Let $\Omega$ be a bounded and smooth domain. Suppose I have a sequence of non-negative functions $u\_n \in L^\infty((0,1)\times \Omega) \cap L^\infty((0,1);L^\infty(\Omega))$ with
$$0 \leq u\_n \leq 1 \quad \text{a.e}$$
and
$$u\_n \rightharpoonup^\* u \quad\text{in $L^\infty((0,1)\times \Omega)$}$$
to some $u$. 


Is it possible to conclude that for a subsequence, $u\_{n\_j}(t) \rightharpoonup^\* u(t)$ in $L^\infty(\Omega)$ for a.e. $t$? The subsequence $\{n\_j\}$ should not depend on the point $t$.


Additional info:


* For each $n$, $u\_n$ is a piecewise constant function, i.e., $u\_n = \sum\_{k=1}^n a\_{kn}\chi\_{I\_{kn}}(t)$ holds for a partition $\{I\_{kn}\}$ and $a\_{kn} \in L^\infty(\Omega)$. 


	+ The sequence $u\_n(t)$ is bounded uniformly in $L^\infty(\Omega)$ for a.e. $t$.



Weak-* convergence in $L^\infty((0,T)\times\Omega)$ implies weak-* convergence in $L^\infty(\Omega)$ for a.e. $t \in (0,T)$?
If I understood correctly the question, this is not true. If $u\_n(t,x)=sin(2πnt)f(x)$ for a fixed $f\neq 0$, then $u\_n \to 0$, $w^\*$ in $(t,x)$ by Riemann Lebesgue, but for fixed $t\neq 0,1/2,1$ does not converge $w^\*$ in $x$. Similarly, there is no way to find a sequence $n\_j$ such that $sin(n\_jt)$ converges a.e. Positivity does not help: take f positive and add 1 to the sinus. 


