Convergence of Radon Nikodym derivatives I apologise in advance if my question is too basic.
Some notation:


*

*$(X,\cal{X})$  denotes a measurable  metric space
where $X$ is a metric space  and
$\cal{X}$ is the associated Borel sigma algebra.

*$B(X)$ is the space of all bounded continuous 
functions defined on $X$.
Let $\{\mu_n\}$ and $\{\nu_n\}$ be sequences of 
probability measures on the above  measurable  space $(X, \mathcal{X})$.
Assume 
that 
each $\mu_n$ is absolutely continuous with respect to 
$\nu_n$, with an density $h_n\in B(X)$.
Suppose that $\mu_n\to \mu$, $\nu_n\to \nu$ 
in the weak  star topology and $h_n$ converges 
to a bounded continuous function $h$.
Question:
 I would like to know if $\mu\ll\nu$. If so, is $h$ the density?  If not, is there some condition in order 
to have $\mu\ll\nu$?
Other information that can be useful
 is that each $\nu_n$  and $\mu_n$ has support in a 
compact subset $K_n\subset X$ which 
increase to $X$, i.e, $X=\bigcup K_n$.
Edit: $h_n$ converges to $h$ uniformly in compacts sets 
 A: I'm going to assume that your space is locally compact (as well as $\sigma$-compact), so that $X$ is the union of a sequence of compact sets where each lies in the interior of the next.
In this case, the answer to your question is yes.
Fix any $g \in B(X)$ with $g \geq 0$. We need $\int_X g \, d\mu_n \to \int_X gh \, d\nu$.
Fix $\varepsilon>0$. Let $M_g:=\sup_{x\in X} g(x)$ and likewise $M_h:=\sup_{x\in X} h(x)$. Let $K \subset X$ be a compact set with $K^\circ$ sufficiently large that
$$ \mu(X \setminus K^\circ) < \frac{\varepsilon}{4M_g} \hspace{4mm} \textrm{and} \hspace{4mm} \nu(X \setminus K^\circ) < \frac{\varepsilon}{4M_gM_h}. $$
Let $N \in \mathbb{N}$ be such that for all $n \geq N$,
$$ \mu_n(X \setminus K^\circ) < \frac{\varepsilon}{4M_g} \ , \hspace{4mm} \max_{x \in K} |h_n(x)-h(x)| < \frac{\varepsilon}{4M_g}  \ , \hspace{4mm} \left| \int_{K^\circ} gh \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| < \frac{\varepsilon}{4}. $$
The third statement is possible since $\nu(\partial K)<\frac{\varepsilon}{4M_gM_h}$ and so $\int_{\partial K} gh \, d\nu < \frac{\varepsilon}{4}$. (This reasoning can be seen by adapting the argument for 3,4$\Rightarrow$5 on p3 of here.)
Then for all $n \geq N$, we have that
\begin{align*}
\Bigg| \int_X g \, d\mu_n & - \int_X gh \, d\nu \Bigg| \\
&\leq \ \left| \int_{K^\circ} gh_n \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \int_{X \setminus K^\circ} g \, d\mu_n \ + \ \int_{X \setminus K^\circ} gh \, d\nu \\
&< \ \left| \int_{K^\circ} gh_n \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \frac{\varepsilon}{4M_g}M_g \ + \ \frac{\varepsilon}{4M_gM_h}M_gM_h \\
&\leq \ \left| \int_{K^\circ} g.\!(h_n - h) \, d\nu_n \right| \ + \ \left| \int_{K^\circ} gh \, d\nu_n - \int_{K^\circ} gh \, d\nu \right| \ + \ \frac{\varepsilon}{2} \\
&< \ M_g.\max_{x \in K} |h_n(x)-h(x)| \ + \ \frac{3\varepsilon}{4} \\
&< \ \varepsilon.
\end{align*}
