Does weak convergence $\mu_n \Rightarrow \mu$ imply convergence of $\int g \, d \mu_n$ for $g$ unbounded (under additional assumptions)?

Question

Let $E$ be a Polish space and let $\mu_n$, $\mu$ be finite non-negative measures on $E$ such that $\mu_n \Rightarrow \mu$, i.e. such that we have the convergence $ \int f \, d \mu_n \to \int f \, d \mu$ for every $f \colon E \to \mathbb{R}$ continuous and bounded. Assume that there exists a continuous function $f \colon E \to [0,\infty)$ (not necessarily bounded) such that $|g(x)| \leqslant f(x)$ for all $x \in E$ and $$ \int f \, d \mu_n \to \int f \, d \mu.$$ The question is: do we have the convergence $$ \int g \, d \mu_n \to \int g \, d \mu?$$ My intuition is that this is false but I could not come up with a counter example. My attempts so far have been to consider $\mu_n = (1-1/n) \delta_{0} + 1/n \delta _{a_n}$ with $a_n \uparrow \infty$ such that $a_n /n \to 0$ and $f(x) = |x|, g(x) = x \sin(x)$ but the convergence holds in this case.

Answer 1

(I assume that $\mu$ and each $\mu_n$ have ben normalized to have mass $1$, and that $g$ is continuous.) We do. Google "Skorokhod's representation theorem" to learn that under the condition you have set, there is a probability space $(\Omega,\mathcal F,\Bbb P)$ and random variables $X, X_1,X_2,\ldots$ defined thereon such that (i) $\mu_n$ is the distribution of $X_n$, (ii) $\mu$ is the distribution of $X$, and (iii) $\lim_nX_n(\omega) = X(\omega)$ for each $\omega\in\Omega$. Consequently, $\int g\,d\mu_n=\Bbb E[g(X_n)]$, etc. The desired conclusion is now seen to follow from a standard variant of the Dominated Convergence Theorem.