null sets

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 $$\limsup t_n=\liminf t_n$$.

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