An mixed weak star convergence problem Let $\Omega\subset \mathbb R^N$ open bounded. Given a sequence of Radon measure $(\mu_n)$ such that $\mu_n\to \mu$ in weak star sense in $\mathcal M_b(\Omega)$ and $\|\mu_n\|\nearrow \|\mu\|$. Also given a function $v\geq 1$ such that $v\in L^1_{\text{loc}}(\Omega)$ and $l.s.c$. Suppose $v$ has properties that there exists a Lipschitz continuous sequence $v_n$ such that $1\leq v_n\leq v$ and $v_n\nearrow v$ for all $x\in\Omega$. Note that $v$ may not bounded above.
Assume there exists a function $u\in C(\Omega)$ such that $u/v\in C_c(\Omega)$.
My question: do we have 
$$
\lim_{n\to\infty}\int_\Omega \frac{u}{v_n}\,d\mu_n = \int_\Omega \frac{u}{v}\,d\mu
$$
My try:
Writing 
$$
\int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu = \int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu_n+\int_\Omega \frac{u}{v}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu
$$
The last two goes to 0 by the definition of weak star convergence. But I don't know how to deal with first two. I was trying to use dominated convergence but it is not obvious... 

I realize that since $u/v\in C_c(\Omega)$ and $v$ is finite a.e., it makes $u$ has compact support and hence $u$, $u/v_n\in C_c(\Omega)$ since $u\in C(\Omega)$. Now, I could write few more steps...
\begin{align*}
\left|\int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu_n\right| &= \left|\int_\Omega u(1/v_n-1/v)\,d\mu_n\right|\\
&=\left|\int_\Omega u(1/v_n-1/v)\,d(\mu_n-\mu+\mu)\right|\\
&\leq \left|\int_\Omega u(1/v_n-1/v)\,d(\mu_n-\mu)\right|+\left|\int_\Omega u(1/v_n-1/v)\,d\mu\right|
\end{align*}
The last one can be done by dominated convergence. But the first one...Maybe I should use the fact that $\|\mu_n\|\to\|\mu\|$?
I know generally I should not hope 
$$
\lim_{n\to\infty}\int_\Omega \frac{u}{v_n}\,d\mu_n = \int_\Omega \frac{u}{v}\,d\mu
$$
since it would require that $u/v_n\to u/v$ uniformly which I don't have. But since I in additional have $0\leq 1/v\leq 1/v_n\leq 1$ and $\|\mu_n\|\to \|\mu\|$, I may expect my result is true.
 A: We will use the version of Fatou's Lemma given here (https://en.wikipedia.org/wiki/Fatou%27s_lemma#Fatou.27s_Lemma_with_Varying_Measures). This states that if we have $\mu_n (A) \to \mu(A)$ for all measurable $A$ and if $(f_n)_n$ is a sequence of nonnegative measurable functions, then
$$
\int \liminf_n f_n \, d\mu \leq \liminf_n \int f_n \, d\mu_n .
$$
In your case, this is satisfied, since you assume $\mu_n \to \mu$ weak star with respect to $M_b$ (and every indicator function is in $M_b$).
But this statement implies a nice version of the dominated convergence theorem: If we have $|f_n| \leq g$ with $\int g \, d\mu_n \to \int g \, d\mu< \infty$ (for example for every bounded measurable function $g$) and $f_n \to f$ pointwise, then $\int f_n \, d\mu_n \to \int f \, d\mu$. Indeed, let $h_n := 2g + |f_n -f|$. Then $h_n \geq 0$ and $h_n \to 2g$ pointwise, so that Fatou's lemma from above yields
$$
\int 2g \, d\mu \leq \liminf_n \int h_n \, d\mu_n = \int 2g \, d\mu - \limsup_n \int |f_n - f| \, d\mu_n 
$$
and thus
$$
\lim_n \int |f_n - f|\, d\mu_n = 0.
$$
Furthermore, if we apply Fatou's lemma with $h_n = g \pm f$ (note $h_n \geq 0$), we get
$$
\int g \pm f \, d\mu \leq \liminf_n \int g \pm f \, d\mu_n  = \int g \, d\mu \pm \lim^\ast \int f_n \, d\mu_n,
$$
with $\lim^\ast = \liminf$ for $\pm = +$ and $\lim^\ast = \limsup$ for $\pm = -$. Hence,
$$
 \int f \, d\mu \leq \liminf \int f_n \, d\mu_n   \leq \limsup \int f_n \, d\mu_n \leq \int f \, d\mu,
$$
i.e. $\inf f_n \, d\mu_n \to \int \int f \, d\mu$. Together with the above, this implies
$$
\int f_n \, d\mu_n \to \int f\, d\mu.
$$
This is the version of the dominated convergence theorem that we will use.
Now, it remains to note (since $v_n \geq 1$) that
$$
\bigg| \frac{u}{v_n}\bigg| \leq \Vert u \Vert_\sup \cdot \chi_K =: g,
$$
where $K$ is the support of $u$. Since all $\mu_n$ are Radon, we have $\int g \, d\mu_n \to \int g \, d\mu < \infty$ ($g$ is bounded measurable). Furthermore, we have pointwise convergence. Now apply the dominated convergence theorem from above.
