Problem on sequence of probability measures

Question

Let $\mu_n$ and $\mu$ be probability measures such that

$$\lim_{n\to \infty} \mu_n(A) = \mu(A)$$ for Borel $A \subset S$ satisfying $\mu(\partial A) =0$ (we call it a $\mu$-continuity set).

I have to prove that for any bounded continuous function $f$

$$\int f d\mu_n \to \int f d\mu$$.

My doubt is that the first line holds only for such special sets $A$. Now, if I have $\mu$ such that there is no such $A$ how to prove ?

Answer 1

Hint:

1. Without loss of generality, we may assume $f \geq 0$ (otherwise: consider $f=f^+-f^-$).
2. It follows from Tonelli's theorem that $$\int f(x) \, d\nu(x) = \int_{(0,\|f\|_{\infty})} \nu(f \geq r) \, dr$$ for any finite measure $\nu$ and $f \geq 0$.
3. Use 2. and the dominated convergence theorem to conclude $$\int f \, d\mu_n \to \int f \, d\mu.$$