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?
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2I think also. If $f \in L^p(\overline \Omega)$ of course $f$ restricts to $L^p(\Omega)$. Conversely, we can choose a representant which is $0$ on $\partial \Omega$. It's a smooth manifold of dimension $n-1$, in particular of measure zero. (Unless I miss something...) – Jul 10 '15 at 08:13
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@N.H. sorry. I missed the smooth manifold part. You are right. – user251257 Jul 10 '15 at 09:23