If $\mu_n(A) \to \mu(A)$ for every $A \in \mathcal A$, then $\mu_n \to\mu$ weakly.

Question

Kolmogorov Prokhorov theorem:

Let $X$ be a metric space and $(\mu_n)$ be a sequence of Borel Probability measure on $X$. Suppose $\mathcal A \subset \mathcal B (X)$ satisfies:

$(1)$ $\mathcal A$ is closed under finite intersection.

$(2)$ Every open set in $X$ equals union of countably many elements of $\mathcal A$.

If $\mu_n(A) \to \mu(A)$ for every $A \in \mathcal A$, then $\mu_n \to\mu$ weakly.

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Need a reference where I can find the proof of the theorem. Or some hints to prove the theorem.

Ref: $\mu_n \to\mu$ weakly, if for every bounded continuous function $f$ on $X$, $$\int_X f d\mu_n \to \int_X d\mu.$$

Answer 1

If $A,B \in \mathcal A$ then $\mu_n (A \cup B) =\mu_n (A)+\mu_n(B)-\mu_n (A \cap B) \to \mu (A)+\mu(B)-\mu (A \cap B)=\mu(A \cup B)$. This extends to finite unions. Let $U$ be any open set and write $U$ as the union of a sequence $\{A_k\}$ from $\mathcal A$. Let $C_k$ be the union of $A_1,A_2,...,A_k$. Then $\mu_n (U) \geq \mu_n(C_k) \to \mu(C_k)$. Hence $\liminf \mu_n(U) \geq \mu(C_k)$ for each $k$. This implies $\liminf \mu_n(U) \geq \mu(U)$. Since $U$ is an arbitrary open set we see that $\mu_n \to \mu$ weakly.