Does weak convergence with uniformly bounded densities imply absolute continuity of the limit? Suppose $(f_n)$ is a sequence of probability density functions on $\mathbb{R}$ such that 
\begin{align*}
f_n &\leq M \text{ for all }n \\
f_n(x) &= 0 \text{ for all } |x| > 1
\end{align*}
Suppose also that there is a Borel probability measure $\mu$ on $\mathbb{R}$ such that
$$
\lim_{n \rightarrow \infty} \int gf_n dx = \int g d\mu \quad \forall g \in C_b(\mathbb{R}),
$$
where $C_b(\mathbb{R})$ is the set of all bounded, continuous, complex-valued functions on $\mathbb{R}$.
(In other words, suppose the measures $\mu_n$ defined by $d\mu_n = f_n dx$ converge to $\mu$ weakly.)
Question: Is $\mu$ absolutely continuous?

One may also prove this via functional analysis. Indeed, for $g \in C^{\infty}([-1,1])$, let us define $$Q(g):= \int g d\mu.$$
By the given condition and Cauchy-Schwarz, we have that $$|Q(g)| = \lim_{n \to \infty} \big| \langle f_n,g \rangle_{L^2} \big|\leq \limsup_{n \to \infty} \|f_n\|_{L^2} \|g\|_{L^2} \leq 2^{1/2}M \|g\|_{L^2},$$ so that $Q$ can be extended to a bounded linear operator from $L^2[-1,1] \to \Bbb R$. By the Riesz representation theorem, we know any bounded linear functional on $L^2$ is given by $\langle f, \cdot \rangle$ for some $f$. Thus there exists $f \in L^2[-1,1]$ such that for all $g \in L^2[-1,1]$ we have $$Q(g) = \int fg$$which then implies that $\mu$ is absolutely continuous with density $f$.
