If $B$ is a Borel set s.t. $\mu(\partial B)=0$ and $\lim_n\mu_n(B)=\mu(B)$ then does $\lim_n\mu(\overline{B}^{1/n}\setminus\overline{B})=0$ hold?

Question

Let $(X, d)$ be a metric space and define $A^t := \{x\in X: d(x, A)<t\}$ for $A\subset X, t > 0$. Let $(\mu_n)_n$ be a sequence of Borel probability measures over $X$ and $\mu$ be a Borel probability measure such that $\lim_{n\to\infty}\mu_n(B) = \mu(B)$ for all Borel sets $B\subset X$ such that $\mu(\partial B) = 0$. Because $\lim_{t\to 0}\nu(B^t) = \lim_{t\to 0}\nu(\overline{B}^t) = \nu(\overline{B})$ for any Borel probability measure $\nu$ over $X$, it is not hard to see that for any fixed $n=1,2,\dots$ it holds that

$$\lim_{t\to 0}\mu_n(\overline{B}^t\setminus \overline{B}) = \mu_n(\varnothing) = 0$$

(Main question:) Can we then conclude that $\lim_{n\to \infty}\mu_n(\overline{B}^{1/n}\setminus \overline{B}) = 0$?

One observation regarding $\mu_n$s behavior is that since $\partial\partial B\subset \partial B$ and $\partial B$ is Borel set, we have that $\lim_{n\to\infty}\mu_n(\partial B) = \mu(\partial B) = 0$. Hence $\lim_{n\to\infty}\mu_n(\overline{B}) = \mu(\overline{B}) = \mu(B)$. Thus we can alternatively look at

$$\left|\mu_n\left(\overline{B}^{1/n}\right) - \mu(\overline{B})\right|$$

(obtained by adding $\pm \mu(\overline{B})$ to the prior limit and using additivity of disjoint sets for measures) and see whether it converges to zero. But I honestly don't know how to move forward with this problem.

Answer 1

Let us denote by $B^{(\varepsilon)}=\{x\in X: d(x,B)\leq \varepsilon\}$. This is a closed set and $\partial B^{(\varepsilon)}\subset\{x\in X: d(x,B)=\varepsilon\}$. Since $\mu$ is a finite measure, $\mu(\partial B^{(\varepsilon)})=0$ for all but countably many $\varepsilon>0$. Let us call $\varepsilon>0$ regular if $\mu(\partial B^{(\varepsilon)})=0$.

Let $\varepsilon>0$ be regular. Then $\partial(B^{(\varepsilon)}\setminus\overline{B})\subset\partial B^{(\varepsilon)}\cup\partial\overline{B}\subset \partial B^{(\varepsilon)}\cup\partial B$, and so $\mu(\partial(B^{(\varepsilon)}\setminus\overline{B}))=0$.

For all $n$ large enough $$ \mu_n(B^{(1/n)}\setminus \overline{B})\leq \mu_n(B^{(\varepsilon)}\setminus\overline{B})$$ Hence $$\limsup_n\mu_n(B^{(1/n)}\setminus \overline{B})\leq \mu(B^{(\varepsilon)}\setminus \overline{B})$$

Taking $\varepsilon\searrow0$ over regular values yields that $\lim_n\mu_n(B^{(1/n)}\setminus \overline{B})=0$.