[proofplan] The proof identifies $D\tilde{u}$ as a Radon measure on $\mathbb{R}^n$ and then computes its total variation. The key computation tests $\tilde{u}$ against $\operatorname{div} \varphi$ for $\varphi \in C_c^\infty(\mathbb{R}^n; \mathbb{R}^n)$: by definition of $\tilde{u} = u\mathbb{1}_\Omega$, the integral over $\mathbb{R}^n$ collapses to an integral over $\Omega$, which a strict-density approximation by smooth functions on $\overline{\Omega}$ converts via the classical divergence theorem into a sum of an interior term involving $Du$ and a boundary term involving the trace $Tu$. This identifies $D\tilde{u}$ as the sum of two mutually singular Radon measures, supported on $\Omega$ and $\partial\Omega$ respectively. The total variation then decomposes additively, yielding the stated formula. [/proofplan]

[step:Test $\tilde{u}$ against the divergence of a smooth compactly supported vector field]For $\varphi \in C_c^\infty(\mathbb{R}^n; \mathbb{R}^n)$ we compute \begin{align*} \int_{\mathbb{R}^n} \tilde{u} \, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega u \, \operatorname{div}\varphi \, d\mathcal{L}^n, \end{align*} since $\tilde{u} = u$ on $\Omega$ and $\tilde{u} = 0$ on $\mathbb{R}^n \setminus \Omega$, while $\partial\Omega$ is $\mathcal{L}^n$-null because $\Omega$ has Lipschitz boundary. The right-hand integrand uses $u \in BV(\Omega) \subseteq L^1(\Omega)$, so the integral is finite.[/step]

[guided]We need to identify $D\tilde{u}$ as a distribution on $\mathbb{R}^n$ and check it is a finite Radon measure. The starting point is the test against $\operatorname{div}\varphi$: By definition of the distributional gradient, $D\tilde{u}(\varphi) = -\int_{\mathbb{R}^n} \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n$ for any $\varphi \in C_c^\infty(\mathbb{R}^n; \mathbb{R}^n)$. Splitting $\mathbb{R}^n$ into the disjoint pieces $\Omega$, $\partial\Omega$, and $\mathbb{R}^n \setminus \overline{\Omega}$: \begin{align*} \int_{\mathbb{R}^n} \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n + \int_{\partial\Omega} \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n + \int_{\mathbb{R}^n \setminus \overline{\Omega}} \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n. \end{align*} The third integral is zero because $\tilde{u} = 0$ on $\mathbb{R}^n \setminus \Omega$. The middle integral is zero because $\partial\Omega$ has $\mathcal{L}^n$-measure zero — this is a property of Lipschitz boundaries (the boundary is locally a Lipschitz graph, hence a countable union of $\mathcal{L}^n$-null sets). On $\Omega$ we have $\tilde{u} = u$, so \begin{align*} \int_{\mathbb{R}^n} \tilde{u}\, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega u\, \operatorname{div}\varphi \, d\mathcal{L}^n. \end{align*} The integral on the right is finite because $u \in L^1(\Omega)$ (since $u \in BV(\Omega)$) and $\operatorname{div}\varphi$ is bounded with compact support.[/guided]

[step:Apply the integration-by-parts formula on $\Omega$ via smooth approximation]By [BV Smooth Approximation](/theorems/3131) there is a sequence $(u_j) \subseteq C^\infty(\overline{\Omega}) \cap BV(\Omega)$ with $u_j \to u$ in $L^1(\Omega)$, $|Du_j|(\Omega) \to |Du|(\Omega)$, and $u_j|_{\partial\Omega} \to Tu$ in $L^1(\partial\Omega; \mathcal{H}^{n-1})$. For each $u_j$, the classical divergence theorem on the Lipschitz domain $\Omega$ gives \begin{align*} \int_\Omega u_j \, \operatorname{div}\varphi \, d\mathcal{L}^n = -\int_\Omega \nabla u_j \cdot \varphi \, d\mathcal{L}^n + \int_{\partial\Omega} u_j \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}, \end{align*} with $\nu_\Omega$ the $\mathcal{H}^{n-1}$-a.e. outward unit normal to $\partial\Omega$ (which exists for Lipschitz $\partial\Omega$). Passing to the limit $j \to \infty$: the left side converges by $L^1$-convergence and boundedness of $\operatorname{div}\varphi$; the second term on the right converges by the $L^1(\partial\Omega; \mathcal{H}^{n-1})$ convergence of $u_j|_{\partial\Omega}$ to $Tu$ and boundedness of $\varphi \cdot \nu_\Omega$; the first term on the right converges to $-\int_\Omega \varphi \cdot d(Du)$ by the strict convergence of $\nabla u_j \, \mathcal{L}^n \to Du$ as Radon measures on $\Omega$. Hence \begin{align*} \int_\Omega u \, \operatorname{div}\varphi \, d\mathcal{L}^n = -\int_\Omega \varphi \cdot d(Du) + \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*}[/step]

[guided]The classical divergence theorem requires $u_j \in C^1(\overline{\Omega})$ (or at least $W^{1,1}$ with sufficient boundary regularity). For $u \in BV(\Omega)$ we proceed by strict-density approximation. By [BV Smooth Approximation](/theorems/3131), there exists a sequence $(u_j) \subseteq C^\infty(\overline{\Omega}) \cap BV(\Omega)$ with the strict properties: \begin{align*} u_j &\to u \quad \text{in } L^1(\Omega), \\ |Du_j|(\Omega) &\to |Du|(\Omega). \end{align*} Combined with [BV Trace Theorem](/theorems/3110), strict convergence implies trace convergence in $L^1(\partial\Omega; \mathcal{H}^{n-1})$: the trace operator $T$ is continuous in the strict topology, so $u_j|_{\partial\Omega} = T u_j \to T u$ in $L^1(\partial\Omega; \mathcal{H}^{n-1})$. For smooth $u_j \in C^\infty(\overline{\Omega})$, the divergence theorem on the Lipschitz domain $\Omega$ asserts: for $\varphi \in C_c^\infty(\mathbb{R}^n; \mathbb{R}^n)$, \begin{align*} \int_\Omega \operatorname{div}(u_j \varphi) \, d\mathcal{L}^n = \int_{\partial\Omega} u_j \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} This is the standard form of the divergence theorem for a Lipschitz domain, where $\nu_\Omega: \partial\Omega \to S^{n-1}$ is the outward unit normal, defined $\mathcal{H}^{n-1}$-a.e. by Rademacher's theorem on the Lipschitz parameterisation of $\partial\Omega$. Expanding $\operatorname{div}(u_j \varphi) = u_j \operatorname{div}\varphi + \nabla u_j \cdot \varphi$: \begin{align*} \int_\Omega u_j \, \operatorname{div}\varphi \, d\mathcal{L}^n + \int_\Omega \nabla u_j \cdot \varphi \, d\mathcal{L}^n = \int_{\partial\Omega} u_j \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} We now pass to the limit $j \to \infty$ in each of the three terms. The first term: $\int_\Omega u_j \operatorname{div}\varphi \, d\mathcal{L}^n \to \int_\Omega u \operatorname{div}\varphi \, d\mathcal{L}^n$ since $u_j \to u$ in $L^1$ and $\operatorname{div}\varphi \in L^\infty(\Omega)$ (compact support, smooth). The second term: $\nabla u_j \cdot \varphi$ tested against the smooth bounded function $\varphi$. We use the strict convergence $\nabla u_j \, \mathcal{L}^n \to Du$ as Radon measures, which is equivalent to convergence in the strict topology. Strict convergence in BV implies weak* convergence in measures with convergence of total masses: \begin{align*} \int_\Omega \nabla u_j \cdot \varphi \, d\mathcal{L}^n \to \int_\Omega \varphi \cdot d(Du). \end{align*} The third term: $u_j|_{\partial\Omega} \to Tu$ in $L^1(\partial\Omega; \mathcal{H}^{n-1})$, so \begin{align*} \int_{\partial\Omega} u_j \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1} \to \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} Combining all three limits: \begin{align*} \int_\Omega u \, \operatorname{div}\varphi \, d\mathcal{L}^n + \int_\Omega \varphi \cdot d(Du) = \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}, \end{align*} which rearranges to the displayed formula.[/guided]

[step:Identify $D\tilde{u}$ as a sum of two mutually singular Radon measures]Combining Steps 1 and 2, \begin{align*} \int_{\mathbb{R}^n} \tilde{u} \, \operatorname{div}\varphi \, d\mathcal{L}^n = -\int_\Omega \varphi \cdot d(Du) + \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} By the definition of the distributional gradient, \begin{align*} D\tilde{u}(\varphi) = -\int_{\mathbb{R}^n} \tilde{u} \, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega \varphi \cdot d(Du) - \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} This identifies $D\tilde{u}$ as the vector-valued Radon measure \begin{align*} D\tilde{u} = (Du)|_\Omega - Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega \quad \text{on } \mathbb{R}^n, \end{align*} where $(Du)|_\Omega$ denotes the extension of $Du$ by zero outside $\Omega$, and $\mathcal{H}^{n-1}\lfloor\partial\Omega$ is the $(n-1)$-dimensional Hausdorff measure restricted to $\partial\Omega$. The two summand measures are supported respectively in $\Omega$ and $\partial\Omega$, which are disjoint, so they are mutually singular as vector-valued Radon measures.[/step]

[guided]We have produced an integration-by-parts identity for $\tilde{u}$ on the whole of $\mathbb{R}^n$. Reading the right-hand side as the action of a Radon measure against $\varphi$, we identify $D\tilde{u}$. The distributional gradient is defined by $D\tilde{u}(\varphi) = -\int \tilde{u} \operatorname{div}\varphi \, d\mathcal{L}^n$. Using the formula from Steps 1–2, \begin{align*} D\tilde{u}(\varphi) = -\int_\Omega u\, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega \varphi \cdot d(Du) - \int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}. \end{align*} The right-hand side is a sum of two integrals against $\varphi$. Each is the action of a vector-valued Radon measure on $\mathbb{R}^n$: - The first, $\int_\Omega \varphi \cdot d(Du)$, is the action of $Du$, regarded as a measure on $\mathbb{R}^n$ with mass concentrated in $\Omega$ (and zero outside $\overline{\Omega}$). - The second, $-\int_{\partial\Omega} Tu \, \varphi \cdot \nu_\Omega \, d\mathcal{H}^{n-1}$, is the action of the vector-valued measure $-Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega$. By the uniqueness of the Radon measure representing a given continuous linear functional on $C_c^\infty$ (a consequence of the Riesz representation theorem), $D\tilde{u}$ is precisely this sum: \begin{align*} D\tilde{u} = (Du)|_\Omega - Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega. \end{align*} The two terms have disjoint supports — $Du$ is supported in $\overline{\Omega}$ but vanishes on the $\mathcal{L}^n$-null Lipschitz boundary $\partial\Omega$ (since $|Du|(\partial\Omega) = 0$ for $u \in BV(\Omega)$ — this follows because $Du$ is a Radon measure on $\Omega$, hence vanishes on $\partial\Omega \cap \Omega = \varnothing$; when extended by zero to $\mathbb{R}^n$, it remains zero on $\partial\Omega$). The boundary term is supported on $\partial\Omega$ exactly. Hence the two summands are mutually singular: there is a Borel partition of $\mathbb{R}^n$ into the interior part $\Omega$ and the boundary part $\partial\Omega$ on which the two measures are concentrated. This is precisely the structure that makes the total variation decompose additively. We use this in the next step.[/guided]

[step:Decompose the total variation along the singular partition]The total variation of a Radon measure $\mu$ that decomposes as $\mu = \mu_1 + \mu_2$ with $\mu_1, \mu_2$ mutually singular satisfies $|\mu|(\mathbb{R}^n) = |\mu_1|(\mathbb{R}^n) + |\mu_2|(\mathbb{R}^n)$. Applying this to $D\tilde{u} = (Du)|_\Omega - Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega$: \begin{align*} |D\tilde{u}|(\mathbb{R}^n) = |(Du)|_\Omega|(\mathbb{R}^n) + |Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega|(\mathbb{R}^n). \end{align*} The first total variation equals $|Du|(\Omega)$ since $(Du)|_\Omega$ is the zero extension of $Du$. For the second, since $|\nu_\Omega| = 1$ $\mathcal{H}^{n-1}$-a.e. on $\partial\Omega$, the total variation of the boundary measure is \begin{align*} |Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega|(\mathbb{R}^n) = \int_{\partial\Omega} |Tu| \, |\nu_\Omega| \, d\mathcal{H}^{n-1} = \int_{\partial\Omega} |Tu| \, d\mathcal{H}^{n-1}. \end{align*} Combining, \begin{align*} |D\tilde{u}|(\mathbb{R}^n) = |Du|(\Omega) + \int_{\partial\Omega} |Tu| \, d\mathcal{H}^{n-1}. \end{align*} Finally, $\tilde{u} \in L^1(\mathbb{R}^n)$ since $\tilde{u} = u$ on $\Omega$ with $u \in L^1(\Omega)$ and $\tilde{u} = 0$ outside; combined with $|D\tilde{u}|(\mathbb{R}^n) < \infty$, this gives $\tilde{u} \in BV(\mathbb{R}^n)$, completing the proof.[/step]

[guided]We exploit the additivity of total variation across mutually singular pieces. Let me state precisely the result we use: if $\mu = \mu_1 + \mu_2$ is a vector-valued Radon measure on $\mathbb{R}^n$ and $\mu_1, \mu_2$ are mutually singular (i.e., supported on disjoint Borel sets $A_1, A_2$), then for every Borel $B \subseteq \mathbb{R}^n$, \begin{align*} |\mu|(B) = |\mu_1|(B \cap A_1) + |\mu_2|(B \cap A_2) = |\mu_1|(B) + |\mu_2|(B). \end{align*} This is a standard consequence of the Hahn–Jordan-type decomposition for vector-valued measures: the total variation is computed pointwise on disjoint supports. In our setting, $\mu_1 = (Du)|_\Omega$ supported in $\overline{\Omega}$ with $|\mu_1|(\partial\Omega) = 0$, and $\mu_2 = -Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega$ supported in $\partial\Omega$. The Borel partition $\mathbb{R}^n = \Omega \cup \partial\Omega \cup (\mathbb{R}^n \setminus \overline{\Omega})$ separates the two supports. Applying the additivity at $B = \mathbb{R}^n$: \begin{align*} |D\tilde{u}|(\mathbb{R}^n) = |(Du)|_\Omega|(\mathbb{R}^n) + |Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega|(\mathbb{R}^n). \end{align*} The first term: $(Du)|_\Omega$ is the zero-extension of $Du$ from $\Omega$ to $\mathbb{R}^n$, so $|(Du)|_\Omega|(\mathbb{R}^n) = |Du|(\Omega)$. The second term: writing the integrand as the scalar $Tu$ times the unit vector field $\nu_\Omega$ on $\partial\Omega$ and the scalar measure $\mathcal{H}^{n-1}\lfloor\partial\Omega$, and using that $|\nu_\Omega(x)| = 1$ for $\mathcal{H}^{n-1}$-a.e. $x \in \partial\Omega$ (Rademacher's theorem on the Lipschitz parameterisation of $\partial\Omega$ gives $\nu_\Omega \in S^{n-1}$ $\mathcal{H}^{n-1}$-a.e.): \begin{align*} |Tu \, \nu_\Omega \, \mathcal{H}^{n-1}\lfloor\partial\Omega|(\mathbb{R}^n) = \int_{\partial\Omega} |Tu(x)\, \nu_\Omega(x)| \, d\mathcal{H}^{n-1}(x) = \int_{\partial\Omega} |Tu(x)| \, d\mathcal{H}^{n-1}(x). \end{align*} Putting the two pieces together: \begin{align*} |D\tilde{u}|(\mathbb{R}^n) = |Du|(\Omega) + \int_{\partial\Omega} |Tu| \, d\mathcal{H}^{n-1}. \end{align*} Finally we verify $\tilde{u} \in BV(\mathbb{R}^n)$. We need $\tilde{u} \in L^1(\mathbb{R}^n)$ and $|D\tilde{u}|(\mathbb{R}^n) < \infty$. The first holds because \begin{align*} \int_{\mathbb{R}^n} |\tilde{u}| \, d\mathcal{L}^n = \int_\Omega |u| \, d\mathcal{L}^n = \|u\|_{L^1(\Omega)} < \infty, \end{align*} and the second holds because $|Du|(\Omega) < \infty$ (by $u \in BV(\Omega)$) and $\int_{\partial\Omega} |Tu| \, d\mathcal{H}^{n-1} < \infty$ (by [BV Trace Theorem](/theorems/3110)). Hence $\tilde{u} \in BV(\mathbb{R}^n)$ with the stated total variation, completing the proof.[/guided]