A sequence of probability measures $\{\mu_n\}$ converges strongly if for each measurable set $A$
$$\mu_{n}(A) \rightarrow \mu(A)$$
(notice that this convergence is different than the total variation)
when does this convergence imply
$$\int f \,d\mu_n\to \int f \,d\mu $$
for every bounded measurable function? I'm interested in the case in which the space is not Polish.
Best Answer
$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{abs}[1]{\left\lvert{#1}\right\rvert}$It always does. I'll use the notation $\mu(f):=\int f\,d\mu$ throughout. Essentially, you can use the fact that all the $\mu$-s are probabilities on $X$ (or the slightly weaker fact that $M=\sup_{n\in\Bbb N}\mu_n(X)$ and $\mu(X)$ are finite).
The result is obvious for simple functions. Now, recall that every bounded measurable function is uniform limit of simple functions. Namely, $$f_n:=\sum_{k=-2n}^{2n-1} \frac kn\cdot 1_{f^{-1}\left[\frac kn,\frac{k+1}n\right)}$$ satisfies $\nrm{f-f_n}_\infty\le\frac1n$ for all $n>\sup_{x\in X}\abs{f(x)}$.
Finally, for any fixed $\delta>0$ consider some simple function $g$ such that $\nrm{f-g}_\infty<\delta$ and observe that, for all $n$, $$\abs{\mu(f)-\mu_n(f)}\le\abs{\mu(f-g)}+\abs{\mu(g)-\mu_n(g)}+\abs{\mu_n(g-f)}\le\\\le (\mu(X)+\mu_n(X))\nrm{f-g}_\infty+\mu_n(g-f)\le (1+M)\delta+\abs{\mu(g)-\mu_n(g)}$$
The last quantity is a sequence converging to $(1+M)\delta$. Therefore, for all $\delta>0$ $$\limsup\limits_{n\to\infty}\abs{\mu(f)-\mu_n(f)}\le (1+M)\delta$$
Therefore $\limsup\limits_{n\to\infty}\abs{\mu(f)-\mu_n(f)}=0$.