# Convergence of Radon Nikodym derivatives

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

• In what sense does $h_n \to h$? Pointwise convergence is not enough. Mar 6, 2017 at 19:27
• This will work as soon as you can guarantee that $h_n\, d\nu_n$ is close to $h\, d\nu_n$, and this will be the case under mild additional assumptions (for example, $\| h_n -h\|_{L^1(d\nu_n)}\to 0$ would give you that these measures are in fact close in norm). Mar 6, 2017 at 19:35
• @Nate Eldredge the convergence is locally compact Mar 6, 2017 at 20:44
• This is a very nicely formulated question: although it initially appears as though some of your assumptions should be redundant, I have had to use every single one for my proof below (except the continuity of $h_n$, and the added statement about the measures having compact support). Mar 7, 2017 at 6:43

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.
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*}
• I suspect it's not actually needed, but I used it to be able to get a compact set $K$ whose interior has measure arbitrarily close to $1$. With only $\sigma$-compactness, I might only be able to get $K$ itself to have measure arbitrarily close to $1$. Mar 9, 2017 at 18:25