Suppose $\Omega\in \mathbb{R}^n$ is open and bounded and $f_n\in L^1(\Omega)$ converge weakly* (as signed measures) to some $f\in L^1$ and pointwise almost everywhere to some $g\in L^1$. Is $f=g$ almost everywhere?
By Egorev we have that there is some subset $I\subset \Omega$ with $|\Omega\backslash I|<\epsilon$ such that a subsequence of the $f_n$ converges uniformly to $g$ when restricted to $I$ and hence also in the norm topology. Therefore it would be enough to show that $\chi_If_n$ still converges weakly* to $\chi_If$, as we then have that $f=g$ on $I$ and by taking a sequence of $\epsilon$'s that go to $0$ and applying this to the corresponding $I$'s, the equality most hold a.e.
The answer is no - this surprises me. On $[0,1]$:
Say $E_n\subset[0,1]$, $0<m(E_n)<1$, and $\sum m(E_n)<\infty$. Let $$f_n(x)=\begin{cases} c_n,&(x\in[0,1]\setminus E_n), \\1,&(x\in E_n)\end{cases},$$where $c_n<0$ is chosen so that $\int_0^1 f_ n=0$. Since $\sum m(E_n)<\infty$ it's clear that $$f_n\to1\, \text{a.e.}$$
BUT now if we require that $E_n$ be the union of $n$ small intervals uniformly distributed in $[0,1]$ we obtain $$\int_{(k-1)/n}^{k/n}f_n=0\quad(k=1,2,\dots,n);$$since $||f_n||_1\le1$ and every $\phi\in C([0,1])$ is uniformly continuous it's not hard to see that $f_n\to0$ weak-*.
Details: Suppose $\phi\in C([0,1])$. Let $\epsilon>0$; choose $\delta>0$ so $|x-y|<\delta$ implies $|\phi(x)-\phi(y)|<\epsilon$. If $n>1/\delta$ then $$\int_0^1\phi(t)f_n(t)=\sum_{k=1}^n\int_{(k-1)/n}^{k/n}(\phi(t)-\phi(k/n))f_n(t)\,dt,$$so $$\left|\int_0^1\phi f_n\right|\le\sum\int_{(k-1)/n}^{k/n}\epsilon|f_n|=\epsilon||f_n||_1\le\epsilon.$$