weak-star limit of a sequence of integrable functions is an integrable function Let $(f_n)_{n\geq1}\subset L^1(\mathbb{R}^d,\mathbb{C})$ be a sequence of Lebesgue integrable functions, and let $g\in L^1(\mathbb{R}^d)$ be a non-negative integrable function be such that
\begin{equation}
\vert f_n(x)\vert\leq g(x)
\end{equation}
holds for every $x\in\mathbb{R}^d$ and every $n\geq 1$.
Since the sequence $(f_n)_n$ is bounded in $L^1(\mathbb{R}^d)\subset \mathcal{M}(\mathbb{R}^d)=C_{0}(\mathbb{R}^d)^*$, there is a finite complex measure $\mu\in\mathcal{M}(\mathbb{R}^d)$ and a subsequence $(n_k)_k$ such that $f_{n_k}\overset{\ast}{\rightharpoonup} \mu$ as $k\rightarrow\infty$, i.e. such that
$$\lim_{k\to\infty}\int_{\mathbb{R}^d}f_{n_k}(x)\varphi(x)dx = \int\varphi(x)d\mu(x)$$
holds for every $\varphi\in C_{0}(\mathbb{R}^d)$, continuous function vanishing at infinity.
How can we deduce in this particular case that $\mu$ is absolutely continuous with respect to Lebesgue measure, or equivalently that that $\mu=fdx$ for some $f\in L^1(\mathbb{R}^{d})$ ?
 A: As I mentioned in my comments, this the statement in the OP follows from the Dunford-Pettis theorem.
First notice that the conditions of the OP imply that the sequence $\{f_n:n\in\mathbb{N}\}$ is uniformly integrable. Indeed,
$$\sup_n\|f_n\|_1\leq \|g\|_1<\infty,$$
and for every $\varepsilon>0$, if $\int_Ag<\varepsilon$, then
$$\sup_n\int_A|f_n|<\varepsilon$$
As the Lebesgue measure is $\sigma$-finite, the Dunford-Pettis theorem implies that $K$ is $\sigma(L_1,L_\infty)$-compact. The Eberlein-Smulain theorem in turn implies that $K$ is sequentially $\sigma(L_1,L_\infty)$ compact. Thus, every subsequence $\{f_n'\}$ of $\{f_n:n\in\mathbb{N}\}$ has a subsequence $\{f_{n''}\}$ that converges to some $f\in L_1$ in in the $\sigma(L_1,L_\infty)$-topology. In particular, for every $g\in C_0(\mathbb{R})$,
$$\lim_{n''}\int f_{n''}g=\int fg$$

There are three key parts to the statement made above.

*

*For any measure space $(\Omega,\mathscr{A},\mu)$, recall that a family $\mathcal{F}\subset L_1(\mu)$ is uniformly integrable if
$$\inf_{g\in L^+_1(\mu)}\sup{f\in\mathcal{F}}\int(|f|-g)_+\,d\mu=0$$
It is not difficult to show that if
$$\sup_{f\in\mathcal{F}}\|f\|_1<\infty,$$
and if there is a $g\in L^+_1(\mu)$ such that for very $\varepsilon>0$ there is $\delta>0$ such that
$$\int_Ag<\delta\qquad\text{implies}\qquad\sup_{f\in\mathcal{F}}\int_A|f|\,d\mu<\varepsilon$$
then $\mathcal{F}$ is uniformly integrable.


*The $\sigma$-finite version of the Dunford-Pettis theorem, which can be proved with slight modifications to the proof of the finite version.

Theorem: (Dunford--Pettis) Suppose $(\Omega,\mathscr{F},\mu)$ is $\sigma$--finite. A subset $K\subset L_1$ is $\sigma(L_1,L_\infty)$-relatively compact iff $K$ is uniformly integrable.



*The Eberlein-Smulain theorem which gives an equivalence between weak compactness and weak sequential compactness


Theorem: Let $X$ be a Banach space with dual space $X^*$. A set $K\subset X$ is $\sigma(X,X^*)$-compact iff  any sequence in $K$ has a $\sigma(X,X^*)$-weakly convergent subsequence in $K$.

