Weak*-convergence to 0 on L^\infty and convergence almost everywhere I am stuck with something standard...
Let $f_n \in L^\infty(\mathbb{R}^d)\cap L^1(\mathbb{R}^d)$, $n\geq1$, be such that
$$
\sup_{n\geq1} \|f_n\|_{L^\infty(\mathbb{R}^d)}<\infty
$$
and
$$
\lim_{n\to\infty} \int_{\mathbb{R}^d}|f_n(x)-f(x)| g(x)\,dx=0, \qquad \text{for all } 0\leq g\in L^1(\mathbb{R}^d).
$$
Is it correct that $f_n(x)\to f(x)$ for a.a. $x\in\mathbb{R}^d$ ?
It can be reformulated as follows: if a sequence from $L^\infty$ converges to 0 in the weak-* topology, does it converge to 0 almost everywhere?
(Update: in the reformulation, I omitted that $f_n\in L^1$ itself, not sure whether this is relevant.)

I think it is not true in general. 
Further more you cannot control all elements of $L^{\infty,*}$ using $L^1$ because $L^1$ it is not reflexive.
Finally you can state that if $(f_n)_n$ converges weak* only if you allow to take a subsequence of it because the unitary ball is weak* compact. ($L^{\infty,*}=L^1$).
