Convergence in a generating algebra $\Rightarrow$ convergence in the weak* topology?

Question

Let $M$ be a compact metric space and $\mathcal{B}$ its Borelian $\sigma$-algebra. Consider $\{\mu_{n}\}_{n\in\mathbb N}$ as a sequence of Borelian probabilities on $M$. Suppose that there exists a Borelian probability $\mu$ on $M$ and a generating algebra $\mathcal{A}$ (i.e. $\mathcal A$ is an algebra and $\sigma(\mathcal{A}) = \mathcal B$) such that $$\mu_n(A)\longrightarrow \mu(A),\ \forall\ A\in \mathcal A\ \text{and}\ \mu(\partial A)=0,\ \forall \ A\in\mathcal{A}. \quad \quad (*)$$

I would like to know if $(*)$ implies that $\mu_n\to\mu$ in the weak* topology, i.e. for every continuous function $f: M \to \mathbb R$ $$\int_M f\ \text{d}\mu_n \longrightarrow \int_M f\ \text{d}\mu. $$

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My attempt

I tried to use the monotone class theorem for functions. I defined the set $$\mathcal H:=\left\{f:M\to\mathbb R;\ f \text{ is bounded, measurable and }\int_M f\ \text{d}\mu_n \longrightarrow \int_M f\ \text{d}\mu\right\}. $$

So if we prove that

1. if $A\in \mathcal A\Rightarrow$ $1_A \in \mathcal H,$
2. if $f,g\in\mathcal H$ $\Rightarrow$ $f+cg \in\mathcal{H}$, for any real number $c$,
3. if $f_n \in \mathcal{H}$ is a sequence of non-negative functions that increase to a bounded function $f$ $\Rightarrow$ $f \in \mathcal{H}$,

holds then, by the monotone class theorem for functions, $\mathcal H$ will all the bounded measurable functions, and we are done. The conditions $1$ and $2$ are obvious to be checked. However, I was not able to conclude the last condition.

Can anyone help me?

Edit: I was thinking and this approach does not make sense since the condition that I am trying to check is a way stronger than convergence in the weak* topology.

Answer 1

Here is a proof if $\mathcal{A}$ consists only of open sets or consists only of closed sets.

First suppose $\mathcal{A}$ consists only of open sets. Let $\nu$ be a weak* limit of $(\mu_n)_n$. Then $\nu(A) \le \liminf_n \mu_n(A) = \mu(A)$ for all $A \in \mathcal{A}$. So $\nu(B) \le \mu(B)$ for all $B \in \mathcal{B}$. Therefore, $\nu = \mu$ (since both are probability measures). Closed sets case is similar.

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The statement is false if $M$ is not compact. Take, for example, $M = \mathbb{R}$ and $\mathcal{A} = \{[x,y] : -\infty < x < y < +\infty\}$, the set of compact intervals. Let $\mu_n = 1_{[n,n+1]}$. Then $\mu_n(A) \to 0$ for each $A \in \mathcal{A}$, but $\mu_n$ does not converge weak* to $0$, since, for example, $\int_{\mathbb{R}} 1d\mu_n = 1$ for each $n$.

Consequently, the approach you outlined in your question probably won't work. It seems that one has to work with a weak* limit of $(\mu_n)_n$, as (partially) done above.