$\DeclareMathOperator\Int{Int}\DeclareMathOperator\Ext{Ext}$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 $\partial^\*E=\partial \tilde{E}$?



If the measure theoretic boundary is closed must it coincide with the topological boundary?
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}


