[proofplan] The identity is the polar decomposition of $D\mathbb{1}_E$ tested against $\phi$. By definition of the distributional gradient of $\mathbb{1}_E \in BV_{\mathrm{loc}}(U)$, the divergence integral $\int_E \operatorname{div} \phi \, d\mathcal{L}^n$ equals $-\int_U \phi \cdot dD\mathbb{1}_E$ for every $\phi \in C^1_c(U; \mathbb{R}^n)$. The substantive content is then [De Giorgi's Structure Theorem](/theorems/599), which provides the polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1} \lfloor \partial^* E$ as an equality of vector-valued Radon measures: the total variation $|D\mathbb{1}_E|$ equals $\mathcal{H}^{n-1} \lfloor \partial^* E$, and the unit Radon-Nikodym derivative is $-\nu_E$ (the minus sign is the convention that makes $\nu_E$ point *outward*). Substituting this expression into the distributional identity and tracking the two minus signs yields the Gauss-Green formula. The proof has two steps: (i) recall the distributional identity for $D\mathbb{1}_E$; (ii) substitute the polar decomposition. [/proofplan]

[step:Apply the distributional definition of $D\mathbb{1}_E$ to $\phi$]The hypothesis "$E$ is a Borel set of finite perimeter in $U$" means $\mathbb{1}_E \in BV(U)$, equivalently that the distributional gradient $D\mathbb{1}_E$ is a finite vector-valued Radon measure on $U$ with total variation $|D\mathbb{1}_E|(U) = P(E; U) < \infty$. By the definition of the distributional gradient, for every $\phi \in C_c^1(U; \mathbb{R}^n)$, \begin{align*} \int_U \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = -\int_U \phi \cdot dD\mathbb{1}_E, \end{align*} where the right-hand integral is the action of the vector-valued Radon measure $D\mathbb{1}_E$ on the continuous compactly supported test field $\phi$. The left-hand integral simplifies because $\mathbb{1}_E = 1$ on $E$ and $0$ off $E$: \begin{align*} \int_U \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_{E \cap U} \operatorname{div} \phi \, d\mathcal{L}^n = \int_E \operatorname{div} \phi \, d\mathcal{L}^n, \end{align*} the second equality because $\phi \in C_c^1(U; \mathbb{R}^n)$ has support contained in $U$, hence $\operatorname{div} \phi$ vanishes on $\mathbb{R}^n \setminus U$. Combining, \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_U \phi \cdot dD\mathbb{1}_E. \end{align*}[/step]

[guided]The identity is the very definition of $D\mathbb{1}_E$ as a distribution. *The distributional gradient $Du$.* For any $u \in L^1_{\mathrm{loc}}(U)$, the distributional gradient $Du$ is the $\mathbb{R}^n$-valued distribution on $U$ defined by \begin{align*} Du(\phi) := -\int_U u \, \operatorname{div} \phi \, d\mathcal{L}^n \quad \text{for every } \phi \in C_c^\infty(U; \mathbb{R}^n). \end{align*} When $u \in BV(U)$, this distribution is represented by a finite vector-valued Radon measure on $U$, also denoted $Du$, in the sense that \begin{align*} Du(\phi) = \int_U \phi \cdot dDu \quad \text{for every } \phi \in C_c(U; \mathbb{R}^n). \end{align*} The class of test fields extends from $C_c^\infty$ to $C_c$ because Radon measures act continuously on $C_c$. *Hypothesis verification.* The hypothesis "$E$ has finite perimeter in $U$" means $\mathbb{1}_E \in BV(U)$ and equivalently $|D\mathbb{1}_E|(U) = P(E; U) < \infty$. So $D\mathbb{1}_E$ is a finite vector-valued Radon measure on $U$, and the displayed integration-by-parts identity \begin{align*} -\int_U \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_U \phi \cdot dD\mathbb{1}_E \end{align*} holds for every $\phi \in C_c^1(U; \mathbb{R}^n)$. The class $C_c^1$ is admissible: $\operatorname{div} \phi \in C_c(U) \subset L^\infty_c(U)$ makes the LHS a well-defined Lebesgue integral, and the RHS is the action of the Radon measure $D\mathbb{1}_E$ on $\phi \in C_c \subseteq C_c$. *Simplifying the LHS.* The function $\mathbb{1}_E(x)$ equals $1$ when $x \in E$ and $0$ otherwise, so \begin{align*} \int_U \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_{E \cap U} \operatorname{div} \phi \, d\mathcal{L}^n. \end{align*} Because $\phi \in C_c^1(U; \mathbb{R}^n)$ has compact support in $U$, the function $\operatorname{div} \phi$ is supported in $U$ — extending it by zero outside $U$ produces a function in $C_c(\mathbb{R}^n)$ supported in $\overline{\operatorname{supp}(\phi)} \subseteq U$. Hence the integrand contributes nothing on $E \setminus U$, and we may write \begin{align*} \int_{E \cap U} \operatorname{div} \phi \, d\mathcal{L}^n = \int_E \operatorname{div} \phi \, d\mathcal{L}^n, \end{align*} the integral on the right being interpreted with $\operatorname{div} \phi$ extended by zero outside $U$. *Combined identity.* Putting these together, \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_U \phi \cdot dD\mathbb{1}_E. \end{align*} This is the half-finished Gauss-Green identity: the divergence integral over $E$ equals (up to a sign) the action of $D\mathbb{1}_E$ on $\phi$. Step 2 will replace $D\mathbb{1}_E$ by its concrete expression in terms of the reduced boundary.[/guided]

[step:Substitute De Giorgi's polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1} \lfloor \partial^* E$]By [De Giorgi's Structure Theorem](/theorems/599), the polar decomposition of the gradient measure $D\mathbb{1}_E$ holds: its total variation is $|D\mathbb{1}_E| = \mathcal{H}^{n-1} \lfloor \partial^* E$ (part (c) of the theorem), and the Radon-Nikodym derivative $dD\mathbb{1}_E / d|D\mathbb{1}_E|$ at $\mathcal{H}^{n-1}$-a.e. $x \in \partial^* E$ equals $-\nu_E(x)$, where $\nu_E(x)$ is the measure-theoretic *outer* unit normal to $E$ at $x$ (part (d)). The minus sign reflects the orientation convention: $\nu_E$ points from $E$ to $E^c$, opposite to the direction in which $\mathbb{1}_E$ "increases." Combining the two parts as vector-valued measures on $U$: \begin{align*} D\mathbb{1}_E = \frac{dD\mathbb{1}_E}{d|D\mathbb{1}_E|} \cdot |D\mathbb{1}_E| = -\nu_E \cdot \mathcal{H}^{n-1} \lfloor \partial^* E. \end{align*} Pairing with $\phi \in C_c^1(U; \mathbb{R}^n)$, \begin{align*} \int_U \phi \cdot dD\mathbb{1}_E = \int_U \phi(x) \cdot (-\nu_E(x)) \, d(\mathcal{H}^{n-1}\lfloor \partial^* E)(x) = -\int_{\partial^* E \cap U} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}. \end{align*} Since $\phi \in C_c^1(U; \mathbb{R}^n)$ has compact support in $U$, the integrand vanishes $\mathcal{H}^{n-1}$-a.e. on $\partial^* E \setminus U$, so we may extend the domain of integration to $\partial^* E$: \begin{align*} \int_{\partial^* E \cap U} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}. \end{align*} Substituting into the identity from Step 1, \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_U \phi \cdot dD\mathbb{1}_E = -\Bigl(-\int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}\Bigr) = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}, \end{align*} which is the Gauss-Green formula. The proof is complete.[/step]

[guided]We use [De Giorgi's Structure Theorem](/theorems/599) to express $D\mathbb{1}_E$ in concrete terms, then substitute into the distributional identity from Step 1. *Polar decomposition of a vector-valued Radon measure.* Any $\mathbb{R}^n$-valued Radon measure $\mu$ on $U$ admits a polar decomposition \begin{align*} \mu = \nu_\mu \cdot |\mu|, \end{align*} where $|\mu|$ is the total variation (a non-negative Radon measure) and $\nu_\mu = d\mu/d|\mu|: U \to S^{n-1}$ is a Borel-measurable unit vector field, defined $|\mu|$-a.e. by the Radon-Nikodym theorem. *Hypothesis verification for [De Giorgi's Structure Theorem](/theorems/599).* The theorem requires $E \subseteq \mathbb{R}^n$ to be a Borel-measurable set of finite perimeter. Both hypotheses are explicit: $E$ is given as a Borel set of finite perimeter in $U$. (Technical point: De Giorgi's theorem in [theorem 599](/theorems/599) is stated for sets of finite perimeter in $\mathbb{R}^n$, but the conclusions are local — the polar decomposition holds inside any open set $U$ where $E$ has finite perimeter, by restricting all measures and densities to $U$.) *Conclusions of De Giorgi.* Part (c) gives $|D\mathbb{1}_E|(A) = \mathcal{H}^{n-1}(A \cap \partial^* E)$ for every Borel $A \subseteq \mathbb{R}^n$, so \begin{align*} |D\mathbb{1}_E| = \mathcal{H}^{n-1}\lfloor \partial^* E \end{align*} as Borel measures. Part (d) gives the Radon-Nikodym derivative $dD\mathbb{1}_E / d|D\mathbb{1}_E| = -\nu_E$ at $\mathcal{H}^{n-1}$-a.e. $x \in \partial^* E$, where $\nu_E$ is the measure-theoretic outer unit normal. The minus sign records the orientation convention: $\nu_E$ points outward (from $E$ to $E^c$), whereas the gradient measure $D\mathbb{1}_E$ "points" from $E^c$ to $E$ (the direction in which $\mathbb{1}_E$ increases). Combining, \begin{align*} D\mathbb{1}_E = (-\nu_E) \cdot (\mathcal{H}^{n-1}\lfloor \partial^* E) = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E. \end{align*} *Action on $\phi$.* For $\phi \in C_c^1(U; \mathbb{R}^n)$, the action of $D\mathbb{1}_E$ on $\phi$ is the integral \begin{align*} \int_U \phi \cdot dD\mathbb{1}_E = \int_U \phi(x) \cdot (-\nu_E(x)) \, d(\mathcal{H}^{n-1}\lfloor \partial^* E)(x) = -\int_{\partial^* E \cap U} \phi(x) \cdot \nu_E(x) \, d\mathcal{H}^{n-1}(x). \end{align*} The notation $\mathcal{H}^{n-1}\lfloor \partial^* E$ means the integration is against $\mathcal{H}^{n-1}$ on the set $\partial^* E$; the further restriction to $\partial^* E \cap U$ comes from $\phi$ being supported in $U$. The integrand $\phi \cdot \nu_E$ is well-defined $\mathcal{H}^{n-1}$-a.e. on $\partial^* E$ because $\nu_E$ is defined $\mathcal{H}^{n-1}$-a.e. on $\partial^* E$ as part of the structure theorem. *Removing the explicit $U$-restriction.* Because $\phi \in C_c^1(U; \mathbb{R}^n)$ has compact support contained in $U$, the integrand $\phi \cdot \nu_E$ vanishes outside the (compact) support of $\phi$, hence in particular vanishes $\mathcal{H}^{n-1}$-a.e. on $\partial^* E \setminus U$. Therefore \begin{align*} \int_{\partial^* E \cap U} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}. \end{align*} *Substitution into Step 1's identity.* From Step 1, \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_U \phi \cdot dD\mathbb{1}_E. \end{align*} Inserting the formula from above: \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\Bigl( -\int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} \Bigr) = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}. \end{align*} The two minus signs cancel. *Conclusion.* For every $\phi \in C_c^1(U; \mathbb{R}^n)$, \begin{align*} \int_E \operatorname{div} \phi \, d\mathcal{L}^n = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}, \end{align*} which is the Gauss-Green theorem for sets of finite perimeter. The proof is complete. *Sign-convention check.* The minus sign in the polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E$ is the only delicate point. To verify it on a model example, take $E := \{x_n < 0\}$ in $\mathbb{R}^n$ (so $U = \mathbb{R}^n$). The classical outer normal is $\nu_E = e_n$ (pointing into $\{x_n > 0\}$, away from $E$). For $\phi = (\phi_1, \ldots, \phi_n) \in C_c^1(\mathbb{R}^n; \mathbb{R}^n)$, the distributional gradient acts as \begin{align*} D\mathbb{1}_E(\phi) = -\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_{\mathbb{R}^{n-1}} \int_{-\infty}^0 \operatorname{div} \phi(x', x_n) \, d\mathcal{L}^1(x_n) \, d\mathcal{L}^{n-1}(x') = -\int_{\mathbb{R}^{n-1}} \phi_n(x', 0) \, d\mathcal{L}^{n-1}(x'), \end{align*} using Fubini and the one-dimensional fundamental theorem of calculus on each fibre (with $\phi_n$ supported in a bounded $x_n$-interval). On the other hand, $\partial^* E = \{x_n = 0\}$ with $\nu_E = e_n$, so \begin{align*} \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} = \int_{\mathbb{R}^{n-1}} \phi(x', 0) \cdot e_n \, d\mathcal{L}^{n-1}(x') = \int_{\mathbb{R}^{n-1}} \phi_n(x', 0) \, d\mathcal{L}^{n-1}(x'). \end{align*} Comparing, $D\mathbb{1}_E(\phi) = -\int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}$, confirming the polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \{x_n = 0\}$. In the Gauss-Green formula the two minus signs cancel, leaving the classical positive sign.[/guided]