I apologise in advance if my question is too basic.


Some notation:


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



Convergence of Radon Nikodym derivatives
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](http://www.ma.utexas.edu/users/gordanz/notes/weak.pdf).)


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\*}


