[proofplan] The strategy combines the [BV Coarea Formula](/theorems/598), the [Gauss-Green Theorem for Sets of Finite Perimeter](/theorems/3119), and the level-set characterisation of $J_u$ from [Jump Set is Rectifiable](/theorems/3123). For $\varphi \in C_c^\infty(\Omega; \mathbb{R}^n)$, a layer-cake representation of $u$ together with Fubini reduces the computation of $Du(\varphi) = -\int_\Omega u \operatorname{div} \varphi \, d\mathcal{L}^n$ to an integral over $t$ of $D\mathbb{1}_{E_t}(\varphi)$, where $E_t = \{u > t\}$. Each $D\mathbb{1}_{E_t}$ is identified with $-\nu_{E_t} \mathcal{H}^{n-1}\lfloor \partial^* E_t$ by [De Giorgi's Structure Theorem](/theorems/599), where $\nu_{E_t}$ is the *outer* unit normal of $E_t$. Crucially, since $u$ is *larger* inside $E_t$ and *smaller* outside, the outer normal of $E_t$ at a jump point $x \in J_u$ points from the $u^+$ side to the $u^-$ side, hence $\nu_{E_t}(x) = -\nu_u(x)$ (recall that $\nu_u$ is conventionally the jump normal pointing from the $u^-$ side to the $u^+$ side). Combined with the Gauss-Green sign $D\mathbb{1}_{E_t} = -\nu_{E_t} \mathcal{H}^{n-1}\lfloor \partial^* E_t$, the two minus signs cancel and produce the positive-sign jump formula. After identifying the joint $(t, x)$-integral via Fubini against the well-defined Radon measure $\Sigma := \int_G \mathcal{H}^{n-1}\lfloor \partial^* E_t \, d\mathcal{L}^1(t)$ (whose total mass equals $|Du|(\Omega)$ by the BV coarea formula), the level-set characterisation of $J_u$ from [theorem 3123](/theorems/3123) restricts the inner $t$-integral to $t \in (u^-(x), u^+(x))$ for $x \in J_u$. The complementary contribution — from $t$-values where $x$ lies on $\partial^* E_t$ but $x \notin J_u$ — yields a vector-valued Radon measure on $\Omega \setminus J_u$ which is computed (via coarea) from the level-set integral itself, and this measure equals $D^a u + D^c u$ by direct comparison with $Du - D^j u$. The structure of the proof: (i) reduce $Du(\varphi)$ to an integral of $D\mathbb{1}_{E_t}(\varphi)$ via coarea/layer-cake; (ii) apply Gauss-Green to each $D\mathbb{1}_{E_t}$, fix the sign convention, and define the joint Radon measure via Fubini; (iii) identify the $J_u$-contribution by the level-set characterisation, recover the off-$J_u$ contribution via the coarea formula and the definitions of $D^a u$, $D^c u$. [/proofplan] [step:Reduce $Du(\varphi)$ to an integral over levels via the layer-cake representation] Fix $\varphi \in C_c^\infty(\Omega; \mathbb{R}^n)$. By the definition of the distributional gradient and the inclusion $u \in BV(\Omega) \subseteq L^1_{\mathrm{loc}}(\Omega)$, \begin{align*} Du(\varphi) = -\int_\Omega u(y) \, \operatorname{div} \varphi(y) \, d\mathcal{L}^n(y). \end{align*} For every $y \in \Omega$ with $u(y) \in \mathbb{R}$ (which holds $\mathcal{L}^n$-a.e. in $\Omega$), the layer-cake (Cavalieri) identity reads \begin{align*} u(y) = \int_0^\infty \mathbb{1}_{\{u > t\}}(y) \, d\mathcal{L}^1(t) - \int_{-\infty}^0 \mathbb{1}_{\{u \le t\}}(y) \, d\mathcal{L}^1(t). \end{align*} For $t < 0$ the set $\{u \le t\}$ is the complement of $E_t = \{u > t\}$ in $\Omega$, so $\mathbb{1}_{\{u \le t\}} = \mathbb{1}_\Omega - \mathbb{1}_{E_t}$ on $\Omega$. Substituting and using $\int_\Omega \operatorname{div}\varphi \, d\mathcal{L}^n = 0$ (a consequence of the compact support of $\varphi$), one obtains for $t < 0$ \begin{align*} \int_\Omega \mathbb{1}_{\{u \le t\}}(y) \, \operatorname{div}\varphi(y) \, d\mathcal{L}^n(y) = -\int_\Omega \mathbb{1}_{E_t}(y) \, \operatorname{div}\varphi(y) \, d\mathcal{L}^n(y). \end{align*} Set $E_t := \{u > t\} \cap \Omega$. The Fubini-Tonelli theorem applies to the joint $(t, y)$-integrand $\mathbb{1}_{E_t}(y) \operatorname{div}\varphi(y)$ on $\mathbb{R} \times \Omega$ because $\varphi \in C_c^\infty$ implies $\operatorname{div}\varphi$ is bounded with compact support $K \subset \Omega$, and the layer-cake integral \begin{align*} \int_K \int_{-\infty}^{+\infty} \mathbb{1}_{E_t}(y) \, d\mathcal{L}^1(t) \, d\mathcal{L}^n(y) = \int_K (u^+(y))_+ \, d\mathcal{L}^n + \cdots \end{align*} (more precisely, the absolute layer-cake decomposition of $|u|$) is finite since $u \in L^1_{\mathrm{loc}}$ and $K$ is compact. Hence \begin{align*} \int_\Omega u(y) \, \operatorname{div} \varphi(y) \, d\mathcal{L}^n(y) = \int_{-\infty}^{+\infty} \int_\Omega \mathbb{1}_{E_t}(y) \, \operatorname{div} \varphi(y) \, d\mathcal{L}^n(y) \, d\mathcal{L}^1(t). \end{align*} Recognising the inner integral as $-D\mathbb{1}_{E_t}(\varphi)$ and substituting yields \begin{align*} Du(\varphi) = \int_{-\infty}^{+\infty} D\mathbb{1}_{E_t}(\varphi) \, d\mathcal{L}^1(t). \end{align*} [guided] The first reduction is to express $Du$ as an integral over levels of $D\mathbb{1}_{E_t}$. The mechanism is the layer-cake representation of $u$ as a superposition of indicators of superlevel sets, fed into the distributional definition of $Du$. *Distributional gradient of $u$.* By definition, for every $\varphi \in C_c^\infty(\Omega; \mathbb{R}^n)$, \begin{align*} Du(\varphi) = -\int_\Omega u(y) \, \operatorname{div} \varphi(y) \, d\mathcal{L}^n(y). \end{align*} This is the distribution-theoretic identity: $D u$ is the gradient distribution, $\varphi$ is the test field, $\operatorname{div} \varphi$ is its divergence. *Layer-cake (Cavalieri) representation.* For any extended real number $r \in \mathbb{R}$, \begin{align*} r = \int_0^\infty \mathbb{1}_{\{r > t\}} \, d\mathcal{L}^1(t) - \int_{-\infty}^0 \mathbb{1}_{\{r \le t\}} \, d\mathcal{L}^1(t). \end{align*} (Check: for $r \ge 0$ the first integral is $r$ and the second is $0$; for $r < 0$ the first is $0$ and the second is $-r$.) Applied pointwise at $y \in \Omega$ with $u(y)$ finite (which holds outside an $\mathcal{L}^n$-null set): \begin{align*} u(y) = \int_0^\infty \mathbb{1}_{E_t}(y) \, d\mathcal{L}^1(t) - \int_{-\infty}^0 \mathbb{1}_{\{u \le t\}}(y) \, d\mathcal{L}^1(t). \end{align*} *Combining the two pieces with $\int \operatorname{div}\varphi = 0$.* For $t < 0$, write $\mathbb{1}_{\{u \le t\}} = \mathbb{1}_\Omega - \mathbb{1}_{E_t}$. Then \begin{align*} \int_\Omega \mathbb{1}_{\{u \le t\}} \operatorname{div}\varphi \, d\mathcal{L}^n = \int_\Omega \operatorname{div}\varphi \, d\mathcal{L}^n - \int_\Omega \mathbb{1}_{E_t} \operatorname{div}\varphi \, d\mathcal{L}^n = -\int_\Omega \mathbb{1}_{E_t} \operatorname{div}\varphi \, d\mathcal{L}^n, \end{align*} using $\int_\Omega \operatorname{div}\varphi \, d\mathcal{L}^n = \int_{\mathbb{R}^n} \operatorname{div}\varphi \, d\mathcal{L}^n = 0$ by the divergence theorem for compactly supported smooth fields. *Fubini-Tonelli justification.* The integrand $(t, y) \mapsto \mathbb{1}_{E_t}(y) \operatorname{div}\varphi(y)$ is jointly Borel measurable on $\mathbb{R} \times \Omega$. Its absolute value is dominated by $\|\operatorname{div}\varphi\|_\infty \mathbb{1}_K(y) \mathbb{1}_{E_t}(y)$, where $K = \operatorname{supp}\varphi \subset \Omega$ is compact. The double integral of this dominator equals \begin{align*} \|\operatorname{div}\varphi\|_\infty \int_K \int_{-\infty}^{+\infty} \mathbb{1}_{E_t}(y) \, d\mathcal{L}^1(t) \, d\mathcal{L}^n(y). \end{align*} The inner integral $\int_{-\infty}^{+\infty} \mathbb{1}_{u(y)>t} \, d\mathcal{L}^1(t)$ is $u(y)_+ + \infty \cdot \mathbb{1}_{u(y)>0}$ — strictly speaking $\mathbb{1}_{u>t}=1$ for *all* $t < u(y)$, so the integral is $+\infty$ for any $y$. The correct dominator is instead $\mathbb{1}_K(y) \cdot \mathbb{1}_{|t| \le |u(y)|+1}(t)$ for the *combined* layer-cake decomposition (positive and negative pieces together), whose double integral equals $\int_K (|u(y)|+1) \, d\mathcal{L}^n(y) < \infty$ since $u \in L^1_{\mathrm{loc}}$ and $K$ is compact. Hence Fubini-Tonelli applies to the combined two-piece integral. Swapping order in each piece and combining via the identity from the previous paragraph: \begin{align*} \int_\Omega u \operatorname{div} \varphi \, d\mathcal{L}^n = \int_{-\infty}^{+\infty} \int_\Omega \mathbb{1}_{E_t}(y) \operatorname{div} \varphi(y) \, d\mathcal{L}^n(y) \, d\mathcal{L}^1(t). \end{align*} *Recognising $D\mathbb{1}_{E_t}$.* The inner integral is $-D\mathbb{1}_{E_t}(\varphi)$ by definition. Hence \begin{align*} \int_\Omega u \operatorname{div} \varphi \, d\mathcal{L}^n = -\int_{-\infty}^{+\infty} D\mathbb{1}_{E_t}(\varphi) \, d\mathcal{L}^1(t), \end{align*} and substituting into $Du(\varphi) = -\int_\Omega u \operatorname{div} \varphi \, d\mathcal{L}^n$: \begin{align*} Du(\varphi) = \int_{-\infty}^{+\infty} D\mathbb{1}_{E_t}(\varphi) \, d\mathcal{L}^1(t). \end{align*} This is the level-set decomposition of $Du$ as a $\mathcal{L}^1$-integral over levels of the gradient measures of the superlevel sets. [/guided] [/step] [step:Apply Gauss-Green to each $D\mathbb{1}_{E_t}$ and fix the sign convention] By the [BV Coarea Formula](/theorems/598), $E_t = \{u > t\}$ has finite perimeter in $\Omega$ for $\mathcal{L}^1$-a.e. $t \in \mathbb{R}$. Define \begin{align*} G := \{ t \in \mathbb{R} : E_t \text{ has finite perimeter in } \Omega \}, \qquad \mathcal{L}^1(\mathbb{R} \setminus G) = 0. \end{align*} For every $t \in G$, the [Gauss-Green Theorem for Sets of Finite Perimeter](/theorems/3119) — using the *outer unit normal* $\nu_{E_t}$ on the reduced boundary $\partial^* E_t$ — gives \begin{align*} \int_{E_t} \operatorname{div}\varphi \, d\mathcal{L}^n = \int_{\partial^* E_t} \varphi \cdot \nu_{E_t} \, d\mathcal{H}^{n-1}. \end{align*} Equivalently, as a vector-valued Radon measure on $\Omega$, \begin{align*} D\mathbb{1}_{E_t} = -\nu_{E_t} \, \mathcal{H}^{n-1} \lfloor \partial^* E_t, \end{align*} so that \begin{align*} D\mathbb{1}_{E_t}(\varphi) = -\int_{\partial^* E_t} \varphi(x) \cdot \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Combining with Step 1 and dropping the $\mathcal{L}^1$-null complement of $G$, \begin{align*} Du(\varphi) = \int_G D\mathbb{1}_{E_t}(\varphi) \, d\mathcal{L}^1(t) = -\int_G \int_{\partial^* E_t} \varphi(x) \cdot \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t). \tag{$\ast$} \end{align*} *Sign convention for $\nu_{E_t}$ versus $\nu_u$.* The outer unit normal $\nu_{E_t}$ at a reduced-boundary point $x \in \partial^* E_t$ points *out* of $E_t = \{u > t\}$, i.e., *into* the complement $\{u \le t\}$. Heuristically: $\nu_{E_t}$ points from larger values of $u$ (inside $E_t$) toward smaller values of $u$ (outside). The convention for the jump normal $\nu_u$ at $x \in J_u$ is that $\nu_u$ points from the $u^-$ side toward the $u^+$ side. Hence at a jump point: \begin{align*} \nu_{E_t}(x) = -\nu_u(x) \qquad \text{for } t \in (u^-(x), u^+(x)) \cap G. \end{align*} This sign — with $\nu_{E_t}$ opposite to $\nu_u$ at jump points — is the source of the positive sign in the final jump formula, after combination with the negative sign in $D\mathbb{1}_{E_t} = -\nu_{E_t} \mathcal{H}^{n-1}\lfloor \partial^* E_t$. [guided] The second reduction uses the [Gauss-Green Theorem for Sets of Finite Perimeter](/theorems/3119) to replace each $D\mathbb{1}_{E_t}(\varphi)$ by an explicit surface integral, and pins down the sign convention relating $\nu_{E_t}$ to the jump normal $\nu_u$. *Hypothesis verification for [theorem 598](/theorems/598).* The [BV Coarea Formula](/theorems/598) requires $u \in BV(\Omega)$ and $\Omega$ open — both given. Conclusion (i): $E_t$ has finite perimeter for $\mathcal{L}^1$-a.e. $t$. The set $G$ is then a Borel co-null subset of $\mathbb{R}$. *Hypothesis verification for [theorem 3119](/theorems/3119).* For $t \in G$, the set $E_t$ is Borel (preimage of $(t, \infty)$ under the Borel function $u$) and has finite perimeter in $\Omega$. Theorem 3119 gives, for every $\varphi \in C_c^1(\Omega; \mathbb{R}^n)$, \begin{align*} \int_{E_t} \operatorname{div} \varphi \, d\mathcal{L}^n = \int_{\partial^* E_t} \varphi \cdot \nu_{E_t} \, d\mathcal{H}^{n-1}, \end{align*} where $\nu_{E_t}$ is the *outer* unit normal of $E_t$ on its reduced boundary. *Equivalent measure-theoretic form.* Recall $D\mathbb{1}_{E_t}(\varphi) := -\int_\Omega \mathbb{1}_{E_t} \operatorname{div}\varphi \, d\mathcal{L}^n = -\int_{E_t} \operatorname{div}\varphi \, d\mathcal{L}^n$. So \begin{align*} D\mathbb{1}_{E_t}(\varphi) = -\int_{\partial^* E_t} \varphi \cdot \nu_{E_t} \, d\mathcal{H}^{n-1}, \end{align*} or equivalently as a vector-valued Radon measure on $\Omega$, \begin{align*} D\mathbb{1}_{E_t} = -\nu_{E_t} \, \mathcal{H}^{n-1} \lfloor \partial^* E_t. \tag{$\dagger$} \end{align*} *Substitution into the Step 1 identity.* Inserting $(\dagger)$ into the level-set integral: \begin{align*} Du(\varphi) = \int_G D\mathbb{1}_{E_t}(\varphi) \, d\mathcal{L}^1(t) = -\int_G \int_{\partial^* E_t} \varphi(x) \cdot \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t). \end{align*} *Sign relation $\nu_{E_t} = -\nu_u$ at jump points.* At a jump point $x \in J_u$ with $\nu_u(x)$ defined and $t \in (u^-(x), u^+(x)) \cap G$, the half-ball blow-up analysis (Step 2 of the proof of [theorem 3123](/theorems/3123)) shows that $E_t$ has the half-space blow-up \begin{align*} \frac{E_t - x}{r} \to \{ y \in \mathbb{R}^n : (y - x) \cdot \nu_u(x) > 0 \}, \quad \text{i.e., the half-space on the } u^+\text{-side of } \nu_u(x). \end{align*} Indeed, on the $u^+$-side of $\nu_u(x)$ the function $u$ has approximate limit $u^+(x) > t$, so the density of $E_t = \{u > t\}$ approaches $1$ on that side; symmetrically on the $u^-$-side it approaches $0$. Hence $E_t$ blows up at $x$ to the half-space whose interior is the $u^+$-side of $\nu_u(x)$. The outer unit normal of this half-space at $x$ points *out* of the $u^+$-side, i.e., *toward* the $u^-$-side, i.e., in the direction $-\nu_u(x)$. So \begin{align*} \nu_{E_t}(x) = -\nu_u(x). \end{align*} This sign relation holds for $\mathcal{H}^{n-1}$-a.e. $x \in J_u$ (where $\nu_u(x)$ is defined) and every $t \in (u^-(x), u^+(x)) \cap G$. *Why this matters.* In $(\dagger)$ the surface measure $D\mathbb{1}_{E_t}$ carries a minus sign relative to $\nu_{E_t}$. Combined with $\nu_{E_t} = -\nu_u$ at jump points, the two minus signs cancel: $D\mathbb{1}_{E_t}$ is *plus* $\nu_u$ times $\mathcal{H}^{n-1}\lfloor \partial^* E_t$ at jump points. Integrating in $t$ over $(u^-(x), u^+(x))$ then produces the positive-sign jump formula $D^j u = (u^+ - u^-) \nu_u \, \mathcal{H}^{n-1}\lfloor J_u$. The next step makes this precise via Fubini against an appropriate Radon measure on $\Omega$. [/guided] [/step] [step:Define the joint Radon measure on $\Omega$ via Fubini] The double integral $(\ast)$ is integration of the vector-valued integrand $-\varphi(x) \cdot \nu_{E_t}(x)$ against the family of measures $\mathcal{H}^{n-1}\lfloor \partial^* E_t$, parameterised by $t \in G$. To apply Fubini, we package this family into a single Radon measure on $\Omega$. *The total perimeter measure $\Sigma$.* Define a positive Radon measure on $\Omega$ by \begin{align*} \Sigma(B) := \int_G \mathcal{H}^{n-1}(\partial^* E_t \cap B) \, d\mathcal{L}^1(t) \quad \text{for every Borel } B \subseteq \Omega. \end{align*} Equivalently, $\Sigma$ is the projection onto $\Omega$ (via Fubini) of the product measure on $G \times \Omega$ whose disintegration in $t$ is $t \mapsto \mathcal{H}^{n-1}\lfloor \partial^* E_t$. By the [BV Coarea Formula](/theorems/598), \begin{align*} \Sigma(\Omega) = \int_G \mathcal{H}^{n-1}(\partial^* E_t) \, d\mathcal{L}^1(t) = \int_G P(E_t; \Omega) \, d\mathcal{L}^1(t) = |Du|(\Omega) < \infty, \end{align*} so $\Sigma$ is a finite Radon measure on $\Omega$. *The vector-valued joint measure $T$.* Define a vector-valued Radon measure $T$ on $\Omega$ by \begin{align*} T(B) := \int_G \int_{\partial^* E_t \cap B} \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t) = \int_G \big( \nu_{E_t} \mathcal{H}^{n-1}\lfloor \partial^* E_t \big)(B) \, d\mathcal{L}^1(t). \end{align*} The total variation $|T|$ is dominated by $\Sigma$ (since $|\nu_{E_t}| = 1$), so $T \ll \Sigma$ with Radon-Nikodym density bounded by $1$ in norm. Moreover, by Fubini-Tonelli applied to the joint $(t, x)$-measurable integrand $\varphi(x) \cdot \nu_{E_t}(x) \mathbb{1}_{\partial^* E_t}(x)$ (whose absolute value is bounded by $\|\varphi\|_\infty \mathbb{1}_{\partial^* E_t}(x)$, jointly integrable against $\mathcal{L}^1 \otimes \mathcal{H}^{n-1}$ on $G \times \Omega$ because $\Sigma(\Omega) < \infty$): \begin{align*} \int_G \int_{\partial^* E_t} \varphi(x) \cdot \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t) = \int_\Omega \varphi(x) \cdot dT(x). \end{align*} Combining with $(\ast)$: \begin{align*} Du(\varphi) = -\int_\Omega \varphi(x) \cdot dT(x). \tag{$\ast\ast$} \end{align*} In other words, $Du = -T$ as vector-valued Radon measures on $\Omega$. [guided] The Fubini step converts the iterated $(t, x)$-integral $(\ast)$ into integration against a single vector-valued Radon measure $T$ on $\Omega$. The role of the BV coarea formula is to certify that the "projected" total perimeter measure $\Sigma$ has finite mass equal to $|Du|(\Omega)$, hence is a legitimate Radon measure. *Construction of $\Sigma$.* For Borel $B \subseteq \Omega$, the function $t \mapsto \mathcal{H}^{n-1}(\partial^* E_t \cap B)$ is Borel measurable in $t$ (a standard fact for the family of reduced boundaries; see the proof of theorem 598). The set function \begin{align*} \Sigma(B) := \int_G \mathcal{H}^{n-1}(\partial^* E_t \cap B) \, d\mathcal{L}^1(t) \end{align*} is countably additive in $B$ by monotone convergence, hence a positive Borel measure on $\Omega$. By the BV coarea formula applied with $B = \Omega$: \begin{align*} \Sigma(\Omega) = \int_G P(E_t; \Omega) \, d\mathcal{L}^1(t) = |Du|(\Omega), \end{align*} which is finite because $u \in BV(\Omega)$. So $\Sigma$ is a finite positive Radon measure on $\Omega$. *Construction of $T$.* The vector-valued counterpart is \begin{align*} T(B) := \int_G \int_{\partial^* E_t \cap B} \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t), \end{align*} well-defined because the inner integral is bounded in norm by $\mathcal{H}^{n-1}(\partial^* E_t \cap B)$ and the outer integration converges absolutely (total variation bounded by $\Sigma(\Omega) = |Du|(\Omega)$). $T$ is a vector-valued Radon measure on $\Omega$ with $|T| \le \Sigma$. *Fubini swap.* The integrand $F(t, x) := \mathbb{1}_{\partial^* E_t}(x) \, \varphi(x) \cdot \nu_{E_t}(x)$ on $G \times \Omega$ is Borel measurable jointly in $(t, x)$. Its absolute value is bounded by $\|\varphi\|_\infty \mathbb{1}_{\partial^* E_t}(x)$. The integral of the dominator over $G \times \Omega$ against $\mathcal{L}^1 \otimes \mathcal{H}^{n-1}$ (more precisely, against the Radon measure on $G \times \Omega$ with disintegration $t \mapsto \mathcal{H}^{n-1}\lfloor \partial^* E_t$) equals $\|\varphi\|_\infty \Sigma(\Omega) = \|\varphi\|_\infty |Du|(\Omega) < \infty$. So Fubini-Tonelli applies, and \begin{align*} \int_G \int_{\partial^* E_t} \varphi(x) \cdot \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t) = \int_\Omega \varphi(x) \cdot dT(x). \end{align*} *Resulting identity.* Combining with $(\ast)$: \begin{align*} Du(\varphi) = -\int_G \int_{\partial^* E_t} \varphi \cdot \nu_{E_t} \, d\mathcal{H}^{n-1} \, d\mathcal{L}^1(t) = -\int_\Omega \varphi \cdot dT, \end{align*} i.e., $Du = -T$ as vector-valued Radon measures. The next step decomposes $T$ on $J_u$ versus off $J_u$, identifying the $J_u$-piece with the jump formula and the off-$J_u$ piece with $-(D^a u + D^c u)$. [/guided] [/step] [step:Identify the $J_u$-contribution to $T$ via the level-set characterisation] We compute $T \lfloor J_u$. The strategy: at $\mathcal{H}^{n-1}$-a.e. $x \in J_u$, the inner $t$-integral defining $T$ is supported on $t \in (u^-(x), u^+(x))$, and $\nu_{E_t}(x) = -\nu_u(x)$ there. Integration in $t$ then produces the factor $u^+(x) - u^-(x)$. Formally, let $x \in J_u$ be a point where $\nu_u(x)$ is defined ($\mathcal{H}^{n-1}$-a.e. on $J_u$, by [theorem 3123](/theorems/3123)) and at which the half-ball density argument applies. [claim:For $\mathcal{H}^{n-1}$-a.e. $x \in J_u$ and $\mathcal{L}^1$-a.e. $t \in \mathbb{R}$, $x \in \partial^* E_t$ if and only if $t \in (u^-(x), u^+(x))$, and in that case $\nu_{E_t}(x) = -\nu_u(x)$] The boundary cases $t = u^\pm(x)$ form an $\mathcal{L}^1$-null set in $\mathbb{R}$ (a fixed two-point set $\{u^-(x), u^+(x)\}$ for each $x$), so they are excluded from the $\mathcal{L}^1$-a.e. statement and need no separate analysis for the Fubini integration. [/claim] [proof] Fix $x \in J_u$ at which $\nu_u(x)$ is defined. *Case $t \in (u^-(x), u^+(x))$:* The half-ball density argument from Step 2 of the proof of [theorem 3123](/theorems/3123) applies. Since $u$ has approximate limit $u^+(x)$ on the $\nu_u(x)$-positive half-ball and $u^-(x)$ on the $\nu_u(x)$-negative half-ball, and $u^-(x) < t < u^+(x)$: - on the positive side, $\Theta(\{u > t\}, x) = 1/2$ (full density restricted to the half-ball); - on the negative side, $\Theta(\{u > t\}, x) = 0$. The blow-up sequence $\mathbb{1}_{(E_t - x)/r}$ converges in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$ to $\mathbb{1}_{\{y \cdot \nu_u(x) > 0\}}$, identifying $x \in \partial^* E_t$ with the outer normal of $E_t$ at $x$ pointing *out of* the $\nu_u(x)$-positive half-space, i.e., $\nu_{E_t}(x) = -\nu_u(x)$. *Case $t < u^-(x)$:* By the definition of $u^-(x)$ (the *largest* threshold $\tau$ with $\Theta^*(\{u < \tau\}, x) = 0$), there exists $\tau$ with $t < \tau < u^-(x)$ and $\Theta^*(\{u < \tau\}, x) = 0$. Since $\{u \le t\} \subseteq \{u < \tau\}$, also $\Theta^*(\{u \le t\}, x) = 0$, so $\Theta^*(E_t^c \cap \Omega, x) = 0$, i.e., $\Theta(E_t, x) = 1$. A density-one point is *not* a reduced-boundary point (the half-space blow-up at a reduced-boundary point has density $1/2$). Hence $x \notin \partial^* E_t$. *Case $t > u^+(x)$:* Symmetric: there exists $\tau$ with $u^+(x) < \tau < t$ and $\Theta^*(\{u > \tau\}, x) = 0$. Since $\{u > t\} \subseteq \{u > \tau\}$, also $\Theta^*(E_t, x) = 0$. So $x \notin \partial^* E_t$. *Boundary cases $t = u^\pm(x)$:* The set of such $t$, for fixed $x$, is the two-point set $\{u^-(x), u^+(x)\}$, which has $\mathcal{L}^1$-measure zero. The Fubini integration in $t$ that defines $T$ ignores $\mathcal{L}^1$-null sets, so these endpoint values do not contribute and need not be classified individually. Combining the three (interior) cases: for fixed $x \in J_u$ with $\nu_u(x)$ defined, the set $\{t \in \mathbb{R} : x \in \partial^* E_t\}$ equals $(u^-(x), u^+(x))$ up to an $\mathcal{L}^1$-null set, and on this interval $\nu_{E_t}(x) = -\nu_u(x)$. [/proof] *Computing $T \lfloor J_u$.* Fix a Borel set $A \subseteq J_u$. By the definition of $T$ and the Claim, \begin{align*} T(A) &= \int_G \int_{\partial^* E_t \cap A} \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t) \\ &= \int_G \int_A \mathbb{1}_{\partial^* E_t}(x) \nu_{E_t}(x) \, d\mathcal{H}^{n-1}(x) \, d\mathcal{L}^1(t). \end{align*} By Fubini (applied as in Step 3 — joint integrability is controlled by $\Sigma(\Omega) = |Du|(\Omega)$), \begin{align*} T(A) = \int_A \left( \int_G \mathbb{1}_{\partial^* E_t}(x) \nu_{E_t}(x) \, d\mathcal{L}^1(t) \right) d\mathcal{H}^{n-1}(x). \end{align*} By the Claim, the inner integral is $-\nu_u(x)$ on the $\mathcal{L}^1$-set $(u^-(x), u^+(x))$ and zero elsewhere, so \begin{align*} \int_G \mathbb{1}_{\partial^* E_t}(x) \nu_{E_t}(x) \, d\mathcal{L}^1(t) = -\nu_u(x) \int_{u^-(x)}^{u^+(x)} d\mathcal{L}^1(t) = -(u^+(x) - u^-(x)) \nu_u(x). \end{align*} Hence \begin{align*} T(A) = -\int_A (u^+(x) - u^-(x)) \nu_u(x) \, d\mathcal{H}^{n-1}(x) \quad \text{for every Borel } A \subseteq J_u. \end{align*} By $(\ast\ast)$, $Du \lfloor J_u = -T \lfloor J_u$, so \begin{align*} Du \lfloor J_u (A) = \int_A (u^+(x) - u^-(x)) \nu_u(x) \, d\mathcal{H}^{n-1}(x). \tag{$\heartsuit$} \end{align*} [guided] We restrict the joint Radon measure $T$ to the jump set $J_u$ and compute its restriction explicitly. *Hypothesis verification for the Claim.* The Claim invokes the half-ball density analysis from Step 2 of the proof of [theorem 3123](/theorems/3123). The hypothesis is $u \in BV(\Omega)$ (which we have) and $x \in J_u$ at which $\nu_u(x)$ is defined ($\mathcal{H}^{n-1}$-a.e. on $J_u$, by theorem 3123). Both hypotheses hold for the relevant points of $J_u$. *Key observation: the boundary cases are $\mathcal{L}^1$-null.* For each fixed $x$, the set $\{t \in \mathbb{R} : t = u^-(x) \text{ or } t = u^+(x)\}$ has at most two points, hence $\mathcal{L}^1$-measure zero. The inner $t$-integral $\int_G \mathbb{1}_{\partial^* E_t}(x) \nu_{E_t}(x) \, d\mathcal{L}^1(t)$ is therefore unchanged by the values at $t = u^\pm(x)$. The case analysis in the Claim need only cover $t \neq u^\pm(x)$, which it does. *Sign tracking.* The Gauss-Green identity $(\dagger)$ has $D\mathbb{1}_{E_t} = -\nu_{E_t} \mathcal{H}^{n-1}\lfloor \partial^* E_t$ (minus sign on the outer normal). The level-set integral identity gives $Du = -T$ where $T$ integrates $\nu_{E_t}$ against $\mathcal{H}^{n-1}\lfloor \partial^* E_t$ (no sign). At a jump point, the Claim gives $\nu_{E_t}(x) = -\nu_u(x)$. Combining: - $T \lfloor J_u (A) = -\int_A (u^+ - u^-) \nu_u \, d\mathcal{H}^{n-1}$ (one minus sign from $\nu_{E_t} = -\nu_u$); - $Du \lfloor J_u = -T \lfloor J_u = +\int (u^+ - u^-) \nu_u \, d\mathcal{H}^{n-1}$ (second minus sign from $Du = -T$). The two minus signs cancel, yielding the positive-sign formula $(\heartsuit)$. *Computation of $T(A)$ for $A \subseteq J_u$.* Apply Fubini to the double integral defining $T(A)$. The integrand is jointly bounded in absolute value by $\mathbb{1}_{\partial^* E_t}(x)$, which is integrable against $\mathcal{L}^1 \otimes \mathcal{H}^{n-1}$ on $G \times A$ with total mass $\Sigma(A) \le \Sigma(\Omega) = |Du|(\Omega) < \infty$. So \begin{align*} T(A) = \int_A \left( \int_G \mathbb{1}_{\partial^* E_t}(x) \nu_{E_t}(x) \, d\mathcal{L}^1(t) \right) d\mathcal{H}^{n-1}(x), \end{align*} and the parenthesised inner integral is, by the Claim, equal to $-(u^+(x) - u^-(x)) \nu_u(x)$ for $\mathcal{H}^{n-1}$-a.e. $x \in J_u$. Integration against $\mathcal{H}^{n-1}\lfloor A$ produces \begin{align*} T(A) = -\int_A (u^+(x) - u^-(x)) \nu_u(x) \, d\mathcal{H}^{n-1}(x). \end{align*} By $(\ast\ast)$, $Du = -T$, so $Du \lfloor J_u (A) = -T(A) = \int_A (u^+ - u^-) \nu_u \, d\mathcal{H}^{n-1}$, which is $(\heartsuit)$. *Vector-measure form.* Equivalently, \begin{align*} Du \lfloor J_u = (u^+ - u^-) \nu_u \, \mathcal{H}^{n-1} \lfloor J_u \end{align*} as a vector-valued Radon measure on $J_u$. [/guided] [/step] [step:Identify the off-$J_u$ contribution as $D^a u + D^c u$] We now compute $T \lfloor (\Omega \setminus J_u)$ and identify it with $-(D^a u + D^c u)$. The key is that $(\ast\ast)$ gives $Du = -T$ as Radon measures on *all* of $\Omega$, so a decomposition of $T$ into $T \lfloor J_u + T \lfloor (\Omega \setminus J_u)$ corresponds to a decomposition of $Du$. By $(\ast\ast)$, $Du = -T$ on $\Omega$. Splitting both sides on $J_u$ versus its complement: \begin{align*} Du \lfloor J_u + Du \lfloor (\Omega \setminus J_u) = -T \lfloor J_u - T \lfloor (\Omega \setminus J_u). \end{align*} By $(\heartsuit)$, $Du \lfloor J_u = -T \lfloor J_u = (u^+ - u^-) \nu_u \, \mathcal{H}^{n-1}\lfloor J_u$. Subtracting, \begin{align*} Du \lfloor (\Omega \setminus J_u) = -T \lfloor (\Omega \setminus J_u). \tag{$\diamondsuit$} \end{align*} It remains to relate $Du \lfloor (\Omega \setminus J_u)$ to $D^a u$ and $D^c u$ — but this is *definitional*, not a citation of the structure theorem statement. The Lebesgue-Radon-Nikodym decomposition of the vector-valued Radon measure $Du$ with respect to $\mathcal{L}^n$ produces an absolutely continuous part $D^a u \ll \mathcal{L}^n$ and a singular part $D^s u \perp \mathcal{L}^n$. The singular part is then split further into atomic-like (concentrated on $\mathcal{H}^{n-1}$-$\sigma$-finite sets) and diffuse parts: \begin{align*} D^j u := D^s u \lfloor J_u, \qquad D^c u := D^s u \lfloor (\Omega \setminus J_u). \end{align*} These are *definitions* — the operative content of the [BV Structure Theorem](/theorems/595) used here is only the existence of the Lebesgue-Radon-Nikodym decomposition $Du = D^a u + D^s u$ (a general measure-theoretic fact, not specific to BV) and the *definition* of $D^j u$ as the restriction of $D^s u$ to $J_u$. With these definitions: - $D^a u + D^c u = D^a u + D^s u \lfloor (\Omega \setminus J_u) = Du \lfloor (\Omega \setminus J_u)$ (since $D^a u \ll \mathcal{L}^n$ and $\mathcal{L}^n(J_u) = 0$, so $D^a u(J_u) = 0$, hence $D^a u = D^a u \lfloor (\Omega \setminus J_u)$). - $D^j u = D^s u \lfloor J_u = Du \lfloor J_u - D^a u \lfloor J_u = Du \lfloor J_u$. Combining with $(\heartsuit)$ and $(\diamondsuit)$: \begin{align*} D^j u = Du \lfloor J_u = (u^+ - u^-) \nu_u \, \mathcal{H}^{n-1}\lfloor J_u, \end{align*} \begin{align*} D^a u + D^c u = Du \lfloor (\Omega \setminus J_u) = -T \lfloor (\Omega \setminus J_u). \end{align*} The first identity is the conclusion. The second is consistent — it is the statement that the off-$J_u$ part of the level-set integral equals the absolutely continuous plus Cantor parts of $Du$, which holds because, by definition, those are precisely the components of $Du$ supported off $J_u$. *Total variation.* Since $|\nu_u(x)| = 1$ at $\mathcal{H}^{n-1}$-a.e. $x \in J_u$ and $u^+(x) > u^-(x)$ on $J_u$ (by the definition of the jump set), \begin{align*} |D^j u| = (u^+ - u^-) \mathcal{H}^{n-1} \lfloor J_u. \end{align*} The proof is complete. [guided] This step closes the off-$J_u$ accounting without invoking the jump-formula content of the [BV Structure Theorem](/theorems/595) (which would be circular). The only structural fact used is the Lebesgue-Radon-Nikodym decomposition of $Du$ with respect to $\mathcal{L}^n$ — a general measure-theoretic identity, not specific to BV — together with the *definition* of $D^j u$ as the restriction of the singular part to $J_u$. *Definitions of $D^a u$, $D^j u$, $D^c u$.* Given the vector-valued Radon measure $Du$ on $\Omega$, the Lebesgue decomposition with respect to $\mathcal{L}^n$ gives $Du = D^a u + D^s u$ with $D^a u \ll \mathcal{L}^n$ and $D^s u \perp \mathcal{L}^n$. Define \begin{align*} D^j u := D^s u \lfloor J_u, \qquad D^c u := D^s u \lfloor (\Omega \setminus J_u). \end{align*} These are definitions, not consequences of theorem 595. (Theorem 595's substantive content is that $D^c u$ vanishes on $\mathcal{H}^{n-1}$-$\sigma$-finite sets — a fact we do not invoke here.) *Restriction of $Du$ to $\Omega \setminus J_u$.* Since $\mathcal{L}^n(J_u) = 0$ and $D^a u \ll \mathcal{L}^n$, $D^a u(J_u) = 0$, so $D^a u = D^a u \lfloor (\Omega \setminus J_u)$. Hence \begin{align*} Du \lfloor (\Omega \setminus J_u) &= D^a u \lfloor (\Omega \setminus J_u) + D^s u \lfloor (\Omega \setminus J_u) \\ &= D^a u + D^c u. \end{align*} *Restriction of $Du$ to $J_u$.* Symmetrically, $Du \lfloor J_u = D^a u \lfloor J_u + D^s u \lfloor J_u = 0 + D^j u = D^j u$. *Combining with the level-set computation.* By $(\heartsuit)$, \begin{align*} D^j u = Du \lfloor J_u = (u^+(x) - u^-(x)) \nu_u(x) \, \mathcal{H}^{n-1} \lfloor J_u. \end{align*} This is the conclusion. The off-$J_u$ identity $D^a u + D^c u = -T \lfloor (\Omega \setminus J_u)$ is a byproduct: it states that the level-set integral, restricted to $\Omega \setminus J_u$, recovers the $D^a u + D^c u$ part of $Du$. Schematically, points $x \notin J_u$ at which $u^-(x) = u^+(x)$ (i.e., points of approximate continuity, which carry the $D^a u$ and $D^c u$ mass) still appear on $\partial^* E_t$ for some $t$, and the level-set integral correctly attributes their contribution. *Total variation $|D^j u|$.* As a vector-valued Radon measure with density $(u^+ - u^-) \nu_u$ relative to $\mathcal{H}^{n-1} \lfloor J_u$, the total variation of $D^j u$ is \begin{align*} |D^j u| = |(u^+ - u^-) \nu_u| \, \mathcal{H}^{n-1} \lfloor J_u = (u^+ - u^-) \mathcal{H}^{n-1} \lfloor J_u, \end{align*} using $|\nu_u(x)| = 1$ (since $\nu_u: J_u \to S^{n-1}$ takes values in the unit sphere) and $u^+(x) > u^-(x)$ on $J_u$ (strict, by the definition of the jump set as the points of $S_u$ where the upper and lower approximate limits differ). *Conclusion.* The jump decomposition formula is established: \begin{align*} D^j u = (u^+ - u^-) \nu_u \, \mathcal{H}^{n-1} \lfloor J_u, \qquad |D^j u| = (u^+ - u^-) \mathcal{H}^{n-1} \lfloor J_u. \end{align*} The proof is complete. [/guided] [/step]