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Let $E\subset \R^n$ be a measurable set. We define the essential boundary of $E$ as$$ \partial^* E := \{x\in\R^n\colon 0 < | E\cap B_\rho(x)| < |B_\rho(x)|,\quad \forall \rho>0\}$$ where $|\cdot|$ is the Lebesgue measure.
Compare the definition of $\partial^* E$ with the definition of the topological boundary $\partial E$ which can be written as$$ \partial E = \{ x \in \R^n \colon \emptyset \subsetneq E\cap B_\rho(x) \subsetneq B_\rho(x),\quad \forall \rho>0\}.$$ Hence one clearly has $\partial^* E\subset \partial E$ .
Notice that the essential boundary does not depend on the Lebesgue representative of the set $E$ , in the sense that if $|E\triangle F|=0$ then $\partial^* E = \partial ^* F$ . For example if $E=\mathbf Q^n\subset \R^n$ is the set of points with rational coordinates, one has $\partial^* E=\emptyset$ while $\partial E=\R^n$ .
Nevertheless one can easily prove that $\partial^*E$ is always a closed set (in the usual sense).
Moreover one has $\mathcal H^{n-1}(\partial^*E\setminus \mathcal F E)=0$ (where $\mathcal F E$ is the reduced boundary and $\mathcal H^{n-1}$ is the $(n-1)$ -dimensional Hausdorff measure) and hence a unit normal vector is defined in $\mathcal H^{n-1}$ -a.e. $x\in\partial^* E$ .
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