Does weak convergence in $L^1$ imply pointwise convergence

Question

Let $u_n$ be a sequence that convergence weakly in $L^1(\Omega)$ to a functions $u$, where $\Omega$ is a subset of $\mathbb{R}^n$. Is it true that
$$u_n(x) \rightarrow u, \ \ \hbox{for} \ \ a.e. x\in \Omega?$$

Answer 1

No. You can even find a sequence $(u_n)$ that converges to $u$ in $L^1$ but s.t. $u_n(x)\to u(x)$ for no $x$. The classical example is $$u_0=\boldsymbol 1_{[0,1/2]},\quad u_1=\boldsymbol 1_{[1/2,1]},\quad u_2=\boldsymbol 1_{[0,1/4]},\quad u_3=\boldsymbol 1_{[1/4,1/2]},$$ $$u_4=\boldsymbol 1_{[1/2,3/4]},\quad u_{5}=\boldsymbol 1_{[3/4,1]},\quad u_6=\boldsymbol 1_{[0,1/8]},\quad u_7=\boldsymbol 1_{[1/8,1/4]},...$$