Show that $\sup_{f\in\mathcal{F}}|\int f(x)d\mu_n-\int f(x)d\mu|\to 0$, for $f\in\mathcal{F}$ equicontinuous, $\mu_n\rightharpoonup\mu$ on compact $X$ If $\mu_n\rightharpoonup\mu$ weakly on a compact metric space $X$, and if $\mathcal{F}$ is a family of equi-continuous functions, show that $\sup_{f\in\mathcal{F}}|\int f(x)d\mu_n-\int f(x)d\mu|\to 0$.
Here are some points that I can get to so far:
I should use Arzela-Ascoli, but I have to show that $\mathcal{F}$ is piecewise bounded, so that it is compact in $C(X)$ under the uniform norm.
Then I argue by contradiction that there exists some $f_n\in\mathcal{F}$ s.t. $|\int f_n(x)d\mu_n-\int f(x)d\mu|\ge\epsilon$. But by compactness of $\mathcal{F}$ and Arzela-Ascoli, there is a uniform convergent subsequence $f_{n_i}\to f$ in uniform norm, which will provide a contradiction.
Here are some points that I am not sure:

*

*I have not used the assumption that $\mu_n$ converges to $\mu$ weakly.


*Does this result still hold if $X$ is complete and separable?
If someone can provide a rather complete version of the proof or point out where I am wrong, I will be grateful. Thanks!
 A: When $X$ is a separable metric space and $\mathcal{F}$ is equicontinuous and uniformly bounded, this is a Theorem of R. Rao.

Theorem: Suppose $(S,d)$ is a separable metric space. Asume $\{\mu_\alpha:\alpha\in D\}$ is a net of positive measures in $\mathscr{B}(X)$ that converges weakly to $\mu$ (another positive Borel measure). If $\mathcal{F}$ is uniformly bounded and equicontinuous, then
$$\lim_\alpha\,\sup\Big\{\Big|\int f d(\mu_\alpha-\mu)\Big|:f\in\mathcal{F}\Big\}=0
$$


*

*Billingsley's book Convergence in Measure has a proof based on the Portmanteau theorem.


*Bogachev's book Measure Theory, Vol. 2 has the version of the theorem as quoted above.

Edit: Without uniform boundedness (besides equicontinuity) the statetment of the OP may fail.  Here is a counterexample:
Suppose $\mu_n=(1+\frac{1}{n})\mu$, where $\mu$ is a positive Borel measure on a compact set $X$ with $0<\mu(X)$. Clearly $\mu\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu_n$.  The collection $\mathcal{F}$ of all positive constant functions is equicontinuous but not bounded. Then $\sup_{f\in\mathcal{F}}|\mu_nf-\mu f|=\frac{1}{n}\mu(X)\sup_{f\in F}|f|=\infty$.

Here is a proof that looks at nets (sequences will be enough if the measures are all probabilities) that I wrote in some paper:
The assumptions on $\mathcal{F}$ imply that $M:=\sup_{f\in\mathcal{F}}\|f\|_u<\infty$;   for any $x\in X$ and $\varepsilon>0$ there is an open ball  $B_x$ centered at $x$ such that $\mu(\partial B_x)=0$ and $|f(x)-f(y)|<\varepsilon$ for all $y\in B_x$ and $f\in \mathcal{F}$. Since $X$ is separable, $X=\bigcup_{n\in\mathbb{N}}B_{x_n}$ for some countable subcollection of balls. Set $A_1=B_{x_1}$, and $A_n=B_{x_n}\setminus\bigcup^{n-1}_{j=1}A_j$ for $n>1$. It follows that $\{A_n:n\in\mathbb{N}\}$ is a pairwise disjoint collection of Borel sets covering $X$ with $\mu(\partial A_n)=0$ for all $n\in\mathbb{N}$. Define
\begin{align*}
  \nu:=\sum_n\mu(A_n)\delta_{x_n},\qquad
  \nu_\alpha:=\sum_n\mu_\alpha(A_n)\delta_{x_n}
\end{align*}
For any $\delta>0$, there is $N\in\mathbb{N}$ large enough such that $\mu\Big(X\setminus\bigcup^N_{n=1}A_n\Big)<\delta$. Since $\partial\Big(X\setminus\bigcup^N_{n=1}A_n\Big)\subset\bigcup^N_{n=1}\partial A_n$, we have that $\lim_\alpha\mu_\alpha\Big(X\setminus\bigcup^N_{n=1}A_n\Big)=\mu\Big(X\setminus\bigcup^N_{n=1}A_n\Big)$. Hence, for any $f\in\mathcal{F}$
\begin{align*}
 \left| \int_X f\,d(\nu_\alpha - \nu)\right| &\leq M \left(\sum^N_{n=1}\Big| \mu_\alpha(A_n)-\mu(A_n)\Big|  + \sum_{n>N}\Big| \mu_\alpha(A_n)-\mu(A_n)\Big|\right)\\
  &\leq M \sum^N_{n=1}|\mu_\alpha(A_n)-\mu(A_n)|+ M\left(\mu_\alpha\Big((X\setminus\bigcup^N_{n=1}A_n\Big) + \mu\Big((X\setminus\bigcup^N_{n=1}A_n\Big)\right)
\end{align*}
Passing to the  limit we obtain that
$\limsup\limits_\alpha\sup_{f\in\mathcal{F}} \left| \int_X f\,d(\nu_\alpha -\nu)\right| \leq 2M\delta$. As $\delta$ may be arbitrarily small, we conclude that
$$
\lim\limits_\alpha\sup\limits_{f\in\mathcal{F}} \left| \int_X f\,d(\nu_\alpha - \nu)\right|=0.\tag{1}\label{discrete-approx-mu}
$$
Since $A_n\subset B_{x_n}$, for any $f\in\mathcal{F}$
\begin{aligned}
  \left|\int_X f\,d(\mu_\alpha-\mu)\right| &\leq \left|\int_X f\,d(\mu_\alpha-\nu_\alpha)\right| + \left|\int_X f\,d(\nu_\alpha-\nu)\right| + \left|\int_X f\,d(\nu-\mu)\right|\\
  &\leq \sum_n\int_{A_n}|f(x)-f(x_n)|(\mu_\alpha+\mu)(dx) +  \left|\int_X f\,d(\nu_\alpha-\nu)\right|\\
  &\leq\varepsilon\big(\mu_\alpha(X)+\mu(X)\big) +  \sup_{f\in\mathcal{F}}\left|\int_X f\,d(\nu_\alpha-\nu)\right|.\label{pre-unif-convergence}
\end{aligned}
From $\eqref{discrete-approx-mu}$ and the fact that $\lim_\alpha\mu_\alpha(X)=\mu(X)$ we get that
$\limsup_\alpha\sup\limits_{f\in\mathcal{F}}\left|\int_X f\,d(\mu_\alpha-\mu)\right|\leq 2\varepsilon\mu(X)$. The conclusion follows by letting $\varepsilon\rightarrow0$.
