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:
If $A$ is compact then $\mu(A)\ge\limsup\mu_n(A)$.
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:
- 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 $\limsup t_n=\liminf t_n$.