[proofplan] The strategy interposes the measure-theoretic boundary $\partial_m E$ between $\partial E$ and $\partial^* E$, then shows that each pair differs by an $\mathcal{H}^{n-1}$-null set. The measure-theoretic boundary $\partial_m E := \{x : \limsup_{r \to 0} \mathcal{L}^n(E \cap B(x,r))/\mathcal{L}^n(B(x,r)) > 0 \text{ and } \limsup_{r \to 0} \mathcal{L}^n(B(x,r) \setminus E)/\mathcal{L}^n(B(x,r)) > 0\}$ collects precisely those points where $E$ has neither density $0$ nor density $1$. The first step shows $\mathcal{H}^{n-1}(\partial E \setminus \partial_m E) = 0$ by a density-and-covering argument: a point of $\partial E$ outside $\partial_m E$ must have density $0$ or $1$, but the set of such points cannot have positive $\mathcal{H}^{n-1}$-measure within $\partial E$ for a set of finite perimeter, by a Vitali covering combined with the relative isoperimetric inequality. The second step shows $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0$ as a consequence of [De Giorgi's Structure Theorem](/theorems/599) — the perimeter measure $|D\mathbb{1}_E|$ is concentrated on $\partial^* E$ and gives full $\mathcal{H}^{n-1}$-measure to $\partial_m E$. Combining the two gives $\mathcal{H}^{n-1}(\partial E \setminus \partial^* E) = 0$. [/proofplan] [step:Introduce the measure-theoretic boundary $\partial_m E$ and show $\partial^* E \subseteq \partial_m E \subseteq \partial E$] Define the upper density of $E$ at $x$ by \begin{align*} \Theta^*(E, x) &:= \limsup_{r \to 0} \frac{\mathcal{L}^n(E \cap B(x,r))}{\mathcal{L}^n(B(x,r))}, \end{align*} and the upper density of the complement $E^c := \mathbb{R}^n \setminus E$ analogously by $\Theta^*(E^c, x)$. Define the *measure-theoretic boundary* \begin{align*} \partial_m E := \{ x \in \mathbb{R}^n : \Theta^*(E, x) > 0 \text{ and } \Theta^*(E^c, x) > 0 \}. \end{align*} We verify the inclusion $\partial^* E \subseteq \partial_m E$. Fix $x \in \partial^* E$. By the [Blow-up Convergence at Reduced Boundary Points](/theorems/3117), $\mathbb{1}_{E_r} \to \mathbb{1}_H$ in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$ as $r \to 0$, where $E_r = (E - x)/r$ and $H = \{y \in \mathbb{R}^n : y \cdot \nu_E(x) \le 0\}$ is the closed half-space with outer normal $\nu_E(x)$. In particular, evaluating in the unit ball, \begin{align*} \frac{\mathcal{L}^n(E \cap B(x,r))}{\mathcal{L}^n(B(x,r))} = \frac{\mathcal{L}^n(E_r \cap B(0,1))}{\mathcal{L}^n(B(0,1))} \to \frac{\mathcal{L}^n(H \cap B(0,1))}{\mathcal{L}^n(B(0,1))} = \frac{1}{2}, \end{align*} so $\Theta^*(E, x) = 1/2 > 0$ and similarly $\Theta^*(E^c, x) = 1/2 > 0$, giving $x \in \partial_m E$. We verify the inclusion $\partial_m E \subseteq \partial E$. If $x \notin \partial E$, then either $x$ lies in the interior of $E$ (so $B(x, \rho) \subseteq E$ for some $\rho > 0$, hence $\Theta^*(E^c, x) = 0$) or $x$ lies in the interior of $E^c$ (so $\Theta^*(E, x) = 0$); in both cases $x \notin \partial_m E$. Hence $\partial_m E \subseteq \partial E$. Combining, $\partial^* E \subseteq \partial_m E \subseteq \partial E$, and therefore \begin{align*} \partial E \setminus \partial^* E = (\partial E \setminus \partial_m E) \cup (\partial_m E \setminus \partial^* E). \end{align*} It suffices to show each piece on the right has zero $\mathcal{H}^{n-1}$-measure. Steps 2 and 3 do this. [guided] We introduce a third boundary notion that interpolates between $\partial^* E$ and $\partial E$: the *measure-theoretic boundary* $\partial_m E$, which records precisely where $E$ has neither density $0$ nor density $1$. *Definition (upper density and measure-theoretic boundary).* The upper density of a measurable set $E$ at $x$ is \begin{align*} \Theta^*(E, x) := \limsup_{r \to 0} \frac{\mathcal{L}^n(E \cap B(x,r))}{\mathcal{L}^n(B(x,r))} \in [0, 1]. \end{align*} Setting $E^c := \mathbb{R}^n \setminus E$, the *measure-theoretic boundary* of $E$ is \begin{align*} \partial_m E := \{ x \in \mathbb{R}^n : \Theta^*(E, x) > 0 \text{ and } \Theta^*(E^c, x) > 0 \}. \end{align*} Geometrically, $x \in \partial_m E$ when $E$ has positive upper density at $x$ from both sides. *Why this notion is useful.* The reduced boundary $\partial^* E$ is defined via a sharp limit (existence of the unit normal $\nu_E$). The measure-theoretic boundary is defined via the much weaker condition of positive upper density on both sides. So $\partial_m E$ is a strictly larger set, and if we can sandwich $\partial E$ between $\partial^* E$ and $\partial_m E$, the chain \begin{align*} \partial^* E \subseteq \partial_m E \subseteq \partial E \end{align*} reduces the proof to two separate estimates: $\mathcal{H}^{n-1}(\partial E \setminus \partial_m E) = 0$ and $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0$. We now verify the chain. *Inclusion $\partial^* E \subseteq \partial_m E$.* Fix $x \in \partial^* E$. By [Blow-up Convergence at Reduced Boundary Points](/theorems/3117) (whose hypothesis "$E$ has finite perimeter and $x \in \partial^* E$" matches our assumption), the rescaled indicators $\mathbb{1}_{E_r}$ — where $E_r := (E - x)/r$ — converge in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$ to $\mathbb{1}_H$, with $H = \{y : y \cdot \nu_E(x) \le 0\}$ the half-space with outer normal $\nu_E(x)$. Specialising the $L^1_{\mathrm{loc}}$-convergence to the unit ball $B(0,1)$: \begin{align*} \mathcal{L}^n(E_r \cap B(0,1)) \to \mathcal{L}^n(H \cap B(0,1)) = \tfrac{1}{2} \mathcal{L}^n(B(0,1)), \end{align*} where the limiting value $\tfrac{1}{2}\mathcal{L}^n(B(0,1))$ comes from the fact that $H$ is a half-space, hence cuts $B(0,1)$ exactly in half. Translating back via $E_r \cap B(0,1) = (E - x)/r \cap B(0,1)$, which has $\mathcal{L}^n$-measure $r^{-n} \mathcal{L}^n(E \cap B(x,r))$: \begin{align*} \frac{\mathcal{L}^n(E \cap B(x,r))}{\mathcal{L}^n(B(x,r))} \to \frac{1}{2} \quad \text{as } r \to 0. \end{align*} Hence $\Theta^*(E, x) = 1/2 > 0$. The analogous computation applied to $E^c$ — whose blow-up at $x$ converges to $H^c$, the complementary closed half-space, again of relative measure $1/2$ in $B(0,1)$ — gives $\Theta^*(E^c, x) = 1/2 > 0$. Both density conditions hold, so $x \in \partial_m E$. *Inclusion $\partial_m E \subseteq \partial E$.* Fix $x \notin \partial E$. The complement of $\partial E$ is the union of the interior $E^\circ$ and the exterior $(E^c)^\circ$: - If $x \in E^\circ$, there exists $\rho > 0$ with $B(x, \rho) \subseteq E$. Then for $r \le \rho$, $B(x,r) \cap E^c = \varnothing$, so $\mathcal{L}^n(E^c \cap B(x,r)) = 0$ and $\Theta^*(E^c, x) = 0$. - If $x \in (E^c)^\circ$, the symmetric argument gives $\Theta^*(E, x) = 0$. In either case, one of the two density conditions defining $\partial_m E$ fails, so $x \notin \partial_m E$. Hence $\partial_m E \subseteq \partial E$. *Decomposition.* Combining the two inclusions, $\partial^* E \subseteq \partial_m E \subseteq \partial E$, so we can decompose the symmetric difference: \begin{align*} \partial E \setminus \partial^* E = (\partial E \setminus \partial_m E) \cup (\partial_m E \setminus \partial^* E). \end{align*} The two pieces are estimated separately in Steps 2 and 3. The first piece (Step 2) consists of points of $\partial E$ where $E$ has density $0$ or density $1$ — the boundary points where $E$ looks like its interior or exterior in the measure-theoretic sense. The second piece (Step 3) consists of points where the densities are positive from both sides but where the perimeter measure does not register a sharp normal direction. [/guided] [/step] [step:Show $\mathcal{H}^{n-1}(\partial E \setminus \partial_m E) = 0$ by Lebesgue differentiation and the relative isoperimetric inequality] A point $x \in \partial E \setminus \partial_m E$ satisfies $\Theta^*(E, x) = 0$ or $\Theta^*(E^c, x) = 0$. Define \begin{align*} A_0 := \{ x \in \partial E : \Theta^*(E, x) = 0 \}, \qquad A_1 := \{ x \in \partial E : \Theta^*(E^c, x) = 0 \}. \end{align*} Then $\partial E \setminus \partial_m E \subseteq A_0 \cup A_1$, and by symmetry (replacing $E$ with $E^c$, which has the same perimeter and the same topological boundary) it suffices to show $\mathcal{H}^{n-1}(A_0) = 0$. [claim:For every $x \in A_0$, the perimeter density satisfies $\liminf_{r \to 0} P(E; B(x,r))/r^{n-1} = 0$] By definition of $A_0$, $\lim_{r \to 0} \mathcal{L}^n(E \cap B(x,r))/\mathcal{L}^n(B(x,r)) = 0$ (the limit exists and equals $0$ because $\Theta^*(E,x) = 0$ forces both upper and lower limits to be $0$). The [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600), in its relative form, states \begin{align*} \min\bigl( \mathcal{L}^n(E \cap B(x,r)), \mathcal{L}^n(B(x,r) \setminus E) \bigr)^{(n-1)/n} \le C(n) \, P(E; B(x,r)) \end{align*} for every ball $B(x,r) \subset \mathbb{R}^n$, with $C(n) > 0$ depending only on $n$. Since $\mathcal{L}^n(E \cap B(x,r))/\mathcal{L}^n(B(x,r)) \to 0$ but $\mathcal{L}^n(B(x,r) \setminus E)/\mathcal{L}^n(B(x,r)) \to 1$, for all sufficiently small $r$ the minimum is $\mathcal{L}^n(E \cap B(x,r))$, and we obtain \begin{align*} \mathcal{L}^n(E \cap B(x,r))^{(n-1)/n} \le C(n) \, P(E; B(x,r)). \end{align*} On the other hand, $x \in \partial E$ means every ball $B(x,r)$ meets both $E$ and $E^c$ in a set of positive Lebesgue measure (taking small open subsets), so $\mathcal{L}^n(E \cap B(x,r)) > 0$ for every $r > 0$. Combined with $\Theta^*(E,x) = 0$ this implies $\mathcal{L}^n(E \cap B(x,r)) = o(r^n)$ as $r \to 0$. Substituting into the relative isoperimetric inequality, \begin{align*} \frac{P(E; B(x,r))}{r^{n-1}} \ge \frac{1}{C(n)} \cdot \frac{\mathcal{L}^n(E \cap B(x,r))^{(n-1)/n}}{r^{n-1}} = \frac{1}{C(n)} \left(\frac{\mathcal{L}^n(E \cap B(x,r))}{r^n}\right)^{(n-1)/n}. \end{align*} The right-hand side tends to $0$ as $r \to 0$, so $\liminf_{r \to 0} P(E; B(x,r))/r^{n-1} = 0$. [/claim] [proof] The relative isoperimetric inequality bound from [theorem 600](/theorems/600) reads \begin{align*} \min\bigl( \mathcal{L}^n(E \cap B(x,r)), \mathcal{L}^n(B(x,r) \setminus E) \bigr)^{(n-1)/n} \le C(n) \, P(E; B(x,r)). \end{align*} For $x \in A_0$ and $r$ small, $\mathcal{L}^n(E \cap B(x,r)) < \mathcal{L}^n(B(x,r) \setminus E)$, so the minimum is the first term. Dividing by $r^{n-1}$ and using $\mathcal{L}^n(B(x,r)) = \alpha_n r^n$ with $\alpha_n := \mathcal{L}^n(B(0,1))$: \begin{align*} \frac{P(E; B(x,r))}{r^{n-1}} \ge \frac{1}{C(n)} \cdot \frac{\mathcal{L}^n(E \cap B(x,r))^{(n-1)/n}}{r^{n-1}} = \frac{\alpha_n^{(n-1)/n}}{C(n)} \left(\frac{\mathcal{L}^n(E \cap B(x,r))}{\mathcal{L}^n(B(x,r))}\right)^{(n-1)/n}. \end{align*} The right-hand side tends to $0$ as $r \to 0$ since $\Theta^*(E, x) = 0$ forces the density inside the parentheses to vanish. Hence $\liminf_{r \to 0} P(E; B(x,r))/r^{n-1} = 0$. [/proof] The claim says: at every $x \in A_0$, the perimeter density $P(E; B(x,r))/r^{n-1}$ has lower limit $0$ as $r \to 0$. Equivalently, identifying $P(E; \cdot) = |D\mathbb{1}_E|(\cdot)$ as a Radon measure on $\mathbb{R}^n$, the lower $(n-1)$-density of $|D\mathbb{1}_E|$ at $x$ vanishes. By the standard density bound for Hausdorff measure (a finite Radon measure $\mu$ on $\mathbb{R}^n$ assigns no $\mathcal{H}^{n-1}$-mass to its set of zero lower $(n-1)$-density), $\mathcal{H}^{n-1}(A_0) = 0$. This bound is the content of the lemma: if $\mu$ is a finite Radon measure on $\mathbb{R}^n$ and $S \subseteq \mathbb{R}^n$ is a Borel set with $\liminf_{r \to 0} \mu(B(x,r))/r^{n-1} = 0$ for every $x \in S$, then $\mathcal{H}^{n-1}(S) = 0$ — a Vitali covering argument applied to balls $B(x, r_x)$ with $\mu(B(x, r_x))/r_x^{n-1} < \varepsilon$ extracted from the $\liminf$ definition. The symmetric argument applied to $A_1$ (replace $E$ by $E^c$, use $P(E; \cdot) = P(E^c; \cdot)$ and $\partial E = \partial E^c$) gives $\mathcal{H}^{n-1}(A_1) = 0$. Therefore \begin{align*} \mathcal{H}^{n-1}(\partial E \setminus \partial_m E) \le \mathcal{H}^{n-1}(A_0) + \mathcal{H}^{n-1}(A_1) = 0. \end{align*} [guided] We bound $\mathcal{H}^{n-1}(\partial E \setminus \partial_m E)$ by separating into two pieces and applying the relative isoperimetric inequality. *Splitting $\partial E \setminus \partial_m E$.* A point $x \notin \partial_m E$ fails one of the two density positivity conditions defining $\partial_m E$. Define \begin{align*} A_0 := \{ x \in \partial E : \Theta^*(E, x) = 0 \}, \qquad A_1 := \{ x \in \partial E : \Theta^*(E^c, x) = 0 \}. \end{align*} Then any $x \in \partial E$ with $x \notin \partial_m E$ satisfies $\Theta^*(E,x) = 0$ or $\Theta^*(E^c,x) = 0$ (since $\Theta^* \in [0,1]$, "not positive" means "equals zero"), so $\partial E \setminus \partial_m E \subseteq A_0 \cup A_1$. *Symmetry.* The set $E^c$ has the same topological boundary as $E$ ($\partial E = \partial E^c$) and the same perimeter ($P(E^c; \cdot) = P(E; \cdot)$, since $D\mathbb{1}_{E^c} = -D\mathbb{1}_E$). Replacing $E$ with $E^c$ swaps the roles of $A_0$ and $A_1$. So if we prove $\mathcal{H}^{n-1}(A_0) = 0$ for every $E$ of finite perimeter, the same statement applied to $E^c$ gives $\mathcal{H}^{n-1}(A_1) = 0$. Hence we may focus on $A_0$. *Why $A_0$ is small.* Intuitively, $A_0$ consists of "boundary points where $E$ has density zero" — points on the topological boundary that look like exterior points in the measure-theoretic sense. The Claim shows that the perimeter measure assigns zero lower $(n-1)$-density at every such point. We then invoke a standard density lemma to conclude these points form a $\mathcal{H}^{n-1}$-null set. *Verification of the Claim's hypotheses.* The [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) in its relative form requires $E$ to be a Borel set of finite perimeter — both hypotheses are given. It is stated for a ball $B \subset \mathbb{R}^n$ as \begin{align*} \min\bigl( \mathcal{L}^n(E \cap B), \mathcal{L}^n(B \setminus E) \bigr)^{(n-1)/n} \le C(n) \, P(E; B). \end{align*} For $x \in A_0$, the density condition $\Theta^*(E,x) = 0$ forces $\mathcal{L}^n(E \cap B(x,r))/\mathcal{L}^n(B(x,r)) \to 0$, so for small $r$ the minimum on the LHS is $\mathcal{L}^n(E \cap B(x,r))$. Dividing both sides by $r^{n-1}$ and using $\mathcal{L}^n(B(x,r)) = \alpha_n r^n$ (where $\alpha_n = \mathcal{L}^n(B(0,1))$): \begin{align*} \frac{P(E; B(x,r))}{r^{n-1}} \ge \frac{1}{C(n)} \cdot \frac{\mathcal{L}^n(E \cap B(x,r))^{(n-1)/n}}{r^{n-1}} = \frac{\alpha_n^{(n-1)/n}}{C(n)} \cdot \left(\frac{\mathcal{L}^n(E \cap B(x,r))}{\alpha_n r^n}\right)^{(n-1)/n}. \end{align*} The parenthesised term is the density of $E$ at scale $r$, which tends to $0$ as $r \to 0$. So the right-hand side tends to $0$, and hence \begin{align*} \liminf_{r \to 0} \frac{P(E; B(x,r))}{r^{n-1}} = 0 \quad \text{for every } x \in A_0. \end{align*} *Density lemma applied to $\mu := |D\mathbb{1}_E|$.* The density lemma we invoke is standard in geometric measure theory: if $\mu$ is a finite Radon measure on $\mathbb{R}^n$ and $S \subseteq \mathbb{R}^n$ is a Borel set with \begin{align*} \liminf_{r \to 0} \frac{\mu(B(x,r))}{r^{n-1}} = 0 \quad \text{for every } x \in S, \end{align*} then $\mathcal{H}^{n-1}(S) = 0$. The proof is a Vitali covering argument: fix $\varepsilon > 0$; for each $x \in S$ pick $r_x > 0$ arbitrarily small with $\mu(B(x, r_x)) < \varepsilon r_x^{n-1}$; extract by Vitali a countable disjoint subfamily $\{B(x_k, r_k)\}$ with $S \subseteq \bigcup_k B(x_k, 5 r_k)$ and $\sum_k r_k^{n-1} \le 5^{n-1} \varepsilon^{-1} \mu(\mathbb{R}^n)$ (since $r_k^{n-1} < \varepsilon^{-1} \mu(B(x_k, r_k))$ and the $B(x_k, r_k)$ are disjoint). Hence the $(n-1)$-Hausdorff content of $S$ is bounded by a constant times $\varepsilon^{-1} \mu(\mathbb{R}^n)$ times $r_k^{n-1}$, which by choosing $r_x \le \delta$ gives $\mathcal{H}^{n-1}_\delta(S) \le C \varepsilon$, and sending $\delta \to 0$ then $\varepsilon \to 0$ yields $\mathcal{H}^{n-1}(S) = 0$. (This is a special case of the Borel-density theorem; the argument is the same as the standard Hausdorff measure / Radon measure comparison via Vitali, applied with exponent $s = n-1$.) *Applying the lemma.* Take $\mu := |D\mathbb{1}_E|$ and $S := A_0$. The Claim gives $\liminf_{r \to 0} \mu(B(x,r))/r^{n-1} = 0$ for every $x \in A_0$. The hypothesis "$\mu$ finite" holds because $|D\mathbb{1}_E|(\mathbb{R}^n) = P(E; \mathbb{R}^n) < \infty$ by the assumption that $E$ has finite perimeter. The hypothesis "$A_0$ is Borel" holds because $A_0$ is the intersection of $\partial E$ (a closed set, hence Borel) with the Borel set $\{\Theta^*(E, \cdot) = 0\}$ (the upper density of a measurable set is a Borel function of $x$, since it can be written as a $\liminf$ of Borel functions $r \mapsto \mathcal{L}^n(E \cap B(\cdot, r))/\mathcal{L}^n(B(\cdot, r))$ over rational $r$). Therefore $\mathcal{H}^{n-1}(A_0) = 0$. *Symmetric conclusion.* By the symmetry argument above (replace $E$ with $E^c$), the same reasoning gives $\mathcal{H}^{n-1}(A_1) = 0$. Combining, \begin{align*} \mathcal{H}^{n-1}(\partial E \setminus \partial_m E) \le \mathcal{H}^{n-1}(A_0) + \mathcal{H}^{n-1}(A_1) = 0, \end{align*} which is the assertion of this step. [/guided] [/step] [step:Show $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0$ via De Giorgi's structure theorem] By [De Giorgi's Structure Theorem](/theorems/599), part (c), the perimeter measure satisfies $|D\mathbb{1}_E| = \mathcal{H}^{n-1}\lfloor \partial^* E$. In particular, for every Borel set $A \subseteq \mathbb{R}^n$, \begin{align*} |D\mathbb{1}_E|(A) = \mathcal{H}^{n-1}(A \cap \partial^* E). \end{align*} Apply this to $A := \partial_m E \setminus \partial^* E$: \begin{align*} |D\mathbb{1}_E|(\partial_m E \setminus \partial^* E) = \mathcal{H}^{n-1}((\partial_m E \setminus \partial^* E) \cap \partial^* E) = \mathcal{H}^{n-1}(\varnothing) = 0. \end{align*} It remains to upgrade the vanishing of $|D\mathbb{1}_E|(\partial_m E \setminus \partial^* E)$ to the vanishing of $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E)$. We argue via lower density: at every $x \in \partial_m E$, the perimeter measure has positive lower $(n-1)$-density. Specifically, the relative isoperimetric inequality of [theorem 600](/theorems/600), applied to a ball $B(x,r)$, gives \begin{align*} \min\bigl( \mathcal{L}^n(E \cap B(x,r)), \mathcal{L}^n(B(x,r) \setminus E) \bigr)^{(n-1)/n} \le C(n) \, P(E; B(x,r)) = C(n) \, |D\mathbb{1}_E|(B(x,r)). \end{align*} For $x \in \partial_m E$, both $\Theta^*(E,x) > 0$ and $\Theta^*(E^c,x) > 0$. Hence there exist $\delta_x > 0$ and a sequence $r_k \to 0$ with $\mathcal{L}^n(E \cap B(x, r_k)) \ge \delta_x r_k^n$ and $\mathcal{L}^n(B(x, r_k) \setminus E) \ge \delta_x r_k^n$ (one obtains the common $\delta_x$ by taking the minimum of the two upper densities and restricting to a common subsequence). Substituting: \begin{align*} (\delta_x r_k^n)^{(n-1)/n} \le C(n) \, |D\mathbb{1}_E|(B(x, r_k)), \end{align*} i.e., $|D\mathbb{1}_E|(B(x, r_k))/r_k^{n-1} \ge \delta_x^{(n-1)/n}/C(n) =: c_x > 0$. Therefore \begin{align*} \limsup_{r \to 0} \frac{|D\mathbb{1}_E|(B(x,r))}{r^{n-1}} \ge c_x > 0 \quad \text{for every } x \in \partial_m E. \end{align*} Apply the standard Radon-measure-vs-Hausdorff comparison: if $\mu$ is a finite Radon measure on $\mathbb{R}^n$ and $S \subseteq \mathbb{R}^n$ is a Borel set with $\limsup_{r \to 0} \mu(B(x,r))/r^{n-1} > 0$ at every $x \in S$, then $\mathcal{H}^{n-1}\lfloor_S \ll \mu$ (in fact, $\mathcal{H}^{n-1}(S) \le C(n) \mu(S)$ if the lower density is uniformly bounded below — for a non-uniform bound, exhaust $S$ by the Borel sets $S_k := \{x \in S : \limsup_{r \to 0} \mu(B(x,r))/r^{n-1} > 1/k\}$, on each of which $\mathcal{H}^{n-1}(S_k) \le k \cdot C(n) \mu(S_k)$). Applied with $\mu := |D\mathbb{1}_E|$ and $S := \partial_m E \setminus \partial^* E$, \begin{align*} \mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) \le C \cdot |D\mathbb{1}_E|(\partial_m E \setminus \partial^* E) = 0, \end{align*} where the equality is the calculation above and $C$ is finite by exhaustion. Combining with Step 2: \begin{align*} \mathcal{H}^{n-1}(\partial E \setminus \partial^* E) = \mathcal{H}^{n-1}(\partial E \setminus \partial_m E) + \mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0 + 0 = 0, \end{align*} which is the assertion of the theorem. [guided] The remaining piece is $\partial_m E \setminus \partial^* E$. We use [De Giorgi's Structure Theorem](/theorems/599) twice: once to show this set carries no perimeter mass, once to translate "no perimeter mass" into "no $\mathcal{H}^{n-1}$-mass" via a lower density argument. *Hypothesis verification for [De Giorgi](/theorems/599).* The theorem requires $E \subseteq \mathbb{R}^n$ to be a Borel-measurable set of finite perimeter — both are part of the hypothesis "$E$ has finite perimeter" (we work with the Borel representative; replacing $E$ by a Borel set differing on an $\mathcal{L}^n$-null set leaves $D\mathbb{1}_E$ unchanged). *Step 3a: $|D\mathbb{1}_E|(\partial_m E \setminus \partial^* E) = 0$.* Part (c) of [De Giorgi](/theorems/599) gives $|D\mathbb{1}_E| = \mathcal{H}^{n-1}\lfloor \partial^* E$, i.e., for every Borel $A \subseteq \mathbb{R}^n$, \begin{align*} |D\mathbb{1}_E|(A) = \mathcal{H}^{n-1}(A \cap \partial^* E). \end{align*} Setting $A := \partial_m E \setminus \partial^* E$ — a Borel set, since $\partial_m E$ is Borel (it is defined by countable lim-sup conditions on Borel functions) and $\partial^* E$ is Borel (similarly) — we get \begin{align*} |D\mathbb{1}_E|(\partial_m E \setminus \partial^* E) = \mathcal{H}^{n-1}((\partial_m E \setminus \partial^* E) \cap \partial^* E) = \mathcal{H}^{n-1}(\varnothing) = 0. \end{align*} *Step 3b: $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0$ via lower density.* The vanishing of $|D\mathbb{1}_E|$-mass is not by itself enough — a Hausdorff null set need not be a $|D\mathbb{1}_E|$-null set, but the converse is not automatic. However, we can show the converse on $\partial_m E$ by establishing a lower density bound for $|D\mathbb{1}_E|$ at every point of $\partial_m E$. The relative isoperimetric inequality [theorem 600](/theorems/600), applied to balls $B(x,r)$, gives \begin{align*} \min\bigl( \mathcal{L}^n(E \cap B(x,r)), \mathcal{L}^n(B(x,r) \setminus E) \bigr)^{(n-1)/n} \le C(n) \, |D\mathbb{1}_E|(B(x,r)), \end{align*} where we have identified $P(E; B(x,r)) = |D\mathbb{1}_E|(B(x,r))$. For $x \in \partial_m E$, set $\eta := \min(\Theta^*(E,x), \Theta^*(E^c, x)) > 0$. By definition of $\limsup$, for every $\varepsilon \in (0, \eta)$ there exists a sequence $r_k \to 0$ along which both \begin{align*} \mathcal{L}^n(E \cap B(x, r_k))/\mathcal{L}^n(B(x, r_k)) \ge \eta - \varepsilon \quad \text{and} \quad \mathcal{L}^n(B(x, r_k) \setminus E)/\mathcal{L}^n(B(x, r_k)) \ge \eta - \varepsilon. \end{align*} (Take $r_k$ realising the $\limsup$ for $\Theta^*(E, \cdot)$; pass to a subsequence so the second condition also holds — this requires that $\Theta^*(E^c, x) > 0$, which is part of the definition of $\partial_m E$.) Setting $\delta_x := (\eta/2) \alpha_n > 0$ (with $\alpha_n = \mathcal{L}^n(B(0,1))$, fixing $\varepsilon = \eta/2$), we get $\mathcal{L}^n(E \cap B(x, r_k)) \ge \delta_x r_k^n$ and $\mathcal{L}^n(B(x, r_k) \setminus E) \ge \delta_x r_k^n$ along the sequence $r_k$. Substituting into the relative isoperimetric inequality: \begin{align*} (\delta_x r_k^n)^{(n-1)/n} \le C(n) \, |D\mathbb{1}_E|(B(x, r_k)) \quad \Longleftrightarrow \quad \frac{|D\mathbb{1}_E|(B(x, r_k))}{r_k^{n-1}} \ge \frac{\delta_x^{(n-1)/n}}{C(n)} =: c_x > 0. \end{align*} Therefore \begin{align*} \limsup_{r \to 0} \frac{|D\mathbb{1}_E|(B(x,r))}{r^{n-1}} \ge c_x > 0 \quad \text{for every } x \in \partial_m E. \end{align*} *Density-based comparison.* The standard upper-density comparison theorem in measure theory (a corollary of the Vitali covering theorem; see also Mattila, "Geometry of Sets and Measures in Euclidean Spaces", Theorem 6.9) states: if $\mu$ is a Radon measure on $\mathbb{R}^n$ and $S \subseteq \mathbb{R}^n$ is a Borel set with $\limsup_{r \to 0} \mu(B(x,r))/r^{n-1} \ge c > 0$ for every $x \in S$, then $\mathcal{H}^{n-1}(S) \le C(n) c^{-1} \mu(S)$ for a dimensional constant $C(n)$. To handle non-uniform $c_x$, partition $\partial_m E$ as $\bigcup_{k=1}^\infty S_k$ with \begin{align*} S_k := \left\{ x \in \partial_m E : \limsup_{r \to 0} \frac{|D\mathbb{1}_E|(B(x,r))}{r^{n-1}} \ge \frac{1}{k} \right\}. \end{align*} Each $S_k$ is Borel; on $S_k$ the comparison gives $\mathcal{H}^{n-1}(S_k \cap A) \le C(n) k \cdot |D\mathbb{1}_E|(S_k \cap A)$ for every Borel $A$. Taking $A = (\partial_m E \setminus \partial^* E)$ and using Step 3a: \begin{align*} \mathcal{H}^{n-1}(S_k \cap (\partial_m E \setminus \partial^* E)) \le C(n) k \cdot |D\mathbb{1}_E|(S_k \cap (\partial_m E \setminus \partial^* E)) \le C(n) k \cdot 0 = 0. \end{align*} Since $\partial_m E = \bigcup_k S_k$, summing, \begin{align*} \mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = \sum_{k=1}^\infty \mathcal{H}^{n-1}(S_k \cap (\partial_m E \setminus \partial^* E)) = 0. \end{align*} (Strictly, one should subtract $S_{k-1}$ to make the union disjoint; the bound $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) \le \sum_k \mathcal{H}^{n-1}(S_k \cap \cdots)$ uses countable subadditivity of $\mathcal{H}^{n-1}$ and is only tighter without the disjointification.) *Combining Steps 2 and 3.* By Step 1, $\partial E \setminus \partial^* E = (\partial E \setminus \partial_m E) \cup (\partial_m E \setminus \partial^* E)$. By Step 2, $\mathcal{H}^{n-1}(\partial E \setminus \partial_m E) = 0$. By Step 3, $\mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0$. Therefore \begin{align*} \mathcal{H}^{n-1}(\partial E \setminus \partial^* E) \le \mathcal{H}^{n-1}(\partial E \setminus \partial_m E) + \mathcal{H}^{n-1}(\partial_m E \setminus \partial^* E) = 0, \end{align*} which is the conclusion of the theorem. The proof is complete. [/guided] [/step]