Suppose that we know that
$ \int{ |f_{n}(x) - f(x)| d\mu(x)} \longrightarrow 0 \qquad (1) $
for every probability measure $\mu \in \mathcal{A}$ in a certain class.
Also, suppose that $\{\nu_{n}\}$ is a sequence of probability measures that converges weakly to $\nu$. Does the convergence of $ \int{ |f_{n}(x) - f(x)| d\mu(x)} \longrightarrow 0 $ for every $\mu \in \mathcal{A}$, imply that (or are there some conditions under which this implies that) $\int{ f_{n}(x) d\nu_{n}(x)} \longrightarrow \int{f(x) d\nu(x)}$?
I believe that it is already known that if $f_{n}(x)$ converges uniformly to $f(x)$ then $\int{ f_{n}(x) d\nu_{n}(x)} \longrightarrow \int{f(x) d\nu(x)}$ but I don't know if the convergence in (1) implies this.
Let $f_n$ be defined on $[0,1]$ with graph consisting of the polygon with vertices $(0,0)$, $(1/n,1)$, $(2/n,0)$, $(1,0)$, and let $f(x) = 0$. For every Borel probability measure $\mu$ on $[0,1]$, $ \int f_n(x) \; d\mu(x) \le \mu((0,2/n)) \to 0$ as $n \to \infty$. But if $\nu_n$ is the point mass at $1/n$, $\nu_n$ converges weakly to the point mass at $0$, while $\int f_n(x) \; d\nu_n(x) = 1$.