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For $1\le p<\infty$, the Lebesgue space $L^p(X)$ consists of equivalence classes modulo equality almost everywhere of measurable functions satisfyingThe space $L^{\infty }(X)$ consists of essentially bounded measurable functions modulo equality almost everywhere, with norm

Let $1\le p<q<\infty$ and put $r=q/p>1$. By Hölder's inequality,Taking $p$th roots givesFor $q=\infty$, directly integrating $|f{|}^p\le \Vert f{\Vert }_{\infty }^p$ gives the same endpoint estimate. Hence $L^q(X)\subseteq L^p(X)$ when $\mu (X)<\infty$, as in Lp inclusion on a finite measure space.

The boundedness in the Sobolev space $H^1(U)$ and the weak sequential compactness of bounded sequences in a reflexive Banach space give a subsequence converging weakly to some $u\in H^1(U)$. For each integer $m\ge 2$, the Rellich-Kondrachov compactness theorem makes $H^1(U)\hookrightarrow L^m(U)$ compact because the dimension is two. Repeated extraction followed by the diagonal argument gives one subsequence $u_{n_k}$ converging strongly to $u$ in every $L^m(U)$ with integral $m\ge 2$.

For any finite real $p\ge 1$, choose an integer $m\ge \max \{2,p\}$. Since $U$ has finite measure, the Lp inclusion on a finite measure space givesThe same subsequence therefore works for every finite $p$.