$\liminf\limits_{n\to\infty} \int fd\mu_n\ge \int fd\mu$? If I have weak convergence of measures $\mu_n\rightharpoonup\mu$ then by definition, for a continuous and bounded function we have $\lim_n \int fd\mu_n=\int fd\mu$ but if we just have that $f$ is continuous and $f\ge 0$, do we have $\liminf_n \int fd\mu_n\ge \int fd\mu$? Can we get rid of the assumptions $f\ge 0$ or $f$ continuous? Any hints are welcomed.
 A: This is more than what you asked but the following  result is seldom taught nowadays in mathematics courses in probability, but still taught in engineering courses in probability (OR, and EE), and  I think it is important to know.
Given a metric space $(S,d)$, denote by $L_b(S)$ the collection of all lower semicontinuous functions that are bounded fro0m below. There is a well know result that states that

Theorem:  Let $(S,d)$ be a metric space.  For any net $\{\mu_\alpha:\alpha\in D\}\subset\mathcal{M}^+(S)$ and  $\mu\in \mathcal{M}^+(S)$, $\mu_\alpha\Rightarrow\mu$ if and only if
$$\begin{align}
\liminf_\alpha\int f\,d\mu_\alpha\geq \int f\,d\mu
\end{align}$$
for all $f\in L_b(S)$.

Proof: Suppose that $\mu_\alpha\Rightarrow\mu$ and let $g\in L_b(S)$ with $g\geq c$.  Then, there is a sequence $g_k$ of bounded Lipschitz functions  such that $c\leq g_k\leq g_{k+1}\nearrow g$. Hence,  for each $k$
$$
\liminf_\alpha\int g\,d\mu_\alpha\geq \liminf_\alpha \int g_k\,d\mu_\alpha =\int g_k\,d\mu.
$$
As $\mu(S)<\infty$,  $\liminf_\alpha\int g\,d\mu_\alpha\geq \int g\,d\mu$ by monotone convergence.
Conversely, suppose $f\in\mathcal{C}_b(S)$. Since  $\mathcal{C}_b(S)\subset L_b(S)$, both $f$ and $-f$ are in $L_b(S)$, so
$$\begin{align}
\liminf_\alpha\int f\,d\mu_\alpha&\geq \int f\,d\mu\\
\liminf_\alpha\int -f\,d\mu_\alpha&\geq \int -f\,d\mu
\end{align}$$
Therefore, $\lim_\alpha \int f\,d\mu_\alpha =\int f\,d\mu$.

In your case, $f\geq0$ continuous, implies that $f\in L_b(S)$.

In my proof of the result, to following result (see for example: Bertsekas, D. P., Shreve, S. E., Optimal Control: The Discrete Time Case, Academic Press, NY, 1978 or Ash, R. Real Analysis and Probability, Academic  Press, San Diego, 1972.) is the key

Lemma: Suppose $(S,d)$ is a metric space. Let $f\in\mathbb{R}^S$ be such that  $-\infty<\inf_{x\in S}f(x)=b$. The function  $f$ is lower semicontinuous if and only if there is a sequence of
bounded Lipschitz continuous functions $f_k$ such that $\inf_{k,x}\{f_k(x)\}\geq b$ an  $f_k\nearrow f$ pointwise.

Proof: Sufficiency is clear since continuous functions are lower semicontinuous, and so is  the  supremum of lower semicontinuous functions.
It suffices to assume that $f\geq0$. For each
$t\geq0$ define
$$g_t(x)=\inf_z\{f(z)+td(x,z)\}$$
Clearly  $0\leq g_s\leq g_t$ whenever $s<t$, and
$g_t(x)\leq f(x)+td(x,x)=f(x)$. Notice
that for all  $x,y\in S$, $f(z)+td(x,z)\leq f(z)+td(y,z)+td(x,y)$;
consequently,  $g_t(x)\leq g_t(y)+td(x,y)$. By symmetry, we obtain
$|g_t(x)-g_t(y)|\leq td(x,y)$, which means that
each $g_t$ is Liptschitz continuous. If  $h=\lim_ng_n$, then
$0\leq h\leq f$. We will show that $h=f$. To that purpose, fix $x\in
S$ and let
$\varepsilon>0$. For each
$n\in\mathbb{N}$, there is $z_n\in S$ such that
$$\begin{align}
g_n(x)+\varepsilon>f(z_n)+nd(x,z_n)\geq nd(x,z_n)\tag{1}\label{lscineq}
\end{align}$$
Since $f(x)\geq g_n(x)$, it follows that $f(x)+\varepsilon>nd(x,z_n)$
for all $n$; hence,  $z_n$ converges to $x$. Since $f$ is lower
semicontinuous, there is $N$ such that for $n\geq N$,
$f(x)-\varepsilon<f(z_n)$. For such $n$, we obtain from \eqref{lscineq} that $g_n(x)>f(x)-2\varepsilon$. Letting $n\rightarrow\infty$ and then
$\varepsilon\searrow0$ shows that  $h=f$. To conclude, notice that
$\{f_n:=g_n\wedge n: n\in\mathbb{Z}_+\}$ is an increasing sequence of nonnegative bounded Lipschitz--continuous functions which converges to $f$.
A: If $f \geq 0$ is continuous, then $g_m := f \wedge m$ is bounded and continuous and $g_m \uparrow f$. By weak convergence and the monotone convergence theorem,
$$\int f d\mu_n \geq \int g_m d\mu_n \to_{n \to \infty} \int g_m d\mu \to_{m \to \infty} \int fd\mu,$$
and the desired result follows by taking the $\liminf$ on the left-hand side.
