[proofplan] The proof has two halves: identifying the precise representative as a quasicontinuous representative, and bounding the capacity of the singular set $S_u$. For the first half, we use a maximal-function argument: the non-centred Hardy–Littlewood maximal operator is bounded on $W^{1,p}$ in the sense that its action on $|\nabla u|$ controls the size of pointwise variations of $u$. This implies that the spherical averages $\fint_{B(x,r)} u\, d\mathcal{L}^n$ converge uniformly outside open sets of arbitrarily small capacity, exhibiting $u^*$ as quasicontinuous. The capacity bound on $S_u$ is then a quantitative restatement of the same maximal-function input: the failure set of the limit definition is contained in level sets of a maximal function, which have small capacity by the Sobolev–Markov inequality applied to a Riesz-potential type expression. The agreement of $u^*$ with $\tilde{u}$ from [Sobolev Quasicontinuous Representative](/theorems/3108) follows from the uniqueness clause of that theorem: both are quasicontinuous representatives of $u$. [/proofplan]

[step:Recall the precise representative and define the maximal function we will control]The precise representative is defined by \begin{align*} u^*: \mathbb{R}^n &\to \mathbb{R} \cup \{\pm\infty\} \\ x &\mapsto \limsup_{r \to 0} \fint_{B(x, r)} u(y)\, d\mathcal{L}^n(y) \quad \text{when the limit exists, else $0$}. \end{align*} By the Lebesgue differentiation theorem applied to $u \in L^p_{\mathrm{loc}} \subseteq L^1_{\mathrm{loc}}$, $u^*(x) = u(x)$ for $\mathcal{L}^n$-a.e. $x$, hence $u^* = u$ $\mathcal{L}^n$-a.e. For the analytic estimates, we use the *non-centred* Hardy–Littlewood maximal function: \begin{align*} M^*g: \mathbb{R}^n &\to [0, \infty] \\ x &\mapsto \sup_{B \ni x} \fint_B |g(y)|\, d\mathcal{L}^n(y), \end{align*} where the supremum is over all open balls $B \subset \mathbb{R}^n$ containing $x$. The non-centred version is comparable to the centred version up to a dimensional constant; we use the non-centred form because it commutes more cleanly with gradient bounds in the form needed below.[/step]

[guided]The precise representative is the natural pointwise representative obtained by averaging on small balls. Where it exists as a true limit (not just a $\limsup$), it captures the "Lebesgue value" of $u$ at the point — the value an absolutely continuous function would take. The Lebesgue differentiation theorem gives, for $u \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ (in particular for $u \in L^p$ with $1 \le p < \infty$, by Hölder and the local finiteness of $\mathcal{L}^n$), \begin{align*} \lim_{r \to 0} \fint_{B(x, r)} u(y)\, d\mathcal{L}^n(y) = u(x) \quad \text{for $\mathcal{L}^n$-a.e. } x \in \mathbb{R}^n. \end{align*} At all such Lebesgue points, $u^*(x) = u(x)$. So $u^* = u$ $\mathcal{L}^n$-a.e. — the precise representative is just one specific representative of the equivalence class of $u$. The quasicontinuity claim is the deeper assertion: $u^*$ is well-behaved at *every* point outside a small set, where "small" is measured by capacity, not Lebesgue measure. We use the non-centred maximal function $M^*$ rather than the centred version $Mg(x) = \sup_{r > 0} \fint_{B(x, r)} |g|\, d\mathcal{L}^n$. The two are pointwise equivalent: $Mg \le M^*g \le 2^n Mg$. The non-centred version has the property that $M^*g$ is automatically lower-semicontinuous (the supremum of an open family of continuous-in-$x$ averages), which avoids issues of needing precise representatives at the maximal-function level. Moreover, the non-centred version interacts well with gradient estimates: pointwise oscillation of $u$ on a ball $B$ is controlled by $M^*(|\nabla u|)$ on a slightly larger ball, where the centring of the ball does not matter.[/guided]

[step:Establish the pointwise oscillation bound via the maximal function of the gradient]We claim that for $u \in W^{1,p}(\mathbb{R}^n)$ and any Lebesgue points $x, y$ of $u$ with $|x - y| = r$, \begin{align*} |u(x) - u(y)| \le C_n \cdot r \cdot \big[(M^*|\nabla u|)(x) + (M^*|\nabla u|)(y)\big], \end{align*} where $C_n > 0$ depends only on the dimension. This is the standard *pointwise gradient bound* for Sobolev functions; we recall its derivation. By a density argument it suffices to prove the inequality for $u \in C^\infty(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$. The integral identity $u(z) - \fint_{B(x, r)} u\, d\mathcal{L}^n$ for $z \in B(x, 2r)$ is bounded, by the standard pointwise Poincaré inequality (sometimes called the Lebesgue–Calderón pointwise Poincaré inequality), by $C_n\, r\, (M^*|\nabla u|)(z)$. Applying this with $z = x$ and $z = y$ and using the triangle inequality on a chain of balls connecting $x$ and $y$ yields the stated inequality.[/step]

[guided]The pointwise gradient inequality is the analytic engine. It says: in a Sobolev function, the difference between values at two nearby points is controlled by the maximal function of the gradient times the distance. The centred bound (a corollary of the Poincaré inequality on balls) reads: for $u \in C^1(\mathbb{R}^n)$ and $x, y \in \mathbb{R}^n$ with $|x - y| = r$, \begin{align*} |u(x) - u(y)| \le C_n\, r\, [M(|\nabla u|)(x) + M(|\nabla u|)(y)], \end{align*} where $C_n$ depends only on the dimension. Since $M \le M^* \le 2^n M$, replacing $M$ by $M^*$ on the right is innocuous and gives the same statement up to a different dimensional constant. The non-centred form has the advantage that the right side is lower-semicontinuous in $x, y$. This is what allows us to derive a quasicontinuity statement for $u$ at *every* Lebesgue point, not just $\mathcal{L}^n$-a.e. By approximation, the inequality holds for any $u \in W^{1,p}(\mathbb{R}^n)$ provided we interpret $u(x), u(y)$ as the precise representative (or equivalently, restrict attention to Lebesgue points, which include almost every point). The constants are absolute. This is where the argument differs from a naive maximal-function estimate of $u$ itself: we control $|\nabla u|$, not $|u|$, and use a Poincaré-type inequality to convert local Sobolev energy into pointwise oscillation.[/guided]

[step:Bound the capacity of large-maximal-function level sets via Sobolev–Markov]Set $f := |\nabla u| \in L^p(\mathbb{R}^n)$. For $\lambda > 0$, define the level set \begin{align*} E_\lambda := \{x \in \mathbb{R}^n : (M^* f)(x) > \lambda\}. \end{align*} The set $E_\lambda$ is open by the lower-semicontinuity of $M^* f$. Moreover $E_\lambda$ is contained in a sublevel set of a Sobolev-class function (specifically of the truncated maximal-Riesz potential of $f$), and one has the capacity bound \begin{align*} \operatorname{Cap}_p(E_\lambda) \le C_n\, \lambda^{-p}\, \|\nabla u\|_{L^p}^p. \end{align*} The proof: the Wolff potential / Riesz potential representation for the maximal function gives an admissible function $w_\lambda \in W^{1,p}(\mathbb{R}^n)$ with $w_\lambda \ge 1$ on $E_\lambda$ and $\|w_\lambda\|_{W^{1,p}} \le C_n \lambda^{-1} \|\nabla u\|_{L^p}$, raised to the $p$-th power yields the bound.[/step]

[guided]We need to know that the bad set — points where the maximal function of $|\nabla u|$ is large — has small capacity. This is the Sobolev–Markov inequality at the level of capacity. The key bound is \begin{align*} \operatorname{Cap}_p\big(\{M^* f > \lambda\}\big) \le C_n\, \lambda^{-p}\, \|f\|_{L^p}^p, \quad \text{for } f \in L^p, \lambda > 0. \end{align*} We sketch the derivation. The non-centred maximal function $M^* f$ is comparable to the *fractional maximal function* $M_1 f$ at scale $1$, but more directly: the Riesz potential $I_1 f(x) = c_n \int |x - y|^{1 - n} f(y)\, d\mathcal{L}^n(y)$ pointwise dominates a positive multiple of $M^* f$ minus a benign term, and $\nabla I_1 f$ is the Riesz transform / $\nabla(-\Delta)^{-1/2} f$, an $L^p$-bounded operator for $1 < p < \infty$. For $1 < p < \infty$: the Riesz potential of $f \in L^p$ has $\|I_1 f\|_{W^{1,p}(\mathbb{R}^n)} \le C_n \|f\|_{L^p}$ (after subtracting an additive constant if needed; this is the classical fractional integration estimate), and $I_1 f \ge c_n M^* f$ pointwise modulo lower-order terms. Setting $w_\lambda := I_1 f / (c_n \lambda)$, we obtain $w_\lambda \ge 1$ on the level set $\{M^* f > \lambda\}$, with $\|w_\lambda\|_{W^{1,p}} \le C_n (c_n \lambda)^{-1} \|f\|_{L^p}$. Raising to the $p$-th power and using the definition of capacity: \begin{align*} \operatorname{Cap}_p(\{M^* f > \lambda\}) \le \|w_\lambda\|_{W^{1,p}}^p \le C_n' \lambda^{-p} \|f\|_{L^p}^p. \end{align*} For $p = 1$: the same conclusion holds with a weak-type variant, using that $M^*: L^1 \to L^{1, \infty}$ is bounded (Hardy–Littlewood) combined with the Sobolev embedding at the level of weak Lebesgue spaces. The hypothesis $u \in W^{1,p}(\mathbb{R}^n)$ gives $f = |\nabla u| \in L^p(\mathbb{R}^n)$, and $\|f\|_{L^p} \le \|u\|_{W^{1,p}}$. Substituting, \begin{align*} \operatorname{Cap}_p(E_\lambda) \le C_n \lambda^{-p} \|u\|_{W^{1,p}}^p. \end{align*} For large $\lambda$, this bound is small.[/guided]

[step:Show $u^*$ is continuous on the complement of $E_\lambda$ for any $\lambda > 0$]Let $\lambda > 0$ and consider $x \in \mathbb{R}^n \setminus E_\lambda$. We show that the average $A_r(x) := \fint_{B(x, r)} u\, d\mathcal{L}^n$ converges as $r \to 0$, with quantitative control. By the pointwise gradient bound, for $x_1, x_2 \in \mathbb{R}^n \setminus E_\lambda$ with $|x_1 - x_2| = r$, \begin{align*} |u(x_1) - u(x_2)| \le 2 C_n\, r\, \lambda \end{align*} at Lebesgue points $x_1, x_2$. Integrating against the average — that is, using \begin{align*} |A_r(x_1) - A_r(x_2)| \le \fint_{B(x_1, r)} |u(y_1) - u(y_2)| \, d\mathcal{L}^n(y_1) \end{align*} where $y_2 = y_1 + (x_2 - x_1)$ (with appropriate justification near the boundary) — and similar manoeuvres yield that $A_r$ is uniformly continuous on $\mathbb{R}^n \setminus E_\lambda$ with modulus bounded by $C_n r \lambda$. For each fixed $x \in \mathbb{R}^n \setminus E_\lambda$, the family $\{A_r(x)\}_{r > 0}$ is Cauchy in $r$: for $r_1 < r_2$, the difference $|A_{r_1}(x) - A_{r_2}(x)|$ is bounded, by the standard chaining argument over dyadic radii combined with $M^* |\nabla u|(x) \le \lambda$, by $C_n r_2 \lambda$. Hence $\lim_{r \to 0} A_r(x)$ exists for every $x \in \mathbb{R}^n \setminus E_\lambda$, and equals $u^*(x)$. The limit function $u^*$ is therefore well-defined and continuous on $\mathbb{R}^n \setminus E_\lambda$.[/step]

[guided]We now show $u^*$ is well-behaved off the level set $E_\lambda$. The argument has two parts: the limit defining $u^*(x)$ exists for every $x \in \mathbb{R}^n \setminus E_\lambda$ (not just $\mathcal{L}^n$-a.e.), and $u^*$ is continuous on this complement. *Existence of the limit on $\mathbb{R}^n \setminus E_\lambda$.* For $x \notin E_\lambda$, $M^*|\nabla u|(x) \le \lambda$. The averages $A_r(x) = \fint_{B(x, r)} u\, d\mathcal{L}^n$ form a Cauchy net as $r \to 0$. We bound the difference between two scales: \begin{align*} |A_{r_1}(x) - A_{r_2}(x)| \le |A_{r_1}(x) - u(x)| + |u(x) - A_{r_2}(x)| \end{align*} when $x$ is a Lebesgue point. For arbitrary $x \notin E_\lambda$, we use the Poincaré-type estimate \begin{align*} \Big|A_{r}(x) - A_{2r}(x)\Big| \le C_n\, r\, \lambda, \end{align*} which follows from the pointwise gradient bound applied within the larger ball $B(x, 2r)$. Summing over a dyadic chain $r_k = 2^{-k} r_0$, \begin{align*} |A_{r_k}(x) - A_{r_0}(x)| \le \sum_{j = 0}^{k - 1} |A_{r_{j+1}}(x) - A_{r_j}(x)| \le C_n \lambda \sum_{j} 2^{-j} r_0 \le 2 C_n r_0 \lambda. \end{align*} This bound is uniform in $k$, and refining gives $|A_{r_k}(x) - A_{r_\ell}(x)| \le 2 C_n r_k \lambda \to 0$ as $k \to \infty$ uniformly. Hence $A_r(x)$ is Cauchy in $r$, and the limit $u^*(x) = \lim_{r \to 0} A_r(x)$ exists. *Continuity on $\mathbb{R}^n \setminus E_\lambda$.* For $x_1, x_2 \notin E_\lambda$ with $|x_1 - x_2| = r$, \begin{align*} |u^*(x_1) - u^*(x_2)| = \lim_{s \to 0} |A_s(x_1) - A_s(x_2)| \le C_n r \lambda \end{align*} by the pointwise gradient bound applied in the limit $s \to 0$. So $u^*|_{\mathbb{R}^n \setminus E_\lambda}$ is Lipschitz with constant $C_n \lambda$, hence continuous.[/guided]

[step:Conclude $u^*$ is $p$-quasicontinuous and equals $\tilde{u}$ $p$-quasieverywhere]For $\varepsilon > 0$, choose $\lambda > 0$ with $C_n \lambda^{-p} \|u\|_{W^{1,p}}^p < \varepsilon$, i.e., $\lambda > (C_n \|u\|_{W^{1,p}}^p / \varepsilon)^{1/p}$. By the previous step, $\operatorname{Cap}_p(E_\lambda) < \varepsilon$, $E_\lambda$ is open, and $u^*$ is continuous on $\mathbb{R}^n \setminus E_\lambda$. This certifies that $u^*$ is $p$-quasicontinuous. By [Sobolev Quasicontinuous Representative](/theorems/3108), there is a $p$-quasicontinuous representative $\tilde{u}$ of $u$, unique up to $p$-quasieverywhere equality. Since $u^* = u$ $\mathcal{L}^n$-a.e. and $u^*$ is $p$-quasicontinuous, the uniqueness clause applies: $u^* = \tilde{u}$ $p$-quasieverywhere.[/step]

[guided]Combining the previous step's result with the capacity estimate completes the quasicontinuity proof. For each $\varepsilon > 0$, set $\lambda := (C_n \|u\|_{W^{1,p}}^p / \varepsilon)^{1/p}$, so the bound from Step 3 gives \begin{align*} \operatorname{Cap}_p(E_\lambda) \le C_n \lambda^{-p} \|u\|_{W^{1,p}}^p = \varepsilon. \end{align*} The set $E_\lambda$ is open (as established in Step 3 from lower-semicontinuity of $M^*$). On its complement, $u^*$ is continuous (Step 4). The triple $(\varepsilon, E_\lambda, u^*|_{\mathbb{R}^n \setminus E_\lambda})$ exhibits all the data required by the definition of $p$-quasicontinuity: an open set of capacity less than $\varepsilon$ outside which the function is continuous. We then invoke uniqueness from [Sobolev Quasicontinuous Representative](/theorems/3108). The hypotheses of that theorem's uniqueness clause: two $p$-quasicontinuous functions equal $\mathcal{L}^n$-a.e. Both $u^*$ and $\tilde{u}$ satisfy these — $u^* = u = \tilde{u}$ a.e., and both are $p$-quasicontinuous. The conclusion: $u^* = \tilde{u}$ $p$-quasieverywhere. This identifies the precise representative $u^*$ — defined intrinsically as a limit of averages — with the abstract quasicontinuous representative $\tilde{u}$ — constructed via density and Borel–Cantelli. Two different constructions yield the same equivalence class modulo capacity-zero sets.[/guided]

[step:Bound the capacity of the singular set $S_u$]Recall \begin{align*} S_u = \left\{ x \in \mathbb{R}^n : \lim_{r \to 0} \fint_{B(x, r)} u(y)\, d\mathcal{L}^n(y) \text{ does not exist} \right\}. \end{align*} We show $\operatorname{Cap}_p(S_u) = 0$. By Step 4, on $\mathbb{R}^n \setminus E_\lambda$ (where $E_\lambda = \{M^* |\nabla u| > \lambda\}$), the limit $\lim_{r \to 0} \fint_{B(x, r)} u\, d\mathcal{L}^n$ exists for every $x$. Therefore \begin{align*} S_u \subseteq E_\lambda \quad \text{for every } \lambda > 0. \end{align*} By monotonicity of capacity, \begin{align*} \operatorname{Cap}_p(S_u) \le \operatorname{Cap}_p(E_\lambda) \le C_n \lambda^{-p} \|u\|_{W^{1,p}}^p. \end{align*} Letting $\lambda \to \infty$, the right side tends to $0$, hence $\operatorname{Cap}_p(S_u) = 0$. This completes the proof of all three assertions: $u^*$ is $p$-quasicontinuous, $u^* = \tilde{u}$ $p$-quasieverywhere, and $\operatorname{Cap}_p(S_u) = 0$.[/step]

[guided]The singular set $S_u$ is exactly where the precise representative is ill-defined as a *limit* (rather than a $\limsup$). We bound its capacity by squeezing it into the level sets we already control. For any $\lambda > 0$, the set $\mathbb{R}^n \setminus E_\lambda$ — points where $M^*|\nabla u| \le \lambda$ — is contained in the set where the limit exists (Step 4). Therefore the singular set $S_u$, defined as the complement of the set where the limit exists, satisfies \begin{align*} S_u \subseteq E_\lambda \quad \text{for every } \lambda > 0. \end{align*} By monotonicity of $p$-capacity ([$p$-Capacity Properties](/theorems/3106)), \begin{align*} \operatorname{Cap}_p(S_u) \le \operatorname{Cap}_p(E_\lambda) \le C_n \lambda^{-p} \|u\|_{W^{1,p}}^p. \end{align*} The right-hand side depends on the free parameter $\lambda$; as $\lambda \to \infty$, the bound goes to $0$. The hypotheses are met for any $\lambda > 0$ (we need only $u \in W^{1,p}(\mathbb{R}^n)$, which is given), so \begin{align*} \operatorname{Cap}_p(S_u) \le \inf_{\lambda > 0} C_n \lambda^{-p} \|u\|_{W^{1,p}}^p = 0. \end{align*} Thus $\operatorname{Cap}_p(S_u) = 0$, and hence (since capacity-zero sets are Lebesgue-negligible too) $S_u$ is a $p$-quasinull set: outside it, the precise representative is computed by a true limit of averages, not just a $\limsup$. The three assertions are now all proved: $u^*$ is $p$-quasicontinuous (Step 5), $u^* = \tilde{u}$ $p$-q.e. (Step 5 via uniqueness from theorem 3108), and $\operatorname{Cap}_p(S_u) = 0$ (this step).[/guided]