# If $\mu_i(A) \to \mu(A)$ for all $A$ with $\mu(\partial A) = 0$, then $\int_E g \mathrm d \mu_i \to \int_E g \mathrm d \mu$

I'm trying to prove below equivalence of weak convergence of finite Borel measures.

Let $$(E, d)$$ be a metric space and $$\mu, \mu_1, \mu_2,\ldots$$ finite Borel measures on $$E$$. Let $$g:E \to \mathbb R$$ be bounded continuous. If $$\mu_i(A) \to \mu(A)$$ for all Borel set $$A \subseteq E$$ with $$\mu(\partial A) = 0$$, then $$\int_E g \mathrm d \mu_i \to \int_E g \mathrm d \mu$$.

Could you verify if my attempt is fine?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Assume $$|g(x)| \le \alpha/2$$ for all $$x\in E$$. Let $$\nu := g_{\sharp} \mu$$ be the induced finite Borel measure on $$\mathbb R$$. Then $$\nu$$ has at most countably many atoms. For each $$m$$, there are $$-\alpha = t_1 < \cdots < t_m = \alpha$$ such that $$\nu (\{t_j\}) = 0, t_{j+1} - t_{j} < 1/m$$, and $$\nu([t_1, t_m]) = \mu(X)$$.
Let $$A_j := g^{-1}([t_j, t_{j+1}))$$. Then $$\overline {A_j} \subseteq g^{-1}([t_j, t_{j+1}])$$ and $$\mathring {A_j} \subseteq g^{-1}((t_j, t_{j+1}))$$. Then $$\partial A_j \subseteq g^{-1} (\{t_j, t_{j+1}\})$$ and thus $$\mu(\partial A_j) \le \nu(\{t_j, t_{j+1}\})=0$$. This implies $$\mu_i(A_j) \to \mu(A_j)$$ as $$i \to \infty$$. Let $$g_m := \sum_{j=1}^{m-1} t_j 1_{A_j}.$$
Notice that $$|g_m (x) - g(x)| \le 1/m$$ for all $$x\in E$$. We have \begin{align} & \left |\int_E g \mathrm d \mu_i - \int_E g \mathrm d \mu \right | \\ = &\left |\int_E (g-g_m) \mathrm d \mu_i + \int_E (g_m-g) \mathrm d \mu+ \left [ \int_E g_m \mathrm d \mu_i - \int_E g_m \mathrm d \mu \right ] \right | \\ \le & \int_E |g-g_m| \mathrm d \mu_i + \int_E |g_m-g| \mathrm d \mu+ \sum_{j=1}^{m-1} \left | \int_{A_j} g_m \mathrm d \mu_i - \int_{A_j} g_m \mathrm d \mu \right | \\ \le &\frac{\mu_i(E)}{m} + \frac{\mu(E)}{m} + \alpha\sum_{j=1}^{m-1} |\mu_i(A_j)-\mu(A_j)|. \end{align}
We first take the limit $$i \to \infty$$ and then $$m \to \infty$$. This completes the proof.