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

  • 3
    $\begingroup$ In what sense does $h_n \to h$? Pointwise convergence is not enough. $\endgroup$ Mar 6, 2017 at 19:27
  • $\begingroup$ 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). $\endgroup$ Mar 6, 2017 at 19:35
  • $\begingroup$ @Nate Eldredge the convergence is locally compact $\endgroup$
    – Eduardo
    Mar 6, 2017 at 20:44
  • 1
    $\begingroup$ 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). $\endgroup$ Mar 7, 2017 at 6:43

1 Answer 1


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*}

  • $\begingroup$ Sorry for the late response, thank you for the really nice the answer. Could you tell me where did you use the locally compact hypotesis? $\endgroup$
    – Eduardo
    Mar 9, 2017 at 18:05
  • $\begingroup$ 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 $1$. $\endgroup$ Mar 9, 2017 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.