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?
 A: 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 $f$.
A: Yes, it is.
Step 1 We have $\mu(A)=0$ for any $A \in \mathcal{B}(\mathbb{R})$ such that $$d(A,[-1,1]) := \inf_{\substack{x \in A \\ y \in [-1,1]}} |x-y|>0.$$ Indeed: Pick a cut-off function $\chi_k \in C_b$ satisfying $\chi(x)=0$ for $|x| \leq 1$, $\chi(x)=1$ for $|x| > 1+1/k$. Since $f_n(x)=0$ for $x \in \mathbb{R} \backslash [-1,1]$ the weak convergence implies
$$\int \chi_k(x) \, d\mu(x) = \lim_{n \to \infty} \int \chi_k(x) \cdot f_n(x) \, dx =0$$
Hence,
$$\mu(A) = \int 1_A \, d\mu \leq \int \chi_k(x) \, d\mu(x) =0$$
for any $K$ such that $d(A,[-1,1]) > 1/k$.
Step 2 By the Radon-Nikodym Theorem, $\mu$ is absolutely continous with respect to the Lebesgue measure $\lambda$, if and only if, $\lambda(A)=0$ implies $\mu(A)=0$ for each $A \in \mathcal{B}(\mathbb{R})$. Since the first part of the proof implies in particular $\mu(\mathbb{R} \backslash [-2,2])=0$, it suffices to show $\lambda(A)=0$ implies $\mu(A)=0$ for $A \subseteq [-2,2]$.
Let $A \subseteq [-2,2]$ be open. We define
$$A_k := \{x \in \mathbb{R}; |x| \leq 2; d(x,A^c) \geq 1/k\}, \qquad k \in \mathbb{N}$$
Obviously, $A_k$ is compact and $A_k \uparrow A$. It is not difficult to show that there exists a sequence $(g_k)_{k \in \mathbb{N}} \subseteq C_C$ such that $1_{A_k} \leq g_k \leq 1_A$ and $g_k \uparrow 1_A$. Therefore, we find by the monotone convergence theorem and the weak convergence
$$\mu(A) = \sup_{k \in \mathbb{N}} \int g_k \, d\mu = \sup_{k \in \mathbb{N}} \limsup_{n \to \infty} \int g_k(x) \cdot f_n(x) \, dx$$
Since $0 \leq f_n \leq M$ and $g_k \uparrow 1_A$, we find
$$\mu(A) \leq \limsup_{n \to \infty} \int 1_A(x) \cdot f_n(x) \, dx \leq M \cdot \lambda(A) \tag{1}$$
Recall that a finite measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is outer regular, i.e.
$$\lambda(B) = \inf\{\lambda(A); A \supseteq B, A \, \text{open}\} \tag{2}$$
By $(1)$ we find that for any $B \in \mathcal{B}(\mathbb{R})$ and $A \supseteq B$ open,
$$\mu(B) \leq \mu(A) \leq M \cdot \lambda(A)$$
Consequently, $(2)$ entails $\mu(B)=0$ for any set $B$ satisfying $\lambda(B)=0$.
Remark Note that the assumption on the uniform boundedness is crucial, see for instance this question.
