[proofplan]
The Gauss-Green theorem $\int_E \operatorname{div} \phi \, d\mathcal{L}^n = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}$ for sets of finite perimeter is a synthesis of two ingredients: the distributional definition of $D\mathbb{1}_E$ and [De Giorgi's Structure Theorem](/theorems/599). The strategy is direct. First, the distributional gradient identity gives $-\int_E \operatorname{div} \phi \, d\mathcal{L}^n = \int_\Omega \phi \cdot dD\mathbb{1}_E$ for every $\phi \in C_c^1(\Omega; \mathbb{R}^n)$, by the very definition of $D\mathbb{1}_E$. Second, De Giorgi's theorem provides the polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E$ as a vector-valued Radon measure (with the sign convention that $\nu_E$ is the *outer* normal). Substituting this decomposition and rearranging the sign yields the Gauss-Green formula. The proof has two short steps: (i) recall the distributional identity; (ii) substitute De Giorgi's polar decomposition.
[/proofplan]
[step:Apply the distributional definition of $D\mathbb{1}_E$ to $\phi$]
Let $E \subseteq \mathbb{R}^n$ have finite perimeter in $\Omega$, so that $\mathbb{1}_E \in BV_{\mathrm{loc}}(\Omega)$ with $|D\mathbb{1}_E|(\Omega) < \infty$. By the definition of the distributional gradient, for every $\phi \in C_c^1(\Omega; \mathbb{R}^n)$,
\begin{align*}
\int_\Omega \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = -\int_\Omega \phi \cdot \, dD\mathbb{1}_E,
\end{align*}
where $D\mathbb{1}_E$ is the vector-valued Radon measure on $\Omega$ whose components $D_i \mathbb{1}_E$ are the distributional partial derivatives of $\mathbb{1}_E$. The left-hand integral simplifies to an integral over $E$:
\begin{align*}
\int_\Omega \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_E \operatorname{div} \phi \, d\mathcal{L}^n.
\end{align*}
Combining,
\begin{align*}
\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_\Omega \phi \cdot \, dD\mathbb{1}_E.
\end{align*}
[guided]
The starting identity is the very definition of the distributional gradient.
*Distributional gradient.* For any locally integrable function $u: \Omega \to \mathbb{R}$, the distributional gradient $Du$ is the $\mathbb{R}^n$-valued distribution defined by
\begin{align*}
Du(\phi) := -\int_\Omega u \, \operatorname{div} \phi \, d\mathcal{L}^n \quad \text{for every } \phi \in C_c^\infty(\Omega; \mathbb{R}^n).
\end{align*}
When $u \in BV_{\mathrm{loc}}(\Omega)$, this distribution is represented by a vector-valued Radon measure on $\Omega$, also denoted $Du$, in the sense that $Du(\phi) = \int_\Omega \phi \cdot dDu$ for $\phi \in C_c(\Omega; \mathbb{R}^n)$.
*Hypothesis verification.* We are given that $E$ has finite perimeter in $\Omega$, which by definition means $\mathbb{1}_E \in BV(\Omega)$, equivalently $|D\mathbb{1}_E|(\Omega) < \infty$. Hence $D\mathbb{1}_E$ is a finite vector-valued Radon measure on $\Omega$, and the identity
\begin{align*}
-\int_\Omega \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_\Omega \phi \cdot dD\mathbb{1}_E
\end{align*}
holds for every $\phi \in C_c^1(\Omega; \mathbb{R}^n)$. (The class $C_c^1$ is admissible because $D\mathbb{1}_E$ is a Radon measure, hence acts on continuous compactly supported test fields by integration; the $C^1$ regularity is needed only for $\operatorname{div} \phi$ to be defined pointwise as an $L^\infty_c$ function.)
*Simplifying the LHS.* Since $\mathbb{1}_E(x) = 1$ when $x \in E$ and $0$ otherwise,
\begin{align*}
\int_\Omega \mathbb{1}_E \, \operatorname{div} \phi \, d\mathcal{L}^n = \int_{E \cap \Omega} \operatorname{div} \phi \, d\mathcal{L}^n = \int_E \operatorname{div} \phi \, d\mathcal{L}^n,
\end{align*}
the last equality holding because $\phi$ has compact support in $\Omega$, so the integrand $\operatorname{div} \phi$ is supported in $\Omega$ and there is no contribution from $E \setminus \Omega$.
*Combined identity.* Putting these together:
\begin{align*}
\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_\Omega \phi \cdot dD\mathbb{1}_E.
\end{align*}
This is the half-finished version of Gauss-Green: the divergence integral over $E$ equals (up to a sign) the action of the gradient measure $D\mathbb{1}_E$ on $\phi$. Step 2 will replace $D\mathbb{1}_E$ by its concrete expression in terms of the reduced boundary.
[/guided]
[/step]
[step:Substitute De Giorgi's polar decomposition $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E$]
[De Giorgi's Structure Theorem](/theorems/599) provides the polar decomposition of the gradient measure: the total variation is $|D\mathbb{1}_E| = \mathcal{H}^{n-1}\lfloor \partial^* E$, and the Radon-Nikodym derivative of $D\mathbb{1}_E$ with respect to its total variation $|D\mathbb{1}_E|$ at $\mathcal{H}^{n-1}$-a.e. $x \in \partial^* E$ is $-\nu_E(x)$, where $\nu_E(x)$ is the measure-theoretic *outer* unit normal to $E$ at $x$. (The minus sign reflects the orientation convention: the gradient of $\mathbb{1}_E$ points from outside to inside, opposite to the outer normal.) Hence as vector-valued measures on $\Omega$,
\begin{align*}
D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E,
\end{align*}
meaning, for every $\phi \in C_c^1(\Omega; \mathbb{R}^n)$,
\begin{align*}
\int_\Omega \phi \cdot dD\mathbb{1}_E = -\int_{\partial^* E \cap \Omega} \phi(x) \cdot \nu_E(x) \, d\mathcal{H}^{n-1}(x).
\end{align*}
Substituting into the identity from Step 1,
\begin{align*}
\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_\Omega \phi \cdot dD\mathbb{1}_E = -\big(-\int_{\partial^* E \cap \Omega} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}\big) = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1},
\end{align*}
the last step using $\partial^* E \cap \Omega = \partial^* E \cap \operatorname{supp}(\phi) \subseteq \partial^* E$ and that $\phi$ vanishes outside $\Omega$. This is the Gauss-Green formula, and the proof is complete.
[guided]
We use [De Giorgi's Structure Theorem](/theorems/599) to express $D\mathbb{1}_E$ in concrete terms, then substitute into the identity from Step 1.
*Polar decomposition of a vector-valued Radon measure.* Any $\mathbb{R}^n$-valued Radon measure $\mu$ on $\Omega$ 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 $\nu_\mu = d\mu / d|\mu|: \Omega \to S^{n-1}$ is a Borel-measurable unit vector field, defined $|\mu|$-a.e. by the Radon-Nikodym theorem.
*Applying De Giorgi to $\mu = D\mathbb{1}_E$.* The hypothesis of [De Giorgi's Structure Theorem](/theorems/599) is that $E \subseteq \mathbb{R}^n$ is a Borel set of finite perimeter, which is given. The theorem yields:
- Part (c): $|D\mathbb{1}_E|(A) = \mathcal{H}^{n-1}(A \cap \partial^* E)$ for every Borel $A$, so $|D\mathbb{1}_E| = \mathcal{H}^{n-1}\lfloor \partial^* E$.
- Part (d): the Radon-Nikodym derivative $dD\mathbb{1}_E / d|D\mathbb{1}_E|$ at $\mathcal{H}^{n-1}$-a.e. $x \in \partial^* E$ is $-\nu_E(x)$ — the minus sign coming from the convention that $\nu_E$ is the *outer* unit normal (the direction in which $\mathbb{1}_E$ decreases, opposite to the gradient direction which records the increase).
Combining the two parts:
\begin{align*}
D\mathbb{1}_E = \frac{dD\mathbb{1}_E}{d|D\mathbb{1}_E|} \cdot |D\mathbb{1}_E| = (-\nu_E) \cdot \big(\mathcal{H}^{n-1}\lfloor \partial^* E\big) = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \partial^* E.
\end{align*}
This is the explicit form of $D\mathbb{1}_E$ as a vector-valued Radon measure on $\Omega$.
*Action on $\phi$.* For $\phi \in C_c^1(\Omega; \mathbb{R}^n)$, the action of $D\mathbb{1}_E$ on $\phi$ is the integral
\begin{align*}
\int_\Omega \phi(x) \cdot dD\mathbb{1}_E(x) = \int_\Omega \phi(x) \cdot (-\nu_E(x)) \, d\big(\mathcal{H}^{n-1}\lfloor \partial^* E\big)(x) = -\int_{\partial^* E \cap \Omega} \phi(x) \cdot \nu_E(x) \, d\mathcal{H}^{n-1}(x).
\end{align*}
The restriction notation $\mathcal{H}^{n-1}\lfloor \partial^* E$ means we integrate against $\mathcal{H}^{n-1}$ over the set $\partial^* E$; the intersection with $\Omega$ comes from the support of $\phi$ being in $\Omega$. (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.)
*Substitution into Step 1's identity.* From Step 1,
\begin{align*}
\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\int_\Omega \phi \cdot dD\mathbb{1}_E.
\end{align*}
Substituting the formula above:
\begin{align*}
\int_E \operatorname{div} \phi \, d\mathcal{L}^n = -\left( -\int_{\partial^* E \cap \Omega} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} \right) = \int_{\partial^* E \cap \Omega} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}.
\end{align*}
*Removing the explicit $\Omega$-restriction.* Because $\phi \in C_c^1(\Omega; \mathbb{R}^n)$ has compact support in $\Omega$, it vanishes outside $\Omega$. Hence
\begin{align*}
\int_{\partial^* E \cap \Omega} \phi \cdot \nu_E \, d\mathcal{H}^{n-1} = \int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1},
\end{align*}
the integrand vanishing $\mathcal{H}^{n-1}$-a.e. on $\partial^* E \setminus \Omega$.
*Conclusion.* We have shown
\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*}
for every $\phi \in C_c^1(\Omega; \mathbb{R}^n)$, which is the Gauss-Green theorem for sets of finite perimeter. The proof is complete.
*Remark on the sign convention.* 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 the convention: take the simple example $E = \{x_n < 0\}$ in $\mathbb{R}^n$. The classical outer normal is $\nu_E = e_n$ (pointing into $\{x_n > 0\}$). The distributional gradient $D\mathbb{1}_E$ is computed against $\phi = (\phi_1, \ldots, \phi_n) \in C_c^1$:
\begin{align*}
D\mathbb{1}_E(\phi) = -\int_{\{x_n < 0\}} \operatorname{div} \phi \, d\mathcal{L}^n = -\int_{\mathbb{R}^{n-1}} \phi_n(x', 0) \, d\mathcal{L}^{n-1}(x') = -\int_{\partial^* E} \phi \cdot \nu_E \, d\mathcal{H}^{n-1}(x'),
\end{align*}
the second equality from integrating in $x_n$ over $(-\infty, 0)$. Therefore $D\mathbb{1}_E = -\nu_E \cdot \mathcal{H}^{n-1}\lfloor \{x_n = 0\}$, confirming the sign. In the Gauss-Green formula, the two minus signs cancel, leaving the classical positive sign $\int_E \operatorname{div} \phi = \int_{\partial^* E} \phi \cdot \nu_E$.
[/guided]
[/step]
