"Let (u\_j) be a bounded sequence from $W^{1,p}(\Omega)$ how to prove that there exists a subsequence such that $u\_j\rightharpoonup u$ in $W^{1,p}\_0(\Omega)$ and $|\nabla u\_j|\rightharpoonup d\mu,$ $|u\_j|^{p^\*}\rightharpoonup d\nu$ weakly\* in the sense of measures."



Question about measure lemma?
Assuming $1 < p < \infty$, $W^{1,p}(\Omega)$ is reflexive, so bounded sets are weakly precompact by Alaoglu's theorem (the weak-\* and weak topologies coincide). Thus $u\_j$ has a subsequence converging weakly to some $u \in W^{1,p}(\Omega)$. Now $W^{1,p}\_0(\Omega)$ is convex and strongly closed in $W^{1,p}(\Omega)$, hence also weakly closed, and therefore $u \in W^{1,p}\_0(\Omega)$.


If $u\_j$ is bounded in $W^{1,p}$ then by definition $|\nabla u\_j|^p$ is bounded in $L^1$. So the measures $|\nabla u\_j|^p \,dm$ are bounded in total variation, i.e. norm-bounded as elements of $C\_0(\Omega)^\*$, which by Riesz representation is the space of Radon measures on $\Omega$. By Alaoglu again, the sequence is weak-\* compact and so a subsequence converges weak-\* to a measure $\mu$.


Likewise, by Sobolev embedding, $u\_j$ is bounded in $L^{p^\*}(\Omega)$ so $|u\_j|^{p^\*}$ is bounded in $L^1(\Omega)$ and a similar argument applies.


