Setwise convergence of measures implies weak convergence under special hypothesis I'm struggling with producing a proof of the following result:
Let $X = \overline{\mathbb{C}}$ be the Riemann sphere, and consider $M(X)$ the space of finite Borel measures on $X$ with norm given by the total variation. Let $\mathcal{H}$ be a basis of open sets for the standard euclidean topology on $\overline{\mathbb{C}}$, such that for any $H \in \mathcal{H}$, $\mu(\overline{H} \setminus H) = \mu(\overline{H} \setminus \text{int}(H)) = 0$ where $\mu \in M(X)$ is some fixed positive measure. Then for any sequence of positive measures $\mu_{n} \in M(X)$ ,, $\mu_{n}(H) \rightarrow \mu(H)$ for all $H \in \mathcal{H}$ implies $\mu_n$ converges weakly to $\mu$.
Supposedly one can proceed by 'uniformly approximating $f \in C(X)$' by linear combinations of indicator functions on the members of $\mathcal{H}$, but I do not see how to accomplish this.
Thank you in advance.
 A: So I think given compactness, the above should imply weak convergence for any basis of open sets at all, regardless of whatever condition we impose on it. My argument is as follows:
First note that by compactness, we have that $\mu_{n}(X),\mu(X)$ are uniformly bounded, say by $C > 0$. Set $\mu_{0} = \mu$. This means that if $\phi_{m} \rightarrow g$ uniformly on $X$, $| \int \phi_m d \mu_{n} - \int g d \mu_{n} | \leq C ||\phi_{m} - g||_{u} \rightarrow_{m} 0$, independent of $n = 0,1,...$
Now WLOG suppose for $f \in C(X)$, we have $0 \leq f \leq 1$, define $\phi_{m} = \frac{1}{2^m} \sum_{j=1}^{2^m -1} \mathbf{1}_{f^{-1}(\frac{j}{2^m},+\infty)} = \sum_{p \geq 1} c^{m}_{p} \mathbf{1}_{U^{m}_{p}}$ for some $c^{m}_{p} > 0$ and $U^{m}_{p} \in \mathcal{H}$.
Then $\int \phi_{m} d \mu_{n} \rightarrow_{n \rightarrow \infty} \int \phi_{m} d \mu$ for all $m$ by hypothesis, however the convergence depends on $m$. Moreover since $f$ is bounded $0 \leq \phi_{m} \uparrow f$ uniformly on $X$.
Let $M$ be such that for $m \geq M$, and some fixed $\epsilon > 0$, we have $|\int \phi_{m} d\mu_{n} - \int f d\mu_{n} | < \frac{\epsilon}{3}$ for all $n$.
Choose $N_0$ such that for $n \geq N_0$, one has $|\int \phi_{M} d\mu_{n} - \int \phi_{M} d \mu | < \frac{\epsilon}{3}$.
Then $|\int f d\mu_{n} - \int f d\mu| \leq |\int f d\mu_{n} - \int \phi_{M} d\mu_{n}| + |\int \phi_{M} d\mu_{n} - \int \phi_{M} d\mu| + |\int \phi_{M} d\mu - \int f d \mu| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} < \epsilon$ whenever $n \geq N_0$, so we have proved $\lim_{n \rightarrow \infty} \int f d\mu_{n} = \int f d\mu$ as desired.
A: I've started writing this before you posted your answer, so I'll post this anyway since it took me some time to write in the hopes it might still benefit someone else.
First, note that $\nu(H)=\int\mathbf{1}_Hd\nu$ for every measure. I'm assuming you want to show that  $\int_X f d\mu_n\to \int_Xfd\mu$ for every $f\in C(X)$.
Notice that if you can find for every $\epsilon>0$ an $f_\epsilon= \sum_{\ell=1}^{N_\epsilon} c_\ell \cdot \mathbf{1}_{G_\ell}$ such that $\Vert f-f_\epsilon\Vert_\infty <\epsilon$, then you can  write
$$\int_X fd\mu-\int_X fd\mu_n= \Big(\int_X fd\mu- \int_X f_\epsilon d\mu \Big)+ \Big(\int_X f_\epsilon d\mu- \int_X f_\epsilon d\mu_n \Big)+ \Big(\int_X f_\epsilon d\mu_n- \int_X f d\mu_n \Big)\leq $$
And we can then deduce that
$$\Big\vert \int_X fd\mu-\int_X fd\mu_n \Big\vert \leq \epsilon(\mu(X)+\mu_n(X))+ \Bigg\vert \int_X f_\epsilon d\mu-\int_X f_\epsilon d\mu_n 
 \Bigg\vert.$$
We can also deduce that
$$ \Bigg\vert \int_X f_\epsilon d\mu-\int_X f_\epsilon d\mu_n 
 \Bigg\vert \leq \sum_{\ell=1}^{N_\epsilon} \vert a_\ell \vert \cdot \vert\mu(G_\ell)-\mu_n(G_\ell)
\vert. $$
We can choose $G_\ell$ from the algebra generated by $\mathcal{H}$ since $X$ is compact and $f$ is uniformly continuous. By the assumption on $\mu$, we can conclude that $\mu_n(G)\to \mu(G)$ for every $G$ in the set algebra generated by $\mathcal{H}$. This last thing does require some work which I skip.
Finally one has to apply the measure convegrence assumption on finitely many sets in $\mathcal{H}$, including $X$ itself, and obtain that
$$ \Big\vert \int_X fd\mu-\int_X fd\mu_n \Big\vert < 4\epsilon. $$
