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?
Best Answer
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.