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In Federer's Theorem, $ \mathcal{H}^{n-1} (\partial ^{m}E \setminus \partial ^{*}E)=0 $, where $E$ is a set of finite perimeter in $ \mathbb R^n $, $\partial ^{e}E$ is the essential boundary of E, and $\partial ^{*}E$ is the reduced boundary of E.

From Maggi's book Prop. 12.19, We know that $\operatorname{spt}(\mu _E) = \{x \in \mathbb R^n \mid 0<|E \cap B(x,r)|< \omega_n r^n, \forall \ r > 0\}$, and by definition, $\partial ^{e}E \subseteq \operatorname{spt} \mu _E$

Now here comes my question, is it true that $ \mathcal{H}^{n-1} (\operatorname{spt} \mu _E \setminus \partial ^{*}E)=0 ?$

I thought about this question for several days, but I could not prove it nor could I give a counterexample. I'm lack of some pictures in mind.

Any idea would be really appreciated.

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  • $\begingroup$ One might want to add that $\mu_E$ denotes $D\mathbf{1}_E$, i.e. the vector-valued Radon measure that is the distributional gradient of the characteristic function $\mathbf{1}_E\in\mathrm{BV}(\mathbb{R}^n)$. $\endgroup$
    – Manfred Sauter
    Commented Mar 30, 2018 at 13:49

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In Maggi's book, Example 12.25, there you have an open set of finite perimeter $E$ in $\mathbb{R}^2$ with $|\text{spt}\, \mu_E|>0$. Since $H^{n-1}(\partial^* E)$ is finite, you have $|\partial^* E|=0$. Thus $|(\text{spt}\, \mu_E)\setminus \partial^*E|>0$, so $H^{n-1}((\text{spt}\, \mu_E)\setminus \partial^*E)= + \infty$

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