Continuity sets are neccesary for weak convergence. Portmanteau theorem In particular if $\mu_n \to \mu$ weakly then $\mu_n(A)\to \mu(A)$ for each continuity set. I want an example to show that the hypothesis of $\mu(\partial(A))=0$ is neccesary. I need to know an example involving probability measures. I tried to contruct an example but I failed. I don't know why this proof is incorrect: $\mu_n(A)=\int{1_Ad\mu_n}\to \int{1_Ad\mu}=\mu(A) $

The problem in your attempt is that $\mathbf 1_A$ is not continuous in general, hence we cannot use the definition of weak convergence.

However, defining $\mu_n:=n\mathbf 1_{(0,1/n]}\lambda$, that is, $\mu_n(A)=n\lambda(A\cap (0,1/n])$, we have that $\mu_n\to\delta_0$ weakly, $\mu_n(\\{0\\})=0$ for each $n$ but $\delta_0(\\{0\\})=1$.