Let $\Omega \subset \mathbb{R}^N$ be an open and bounded domain with $C^1$ boundary . It is well known that $W^{1,2}(\Omega)$ is compactly embedded in $L^{p}(\Omega)$ for any $1 \leq p < 2^*$. However, in $\mathbb{R}^N$, this embedding is never compact. I would like to know if there's some result like this for sets with finite Lebesgue measure. Actually, my professor said that $W^{1,2}(\Omega)$ is compactly embedded in $L^{s}(\Omega)$ for $2 \leq p < 2^*$, if $\Omega$ has finite Lebesgue measure.
What I tried: Denote by $\mathcal{L}^N$ the $N-$dimensional Lebesgue measure. Let $\Omega \subset \mathbb{R}^N$ such that $\mathcal{L}^N(\Omega) < \infty$. From Folland, Real Analysis: Modern Techniques and Their Applications, theorem 2.40, p. 70, we have $$ \mathcal{L}^N(\Omega) = \sup\{\mathcal{L}^N(K) : K \subset \Omega \text{ is compact}\}. $$ So, given $\delta > 0$, there's a compact $K_\delta \subset \Omega$ such that $\mathcal{L}^N(\Omega) - \delta < \mathcal{L}^N(K_\delta)$. So, $$ \mathcal{L}^N(\Omega) - \mathcal{L}^N(K_\delta) < \delta. $$ Yet, note that $$ \mathcal{L}^N(\Omega) = \mathcal{L}^N(\Omega\backslash K_\delta \cup K_\delta) = \mathcal{L}^N(\Omega\backslash K_\delta) + \mathcal{L}^N(K_\delta) \implies \mathcal{L}^N(\Omega\backslash K_\delta) = \mathcal{L}^N(\Omega) - \mathcal{L}^N(K_\delta) $$ Therefore, $\mathcal{L}^N(\Omega\backslash K_\delta) < \delta$.
As I want to show that $W^{1,2}(\Omega)$ is compactly embedded in $L^{s}(\Omega)$ for $2 \leq p < 2^*$ and I already know that $W^{1,2}(\Omega)$ is continuously embedded in $L^{s}(\Omega)$ for $2 \leq p \leq 2^*$, by an interpolation argument it is enough to prove that $W^{1,2}(\Omega)$ is compactly embedded in $L^{2}(\Omega)$. So, let $\{u_n\}_{n \in \mathbb{N}}$ a bounded sequence in $W^{1,2}(\Omega)$. I have to show that this sequence has a convergente subsequence in $L^2(\Omega)$. As $W^{1,2}(\Omega)$ is a reflexive Banach space, there's $u \in W^{1,2}(\Omega)$ such that $u_n \rightharpoonup u$ in $W^{1,2}(\Omega)$.
In general, from Folland, Corollary 3.6, p. 89, if $f \in L^1(\Omega)$ given $\epsilon > 0$, there's an $\delta > 0$ (depending on f, I guees) such that $|\int_{E} f(x) dx| < \epsilon$ whenever $\mathcal{L}^N(E) < \delta$.
As $u_n, u \in L^2(\Omega)$, we have $f_n := |u_n - u|^2 \in L^{1}(\Omega)$. So, given $\epsilon > 0$, there's some $\delta_n > 0$ such that $|\int_{E} f_n(x) dx| < \epsilon$ whenever $\mathcal{L}^N(E) < \delta_n$. On the other hand, by the initial discussion, given $\delta_n > 0$, there's a compact $K_n \subset \Omega$ such that $\mathcal{L}^N(\Omega\backslash K_n) < \delta_n$. Then, $|\int_{\Omega\backslash K_n} f_n| < \epsilon$. Therefore, $$ \int_{\Omega} |u_n - u|^2 dx = \int_{\Omega} f_n dx = \int_{\Omega\backslash K_n} f_n dx + \int_{K_n} f_n dx < \epsilon + \int_{K_n} f_n dx. $$ But how to control the term $\int_{K_n} f_n dx$ ? I know how to control $\int_{B} |u_n - u|^2 dx$ for a ball $B$ fixed, as $W^{1,2}(B)$ is compactly embedded into $L^2(B)$.
Any hints, modification of my argument or references would be very welcome :-)