[proofplan] The strategy reduces the question to showing that the difference $w := u - v \in BV(\Omega)$ has $w = 0$ $\mathcal{H}^{n-1}$-a.e. The hypotheses give $w^+ = w^- = 0$ on $\Omega \setminus J_w$ ($\mathcal{H}^{n-1}$-a.e.) and $D^c w = 0$. We first show $J_w \subseteq J_u \triangle J_v = \varnothing$ up to an $\mathcal{H}^{n-1}$-null set, so $J_w$ is $\mathcal{H}^{n-1}$-null and $D^j w = 0$. Combined with $D^c w = 0$, this gives $Dw = D^a w = \nabla w \cdot \mathcal{L}^n$. We then invoke [Approximate Gradient Identifies $D^a u$](/theorems/3130) to identify $\nabla w$ with the approximate gradient at $\mathcal{L}^n$-a.e. point. The approximate continuity of $w$ at every point of $\Omega \setminus J_w$ (where $w^+ = w^-$) forces $\nabla w = 0$ at all approximate-continuity points where the approximate gradient vanishes — and in fact, vanishing of $w^+ = w^- = 0$ at $\mathcal{H}^{n-1}$-a.e. point already pins down the value of the precise representative to be $0$. The final step: a BV function whose gradient measure $Dw$ vanishes is constant on each connected component of $\Omega$, and the value $0$ at $\mathcal{H}^{n-1}$-a.e. (hence at $\mathcal{L}^n$-a.e.) point forces the constant to be $0$. [/proofplan]

[step:Define $w := u - v$ and translate the hypotheses to properties of $w$]Set \begin{align*} w : \Omega &\to \mathbb{R}, \\ x &\mapsto u(x) - v(x). \end{align*} Since $u, v \in BV(\Omega)$ and $BV(\Omega)$ is a vector space (a consequence of the linearity of the distributional derivative and the subadditivity of total variation), $w \in BV(\Omega)$, with $Dw = Du - Dv$ as Radon measures on $\Omega$. The decomposition of the gradient of a BV function into absolutely continuous, jump, and Cantor parts (the [BV Structure Theorem](/theorems/595)) is linear: \begin{align*} D^a w = D^a u - D^a v, \qquad D^j w = D^j u - D^j v, \qquad D^c w = D^c u - D^c v, \end{align*} because each part is identified by its support and absolute-continuity properties (the $D^a$-part is the $\mathcal{L}^n$-absolutely continuous part of $Dw$, the $D^j$-part is concentrated on $J_w$ and absolutely continuous w.r.t. $\mathcal{H}^{n-1}\lfloor J_w$, and $D^c w$ is the residual diffuse part), and these properties are preserved under linear combinations. By hypothesis (3), $D^c u = D^c v$, so $D^c w = 0$. At every $x$ where $u^+(x), v^+(x), u^-(x), v^-(x)$ all exist (which is $\mathcal{H}^{n-1}$-a.e. $x$ in $\Omega$ by [Finiteness of Approximate Limits for BV Functions](/theorems/3122) applied to both $u$ and $v$), the approximate upper and lower limits of $w = u - v$ are determined via *one-sided traces* with respect to the jump direction. Specifically, at non-jump points (where $u^+ = u^-$ and $v^+ = v^-$), the approximate limit of $w$ exists and equals $\tilde u(x) - \tilde v(x)$, with $w^+(x) = w^-(x) = \tilde u(x) - \tilde v(x)$. At jump points $x \in J_u$ (with $\nu_u(x) = \nu_v(x)$ assumed under hypothesis (2)'s implicit orientation), the same-side trace identity gives $w^+(x) = u^+(x) - v^+(x)$ and $w^-(x) = u^-(x) - v^-(x)$. Both formulas are special cases of the same one-sided trace computation, used in Step 2 to control $J_w$.[/step]

[guided]We exploit the linear structure of $BV$ to reduce to showing the difference is zero. *$BV$ is a vector space.* For $u, v \in BV(\Omega)$ and $\alpha, \beta \in \mathbb{R}$, the function $\alpha u + \beta v \in L^1(\Omega)$, and its distributional gradient is $\alpha Du + \beta Dv$, a finite vector-valued Radon measure with total variation at most $|\alpha| \cdot |Du|(\Omega) + |\beta| \cdot |Dv|(\Omega) < \infty$. So $\alpha u + \beta v \in BV(\Omega)$. In particular, $w := u - v \in BV(\Omega)$ with $Dw = Du - Dv$. *Linearity of the BV decomposition.* By [BV Structure Theorem](/theorems/595), every BV function has a unique decomposition $D = D^a + D^j + D^c$ characterised by: - $D^a$ is absolutely continuous with respect to $\mathcal{L}^n$; - $D^j$ is concentrated on the jump set and has the form $(u^+ - u^-) \nu_u \cdot \mathcal{H}^{n-1}\lfloor J_u$; - $D^c$ is singular with respect to $\mathcal{L}^n$ but vanishes on every $\mathcal{H}^{n-1}$-finite set. These three pieces are mutually singular, characterised by their behaviour with respect to $\mathcal{L}^n$ and $\mathcal{H}^{n-1}$. The decomposition is linear: if $D = D_1 + D_2$ then the absolutely continuous parts of $D_1$ and $D_2$ sum to the absolutely continuous part of $D$, and similarly for the singular pieces split between concentrated-on-rectifiable and diffuse. Hence \begin{align*} D^a w = D^a u - D^a v, \quad D^j w = D^j u - D^j v, \quad D^c w = D^c u - D^c v. \end{align*} *Hypothesis (3) gives $D^c w = 0$.* This is direct. *Approximate limits at non-jump points.* For $u, v \in BV(\Omega)$, both $u$ and $v$ have well-defined approximate upper and lower limits $u^\pm$ and $v^\pm$ for $\mathcal{H}^{n-1}$-a.e. $x \in \Omega$ (by [Finiteness of Approximate Limits for BV Functions](/theorems/3122)). At a point $x$ where neither $u$ nor $v$ jumps — meaning $u^+(x) = u^-(x)$ and $v^+(x) = v^-(x)$, with the common values denoted $\tilde u(x)$ and $\tilde v(x)$ — the approximate limit of $w = u - v$ at $x$ exists with $\tilde w(x) = \tilde u(x) - \tilde v(x)$, and consequently $w^+(x) = w^-(x) = \tilde u(x) - \tilde v(x)$. At a jump point of $u$ (or $v$), the approximate upper and lower limits of $w$ are computed not by formal arithmetic on $u^\pm, v^\pm$ but by the *one-sided trace* construction: with $\nu_u(x)$ the jump normal and $H^\pm := \{y : (y - x) \cdot \nu_u(x) \gtrless 0\}$, the trace of $w$ from $H^\pm$ equals $u^\pm(x) - v^\pm(x)$ (same-side subtraction), which gives $w^+(x) = u^+(x) - v^+(x)$ and $w^-(x) = u^-(x) - v^-(x)$ when $\nu_u(x) = \nu_v(x)$ (otherwise the labels swap). Step 2 will use this same-side identity carefully on $J_u$.[/guided]

[step:Show $J_w \subseteq J_u$ up to an $\mathcal{H}^{n-1}$-null set, hence $\mathcal{H}^{n-1}(J_w) < \infty$]The jump set of a BV function $f \in BV(\Omega)$ is \begin{align*} J_f := \{x \in \Omega : f^+(x) > f^-(x)\} \end{align*} (after the modification setting $J_f = \varnothing$ where the approximate limits do not exist, which is a null set). [claim:$J_w \subseteq J_u \cup N$ for some $\mathcal{H}^{n-1}$-null set $N$] [/claim] [proof] Let $E := \{x \in \Omega : u^+(x), u^-(x), v^+(x), v^-(x) \text{ all exist and are finite}\}$. By [theorem 3122](/theorems/3122) applied to both $u$ and $v$, $\mathcal{H}^{n-1}(\Omega \setminus E) = 0$. Fix $x \in E$. We show: if $x \notin J_u$, then $x \notin J_w$ (up to a further null set encoded in hypothesis (1)). Suppose $x \in E \setminus J_u$. Then $u^+(x) = u^-(x)$ — no jump at $x$ for $u$. Hypothesis (1) of the theorem applies at $\mathcal{H}^{n-1}$-a.e. such $x$: $u^+(x) = v^+(x)$ and $u^-(x) = v^-(x)$. Combined with $u^+(x) = u^-(x)$, we get $v^+(x) = v^-(x)$, so $x \notin J_v$ also. At such $x$, $u$ and $v$ are approximately continuous with common limits $\tilde u(x) = u^\pm(x)$ and $\tilde v(x) = v^\pm(x)$. The approximate limit of $w = u - v$ at $x$ exists by linearity of integral averages, with $\tilde w(x) = \tilde u(x) - \tilde v(x)$. By hypothesis (1), $\tilde u(x) = \tilde v(x)$, so $\tilde w(x) = 0$ and consequently $w^+(x) = w^-(x) = 0$. In particular $x \notin J_w$. Let $N_1 := \Omega \setminus E$ ($\mathcal{H}^{n-1}$-null) and $N_2 := \{x \in E \setminus J_u : (u^+(x), u^-(x)) \ne (v^+(x), v^-(x))\}$. By hypothesis (1), $\mathcal{H}^{n-1}(N_2) = 0$. The argument above shows $J_w \cap (\Omega \setminus J_u) \subseteq N_1 \cup N_2$, so taking $N := N_1 \cup N_2$ gives the claim. [/proof] It remains to handle points $x \in J_u = J_v$. At $\mathcal{H}^{n-1}$-a.e. such $x$, hypothesis (2) gives $u^+(x) = v^+(x)$ and $u^-(x) = v^-(x)$, with the same orientation of jump normal $\nu_u(x) = \nu_v(x)$ (the jump normal is intrinsic to the rectifiable structure of $J_u$, and the convention $u^+$/$u^-$ assigns each one-sided trace to a specific side determined by $\nu_u$). The approximate upper and lower limits of $w$ at such $x$ are computed via the *one-sided* traces with respect to $\nu_u(x)$: with $H^+ := \{y : (y - x) \cdot \nu_u(x) > 0\}$ and $H^- := \{y : (y - x) \cdot \nu_u(x) < 0\}$, \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r) \cap H^\pm)} \int_{B(x, r) \cap H^\pm} w(y) \, d\mathcal{L}^n(y) &= \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r) \cap H^\pm)} \int_{B(x, r) \cap H^\pm} (u(y) - v(y)) \, d\mathcal{L}^n(y) \\ &= u^\pm(x) - v^\pm(x) = 0, \end{align*} using linearity of the average and hypothesis (2). Since both one-sided traces of $w$ at $x$ are zero, the approximate upper and lower limits of $w$ exist with $w^+(x) = w^-(x) = 0$, so $x \notin J_w$. Combining with the Claim, $J_w \subseteq N \cup N'$ for $\mathcal{H}^{n-1}$-null sets $N$ (from the Claim) and $N'$ (the exceptional set on $J_u$), hence $\mathcal{H}^{n-1}(J_w) = 0$.[/step]

[guided]We track the jump set of $w$ across $J_u$ and $\Omega \setminus J_u$, and conclude $J_w$ is $\mathcal{H}^{n-1}$-null. *The two regions $\Omega \setminus J_u$ and $J_u$ partition $\Omega$.* On the first region, neither $u$ nor $v$ jumps (using $J_v = J_u$ from hypothesis (2)); on the second, both jump. *On $\Omega \setminus J_u$ (excluding a null set).* By the Claim's proof: if $x \in \Omega \setminus J_u$ and the four approximate limits $u^\pm, v^\pm$ exist at $x$ and satisfy $u^\pm(x) = v^\pm(x)$ (hypothesis (1)), then $w^+(x) = w^-(x) = 0$, so $x \notin J_w$. The exceptional set is the union of two $\mathcal{H}^{n-1}$-null sets and is itself null. *On $J_u$ (excluding a null set).* The approximate upper and lower limits of $w$ at jump points must be computed via the *one-sided* trace structure of $u$ and $v$ at jump points, not by formal arithmetic on $u^\pm$ and $v^\pm$. *One-sided traces at jump points.* At $\mathcal{H}^{n-1}$-a.e. $x \in J_u$, the function $u$ has well-defined "trace from each side" with respect to the jump direction $\nu_u(x)$: with $H^+ := \{y : (y - x) \cdot \nu_u(x) > 0\}$ and $H^- := \{y : (y - x) \cdot \nu_u(x) < 0\}$, \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r) \cap H^+)} \int_{B(x, r) \cap H^+} u \, d\mathcal{L}^n &= u^+(x), \\ \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r) \cap H^-)} \int_{B(x, r) \cap H^-} u \, d\mathcal{L}^n &= u^-(x). \end{align*} These one-sided traces are precisely $u^\pm(x)$ — that's the geometric meaning of the approximate upper and lower limits at jump points. Similarly for $v$. *Computing $w^\pm$ at $x \in J_u$.* Since $J_u = J_v$ and the jump normals $\nu_u(x), \nu_v(x)$ are determined by the rectifiable structure of the common jump set, $\nu_u(x) = \pm \nu_v(x)$ for $\mathcal{H}^{n-1}$-a.e. $x \in J_u$. The convention is that $\nu_u$ points from the "$-$" side (where $u^-$ is the trace) to the "$+$" side (where $u^+$ is the trace). Hypothesis (2) — $u^\pm = v^\pm$ — implicitly assumes the same orientation, so $\nu_v(x) = \nu_u(x)$ at $\mathcal{H}^{n-1}$-a.e. $x \in J_u$. The one-sided trace of $w = u - v$ on $H^+$ then equals \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r) \cap H^+)} \int_{B(x, r) \cap H^+} (u - v) \, d\mathcal{L}^n = u^+(x) - v^+(x) = 0 \end{align*} (using hypothesis (2)). Similarly on $H^-$, the trace of $w$ is $u^-(x) - v^-(x) = 0$. So both one-sided traces of $w$ at $x$ are zero, meaning the approximate upper and lower limits $w^+(x) = w^-(x) = 0$ — there is no jump of $w$ at $x$. *Why same-side subtraction is correct.* The approximate limits of $u - v$ are limits of integrals of $u - v$, which split into $u$-integrals minus $v$-integrals on the *same* one-sided neighbourhood. The trace of $w = u - v$ from the "$+$" side is $u^+(x) - v^+(x)$ (both traces from the same side), and the trace from the "$-$" side is $u^-(x) - v^-(x)$. At a non-jump point of $u$ (where $u^+ = u^-$), this reduces to $u(x) - v(x)$; at a jump point, the sides matter, and the relevant quantities are $u^+(x) - v^+(x)$ and $u^-(x) - v^-(x)$, both zero by hypothesis (2). *Conclusion of Step 2.* Combining the analysis on $\Omega \setminus J_u$ (via the Claim) and on $J_u$ (via the one-sided trace argument), $J_w \subseteq N$ for some $\mathcal{H}^{n-1}$-null set $N$. So $\mathcal{H}^{n-1}(J_w) = 0$, and consequently $D^j w = 0$ as a measure (by [Jump Decomposition Formula](/theorems/3124), $|D^j w|$ is concentrated on $J_w$ which has $\mathcal{H}^{n-1}$-measure zero, hence $|D^j w| = 0$).[/guided]

[step:Conclude $Dw = D^a w$ and identify $D^a w = \nabla w \cdot \mathcal{L}^n$ with $\nabla w = 0$ a.e.]By Step 2, $D^j w = 0$, and by hypothesis (3), $D^c w = 0$. Hence \begin{align*} Dw = D^a w = \nabla w \cdot \mathcal{L}^n, \end{align*} where $\nabla w \in L^1(\Omega; \mathbb{R}^n)$ is the Radon-Nikodym density of $D^a w$ with respect to $\mathcal{L}^n$. By [Approximate Gradient Identifies $D^a u$](/theorems/3130), the Radon-Nikodym density $\nabla w$ coincides $\mathcal{L}^n$-a.e. with the *approximate gradient* of $w$, defined as the unique vector $\nabla w(x) \in \mathbb{R}^n$ (when it exists) such that \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \frac{|w(y) - \tilde w(x) - \nabla w(x) \cdot (y - x)|}{r} \, d\mathcal{L}^n(y) = 0, \end{align*} where $\tilde w(x)$ is the approximate limit of $w$ at $x$. We claim $\nabla w(x) = 0$ for $\mathcal{L}^n$-a.e. $x \in \Omega$. Indeed: by Step 2, $J_w$ is $\mathcal{H}^{n-1}$-null, hence $\mathcal{L}^n(J_w) = 0$ (since $\mathcal{H}^{n-1}$-null implies Hausdorff-dimension at most $n - 1$, hence $\mathcal{L}^n$-null). At every $x \in \Omega \setminus J_w$ where the approximate limits $w^+(x) = w^-(x)$ exist (which is $\mathcal{H}^{n-1}$-a.e. by [theorem 3122](/theorems/3122), in particular $\mathcal{L}^n$-a.e.), the approximate limit $\tilde w(x) := w^+(x) = w^-(x)$ is well-defined. Hypothesis (1) — $u^\pm = v^\pm$ on $\Omega \setminus J_u$ at $\mathcal{H}^{n-1}$-a.e. point — combined with the Step-1 identities $w^\pm = u^\pm - v^\mp$ on the non-jump region (where $u^+ = u^-$ and $v^+ = v^-$, so this reduces to the pointwise difference) gives $w^\pm(x) = 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$. By Step 2, $J_w \subseteq J_u \cup N$ ($\mathcal{H}^{n-1}$-null $N$), and on $J_u \setminus J_w$ the analogous calculation (using hypothesis (2)) gives $w^\pm = 0$. So $\tilde w(x) = 0$ at $\mathcal{H}^{n-1}$-a.e. $x$ in $\Omega \setminus J_w$, hence at $\mathcal{L}^n$-a.e. $x \in \Omega$. The function $\tilde w$ extends to $w$ at almost every point in $L^1$-equivalence sense. [claim:$\nabla w = 0$ $\mathcal{L}^n$-a.e. on $\Omega$] [/claim] [proof] By the identity $\tilde w \equiv 0$ $\mathcal{L}^n$-a.e., the function $w(y) - \tilde w(x) = w(y)$ at every $x$ where $\tilde w(x) = 0$. The defining limit of the approximate gradient becomes \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \frac{|w(y) - \nabla w(x) \cdot (y - x)|}{r} \, d\mathcal{L}^n(y) = 0. \end{align*} The function $\tilde w \equiv 0$ in particular has $\tilde w(y) = 0$ for $\mathcal{L}^n$-a.e. $y$, so $w(y) = 0$ for $\mathcal{L}^n$-a.e. $y$. Substituting into the limit: \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \frac{|0 - \nabla w(x) \cdot (y - x)|}{r} \, d\mathcal{L}^n(y) = |\nabla w(x)| \cdot c_n, \end{align*} where $c_n := \frac{1}{\mathcal{L}^n(B(0, 1))} \int_{B(0, 1)} |z_1| \, d\mathcal{L}^n(z) > 0$ is a positive dimensional constant (using the substitution $z = (y - x)/r$ and the fact that the integral of $|\xi \cdot z|$ over $B(0, 1)$ equals $|\xi| \cdot c_n$ by rotational symmetry). For this limit to equal $0$ at $\mathcal{L}^n$-a.e. $x$, we must have $|\nabla w(x)| \cdot c_n = 0$, hence $\nabla w(x) = 0$. [/proof] So $D^a w = \nabla w \cdot \mathcal{L}^n = 0$, hence $Dw = 0$.[/step]

[guided]We use the BV decomposition to reduce the question to: a BV function with zero distributional gradient is constant. *Reducing to $Dw = 0$.* By Step 2, $D^j w = 0$. By hypothesis (3), $D^c w = 0$. So \begin{align*} Dw = D^a w = \nabla w \cdot \mathcal{L}^n, \end{align*} where $\nabla w \in L^1(\Omega; \mathbb{R}^n)$ is the Radon-Nikodym density. *Identifying $\nabla w$ with the approximate gradient.* By [theorem 3130](/theorems/3130), at $\mathcal{L}^n$-a.e. $x \in \Omega$, the approximate gradient of $w$ at $x$ exists and equals the Radon-Nikodym density $\nabla w(x)$. *Constraint on $w$ at $\mathcal{L}^n$-a.e. point.* Hypothesis (1) gives $w^\pm = 0$ on $\Omega \setminus J_u$ at $\mathcal{H}^{n-1}$-a.e. point (using the on-region calculation $w^\pm = u^\pm - v^\pm = 0$). Hypothesis (2) gives the same on $J_u$ via the one-sided trace argument of Step 2. So the approximate limit $\tilde w(x)$ exists and equals zero at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_w$, where $J_w$ is itself $\mathcal{H}^{n-1}$-null. In particular, at $\mathcal{L}^n$-a.e. $x \in \Omega$, $\tilde w(x) = 0$, and since the approximate limit is the integral average limit, $w(y) = \tilde w(x) = 0$ for $\mathcal{L}^n$-a.e. $y$ in any small neighbourhood of $\mathcal{L}^n$-a.e. $x$. By Lebesgue density, this propagates: $w(y) = 0$ for $\mathcal{L}^n$-a.e. $y \in \Omega$. *Conclusion: $\nabla w = 0$.* The defining condition for the approximate gradient is \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \frac{|w(y) - \tilde w(x) - \nabla w(x) \cdot (y - x)|}{r} \, d\mathcal{L}^n(y) = 0. \end{align*} Substituting $w(y) = 0$ a.e. and $\tilde w(x) = 0$: \begin{align*} \lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} \frac{|\nabla w(x) \cdot (y - x)|}{r} \, d\mathcal{L}^n(y) = 0. \end{align*} The integrand is $|\nabla w(x) \cdot z| / 1$ in the rescaled variable $z = (y - x)/r$ on $B(0, 1)$; by rotational symmetry of $\mathcal{L}^n$ on the ball, the average equals $|\nabla w(x)| \cdot c_n$ for a positive dimensional constant $c_n > 0$ (the average of $|z_1|$ on $B(0, 1)$). For the limit to be zero, $|\nabla w(x)| = 0$. So $\nabla w = 0$ $\mathcal{L}^n$-a.e., and $D^a w = 0$. *Final consolidation.* All three pieces of $Dw$ vanish: \begin{align*} Dw = D^a w + D^j w + D^c w = 0 + 0 + 0 = 0. \end{align*}[/guided]

[step:Use $Dw = 0$ to conclude $w \equiv 0$ $\mathcal{H}^{n-1}$-a.e.]We have $Dw = 0$ as a Radon measure on $\Omega$. By a standard distributional argument: if $w \in L^1(\Omega)$ and $Dw = 0$, then $w$ is constant on each connected component of $\Omega$. Indeed, for any $\varphi \in C_c^\infty(\Omega)$ and any $i$, $\int_\Omega w \, \partial_{x_i}\varphi \, d\mathcal{L}^n = -\int_\Omega \varphi \, d(\partial_i w) = 0$. So $w$ is a distribution whose distributional gradient vanishes, hence $w$ is constant on each connected component of $\Omega$. It remains to identify the constant. From Step 3, $\tilde w(x) = 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega$. The precise representative $\tilde w$ agrees with $w$ at $\mathcal{L}^n$-a.e. $x$ (this is part of the definition of the approximate limit / Lebesgue point). Hence $w = 0$ $\mathcal{L}^n$-a.e. on $\Omega$. Combining: on each connected component, $w$ is constant ($\mathcal{L}^n$-a.e.) and that constant is $0$ (by the $\mathcal{H}^{n-1}$-a.e. or $\mathcal{L}^n$-a.e. evaluation just established). So $w = 0$ at $\mathcal{L}^n$-a.e. (and $\mathcal{H}^{n-1}$-a.e.) point of $\Omega$, i.e., $u = v$ $\mathcal{H}^{n-1}$-a.e. on $\Omega$.[/step]

[guided]We translate $Dw = 0$ into the pointwise conclusion. *$Dw = 0 \Rightarrow w$ is locally constant.* The distributional partial derivatives $\partial_i w$ vanish for each $i$. For any $\varphi \in C_c^\infty(\Omega)$: \begin{align*} \int_\Omega w(x) \, \partial_{x_i}\varphi(x) \, d\mathcal{L}^n(x) = -\partial_i w(\varphi) = 0, \end{align*} since $\partial_i w$ is the zero measure. By the standard result that an $L^1_{\mathrm{loc}}$ function with vanishing distributional gradient is locally constant (proved via convolution with mollifiers: $(w * \eta_\varepsilon)$ is smooth with $\nabla(w * \eta_\varepsilon) = w * \nabla \eta_\varepsilon$; a calculation using $Dw = 0$ shows this gradient vanishes, hence $w * \eta_\varepsilon$ is constant on every connected open subset of its domain; passing to the limit $\varepsilon \to 0$, $w$ itself is constant on each connected component of $\Omega$ at $\mathcal{L}^n$-a.e. point). *Pinpointing the constant.* From Step 3, $\tilde w(x) = 0$ at $\mathcal{H}^{n-1}$-a.e. $x$, hence at $\mathcal{L}^n$-a.e. $x$. Since the precise representative $\tilde w$ equals $w$ at every Lebesgue point — and Lebesgue points have full $\mathcal{L}^n$-measure by the [Lebesgue Differentiation Theorem](/theorems/4) applied to $w \in L^1_{\mathrm{loc}}(\Omega)$ — $w(x) = 0$ at $\mathcal{L}^n$-a.e. $x$. Combined with locally constant, $w \equiv 0$ on each connected component (where the constant must be $0$). *Conclusion: $u = v$ $\mathcal{H}^{n-1}$-a.e.* The function $w = u - v$ vanishes $\mathcal{L}^n$-a.e., hence $u = v$ $\mathcal{L}^n$-a.e. on $\Omega$. The stronger statement $u = v$ $\mathcal{H}^{n-1}$-a.e. follows from the $\mathcal{H}^{n-1}$-a.e. vanishing of the approximate limit $\tilde w$: at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_w$, the values $w^\pm(x)$ exist and equal zero, so $u^\pm(x) = v^\pm(x)$, and in particular the precise representatives agree $\mathcal{H}^{n-1}$-a.e. The "pointwise data" — the approximate upper and lower limits — are themselves equal $\mathcal{H}^{n-1}$-a.e., hence $u$ and $v$ define the same BV equivalence class up to an $\mathcal{H}^{n-1}$-null set, which is the strongest form of "$u = v$ $\mathcal{H}^{n-1}$-a.e." in this context.[/guided]