Do we necessarily have that $\int g\,d\mu_n \to \int_0^1 g\,dx$?

Question

Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $[0, 1]$. Suppose $\mu_n$ are finite measures on $([0, 1], \mathcal{B})$ such that $\int f\,d\mu_n \to \int_0^1 f\,dx$ whenever $f$ is a real-valued continuous function on $[0, 1]$. Suppose that $g$ is a bounded measurable function such that the set of discontinuities of $g$ has measure $0$. Do we have that$$\int g\,d\mu_n \to \int_0^1 g\,dx?$$

Answer 1

Since $\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$ for all continuous bounded funtions $f$, we know that $\mu_n$ converges weakly to $\lambda|_{[0,1]}$. Now it follows directly from the portmanteau theorem that $$\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$$ for all bounded measurable functions $f:[0,1] \to \mathbb{R}$ with $\lambda(D_f)=0$ where $D_f$ denotes the set of discontinuity points of $f$.

For a proof see e.g. this article, Theorem 1.20.