Is it true that $L^p(\Omega) = L^p(\bar {\Omega})$, where let us say $\Omega$ is a bounded domain of $\mathbb{R}^n$ with smooth boundary? I think it is true, because $\partial \bar {\Omega}$ has measure zero. Am I correct?
2026-05-05 01:17:44.1777943864
Is $L^p(\Omega) = L^p(\bar {\Omega})$?
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