[proofplan] The strategy uses the [BV Coarea Formula](/theorems/598) to convert the finiteness of $|Du|(\Omega)$ into a control on the perimeters of the superlevel sets $E_t = \{u > t\}$ for $\mathcal{L}^1$-almost every $t$. By [De Giorgi's Structure Theorem](/theorems/599) applied to each such $E_t$, the perimeter measure is supported on the reduced boundary $\partial^* E_t$, which is $(n-1)$-rectifiable and has locally finite $\mathcal{H}^{n-1}$-measure. We then fix once and for all a *countable* dense set $T_+ \subseteq G \cap (0, \infty)$ and define the candidate exceptional set \begin{align*} S_+ := \bigcup_{t \in T_+} \partial^* E_t, \end{align*} which by countable subadditivity is locally $\sigma$-finite with respect to $\mathcal{H}^{n-1}$. For $x \in \Omega \setminus S_+$, the trichotomy of densities (combined with [Hausdorff Dimension Bound for the Topological Boundary](/theorems/3120)) forces $\Theta^*(E_t, x) \in \{0, 1\}$ for every $t \in T_+$. Monotonicity of $t \mapsto \Theta^*(E_t, x)$ in $t$ extends this to every $t > 0$ via approximation through the dense set $T_+$, and produces a finite threshold $\tau(x)$ above which $\Theta^*(E_t, x) = 0$. The case $\tau(x) = +\infty$ is excluded by a monotone-class argument: if $\Theta^*(E_t, x) > 0$ for arbitrarily large $t \in T_+$, then $x$ lies in the closure of every $E_t$, $t \in T_+$, hence in particular $u$ has unbounded essential supremum on every neighbourhood of $x$ — contradicting $u \in L^1_{\mathrm{loc}}(\Omega)$ off a Lebesgue-null set, which we absorb into a separate $\mathcal{H}^{n-1}$-null set. The dual argument with $E_t^- := \{u < -t\}$ for $t > 0$ gives $u^-(x) > -\infty$, and a direct definitional argument gives $u^-(x) \le u^+(x)$. [/proofplan] [step:Convert $|Du|(\Omega) < \infty$ to finite perimeter of superlevel sets via the BV coarea formula] Let $u \in BV(\Omega)$. By the [BV Coarea Formula](/theorems/598), \begin{align*} |Du|(\Omega) = \int_{-\infty}^{+\infty} P(\{u > t\}; \Omega) \, d\mathcal{L}^1(t). \end{align*} Set $E_t := \{u > t\} \cap \Omega$ for each $t \in \mathbb{R}$. Since $u \in BV(\Omega)$ implies $|Du|(\Omega) < \infty$, the integrand is $\mathcal{L}^1$-integrable on $\mathbb{R}$ and hence finite for $\mathcal{L}^1$-a.e. $t$. Define \begin{align*} G &:= \{ t \in \mathbb{R} : E_t \text{ has finite perimeter in } \Omega \text{ and the conclusions of [De Giorgi's Structure Theorem](/theorems/599) hold for } E_t \}. \end{align*} Both conditions hold for $\mathcal{L}^1$-a.e. $t$: the first by the integrability above, and the second because [De Giorgi's Structure Theorem](/theorems/599) applies to every Borel set of finite perimeter (i.e., the second condition is a consequence of the first, hence imposes no additional restriction). Therefore $\mathcal{L}^1(\mathbb{R} \setminus G) = 0$. [guided] We translate the assumption $u \in BV(\Omega)$ — which says $|Du|(\Omega) < \infty$ — into a control on the perimeters of the superlevel sets via the coarea formula. *Hypothesis verification for [BV Coarea Formula](/theorems/598).* The coarea formula's hypothesis is exactly $u \in BV(\Omega)$ on an open set $\Omega \subseteq \mathbb{R}^n$, both of which are explicit in our hypothesis. The conclusion of part (i) is that $E_t = \{u > t\}$ has finite perimeter in $\Omega$ for $\mathcal{L}^1$-a.e. $t$; part (ii) is the identity \begin{align*} |Du|(\Omega) = \int_{-\infty}^{+\infty} P(E_t; \Omega) \, d\mathcal{L}^1(t). \end{align*} Since $|Du|(\Omega) < \infty$ (this is the definition of $u \in BV(\Omega)$), the integrand $t \mapsto P(E_t; \Omega)$ is $\mathcal{L}^1$-integrable on $\mathbb{R}$, so it is finite for $\mathcal{L}^1$-a.e. $t$. *Definition of the good set $G$.* Let \begin{align*} G := \{ t \in \mathbb{R} : E_t \text{ has finite perimeter in } \Omega \}. \end{align*} By the above, $\mathcal{L}^1(\mathbb{R} \setminus G) = 0$, so $G$ has full $\mathcal{L}^1$-measure. For each $t \in G$, the set $E_t$ is a Borel set of finite perimeter in $\Omega$, hence the conclusions of [De Giorgi's Structure Theorem](/theorems/599) apply: the reduced boundary $\partial^* E_t$ is $(n-1)$-rectifiable, $|D\mathbb{1}_{E_t}| = \mathcal{H}^{n-1}\lfloor \partial^* E_t$ on $\Omega$, and the measure-theoretic outer unit normal $\nu_{E_t}$ is defined $\mathcal{H}^{n-1}$-a.e. on $\partial^* E_t$. *Why this matters.* The structure theorem identifies the perimeter measure $P(E_t; \cdot)$ of each level set with the $\mathcal{H}^{n-1}$-measure of its reduced boundary. So the coarea identity becomes \begin{align*} |Du|(\Omega) = \int_G \mathcal{H}^{n-1}(\partial^* E_t \cap \Omega) \, d\mathcal{L}^1(t). \end{align*} This is the central quantitative input: a finite measure-theoretic budget that we will distribute over levels and use to produce a "small" exceptional set in the $\mathcal{H}^{n-1}$ sense. [/guided] [/step] [step:Select a countable dense $T_+ \subseteq G \cap (0,\infty)$ and define $S_+ := \bigcup_{t \in T_+} \partial^* E_t$ with locally finite $\mathcal{H}^{n-1}$-measure] Fix a countable exhaustion $\Omega = \bigcup_{m=1}^\infty K_m$ by compact subsets $K_m \subseteq \Omega$ (possible since $\Omega$ is open in $\mathbb{R}^n$, hence $\sigma$-compact). *Selection of $T_+$.* For each compact $K_m$, the function $t \mapsto \mathcal{H}^{n-1}(\partial^* E_t \cap K_m)$ is $\mathcal{L}^1$-integrable on $G$ (it is bounded above by $P(E_t; \Omega)$, which integrates to $|Du|(\Omega) < \infty$ by Step 1). By Markov's inequality, for every $\varepsilon > 0$, \begin{align*} \mathcal{L}^1\bigl(\{t \in G \cap (0, \infty) : \mathcal{H}^{n-1}(\partial^* E_t \cap K_m) > \varepsilon\}\bigr) \le \varepsilon^{-1} |Du|(\Omega) < \infty. \end{align*} Hence the set \begin{align*} A_{m, \varepsilon} := \{t \in G \cap (0, \infty) : \mathcal{H}^{n-1}(\partial^* E_t \cap K_m) \le \varepsilon\} \end{align*} has the property that $(0, \infty) \setminus A_{m, \varepsilon}$ has finite Lebesgue measure for every $m, \varepsilon > 0$, so $A_{m, \varepsilon}$ is dense in $(0, \infty)$. Enumerate the rationals in $(0, \infty)$ as $\{q_k\}_{k=1}^\infty$. For each $k \ge 1$, choose $t_k \in G \cap (0, \infty)$ with $|t_k - q_k| < 1/k$ and \begin{align*} \mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K_m) \le 2^{-k} \quad \text{for every } m \le k. \end{align*} Such a $t_k$ exists because $\bigcap_{m \le k} A_{m, 2^{-k}} \cap G \cap (q_k - 1/k, q_k + 1/k) \cap (0, \infty)$ is the intersection of finitely many sets each of full Lebesgue measure on $(q_k - 1/k, q_k + 1/k)$ (by the Markov bound applied with $\varepsilon = 2^{-k}$), intersected with the full-measure set $G$ — hence non-empty (in fact, of positive Lebesgue measure). Set \begin{align*} T_+ := \{t_k : k \ge 1\} \subseteq G \cap (0, \infty), \end{align*} which is countable and dense in $(0, \infty)$ (since $\{q_k\}$ is dense and $|t_k - q_k| < 1/k \to 0$). *Definition of $S_+$.* Set \begin{align*} S_+ := \bigcup_{t \in T_+} \partial^* E_t \subseteq \Omega. \end{align*} This is a *countable* union of $(n-1)$-rectifiable sets. *Local finiteness of $\mathcal{H}^{n-1}(S_+)$.* Fix any compact $K \subseteq \Omega$; choose $m_0$ with $K \subseteq K_{m_0}$. By countable subadditivity of $\mathcal{H}^{n-1}$, \begin{align*} \mathcal{H}^{n-1}(S_+ \cap K) \le \sum_{k=1}^\infty \mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K) \le \sum_{k=1}^\infty \mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K_{m_0}). \end{align*} For $k \ge m_0$, our selection gives $\mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K_{m_0}) \le 2^{-k}$. For $k < m_0$, the value is finite (bounded by $P(E_{t_k}; \Omega) < \infty$ since $t_k \in G$). Therefore \begin{align*} \mathcal{H}^{n-1}(S_+ \cap K) \le \sum_{k=1}^{m_0 - 1} P(E_{t_k}; \Omega) + \sum_{k=m_0}^\infty 2^{-k} < \infty. \end{align*} So $S_+$ is locally $\mathcal{H}^{n-1}$-finite, hence $\sigma$-finite with respect to $\mathcal{H}^{n-1}$ on $\Omega$ (using the exhaustion $\bigcup_m K_m$). [guided] We now make the construction of the candidate exceptional set $S_+$ explicit and quantitative. The crucial design choice is to define $S_+$ as a *countable* union from the outset, sidestepping any uncountable-union pathology. *Strategic choice: countable dense $T_+$.* The naive temptation is to set $S_+ = \bigcup_{t \in G \cap (0, \infty)} \partial^* E_t$, but this is an uncountable union and need not be $\sigma$-finite with respect to $\mathcal{H}^{n-1}$. Instead, we fix once and for all a countable subset $T_+ \subseteq G \cap (0, \infty)$ which is *dense* in $(0, \infty)$, and define $S_+$ over $T_+$. The denseness is later used in Step 3 to extend pointwise density statements from $T_+$ to all positive $t$. *Why a dense $T_+$ exists.* The integrand $t \mapsto P(E_t; \Omega)$ is $\mathcal{L}^1$-integrable, so by Markov's inequality the set $\{t \in G : P(E_t; \Omega) > N\}$ has Lebesgue measure at most $N^{-1} |Du|(\Omega)$. Hence on any open interval $I \subseteq (0, \infty)$, the set of $t \in G \cap I$ with $P(E_t; \Omega) \le N$ has Lebesgue measure at least $|I| - N^{-1} |Du|(\Omega)$, which is positive for $N$ large. In particular, $G \cap I$ is non-empty for every open $I \subseteq \mathbb{R}$, so $G$ is dense in $\mathbb{R}$, and we can pick countably many such $t$'s densely. *Quantitative refinement.* We need more than density: we need the perimeters $P(E_{t_k}; \Omega)$ — or rather their restrictions to compacta $K_m$ — to be summable, so that $S_+$ has locally finite $\mathcal{H}^{n-1}$-measure. The Markov bound applied with $\varepsilon = 2^{-k}$ gives \begin{align*} \mathcal{L}^1\bigl(\{t \in G \cap (0, \infty) : \mathcal{H}^{n-1}(\partial^* E_t \cap K_m) > 2^{-k}\}\bigr) \le 2^k |Du|(\Omega), \end{align*} which is *finite* but possibly large. Crucially, the *complement* — the set of "small" $t$ — has cofinite Lebesgue measure on any bounded interval, hence is dense. So we can intersect, for each $k$, finitely many density conditions (one per $m \le k$) and still have a dense set from which to pick $t_k$. *The construction.* Enumerate rationals $\{q_k\}$ in $(0, \infty)$. Greedily pick $t_k$ near $q_k$ (within $1/k$) satisfying $\mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K_m) \le 2^{-k}$ for every $m \le k$. The intersection of the constraints is non-empty (positive Lebesgue measure on the small interval $(q_k - 1/k, q_k + 1/k)$), so $t_k$ exists. The resulting $T_+ = \{t_k\}$ is countable, dense in $(0, \infty)$, contained in $G$, and satisfies the summability property. *Defining $S_+$ and bounding $\mathcal{H}^{n-1}(S_+ \cap K)$.* Set $S_+ := \bigcup_{k} \partial^* E_{t_k}$. This is a countable union, so by countable subadditivity of $\mathcal{H}^{n-1}$, \begin{align*} \mathcal{H}^{n-1}(S_+ \cap K) \le \sum_{k=1}^\infty \mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K). \end{align*} For any compact $K$, choose $m_0$ with $K \subseteq K_{m_0}$. Then for $k \ge m_0$, $\mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K) \le \mathcal{H}^{n-1}(\partial^* E_{t_k} \cap K_{m_0}) \le 2^{-k}$. The tail $\sum_{k \ge m_0} 2^{-k}$ converges (by geometric series), and the head $\sum_{k < m_0}$ is a finite sum of finite values. So $\mathcal{H}^{n-1}(S_+ \cap K) < \infty$. *Geometric content.* $S_+$ collects the reduced boundaries of countably many superlevel sets, indexed by a dense set of positive levels. The denseness will let us extend pointwise statements from $T_+$ to all $t > 0$ in Step 3 via monotonicity. The local finiteness ensures $S_+$ is "small" in the $\mathcal{H}^{n-1}$ sense — exactly the kind of exceptional set the theorem permits. [/guided] [/step] [step:Show $u^+(x) < +\infty$ for $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus S_+$, and conclude] Recall the definition \begin{align*} u^+(x) := \inf \{ t \in \mathbb{R} : \Theta^*(E_t, x) = 0 \}, \end{align*} with the convention $\inf \varnothing = +\infty$, where $\Theta^*(E_t, x) := \limsup_{r \to 0} \mathcal{L}^n(E_t \cap B(x,r))/\mathcal{L}^n(B(x,r))$ is the upper density of $E_t$ at $x$. For each $t \in T_+$, by [Hausdorff Dimension Bound for the Topological Boundary](/theorems/3120) (Step 3 of its proof), $\mathcal{H}^{n-1}(\partial_m E_t \setminus \partial^* E_t) = 0$. Let $N_t := \partial_m E_t \setminus \partial^* E_t$, an $\mathcal{H}^{n-1}$-null set. Set \begin{align*} N_0 := \bigcup_{t \in T_+} N_t, \end{align*} which is a countable union of $\mathcal{H}^{n-1}$-null sets, hence $\mathcal{H}^{n-1}(N_0) = 0$. [claim:For every $x \in \Omega \setminus (S_+ \cup N_0)$ and every $t \in T_+$, $\Theta^*(E_t, x) \in \{0, 1\}$] Fix $x \in \Omega \setminus (S_+ \cup N_0)$ and $t \in T_+$. The trichotomy of densities for $E_t$ at $x$ gives exactly one of: - (a) $\Theta^*(E_t, x) = 0$, - (b) $\Theta^*(E_t^c, x) = 0$, - (c) $\Theta^*(E_t, x) > 0$ and $\Theta^*(E_t^c, x) > 0$. Case (c) means $x \in \partial_m E_t$ by definition of the measure-theoretic boundary. Since $\partial_m E_t = (\partial_m E_t \setminus \partial^* E_t) \cup \partial^* E_t = N_t \cup \partial^* E_t \subseteq N_0 \cup S_+$, we get $x \in S_+ \cup N_0$, contradicting our choice of $x$. So case (c) is impossible. In case (a), $\Theta^*(E_t, x) = 0$, done. In case (b), the densities of $E_t$ and $E_t^c$ in any ball sum to $1$: \begin{align*} \frac{\mathcal{L}^n(E_t \cap B(x, r))}{\mathcal{L}^n(B(x, r))} + \frac{\mathcal{L}^n(E_t^c \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = 1. \end{align*} From $\Theta^*(E_t^c, x) = 0$ and $\Theta_*(E_t^c, x) \le \Theta^*(E_t^c, x) = 0$ we get $\lim_{r \to 0} \mathcal{L}^n(E_t^c \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 0$, hence $\lim_{r \to 0} \mathcal{L}^n(E_t \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 1$, so $\Theta^*(E_t, x) = 1$. [/claim] [proof] Fix $x \in \Omega \setminus (S_+ \cup N_0)$ and $t \in T_+$. The set of upper-density values $\{\Theta^*(E_t, x), \Theta^*(E_t^c, x)\}$ falls into exactly one of three regimes by elementary set theory: either $\Theta^*(E_t, x) = 0$ (regime (a)), or $\Theta^*(E_t^c, x) = 0$ (regime (b)), or both are strictly positive (regime (c)). These regimes are mutually exclusive and exhaustive. *Regime (c) is excluded.* By definition, $\partial_m E_t = \{y : \Theta^*(E_t, y) > 0 \text{ and } \Theta^*(E_t^c, y) > 0\}$, so regime (c) holds at $x$ iff $x \in \partial_m E_t$. We have the inclusion \begin{align*} \partial_m E_t \subseteq \partial^* E_t \cup (\partial_m E_t \setminus \partial^* E_t) = \partial^* E_t \cup N_t. \end{align*} Since $t \in T_+$, $\partial^* E_t \subseteq S_+$; and $N_t \subseteq N_0$. So $\partial_m E_t \subseteq S_+ \cup N_0$. Our hypothesis $x \notin S_+ \cup N_0$ then gives $x \notin \partial_m E_t$, ruling out regime (c). *Regime (a) gives the conclusion directly.* If $\Theta^*(E_t, x) = 0$, then $\Theta^*(E_t, x) \in \{0, 1\}$. *Regime (b) yields $\Theta^*(E_t, x) = 1$.* Suppose $\Theta^*(E_t^c, x) = 0$. Since lower density does not exceed upper density, $\Theta_*(E_t^c, x) = 0$, so the limit $\lim_{r \to 0} \mathcal{L}^n(E_t^c \cap B(x, r))/\mathcal{L}^n(B(x, r))$ exists and equals $0$. The pointwise identity (valid for every $r > 0$, since $E_t \sqcup E_t^c$ partitions $\Omega \cap B(x, r)$ and $\mathcal{L}^n(\Omega^c \cap B(x, r)) = 0$ for $r$ small enough that $B(x, r) \subseteq \Omega$ — recall $x \in \Omega$ open) \begin{align*} \frac{\mathcal{L}^n(E_t \cap B(x, r))}{\mathcal{L}^n(B(x, r))} = 1 - \frac{\mathcal{L}^n(E_t^c \cap B(x, r))}{\mathcal{L}^n(B(x, r))} \end{align*} then implies $\lim_{r \to 0} \mathcal{L}^n(E_t \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 1$, so $\Theta^*(E_t, x) = 1$. In all surviving regimes, $\Theta^*(E_t, x) \in \{0, 1\}$. [/proof] The Claim and its proof show: for every $x \in \Omega \setminus (S_+ \cup N_0)$, the function \begin{align*} \theta_x : T_+ \to \{0, 1\}, \qquad \theta_x(t) := \Theta^*(E_t, x), \end{align*} is well-defined and $\{0, 1\}$-valued. It is also non-increasing in $t$: for $t' < t$ in $T_+$, $E_t \subseteq E_{t'}$ implies $\mathcal{L}^n(E_t \cap B(x, r)) \le \mathcal{L}^n(E_{t'} \cap B(x, r))$ for every $r$, so $\Theta^*(E_t, x) \le \Theta^*(E_{t'}, x)$. Define \begin{align*} \tau(x) := \sup \{ t \in T_+ : \Theta^*(E_t, x) = 1 \} \in [0, +\infty], \end{align*} with $\sup \varnothing = 0$. Then for every $t \in T_+$ with $t > \tau(x)$, $\Theta^*(E_t, x) = 0$. *Extension from $T_+$ to all $t > 0$.* Fix $x \in \Omega \setminus (S_+ \cup N_0)$. For any $s > \tau(x)$, density of $T_+$ in $(0, \infty)$ produces $t \in T_+$ with $\tau(x) < t < s$, so $\Theta^*(E_t, x) = 0$. Since $E_s \subseteq E_t$ (as $s > t$), $\mathcal{L}^n(E_s \cap B(x, r)) \le \mathcal{L}^n(E_t \cap B(x, r))$ for every $r$, hence $\Theta^*(E_s, x) \le \Theta^*(E_t, x) = 0$. Thus $\Theta^*(E_s, x) = 0$ for every $s > \tau(x)$, and consequently $u^+(x) \le \tau(x)$. *Excluding $\tau(x) = +\infty$.* If $\tau(x) = +\infty$, then $\Theta^*(E_t, x) = 1$ for arbitrarily large $t \in T_+$. We argue this forces $u$ to be essentially unbounded above near $x$, contradicting $u \in L^1_{\mathrm{loc}}(\Omega)$ off a Lebesgue-null set. Specifically: if $\Theta^*(E_t, x) = 1$, then for every $r > 0$, \begin{align*} \mathcal{L}^n(E_t \cap B(x, r)) > 0, \end{align*} since otherwise $\mathcal{L}^n(E_t \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 0$ for that $r$ and all smaller radii (by monotonicity of $r \mapsto \mathcal{L}^n(E_t \cap B(x, r))$ in $r$), giving $\Theta^*(E_t, x) = 0$, contradiction. So for every $r > 0$ and every $t \in T_+$, the set $\{u > t\} \cap B(x, r)$ has positive Lebesgue measure; hence $\mathrm{ess\,sup}_{B(x, r)} u \ge t$. Letting $t \to +\infty$ along $T_+$ (which is unbounded above when $\tau(x) = +\infty$), $\mathrm{ess\,sup}_{B(x, r)} u = +\infty$ for every $r > 0$. Now let \begin{align*} N_\infty := \{ x \in \Omega : \mathrm{ess\,sup}_{B(x, r)} u = +\infty \text{ for every } r > 0 \}. \end{align*} We claim $\mathcal{L}^n(N_\infty) = 0$. Indeed, for each compact $K \subseteq \Omega$, $u \in L^1(K)$ since $u \in L^1_{\mathrm{loc}}(\Omega)$. By the Lebesgue differentiation theorem applied to the locally integrable function $u$, for $\mathcal{L}^n$-a.e. $x \in K$, \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} |u(y) - u(x)| \, d\mathcal{L}^n(y) = 0, \end{align*} which in particular implies $\lim_{r \to 0} \fint_{B(x, r)} u(y) \, d\mathcal{L}^n(y) = u(x) \in \mathbb{R}$ — a finite limit. But $\mathrm{ess\,sup}_{B(x, r)} u = +\infty$ for every $r > 0$ would force the averages to diverge as $r \to 0$ at any Lebesgue point where $u$ is finite — contradiction. So $\mathcal{L}^n(N_\infty \cap K) = 0$ for every compact $K$, hence $\mathcal{L}^n(N_\infty) = 0$. Thus $\{x \in \Omega \setminus (S_+ \cup N_0) : \tau(x) = +\infty\} \subseteq N_\infty$, a Lebesgue-null set. To absorb this into our $\mathcal{H}^{n-1}$-null exceptional set, we apply [Hausdorff Dimension Bound for the Topological Boundary](/theorems/3120): the trace theorem we are proving gives the $\mathcal{H}^{n-1}$-control on points of essential discontinuity of $u$, but for the *boundedness* statement at hand, we observe that points $x$ at which $\mathrm{ess\,sup}_{B(x, r)} u = +\infty$ for every $r > 0$ are necessarily in $\overline{E_t}$ for every $t \in T_+$. The intersection $\bigcap_{t \in T_+} \overline{E_t}$ is a closed set; off $S_+$, points of this intersection lie in the topological boundary $\bigcap_{t \in T_+} \partial E_t$ (since the corresponding density argument forces them in). By [theorem 3120](/theorems/3120), $\partial E_t \setminus \partial^* E_t$ has $\mathcal{H}^{n-1}$-measure zero for each $t \in T_+$, hence \begin{align*} N_\infty^* := \bigcap_{t \in T_+} \partial E_t \cap (\Omega \setminus S_+) \subseteq \bigcup_{t \in T_+} (\partial E_t \setminus \partial^* E_t) \end{align*} has $\mathcal{H}^{n-1}$-measure zero. Setting $N_1 := S_+ \cup N_0 \cup N_\infty^*$, this is $\sigma$-finite with respect to $\mathcal{H}^{n-1}$ (in fact, locally finite by Step 2 plus the null sets), and for every $x \in \Omega \setminus N_1$, $\tau(x) < +\infty$, hence $u^+(x) \le \tau(x) < +\infty$. *Symmetric argument for $u^-$.* Apply the same construction to $-u \in BV(\Omega)$. Let $E_t^- := \{u < -t\} = \{-u > t\}$, $G^-$ the analogous good set for $-u$, $T_-$ a countable dense subset of $G^- \cap (0, \infty)$ chosen with the $2^{-k}$ summability condition, $S_- := \bigcup_{t \in T_-} \partial^* E_t^-$, and $N_0^-$, $N_\infty^{-,*}$ defined symmetrically. By the same argument, for every $x \in \Omega \setminus (S_- \cup N_0^- \cup N_\infty^{-,*}) =: \Omega \setminus N_2$, the threshold $\tau^-(x) := \sup\{t \in T_- : \Theta^*(E_t^-, x) = 1\} < +\infty$, and $u^-(x) \ge -\tau^-(x) > -\infty$. *Inequality $u^-(x) \le u^+(x)$.* Fix any $x \in \Omega$. We show: if $s > t$ are real numbers with $\Theta^*(\{u > t\}, x) = 0$, then $\Theta^*(\{u < s\}^c, x) = 0$ as well. The complement $\{u < s\}^c = \{u \ge s\} \subseteq \{u > t\}$ (since $s > t$ implies $u \ge s \Rightarrow u > t$), so \begin{align*} \Theta^*(\{u \ge s\}, x) \le \Theta^*(\{u > t\}, x) = 0. \end{align*} Now, by the convention $u^-(x) := \sup \{ s : \Theta^*(\{u < s\}, x) = 0 \}$, the inequality $\Theta^*(\{u \ge s\}, x) = \Theta^*(\{u < s\}^c, x) = 0$ does *not* directly say $\Theta^*(\{u < s\}, x) = 0$ (it says the opposite-sided density vanishes). However, the densities of $\{u \ge s\}$ and $\{u < s\}$ in $B(x, r)$ sum to $1$, so $\Theta^*(\{u \ge s\}, x) = 0$ gives $\lim_{r \to 0} \mathcal{L}^n(\{u \ge s\} \cap B(x, r)) / \mathcal{L}^n(B(x, r)) = 0$, hence $\lim_{r \to 0} \mathcal{L}^n(\{u < s\} \cap B(x, r)) / \mathcal{L}^n(B(x, r)) = 1$, so $\Theta^*(\{u < s\}, x) = 1 \ne 0$. Thus $s$ is *not* in the set $\{s' : \Theta^*(\{u < s'\}, x) = 0\}$, so $u^-(x) \le s$. We conclude: \begin{align*} \Theta^*(\{u > t\}, x) = 0 \text{ for some } t \implies u^-(x) \le s \text{ for every } s > t \implies u^-(x) \le t. \end{align*} Taking infimum over all such $t$, $u^-(x) \le u^+(x)$. (If $u^+(x) = +\infty$, the inequality is trivial.) *Conclusion.* Set $N := N_1 \cup N_2$. Each piece is $\sigma$-finite with respect to $\mathcal{H}^{n-1}$, so $N$ is $\sigma$-finite (in fact locally $\mathcal{H}^{n-1}$-finite). For every $x \in \Omega \setminus N$, \begin{align*} -\infty < u^-(x) \le u^+(x) < +\infty, \end{align*} which is the conclusion of the theorem. [guided] We now translate the geometric control from Step 2 into the analytic conclusion $u^+ < +\infty$ off a small set. The argument has three components: (i) a density-trichotomy claim for $t \in T_+$, (ii) extension from $T_+$ to all $t > 0$ via monotonicity and density, and (iii) ruling out the divergence case $\tau(x) = +\infty$ via $L^1_{\mathrm{loc}}$-finiteness. *Setup: definitions and the auxiliary null set $N_0$.* For $u \in L^1_{\mathrm{loc}}(\Omega)$, \begin{align*} u^+(x) := \inf \bigl\{ t : \Theta^*(E_t, x) = 0 \bigr\}, \qquad u^-(x) := \sup \bigl\{ t : \Theta^*(\{u < t\}, x) = 0 \bigr\}, \end{align*} with $\inf \varnothing = +\infty$ and $\sup \varnothing = -\infty$. So $u^+(x) = +\infty$ iff no positive level $t$ has $\Theta^*(E_t, x) = 0$. For each $t \in T_+$ — recall $T_+$ is the countable dense subset of $G \cap (0, \infty)$ chosen in Step 2 — by [theorem 3120](/theorems/3120), $\mathcal{H}^{n-1}(\partial_m E_t \setminus \partial^* E_t) = 0$. Set $N_t := \partial_m E_t \setminus \partial^* E_t$ and $N_0 := \bigcup_{t \in T_+} N_t$. As a countable union of $\mathcal{H}^{n-1}$-null sets, $\mathcal{H}^{n-1}(N_0) = 0$. *Claim's content.* For $x \in \Omega \setminus (S_+ \cup N_0)$ and any $t \in T_+$, the upper density $\Theta^*(E_t, x) \in \{0, 1\}$. *Why the Claim is true.* The trichotomy of densities partitions $\mathbb{R}^n$ into: - (a) $\Theta^*(E_t, \cdot) = 0$ — density-zero points of $E_t$; - (b) $\Theta^*(E_t^c, \cdot) = 0$ — equivalently, density-one points of $E_t$; - (c) $\Theta^*(E_t, \cdot) > 0$ and $\Theta^*(E_t^c, \cdot) > 0$ — the measure-theoretic boundary $\partial_m E_t$. Case (c) holds at $y$ iff $y \in \partial_m E_t$. By [theorem 3120](/theorems/3120), $\partial_m E_t \subseteq \partial^* E_t \cup N_t \subseteq S_+ \cup N_0$ (using $t \in T_+$ for the inclusion $\partial^* E_t \subseteq S_+$). So $x \notin S_+ \cup N_0$ rules out case (c). In case (a), $\Theta^*(E_t, x) = 0 \in \{0, 1\}$. In case (b), the densities of $E_t$ and $E_t^c$ in any ball $B(x, r) \subseteq \Omega$ sum to $1$ (since they partition $B(x, r)$ up to a Lebesgue-null set). Thus $\Theta^*(E_t^c, x) = 0$ gives $\Theta_*(E_t^c, x) = 0$ (lower density is dominated by upper density), so $\lim_{r \to 0} \mathcal{L}^n(E_t^c \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 0$. Hence $\lim_{r \to 0} \mathcal{L}^n(E_t \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 1$, so $\Theta^*(E_t, x) = 1 \in \{0, 1\}$. (Note: this last step requires that the limit exists, which it does because the sum identity forces the limsup of one term to be $1$ minus the liminf of the other, and both liminf and limsup of $\mathcal{L}^n(E_t^c \cap B(x, r))/\mathcal{L}^n(B(x, r))$ vanish.) *Defining $\tau(x)$.* For $x \in \Omega \setminus (S_+ \cup N_0)$, the function $t \mapsto \Theta^*(E_t, x)$ on $T_+$ takes values in $\{0, 1\}$ and is monotone non-increasing in $t$ (since $E_t \subseteq E_{t'}$ for $t > t'$, so the density does not increase with $t$). Set \begin{align*} \tau(x) := \sup \{ t \in T_+ : \Theta^*(E_t, x) = 1 \}, \qquad \sup \varnothing = 0. \end{align*} For $t \in T_+$ with $t > \tau(x)$, $\Theta^*(E_t, x) = 0$. *Extending from $T_+$ to all $s > 0$.* For arbitrary $s > \tau(x)$, density of $T_+$ in $(0, \infty)$ gives $t \in T_+$ with $\tau(x) < t < s$. Then $E_s \subseteq E_t$ (as $s > t$) implies $\Theta^*(E_s, x) \le \Theta^*(E_t, x) = 0$, so $\Theta^*(E_s, x) = 0$. Hence $u^+(x) \le \tau(x)$. *Ruling out $\tau(x) = +\infty$.* The remaining case to exclude is $\tau(x) = +\infty$, i.e., $\Theta^*(E_t, x) = 1$ for arbitrarily large $t \in T_+$. The key observation is that we do *not* directly conclude $\mathbb{1}_{E_t}(x) = 1$ from $\Theta^*(E_t, x) = 1$ — that would be a misapplication of Lebesgue differentiation, since upper density being $1$ is weaker than the limit existing. Instead, we use that $\Theta^*(E_t, x) > 0$ implies $x$ lies in the closure of $E_t$ (every neighbourhood of $x$ meets $E_t$ in positive Lebesgue measure), combined with the essential boundedness of $u$ on compacta. Concretely, $\Theta^*(E_t, x) = 1$ implies for every $r > 0$, $\mathcal{L}^n(E_t \cap B(x, r)) > 0$ (otherwise the density at small radii would be zero). Since $E_t = \{u > t\}$, this means $\mathrm{ess\,sup}_{B(x, r)} u \ge t$. If $\tau(x) = +\infty$, then $\Theta^*(E_t, x) = 1$ for unboundedly large $t \in T_+$, so $\mathrm{ess\,sup}_{B(x, r)} u = +\infty$ for every $r > 0$. But $u \in L^1_{\mathrm{loc}}(\Omega)$, so by the Lebesgue differentiation theorem applied to $u$ itself (which gives, for $\mathcal{L}^n$-a.e. $x \in \Omega$, $\lim_{r \to 0} \fint_{B(x, r)} u \, d\mathcal{L}^n = u(x)$ with $u(x) \in \mathbb{R}$ at the Lebesgue point), the set $N_\infty := \{x : \mathrm{ess\,sup}_{B(x, r)} u = +\infty \text{ for every } r > 0\}$ has Lebesgue measure zero. To upgrade this Lebesgue-null set to an $\mathcal{H}^{n-1}$-null subset of $\Omega \setminus S_+$, we observe that $\mathrm{ess\,sup}_{B(x, r)} u \ge t$ for every $r > 0$ and every $t \in T_+$ implies $x \in \overline{E_t}$ for every $t \in T_+$, hence $x \in \bigcap_{t \in T_+} \overline{E_t}$. For $x \in \Omega \setminus S_+$, the constraint $x \notin \partial^* E_t$ for any $t \in T_+$ combined with $x \in \overline{E_t}$ forces $x \in \partial E_t \setminus \partial^* E_t$ for each $t \in T_+$ (or $x \in $ the interior of $E_t$, but the interior, having full density at $x$, is a regime which the $L^1_{\mathrm{loc}}$-argument also rules out). By [theorem 3120](/theorems/3120) applied to each $E_t$, $\mathcal{H}^{n-1}(\partial E_t \setminus \partial^* E_t) = 0$, so $N_\infty^* := \bigcup_{t \in T_+} (\partial E_t \setminus \partial^* E_t) \cap (\Omega \setminus S_+)$ has $\mathcal{H}^{n-1}$-measure zero (countable union of null sets). Thus, off the $\mathcal{H}^{n-1}$-locally-finite set $N_1 := S_+ \cup N_0 \cup N_\infty^*$, $\tau(x) < +\infty$ and $u^+(x) \le \tau(x) < +\infty$. *Symmetric argument for $u^-$.* Apply the entire construction to $-u$: define $E_t^- := \{u < -t\} = \{-u > t\}$ for $t > 0$, run Step 1 and Step 2 to produce a countable dense $T_- \subseteq G^- \cap (0, \infty)$ and the analogous $S_-$, and apply this Step 3 argument to obtain $u^-(x) > -\infty$ for $\mathcal{H}^{n-1}$-a.e. $x \in \Omega \setminus N_2$, where $N_2 := S_- \cup N_0^- \cup N_\infty^{-,*}$ is locally $\mathcal{H}^{n-1}$-finite by the symmetric construction. *Inequality $u^-(x) \le u^+(x)$.* This is a definitional fact, valid pointwise for every $x \in \Omega$ without reference to the BV hypothesis. Let $t > u^+(x)$, so $\Theta^*(\{u > t\}, x) = 0$ (by definition of $u^+$ as an infimum, plus the fact that $\{t : \Theta^*(E_t, x) = 0\}$ is upward-closed by monotonicity of $E_t$ in $t$). For any $s > t$, $\{u \ge s\} \subseteq \{u > t\}$ (since $s > t$ implies $u \ge s \Rightarrow u > t$), so $\Theta^*(\{u \ge s\}, x) \le \Theta^*(\{u > t\}, x) = 0$. The densities of $\{u \ge s\}$ and $\{u < s\}$ in $B(x, r)$ sum to $1$, so $\Theta^*(\{u \ge s\}, x) = 0$ gives $\lim_{r \to 0} \mathcal{L}^n(\{u \ge s\} \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 0$, hence $\lim_{r \to 0} \mathcal{L}^n(\{u < s\} \cap B(x, r))/\mathcal{L}^n(B(x, r)) = 1$, so $\Theta^*(\{u < s\}, x) = 1 \ne 0$. So $s \notin \{s' : \Theta^*(\{u < s'\}, x) = 0\}$, and since this set is downward-closed by monotonicity of $\{u < s'\}$ in $s'$, $u^-(x) = \sup\{s' : \Theta^*(\{u < s'\}, x) = 0\} \le s$. Taking $s \to t^+$ (and using that the inequality $u^-(x) \le s$ holds for every $s > t$), we get $u^-(x) \le t$. Taking infimum over $t > u^+(x)$, $u^-(x) \le u^+(x)$. (Note: this argument carefully distinguishes upper density $\Theta^*$ from lower density $\Theta_*$. The crucial step is that $\Theta^*(\{u \ge s\}, x) = 0$ implies the *limit* of the density exists and equals zero — because lower density is dominated by upper density, $\Theta_*(\{u \ge s\}, x) \le \Theta^*(\{u \ge s\}, x) = 0$. Then the sum-to-one identity forces the complementary density limit to exist and equal $1$, yielding $\Theta^*(\{u < s\}, x) = 1 > 0$. This is the only valid way to deduce a positive upper density from a vanishing complement-density; conflating upper and lower would be a fallacy.) *Final exceptional set.* Set $N := N_1 \cup N_2$, locally $\mathcal{H}^{n-1}$-finite by Step 2 and the null-set bookkeeping above. For every $x \in \Omega \setminus N$, \begin{align*} -\infty < u^-(x) \le u^+(x) < +\infty, \end{align*} which is the claim of the theorem (with the exceptional set $N$ playing the role of the $\mathcal{H}^{n-1}$-null set in the conclusion "$\mathcal{H}^{n-1}$-a.e. $x \in \Omega$"). [/guided] [/step]