$L^p_0(\Omega)\cap L^p(\Omega)$ dense in $L^p(\Omega)$ when $m(\Omega)=\infty$? Let $1\leq p < \infty$ and $\Omega\subset\mathbb{R}^n$ be a measurable (Lebesgue) set. I know that $L_0^p(\Omega)\cap L^p(\Omega)$ is dense in $L^p(\Omega)$ when $m(\Omega)$ is finite. For the proof I used the absolute continuity of the integral and the fact that $\Omega$ can be approximated with a compact set.
Now I'm wondering if the claim is true when $m(\Omega)=\infty$. Obviously, I cannot use the same proof that I used in the case where $m(\Omega)<\infty$. If it is true, could you give a hint from where to start?

Sure it is. Since $\mathbb{R}^n$ is $\sigma$-finite. Intersect your $\Omega$ with sets of the form $\{n<|x|<n+1\}$ and approximate on each annulus with error at most $\epsilon/2^n$, then you will see the total error will be less than $\epsilon$. 
