Generally we know that $W^{1,p}(\Omega)$ is bigger then $W^{1,p}_0(\Omega)$ for arbitrary $\Omega\subset \mathbb R^N$ and also we have $W_0^{1,p}(\mathbb R^N)=W^{1,p}(\mathbb R^N)$. Today I found on H. Brezis book a remark states that for domain $\Omega$, if $\mathbb R^N\setminus \Omega$ is "sufficiently thin" and $p<N$, then $W_0^{1,p}(\Omega)= W^{1,p}(\Omega)$.
What does he means by "sufficiently thin"? And how we prove this remark? I don't have an idea to get start...
Also, it gives an example. By defining $\Omega:=\mathbb R^N\setminus\{0\}$ and $N\geq 2$, one can prove that $H_0^1(\Omega)=H^1(\Omega)$.
I tried to prove this example by using approximation. But I think, since $C_c^\infty(\Omega)\neq C^\infty (\Omega)$, why should we have $H_0^1(\Omega)=H^1(\Omega)$?
Any help is really welcome. Thank you!