If the measure theoretic boundary is closed must it coincide with the topological boundary? $\DeclareMathOperator\Int{Int}\DeclareMathOperator\Ext{Ext}$Suppose $E\subset\mathbb{R}^n$ is a set of finite perimeter and suppose that the measure theoretic boundary $\partial^*E=\mathbb{R}^n\setminus(\Int(E)\cup \Ext(E))$ is closed, where $\Int(E)=\{x\in\mathbb{R}^n\,:\, \lim_{r\to 0}\frac{\mathscr{L}^n(E\cap B_r(x))}{\omega_n r^n}=1\}$ and $\Ext(E)=\{x\in\mathbb{R}^n\,:\,\lim_{r\to 0}\frac{\mathscr{L}^n(E\cap B_r(x))}{\omega_n r^n}=0\}$. Is it true that then there is a Borel set $\tilde{E}$ with $\mathscr{L}^n(\tilde{E}\mathbin\Delta E)=0$ and $\partial^*E=\partial \tilde{E}$?
 A: Yes, this is essentially true, if you allow the conclusion to hold up to sets of measure zero. You can even choose $\widetilde{E} $ to be open. See Lemma 6.2 of De Rosa, Kolasiński, and Santilli - Uniqueness of Critical Points of the Anisotropic Isoperimetric Problem for Finite Perimeter Sets.
A: This is indeed true; you can find the answer to your question in Francesco Maggi's book for example, by combining Proposition 12.9 and Remark 15.3. I'll paraphrase the argument.
Let $E \subset \mathbf{R}^n$ be a Caccioppoli set, and $\mu_E$ be its Gauss–Green measure. Then $\partial^* E \subset \operatorname{spt} \mu_E$, and as the support is closed, one has
\begin{equation}
\overline{\partial^* E} = \operatorname{spt} \mu_E.
\end{equation}
Now, said proposition gives the existence of a Borel set $\tilde{E}$ so that $\mathcal{H}^n(E \triangle \tilde{E}) = 0$ and $\partial \tilde{E} = \operatorname{spt} \mu_{\tilde{E}}$. But then
\begin{equation}
\partial \tilde{E} = \operatorname{spt} \mu_{\tilde{E}} = \operatorname{spt} \mu_E = \overline{\partial^* E} = \partial^* E.
\end{equation}
A: I have found that an answer to my question is contained in Maggi's book "Sets of finite perimeter and geometric variational problems" at page 127 (Proposition $12.19$). This is in the same spirit of the result Longyearbyen referred to in his answer but is more precise since it gives a true equality, not one that holds up to a set of measure zero (using that, as proven in Remark $15.3$ of the same book, $\overline{\partial^*E}=spt(\mu_{ E})$ ).
