3
$\begingroup$

Let $(\mu_n)$ be a sequence of probability measures on $\mathbb{R}$ that converges weakly to a probability measure $\mu$ on $\mathbb{R}$. So $$ \lim \int hd\mu_n = \int h d\mu $$ whenever $h$ is a bounded, continuous function defined on $\mathbb{R}$.

Suppose each $\mu_n$ is absolutely continuous with respect to Lebesgue measure $\lambda$. So there are $f_n \in L^1(\lambda)$ with $f_n \geq 0$ such that $$ \int h d\mu_n = \int h f_n d\lambda $$ whenever $h$ is Lebesgue measurable and non-negative.

What are some techniques for determining whether $\mu$ is absolutely continuous with respect to Lebesgue measure?

For example, what information about $(\mu_n)$ would be sufficient to conclude that $\mu$ is absolutely continuous?

$\endgroup$
6
$\begingroup$

In general, the weak limit is not absolutely continuous with respect to the Lebesgue measure, see for instance this example. Roughly speaking, the sequence $(f_n)_{n \in \mathbb{N}}$ of probability densities should not explode and move to much mass to $\pm \infty$.

A sufficient condition is the following: Suppose that there exists a constant $M>0$ and a compact set $K \subseteq \mathbb{R}$ such that

$$|f_n(x)| \leq M \qquad \text{for all} \, x \in \mathbb{R}, n \in \mathbb{N}$$

and

$$f_n(x) = 0 \qquad \text{for all} \, x \in \mathbb{R} \backslash K, n \in \mathbb{N}$$

Then $\mu$ is absolutely continuous with respect to the Lebesgue measure $\lambda$. A proof can be found here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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