[proofplan] The strategy combines the definition of the approximate upper and lower limits with the structure of $\Omega \setminus J_u$ to show that, off the jump set, the approximate limit of $u$ exists and the average of $|u - \tilde u(x)|$ over small balls vanishes. We work with two key tools: [Finiteness of Approximate Limits for BV Functions](/theorems/3122), which establishes the existence and finiteness of $u^\pm$ at $\mathcal{H}^{n-1}$-a.e. point; and the trichotomy of upper densities for the superlevel sets $E_t = \{u > t\}$, which forces $\Theta^*(E_t, x) \in \{0, 1\}$ off the measure-theoretic boundary $\partial_m E_t$. The proof has three steps: (i) verify that $\tilde u$ is well-defined $\mathcal{H}^{n-1}$-a.e. on $\Omega \setminus J_u$; (ii) show that for such $x$, the upper density of $\{u > t\}$ at $x$ is the indicator $\mathbb{1}_{t < \tilde u(x)}$; (iii) use this density characterisation, together with the layer-cake representation of $|u - \tilde u(x)|$, to compute the average on small balls and verify it vanishes. The Borel measurability and the $\mathcal{L}^n$-a.e. agreement with $u$ follow from the Lebesgue differentiation theorem and the definition of the approximate limit. [/proofplan] [step:Set up the precise representative and identify it on $\Omega \setminus J_u$] Recall the definitions: for $u \in BV(\Omega) \subseteq L^1_{\mathrm{loc}}(\Omega)$, define \begin{align*} u^+(x) &:= \inf \bigl\{ t \in \mathbb{R} : \Theta^*(\{u > t\}, x) = 0 \bigr\}, \\ u^-(x) &:= \sup \bigl\{ t \in \mathbb{R} : \Theta^*(\{u < t\}, x) = 0 \bigr\}, \end{align*} with $\Theta^*(E, x) := \limsup_{r \to 0} \mathcal{L}^n(E \cap B(x, r))/\mathcal{L}^n(B(x, r))$ the upper Lebesgue density of a Borel set $E \subseteq \mathbb{R}^n$ at $x$, and the conventions $\inf \varnothing = +\infty$, $\sup \varnothing = -\infty$. The jump set of $u$ is \begin{align*} J_u := \{ x \in \Omega : u^-(x) < u^+(x) \} \end{align*} (modulo an $\mathcal{H}^{n-1}$-null set where $u^\pm$ may not exist). By [Finiteness of Approximate Limits for BV Functions](/theorems/3122), there is an $\mathcal{H}^{n-1}$-null set $N_0 \subseteq \Omega$ such that \begin{align*} -\infty < u^-(x) \le u^+(x) < +\infty \quad \text{for every } x \in \Omega \setminus N_0. \end{align*} For $x \in \Omega \setminus (J_u \cup N_0)$, $u^+(x) = u^-(x) \in \mathbb{R}$. Define \begin{align*} \tilde u : \Omega \setminus (J_u \cup N_0) &\to \mathbb{R}, \\ x &\mapsto u^+(x) = u^-(x). \end{align*} This is well-defined and finite on $\Omega \setminus (J_u \cup N_0)$. Since $\mathcal{H}^{n-1}(N_0) = 0$, the precise representative is defined $\mathcal{H}^{n-1}$-a.e. on $\Omega \setminus J_u$. [guided] We make the precise representative $\tilde u$ explicit and describe the structure of points where it is defined. *Approximate upper and lower limits.* For $u \in L^1_{\mathrm{loc}}(\Omega)$, the approximate upper limit $u^+(x)$ at $x \in \Omega$ is the infimum of values $t$ at which the superlevel set $\{u > t\}$ has zero upper density at $x$. Geometrically: $u^+(x) \le t$ means "in a measure-theoretic sense, $u$ is at most $t$ on most of any small ball around $x$". The lower limit $u^-(x)$ is the dual notion. The convention $\inf \varnothing = +\infty$ handles points where $u$ is essentially unbounded above near $x$; $\sup \varnothing = -\infty$ does the dual. *Hypothesis verification for [theorem 3122](/theorems/3122).* The theorem requires $u \in BV(\Omega)$, exactly our hypothesis. Its conclusion is the existence of an $\mathcal{H}^{n-1}$-null set $N_0$ outside which $-\infty < u^-(x) \le u^+(x) < +\infty$. So at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega$, the approximate upper and lower limits exist and are finite real numbers. *Definition of $\tilde u$.* On $\Omega \setminus (J_u \cup N_0)$, the inequality $u^-(x) \le u^+(x)$ is an equality (by definition of $J_u$), so the common value \begin{align*} \tilde u(x) := u^+(x) = u^-(x) \in \mathbb{R} \end{align*} is unambiguously defined and finite. The set $J_u \cup N_0$ where $\tilde u$ is undefined is the union of $J_u$ (where the limits differ) and a $\mathcal{H}^{n-1}$-null set; in particular, $\tilde u$ is defined at every $x \in \Omega \setminus J_u$ outside an $\mathcal{H}^{n-1}$-null subset. *Why this matters.* The theorem's conclusion is about the asymptotic behaviour of $u$ near $\tilde u(x)$. Step 2 will identify $\tilde u(x)$ with the limit of integral averages $\fint_{B(x, r)} u \, d\mathcal{L}^n$ via a density argument, and Step 3 will upgrade this to the $L^1$-average of $|u - \tilde u(x)|$. [/guided] [/step] [step:For $x \in \Omega \setminus (J_u \cup N_0)$, the upper density of $\{u > t\}$ at $x$ is $\mathbb{1}_{\{t < \tilde u(x)\}}$ for $\mathcal{L}^1$-a.e. $t$] Fix $x \in \Omega \setminus (J_u \cup N_0)$, with $\tilde u(x) = u^+(x) = u^-(x) =: a$. [claim:For every $t > a$, $\Theta^*(\{u > t\}, x) = 0$. For every $t < a$, the lower density satisfies $\Theta_*(\{u > t\}, x) = 1$ — equivalently, $\Theta^*(\{u \le t\}, x) = 0$] [/claim] [proof] *Upper bound, $t > a$.* By definition of $u^+(x) = a$ as an infimum, the set $\{t : \Theta^*(\{u > t\}, x) = 0\}$ contains every $t > a$. Indeed, $u^+(x) = a$ means for every $\varepsilon > 0$ there exists $t_\varepsilon < a + \varepsilon$ with $\Theta^*(\{u > t_\varepsilon\}, x) = 0$. By monotonicity of $\{u > t\}$ in $t$ (the set shrinks as $t$ increases), $\{u > t\} \subseteq \{u > t_\varepsilon\}$ for $t > t_\varepsilon$, so \begin{align*} \Theta^*(\{u > t\}, x) \le \Theta^*(\{u > t_\varepsilon\}, x) = 0 \end{align*} for every $t > t_\varepsilon$. Taking $\varepsilon \to 0$ shows $\Theta^*(\{u > t\}, x) = 0$ for every $t > a$. *Lower bound, $t < a$.* By the dual definition $u^-(x) = a$ as a supremum, the set $\{t : \Theta^*(\{u < t\}, x) = 0\}$ contains every $t < a$. Specifically, $u^-(x) = a$ means for every $\varepsilon > 0$ there exists $s_\varepsilon > a - \varepsilon$ with $\Theta^*(\{u < s_\varepsilon\}, x) = 0$. By monotonicity, $\{u < t\} \subseteq \{u < s_\varepsilon\}$ for $t < s_\varepsilon$, so $\Theta^*(\{u < t\}, x) = 0$ for every $t < a$. The set $\{u \le t\}$ differs from $\{u < t\}$ by the level set $\{u = t\}$. By Fubini-Tonelli applied to the indicator $\mathbb{1}_{\{(y, s) : u(y) = s\}}$ (whose slice in $s$ at fixed $y$ is the singleton $\{u(y)\}$, hence $\mathcal{L}^1$-null), $\int_\mathbb{R} \mathcal{L}^n(\{u = s\}) \, d\mathcal{L}^1(s) = 0$, so $\mathcal{L}^n(\{u = s\}) = 0$ for $\mathcal{L}^1$-a.e. $s \in \mathbb{R}$. Choose $t < a$ with $\mathcal{L}^n(\{u = t\}) = 0$, so $\{u \le t\} = \{u < t\}$ up to a Lebesgue-null set, and $\Theta^*(\{u \le t\}, x) = \Theta^*(\{u < t\}, x) = 0$ for every such $t$. For $\mathcal{L}^1$-a.e. $t < a$, the densities of $\{u > t\}$ and $\{u \le t\}$ in any ball $B(x, r) \subseteq \Omega$ sum to $1$: \begin{align*} \frac{\mathcal{L}^n(\{u > t\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} + \frac{\mathcal{L}^n(\{u \le t\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = 1. \end{align*} Taking limits as $r \to 0$: $\Theta^*(\{u \le t\}, x) = 0$ implies $\Theta_*(\{u \le t\}, x) \le \Theta^*(\{u \le t\}, x) = 0$, so the actual density limit of $\{u \le t\}$ at $x$ exists and equals $0$. Subtracting from $1$, the density limit of $\{u > t\}$ at $x$ exists and equals $1$. So $\Theta^*(\{u > t\}, x) = 1$ (in fact the full density limit exists and equals $1$). [/proof] The Claim says: for $\mathcal{L}^1$-a.e. $t \in \mathbb{R}$, \begin{align*} \Theta(\{u > t\}, x) := \lim_{r \to 0} \frac{\mathcal{L}^n(\{u > t\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = \mathbb{1}_{\{t < \tilde u(x)\}}. \end{align*} The exceptional $t$-set has $\mathcal{L}^1$-measure zero (consisting of $t = a$, where the limit may not be sharply $0$ or $1$, plus the at-most-countable set $\{t : \mathcal{L}^n(\{u = t\}) > 0\}$). [guided] We translate the definition of $u^+(x) = u^-(x) = a$ into a complete description of the upper density of every superlevel set $\{u > t\}$ at $x$. *Strategy.* For $t > a$, the set $\{u > t\}$ has zero upper density at $x$ — directly from the infimum definition of $u^+(x)$. For $t < a$, the dual statement gives zero upper density of $\{u < t\}$ at $x$, which by the sum-to-one identity for densities upgrades to upper density $1$ for $\{u > t\}$ (or its near-equivalent $\{u \ge t\}$). *The equality of $\{u < t\}$ and $\{u \le t\}$ for a.e. $t$.* The two sets differ by the level set $\{u = t\}$. Apply Fubini-Tonelli to the indicator $\mathbb{1}_{\{(y, t) \in \Omega \times \mathbb{R} : u(y) = t\}}$, which is $(\mathcal{L}^n \otimes \mathcal{L}^1)$-measurable since $u$ is measurable. Integrating in $t$ first: for each fixed $y \in \Omega$, the slice $\{t \in \mathbb{R} : u(y) = t\} = \{u(y)\}$ is a singleton, hence $\mathcal{L}^1$-null. So \begin{align*} (\mathcal{L}^n \otimes \mathcal{L}^1)\bigl( \{(y, t) : u(y) = t\} \bigr) = \int_\Omega \mathcal{L}^1(\{u(y)\}) \, d\mathcal{L}^n(y) = 0. \end{align*} Now integrate in the other order: for each fixed $t$, the slice is $\{y \in \Omega : u(y) = t\} = \{u = t\}$, so \begin{align*} \int_\mathbb{R} \mathcal{L}^n(\{u = t\}) \, d\mathcal{L}^1(t) = (\mathcal{L}^n \otimes \mathcal{L}^1)\bigl( \{(y, t) : u(y) = t\} \bigr) = 0. \end{align*} Since the integrand is non-negative, $\mathcal{L}^n(\{u = t\}) = 0$ for $\mathcal{L}^1$-a.e. $t \in \mathbb{R}$, and $\{u < t\}$ and $\{u \le t\}$ coincide modulo $\mathcal{L}^n$-null sets for $\mathcal{L}^1$-a.e. $t$. *Density limit for $t < a$.* For $\mathcal{L}^1$-a.e. $t < a$ with $\mathcal{L}^n(\{u = t\}) = 0$: \begin{align*} \Theta^*(\{u \le t\}, x) = \Theta^*(\{u < t\}, x) = 0, \end{align*} where the first equality is by the null-set agreement and the second comes from the supremum definition of $u^-(x) = a$. Since $\Theta_* \le \Theta^*$, the lower density also vanishes, so the limit exists and is zero. The sum-to-one identity \begin{align*} \frac{\mathcal{L}^n(\{u > t\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} + \frac{\mathcal{L}^n(\{u \le t\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = 1 \end{align*} then forces the limit of $\mathcal{L}^n(\{u > t\} \cap B(x, r))/\mathcal{L}^n(B(x, r))$ to exist and equal $1$. *Density limit for $t > a$.* Direct from $u^+(x) = a$: the infimum gives $t \in \{t' : \Theta^*(\{u > t'\}, x) = 0\}$ for every $t > a$. Since $\Theta^* \ge \Theta_* \ge 0$, the limit exists and is $0$. *Borderline $t = a$.* The behaviour at $t = a$ is not determined by the definition of $u^\pm$; this is at most one point, of $\mathcal{L}^1$-measure zero, hence ignorable in the integration steps to follow. *Summary of density structure.* For $\mathcal{L}^1$-a.e. $t$, \begin{align*} \Theta(\{u > t\}, x) = \begin{cases} 1 & t < \tilde u(x), \\ 0 & t > \tilde u(x). \end{cases} \end{align*} This is exactly the density structure of $\mathbb{1}_{(-\infty, \tilde u(x))}$ — in other words, the values of $u$ on small balls around $x$ are concentrated at $\tilde u(x)$ in the measure-theoretic sense. [/guided] [/step] [step:Compute $\lim_{r \to 0} \frac{1}{\mathcal{L}^n(B(x, r))} \int_{B(x, r)} |u(y) - \tilde u(x)| \, d\mathcal{L}^n(y) = 0$] Fix $x \in \Omega \setminus (J_u \cup N_0)$ with $\tilde u(x) = a$, and fix $r_0 > 0$ small enough that $B(x, r_0) \subseteq \Omega$. We split the average via the triangle inequality: \begin{align*} \fint_{B(x, r)} |u(y) - a| \, d\mathcal{L}^n(y) \le \fint_{B(x, r)} |u(y) - u_{B(x, r)}| \, d\mathcal{L}^n(y) + |u_{B(x, r)} - a|, \end{align*} where $u_{B(x, r)} := \fint_{B(x, r)} u \, d\mathcal{L}^n$ and $\fint$ denotes the integral average. We show both right-hand-side terms tend to zero as $r \to 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$. [claim:The oscillation $\fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \to 0$ as $r \to 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$] [/claim] [proof] By [BV Poincaré Inequality on Balls](/theorems/3116) applied to $u \in BV(\Omega)$ on $B(x, r) \subseteq \Omega$ (the hypothesis $u \in BV$ on the ball follows from $u \in BV(\Omega)$ since $B(x, r) \subseteq \Omega$, and the open-ball Poincaré inequality requires only this), there is a dimensional constant $C(n) > 0$ such that \begin{align*} \fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \le C(n) \cdot \frac{|Du|(B(x, r))}{\omega_n r^{n-1}}, \end{align*} where $\omega_n := \mathcal{L}^n(B(0, 1))$. By [BV Structure Theorem](/theorems/595), $|Du| = |\nabla u| \mathcal{L}^n + |D^j u| + |D^c u|$ as measures on $\Omega$, where the three pieces are mutually singular. Each piece has zero upper $(n-1)$-density at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$: - *The absolutely continuous part.* $|\nabla u| \mathcal{L}^n(B(x, r)) = \int_{B(x, r)} |\nabla u| \, d\mathcal{L}^n = \omega_n r^n \cdot \fint_{B(x, r)} |\nabla u| \, d\mathcal{L}^n$, so $|\nabla u| \mathcal{L}^n(B(x, r))/r^{n-1} = \omega_n r \cdot \fint_{B(x, r)} |\nabla u| \, d\mathcal{L}^n \to 0$ at $\mathcal{L}^n$-a.e. $x$ (the average converges to $|\nabla u(x)| < \infty$ at Lebesgue points of $|\nabla u| \in L^1_\mathrm{loc}(\Omega)$). - *The jump part.* By [Jump Decomposition Formula](/theorems/3124), $|D^j u| = (u^+ - u^-) \mathcal{H}^{n-1} \lfloor J_u$ is concentrated on $J_u$. By definition of concentration, $|D^j u|(\Omega \setminus J_u) = 0$, so for any Borel set $B$, $|D^j u|(B) = |D^j u|(B \cap J_u) \le \int_{B \cap J_u} (u^+ - u^-) \, d\mathcal{H}^{n-1}$. In particular, $|D^j u|$ vanishes on every Borel set $B$ with $\mathcal{H}^{n-1}(B \cap J_u) = 0$. At $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$, the upper $(n-1)$-density $\Theta^{*, n-1}(|D^j u|, x) := \limsup_{r \to 0} |D^j u|(B(x, r)) / (\omega_{n-1} r^{n-1})$ vanishes by the standard density theorem for Borel measures concentrated on $\mathcal{H}^{n-1}$-rectifiable sets (cf. the structure of $|Du|$ in [BV Structure Theorem](/theorems/595)): the set where the upper density is positive has $\mathcal{H}^{n-1}$-mass controlled by the measure restricted to its support, and intersects $\Omega \setminus J_u$ only on an $\mathcal{H}^{n-1}$-null set. - *The Cantor part.* By [BV Structure Theorem](/theorems/595), $|D^c u|$ is singular with respect to $\mathcal{L}^n$ and additionally diffuse: $|D^c u|(B) = 0$ on every Borel set $B$ with $\mathcal{H}^{n-1}(B) < \infty$ (cf. [De Giorgi's Structure Theorem](/theorems/599)). By a standard density theorem for measures vanishing on $\mathcal{H}^{n-1}$-finite sets (the $5r$-covering / Vitali argument applied to the upper-density level sets), the upper $(n-1)$-density $\Theta^{*, n-1}(|D^c u|, x)$ vanishes at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega$ — hence in particular at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$. Combining, $|Du|(B(x, r))/r^{n-1} \to 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$. By the BV-Poincaré bound, the oscillation tends to zero at the same set of $x$. [/proof] [claim:The average $u_{B(x, r)} \to a$ as $r \to 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$ with $\tilde u(x) = a$] [/claim] [proof] Combine Step 2 (convergence in measure of $u$ to $a$ on $B(x, r)$) with the first claim (oscillation $\fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \to 0$). *Convergence in measure.* By Step 2, for $\mathcal{L}^1$-a.e. $\delta > 0$ both $D_+^r(a + \delta) \to 0$ and $D_-^r(a - \delta) \to 0$ as $r \to 0$, where $D_+^r(t) := \mathcal{L}^n(\{u > t\} \cap B(x, r))/\mathcal{L}^n(B(x, r))$ and $D_-^r(t) := \mathcal{L}^n(\{u < t\} \cap B(x, r))/\mathcal{L}^n(B(x, r))$. Since $\{|u - a| > \delta\} = \{u > a + \delta\} \cup \{u < a - \delta\}$, \begin{align*} \frac{\mathcal{L}^n(\{|u - a| > \delta\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} \to 0 \quad (r \to 0) \end{align*} for $\mathcal{L}^1$-a.e. $\delta > 0$. *Bootstrapping to convergence of the average.* Suppose for contradiction that $u_{B(x, r)} \not\to a$ as $r \to 0$. Then there exist $\delta_0 > 0$ and $r_k \downarrow 0$ with $|u_{B(x, r_k)} - a| \ge 2 \delta_0$. Pick $\delta \in (0, \delta_0)$ such that the convergence in measure above holds with this $\delta$. For every $y$ with $|u(y) - u_{B(x, r_k)}| < \delta$, the reverse triangle inequality gives $|u(y) - a| \ge |u_{B(x, r_k)} - a| - |u(y) - u_{B(x, r_k)}| > 2\delta_0 - \delta > \delta$. Hence \begin{align*} \{y \in B(x, r_k) : |u(y) - u_{B(x, r_k)}| < \delta\} \subseteq \{y \in B(x, r_k) : |u(y) - a| > \delta\}, \end{align*} so by Markov's inequality applied to the oscillation, \begin{align*} \frac{\mathcal{L}^n(\{|u - a| > \delta\} \cap B(x, r_k))}{\mathcal{L}^n(B(x, r_k))} \ge 1 - \frac{\mathcal{L}^n(\{|u - u_{B(x, r_k)}| \ge \delta\} \cap B(x, r_k))}{\mathcal{L}^n(B(x, r_k))} \ge 1 - \frac{1}{\delta} \fint_{B(x, r_k)} |u - u_{B(x, r_k)}| \, d\mathcal{L}^n. \end{align*} By the first claim, the rightmost average tends to $0$, so the left-hand side has lower limit $\ge 1$ — contradicting convergence in measure. Hence $u_{B(x, r)} \to a$ as $r \to 0$. [/proof] Combining the two claims: at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$, \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} |u(y) - \tilde u(x)| \, d\mathcal{L}^n(y) \le \lim_{r \to 0} \fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n + \lim_{r \to 0} |u_{B(x, r)} - a| = 0 + 0 = 0. \end{align*} [guided] We compute the average $\fint_{B(x, r)} |u - \tilde u(x)| \, d\mathcal{L}^n$ as $r \to 0$ and verify it vanishes, using both Step 2's density structure and the BV-Poincaré inequality. *Decomposition via the average $u_{B(x, r)}$.* The triangle inequality gives \begin{align*} \fint_{B(x, r)} |u - a| \, d\mathcal{L}^n \le \fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n + |u_{B(x, r)} - a|, \end{align*} where $u_{B(x, r)} := \fint_{B(x, r)} u \, d\mathcal{L}^n$. Both right-hand-side terms must tend to zero. *Average tends to $a$: $u_{B(x, r)} \to a$ as $r \to 0$.* Combine Step 2's convergence in measure with the BV-Poincaré oscillation bound. By Step 2, for $\mathcal{L}^1$-a.e. $\delta > 0$, \begin{align*} \frac{\mathcal{L}^n(\{|u - a| > \delta\} \cap B(x, r))}{\mathcal{L}^n(B(x, r))} \to 0 \quad (r \to 0), \end{align*} i.e., $u \to a$ in $\mathcal{L}^n$-measure on $B(x, r)$. Suppose $u_{B(x, r_k)} \not\to a$ along some sequence $r_k \downarrow 0$, so $|u_{B(x, r_k)} - a| \ge 2 \delta_0 > 0$ for some $\delta_0$. Pick $\delta \in (0, \delta_0)$ in the convergence-in-measure exceptional-set complement. By the reverse triangle inequality, \begin{align*} \{|u - u_{B(x, r_k)}| < \delta\} \cap B(x, r_k) \subseteq \{|u - a| > \delta\} \cap B(x, r_k), \end{align*} so \begin{align*} \frac{\mathcal{L}^n(\{|u - a| > \delta\} \cap B(x, r_k))}{\mathcal{L}^n(B(x, r_k))} \ge 1 - \frac{1}{\delta} \fint_{B(x, r_k)} |u - u_{B(x, r_k)}| \, d\mathcal{L}^n. \end{align*} The right-hand side $\to 1$ by the BV-Poincaré oscillation bound (proven below), contradicting convergence in measure. Hence $u_{B(x, r)} \to a$. *Oscillation tends to zero: $\fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \to 0$.* By [BV Poincaré Inequality on Balls](/theorems/3116) applied to $u \in BV(\Omega)$ on $B(x, r) \subseteq \Omega$, \begin{align*} \fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \le C(n) \cdot r \cdot \frac{|Du|(B(x, r))}{\mathcal{L}^n(B(x, r))} = C(n) \cdot \frac{|Du|(B(x, r))}{\omega_n r^{n-1}}, \end{align*} where the constant $C(n)$ depends only on the dimension and $\omega_n = \mathcal{L}^n(B(0, 1))$. The quantity $|Du|(B(x, r))/(\omega_{n-1} r^{n-1})$ is the upper $(n-1)$-density of $|Du|$ at $x$ (up to the constant $\omega_{n-1}$). By the structure of BV functions, at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$ the upper $(n-1)$-density of $|Du|$ vanishes. The argument splits along the decomposition $|Du| = |\nabla u| \mathcal{L}^n + |D^j u| + |D^c u|$ from [BV Structure Theorem](/theorems/595): the $\mathcal{L}^n$-piece contributes $O(r) \to 0$ at Lebesgue points of $|\nabla u|$, the jump piece $|D^j u|$ is concentrated on $J_u$ and contributes zero density off $J_u$ in an $\mathcal{H}^{n-1}$-a.e. sense, and the Cantor piece $|D^c u|$ is diffuse and contributes zero $(n-1)$-density $\mathcal{H}^{n-1}$-a.e. The three quantitative bounds are derived in turn below. Combining, the upper $(n-1)$-density of $|Du|$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$ is zero, so $|Du|(B(x, r))/r^{n-1} \to 0$ as $r \to 0$. By the BV-Poincaré bound, \begin{align*} \fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \le \frac{C(n)}{\omega_n} \cdot \frac{|Du|(B(x, r))}{r^{n-1}} \to 0. \end{align*} *Combining.* Both terms tend to zero, so \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} |u(y) - \tilde u(x)| \, d\mathcal{L}^n(y) = 0, \end{align*} which is exactly the statement of approximate continuity of $\tilde u$ at $x$. *Where this fails.* The argument fails on $J_u$ (where $u^+ \ne u^-$, so $\tilde u(x)$ is undefined), and on the $\mathcal{H}^{n-1}$-null set $N_0$ where the structure theorem's exceptional set lives, plus the $\mathcal{H}^{n-1}$-null set where $|Du|/r^{n-1}$ has positive upper density. The union of these is an $\mathcal{H}^{n-1}$-null subset of $\Omega \setminus J_u$, off which approximate continuity holds. *Upper $(n-1)$-density of the absolutely continuous part vanishes at $\mathcal{L}^n$-a.e. $x$.* Since $|\nabla u| \in L^1(\Omega)$, by the [Lebesgue Differentiation Theorem](/theorems/4) the average $\fint_{B(x, r)} |\nabla u| \, d\mathcal{L}^n$ converges to $|\nabla u(x)|$ at $\mathcal{L}^n$-a.e. $x \in \Omega$. The corresponding $(n-1)$-density satisfies \begin{align*} \frac{|\nabla u| \mathcal{L}^n(B(x, r))}{r^{n-1}} = \frac{\int_{B(x, r)} |\nabla u| \, d\mathcal{L}^n}{r^{n-1}} = \omega_n r \cdot \fint_{B(x, r)} |\nabla u| \, d\mathcal{L}^n. \end{align*} As $r \to 0$, the right-hand side is $\omega_n r \cdot |\nabla u(x)| + o(r) \to 0$. So $|\nabla u| \mathcal{L}^n$ has zero upper $(n-1)$-density at $\mathcal{L}^n$-a.e. $x$, and since $\mathcal{H}^{n-1}$-a.e. is a finer notion than $\mathcal{L}^n$-a.e., this also holds at $\mathcal{H}^{n-1}$-a.e. $x$. *Upper $(n-1)$-density of the jump part vanishes at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$.* By [Jump Decomposition Formula](/theorems/3124), $|D^j u| = (u^+ - u^-) \mathcal{H}^{n-1} \lfloor J_u$, a Borel measure absolutely continuous with respect to $\mathcal{H}^{n-1} \lfloor J_u$ with locally finite $(n-1)$-mass. For such measures, a standard density theorem holds: writing $E_\alpha := \{x \in \Omega : \Theta^{*, n-1}(|D^j u|, x) > \alpha\}$ for $\alpha > 0$, the $5r$-covering lemma applied to balls $\{B(x, r_x)\}_{x \in E_\alpha}$ with $|D^j u|(B(x, r_x)) > \alpha \omega_{n-1} r_x^{n-1}$ gives $\mathcal{H}^{n-1}(E_\alpha) \le C |D^j u|(\Omega) / \alpha$ for a dimensional constant $C$. Restricted to $E_\alpha \cap (\Omega \setminus J_u)$, the right-hand side is replaced by $|D^j u|(\Omega \setminus J_u) = 0$, so $\mathcal{H}^{n-1}(E_\alpha \cap (\Omega \setminus J_u)) = 0$. Taking $\alpha = 1/k$ and unioning, the set where $\Theta^{*, n-1}(|D^j u|, \cdot) > 0$ has $\mathcal{H}^{n-1}$-measure zero in $\Omega \setminus J_u$. *Upper $(n-1)$-density of the Cantor part vanishes off an $\mathcal{H}^{n-1}$-null set.* By [BV Structure Theorem](/theorems/595) (cf. [De Giorgi's Structure Theorem](/theorems/599)), $|D^c u|$ vanishes on every Borel set $B$ with $\mathcal{H}^{n-1}(B) < \infty$ — this is the "diffuse" property, which contrasts with $|D^j u|$'s concentration on the rectifiable jump set. Apply the same density theorem: $E_\alpha := \{x \in \Omega : \Theta^{*, n-1}(|D^c u|, x) > \alpha\}$ has $\mathcal{H}^{n-1}(E_\alpha) \le C |D^c u|(\Omega) / \alpha < \infty$ by the $5r$-covering bound. But then $|D^c u|(E_\alpha) = 0$ by the diffuse property. The same covering bound applied with $|D^c u|$-restricted to $E_\alpha$ now yields $\mathcal{H}^{n-1}(E_\alpha) \le C |D^c u|(E_\alpha) / \alpha = 0$. So $\Theta^{*, n-1}(|D^c u|, \cdot) = 0$ at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega$. *Putting the three pieces together.* At $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$: \begin{align*} \frac{|Du|(B(x, r))}{r^{n-1}} = \frac{|\nabla u| \mathcal{L}^n(B(x, r)) + |D^j u|(B(x, r)) + |D^c u|(B(x, r))}{r^{n-1}} \to 0 + 0 + 0 = 0. \end{align*} Combined with the BV-Poincaré bound, the oscillation $\fint_{B(x, r)} |u - u_{B(x, r)}| \, d\mathcal{L}^n \to 0$. *Why the BV-Poincaré inequality is the right tool.* The classical Poincaré inequality on balls for Sobolev functions involves the gradient $\nabla u$. For BV functions, the gradient is replaced by the variation measure $|Du|$, with the same scaling: the $(n-1)$-density of $|Du|$ at $x$ controls the oscillation of $u$ on small balls. This is geometrically natural: BV functions can have jumps, but the size of jumps near $x$ is controlled by $|Du|$ near $x$, and far from any jump, $|Du|$ has small $(n-1)$-density. [/guided] [/step] [step:Verify Borel measurability of $\tilde u$ and the $\mathcal{L}^n$-a.e. agreement $\tilde u = u$] *Borel measurability.* The functions $u^+$ and $u^-$ are Borel-measurable on $\Omega$. To see this for $u^+$: for each $\lambda \in \mathbb{R}$, \begin{align*} \{u^+ < \lambda\} = \bigcup_{t < \lambda, t \in \mathbb{Q}} \{x : \Theta^*(\{u > t\}, x) = 0\}. \end{align*} Each set $\{x : \Theta^*(\{u > t\}, x) = 0\}$ is Borel (being a level set of a function defined as a $\limsup$ of integrals against the Borel-measurable indicator $\mathbb{1}_{\{u > t\}}$, divided by the Borel function $r \mapsto \mathcal{L}^n(B(x, r))$). The countable union is therefore Borel, and Borel measurability of $u^+$ follows. The argument for $u^-$ is symmetric. The set $\Omega \setminus (J_u \cup N_0) = \{u^+ = u^-\} \setminus N_0$ is Borel (as the difference of Borel sets, since $\{u^+ = u^-\} = \{u^+ - u^- = 0\}$ is Borel and $N_0$ is contained in a Borel $\mathcal{H}^{n-1}$-null set). On this Borel set, $\tilde u = u^+$ is Borel-measurable. *Agreement $\tilde u = u$ $\mathcal{L}^n$-a.e.* By the Lebesgue differentiation theorem applied to $u \in L^1_{\mathrm{loc}}(\Omega)$, at $\mathcal{L}^n$-a.e. $x \in \Omega$, \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} |u(y) - u(x)| \, d\mathcal{L}^n(y) = 0, \end{align*} i.e., $x$ is a Lebesgue point of $u$ with $\lim_{r \to 0} u_{B(x, r)} = u(x)$. By Step 3, at $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$ (and hence at $\mathcal{L}^n$-a.e. $x \in \Omega \setminus J_u$, since $\mathcal{H}^{n-1}$-a.e. is a stronger conclusion than $\mathcal{L}^n$-a.e. when restricted to subsets of $\mathbb{R}^n$ — both are negligible compared to $\mathcal{L}^n$), \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} |u - \tilde u(x)| \, d\mathcal{L}^n = 0, \end{align*} which forces $\tilde u(x) = \lim_r u_{B(x, r)} = u(x)$ at every $x$ that is simultaneously a Lebesgue point of $u$ and a point of approximate continuity of $\tilde u$. The set of Lebesgue points has full $\mathcal{L}^n$-measure. The complement of $\Omega \setminus (J_u \cup N_0)$ in $\Omega$ is $J_u \cup N_0$. Since $J_u$ is $\mathcal{H}^{n-1}$-$\sigma$-finite (by [BV Structure Theorem](/theorems/595)), $\mathcal{L}^n(J_u) = 0$ (as $\mathcal{H}^{n-1}$-$\sigma$-finite sets in $\mathbb{R}^n$ have Lebesgue measure zero); and $\mathcal{H}^{n-1}(N_0) = 0$ implies $\mathcal{L}^n(N_0) = 0$. So $\Omega \setminus (J_u \cup N_0)$ has full $\mathcal{L}^n$-measure, and on the intersection with the Lebesgue point set (still full $\mathcal{L}^n$-measure), $\tilde u(x) = u(x)$. This concludes the proof: the precise representative $\tilde u$ is approximately continuous at $\mathcal{H}^{n-1}$-a.e. point of $\Omega \setminus J_u$, is a Borel function, and agrees with $u$ at $\mathcal{L}^n$-a.e. point of $\Omega$. [guided] We close out by addressing the two remaining pieces of the conclusion: Borel measurability of $\tilde u$ and pointwise agreement with $u$ at $\mathcal{L}^n$-a.e. point. *Borel measurability.* The function $u^+$ is defined as an infimum of $t$ over a Borel-measurable set-condition $\Theta^*(\{u > t\}, x) = 0$. The condition $\Theta^*(E, x) = 0$ for a fixed Borel set $E$ is the vanishing of the function $x \mapsto \limsup_{r \to 0} \mathcal{L}^n(E \cap B(x, r))/\mathcal{L}^n(B(x, r))$, which is Borel-measurable in $x$ (it is a $\limsup$ of Borel-measurable functions $x \mapsto \mathcal{L}^n(E \cap B(x, r))/\mathcal{L}^n(B(x, r))$, indexed by countable rational $r > 0$, since the function $x \mapsto \mathcal{L}^n(E \cap B(x, r))$ is continuous in $x$ for fixed $r$ — by the dominated convergence theorem applied to the indicator $\mathbb{1}_{E \cap B(\cdot, r)}$). Hence $\{x : \Theta^*(\{u > t\}, x) = 0\}$ is Borel for every $t$. For any $\lambda \in \mathbb{R}$, \begin{align*} \{u^+ < \lambda\} = \bigcup_{t \in \mathbb{Q}, t < \lambda} \{x : \Theta^*(\{u > t\}, x) = 0\} \end{align*} is a countable union of Borel sets, hence Borel. So $u^+$ is Borel-measurable. Symmetrically $u^-$ is Borel-measurable, and $\tilde u = u^+$ on the Borel set $\{u^+ = u^-\}$, undefined elsewhere. After Borel-extending $\tilde u$ by setting it equal to (say) $0$ on the complement, $\tilde u: \Omega \to \mathbb{R}$ is Borel-measurable. *Agreement $\tilde u = u$ $\mathcal{L}^n$-a.e.* By [Lebesgue Differentiation Theorem](/theorems/4) applied to $u \in L^1_{\mathrm{loc}}(\Omega)$, $\mathcal{L}^n$-a.e. $x \in \Omega$ is a Lebesgue point: $\fint_{B(x, r)} |u(y) - u(x)| \, d\mathcal{L}^n(y) \to 0$ as $r \to 0$, hence $u_{B(x, r)} \to u(x)$. By Step 3, $\fint_{B(x, r)} |u(y) - \tilde u(x)| \, d\mathcal{L}^n(y) \to 0$ at $\mathcal{H}^{n-1}$-a.e. (hence $\mathcal{L}^n$-a.e.) $x \in \Omega \setminus J_u$. Triangle inequality: \begin{align*} |u(x) - \tilde u(x)| \le \limsup_{r \to 0} \fint_{B(x, r)} |u - u(x)| \, d\mathcal{L}^n + \limsup_{r \to 0} \fint_{B(x, r)} |u - \tilde u(x)| \, d\mathcal{L}^n = 0 \end{align*} at every $x$ that is both a Lebesgue point of $u$ and a point of approximate continuity of $\tilde u$. The intersection of the two full-measure sets has full $\mathcal{L}^n$-measure (Lebesgue points: full $\mathcal{L}^n$-measure; approximate-continuity points: $\mathcal{L}^n$-a.e. on $\Omega \setminus J_u$, which itself has full $\mathcal{L}^n$-measure since $J_u$ is $\mathcal{H}^{n-1}$-$\sigma$-finite, hence $\mathcal{L}^n$-null). So $\tilde u(x) = u(x)$ at $\mathcal{L}^n$-a.e. $x \in \Omega$. *Final summary.* We have shown: - $\tilde u$ is well-defined and Borel-measurable on $\Omega \setminus (J_u \cup N_0)$, and admits a Borel extension to all of $\Omega$. - At $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus J_u$, $\tilde u$ is approximately continuous: $\fint_{B(x, r)} |u - \tilde u(x)| \, d\mathcal{L}^n \to 0$. - $\tilde u = u$ at $\mathcal{L}^n$-a.e. $x \in \Omega$. [/guided] [/step]