When is $L^2(\Omega) = L^2(\bar \Omega)$?

Question

Let $\Omega \subset \mathbb{R}^n$ be a subset.

Is it always true that whenever $L^2(\Omega)$ is well defined that $L^2(\Omega) = L^2(\bar \Omega)$, where $\bar \Omega$, where $\bar \Omega$ means the closure of $\Omega$?

It seems true when eg. $\Omega$ is a smooth bounded domain since its closure just includes its boundary which is a null set. However, what if $\Omega$ is not as nice?

Answer 1

This is true if and only if $\bar\Omega \setminus \Omega$ has measure zero. One extreme case is the following: $\Omega$ is a countable, dense subset. Then, $$\{0\} = L^2(\Omega) \ne L^2(\bar\Omega) = L^2(\mathbb R^n).$$