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. 



Is it true that $ \mathcal{H}^{n-1} (\operatorname{spt} \mu _E \setminus \partial ^{*}E)=0$?
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 $H^{n-1}((\text{spt}\, \mu\_E)\setminus \partial^\*E)= + \infty$


