Let $X\subset\mathbb{R}^d$ be compact.
Given sequences of real-vauled random variables $f_n\to f$
and positive radon measures $\mu_n\to \mu$ both converging weakly for $n\to\infty$.
Under which further conditions can we deduce that
$$\lim_{n\to\infty}\int_X f_nd\mu_n =\int_X f d\mu ?$$
Best Answer
Claim: the conclusion holds for every sequence $(\mu_n)$ converging weakly to $\mu$ iff $f_n \to f$ uniformly.
Since $\mu$ is Radon and $X$ is compact, $\mu$ is a finite measure. Since $\mu_n(X) \to \mu(X)$ it follows that $sup_n \mu_n(X)<\infty$. So, if $f_n \to f$ uniformly then $\int f_n d\mu_n -\int fd\mu_n \to 0$ from which the conclusion follow easily.
Now suppose the conclusion holds for every sequence $(\mu_n)$ converging weakly to $\mu$. Let $x_n \to x$ and $\mu_n=\delta_{x_n}, \mu =\delta_x$. Then $\mu_n \to \mu$ weakly so $f(x_n)=\int f_n d\mu_n \to \int f d\mu=f(x)$. Since $X$ is compact the statement $f(x_n) \to f(x)$ whenever $x_n \to x$ is equivalent to uniform convergence of $f_n$ to $f$.