Remark below the proof of Rellich-Kondrachov compactness theorem The following if a remark left at the end of the proof of the RK compactness theorem p.274 Evans.

Here I think I got through everything more or less, but have trouble proving the final claim
$$ W_0^{1,p}(U)\subset\subset L^q(U)$$
even if $\partial U$ is not $C^1$.
Appreciate for anyone can help.
 A: You can let $U'$ be an open set containing $U$ so that $\partial U'$ is $C^1$. Any function $u \in W^{1,p}_0(U)$ extends naturally to a function in $W^{1,p}_0(U')$ without changing its norm: just define it to be zero outside $U$. Use the fact that $W_0^{1,p}(U')\subset\subset L^q(U')$ to conclude that a bounded sequence in $W^{1,p}_0(U)$, once extended, has a limit (along a subsequence) in $L^p(U')$. Standard techniques will show you that the limit vanishes almost everywhere outside of $U$ so that the convergence is in fact in $L^q(U)$.

Added in response to comment: If $\{u_k\}$ is a bounded sequence in $W_0^{1,p}(U)$, extend the domain of each $u_k$ to $U'$ by defining $u_k(x) = 0$ if $x \in U' \setminus U$. Then there is a subsequence $\{u_{k_j}\}$ that converges in $L^q(U')$ to some function $v$.  Then
$$\int_{U' \setminus U} |v|^q \, dx = \int_{U' \setminus U} |v - u_{k_j}|^q \, dx \le \int_{U'} |v - u_{k_j}|^q \, dx$$ for all $j$. Since the last term tends toward $0$ you conclude that $$\int_{U' \setminus U} |v|^q \, dx = 0$$ so that $v$ vanishes (a.e.) outside $U$.
