[proofplan] We construct $\tilde{u}$ as the uniform-on-good-sets limit of a rapidly convergent sequence of smooth approximations. The density of $C^\infty(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$ in $W^{1,p}(\mathbb{R}^n)$ supplies a sequence $u_k \to u$ in $W^{1,p}$ at the geometric rate $\|u_k - u\|_{W^{1,p}} < 2^{-k-1}$, hence $\|u_{k+1} - u_k\|_{W^{1,p}} < \tfrac{3}{4}\, 2^{-k}$. Applying the Sobolev–Markov inequality for $p$-capacity at threshold $\lambda_k := 2^{-k(p-1)/p} \cdot 2^{-1/p}$ gives $\operatorname{Cap}_p(\{|u_{k+1} - u_k| > \lambda_k\}) < 2^{-k}$, and these capacities sum. A Borel–Cantelli argument on capacity then exhibits, for each $\varepsilon > 0$, an open set $H_m$ of capacity less than $\varepsilon$ outside which $u_k$ converges uniformly to a continuous limit. The uniqueness statement reduces to showing that two $p$-quasicontinuous functions agreeing $\mathcal{L}^n$-a.e. agree $p$-quasieverywhere — this follows from the fact that capacity-small sets are also Lebesgue-small (via the Sobolev embedding theorem), so the closed set on which both representatives are continuous fills positive Lebesgue volume in every ball, allowing a continuity argument to detect any non-trivial gap. We treat the case $1 < p < n$ in the body and indicate the modifications for $p = 1$, $p = n$, and $p > n$ where they arise. [/proofplan]

[step:Approximate $u$ by smooth functions at geometric rate in $W^{1,p}$]By the density of $C^\infty(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$ in $W^{1,p}(\mathbb{R}^n)$ (a global version of [Meyers–Serrin](/theorems/58)), there exists a sequence $(u_k)_{k \ge 1}$ with $u_k \in C^\infty(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$ and \begin{align*} \|u_k - u\|_{W^{1,p}(\mathbb{R}^n)} < 2^{-k-1}. \end{align*} This rate is fixed once and used throughout the proof. By the triangle inequality, for $k \ge 1$, \begin{align*} \|u_{k+1} - u_k\|_{W^{1,p}(\mathbb{R}^n)} \le \|u_{k+1} - u\|_{W^{1,p}} + \|u - u_k\|_{W^{1,p}} < 2^{-k-2} + 2^{-k-1} = \tfrac{3}{4}\, 2^{-k}. \end{align*}[/step]

[guided]The strategy is to convert $W^{1,p}$ convergence — which is at the level of integrals — into pointwise convergence outside a small (in capacity) bad set. The conversion mechanism is a Borel–Cantelli–style argument applied to the level sets where consecutive approximations differ noticeably. For this to work, we need approximations that converge fast enough that the level-set capacities form a summable series. Density gives convergence; to get geometric convergence we extract a rapidly convergent subsequence. **Hypotheses of global Meyers–Serrin density.** $\mathbb{R}^n$ is open and $1 \le p < \infty$. Both are given. The conclusion: $C^\infty(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$ is dense in $W^{1,p}(\mathbb{R}^n)$. Density yields, for each prescribed $\varepsilon_k = 2^{-k-1}$, a smooth element within $W^{1,p}$-distance $\varepsilon_k$ of $u$; we choose one and call it $u_k$. **The triangle bound and its strict factor $3/4$.** We have \begin{align*} \|u_{k+1} - u_k\|_{W^{1,p}} \le \|u_{k+1} - u\|_{W^{1,p}} + \|u - u_k\|_{W^{1,p}} < 2^{-k-2} + 2^{-k-1} = \tfrac{3}{4}\, 2^{-k}. \end{align*} The factor $3/4$ leaves headroom in subsequent estimates and is what powers the strict inequality $\operatorname{Cap}_p(G_k) < 2^{-k}$ in the next step. The choice of constant $2^{-k-1}$ (rather than $2^{-k}$) is precisely tuned to give the clean strict bound $\tfrac{3}{4}\, 2^{-k}$ on the consecutive differences. This is the only approximation rate used in the proof; it is not re-picked later.[/guided]

[step:Bound the capacity of the level set $G_k$ by $2^{-k}$]Set \begin{align*} \lambda_k := \tfrac{3}{4}\, 2^{-k}\, \cdot 2^{k/p} = \tfrac{3}{4}\, 2^{-k(p-1)/p},\qquad G_k := \{x \in \mathbb{R}^n : |u_{k+1}(x) - u_k(x)| > \lambda_k\}. \end{align*} The set $G_k$ is open since $u_{k+1} - u_k \in C^\infty(\mathbb{R}^n)$ is continuous. The function $w_k := |u_{k+1} - u_k|/\lambda_k \in W^{1,p}(\mathbb{R}^n)$ satisfies $w_k > 1$ on $G_k$ pointwise and $w_k \in W^{1,p}$ with $\|w_k\|_{W^{1,p}} = \lambda_k^{-1}\,\|u_{k+1} - u_k\|_{W^{1,p}}$. So $w_k$ is admissible in the definition of $\operatorname{Cap}_p(G_k)$ (cf. [$p$-Capacity Properties](/theorems/3106)): \begin{align*} \operatorname{Cap}_p(G_k) \le \|w_k\|_{W^{1,p}}^p = \lambda_k^{-p}\, \|u_{k+1} - u_k\|_{W^{1,p}}^p < \lambda_k^{-p}\,(\tfrac{3}{4})^p\, 2^{-kp}. \end{align*} Substituting $\lambda_k^{-p} = (\tfrac{4}{3})^p\, 2^{k(p-1)}$: \begin{align*} \operatorname{Cap}_p(G_k) < (\tfrac{4}{3})^p \cdot 2^{k(p-1)} \cdot (\tfrac{3}{4})^p \cdot 2^{-kp} = 2^{-k}. \end{align*} This is the single capacity bound used throughout: $\operatorname{Cap}_p(G_k) < 2^{-k}$, geometrically summable for any $p \ge 1$. The threshold $\lambda_k$ also satisfies $\sum_k \lambda_k < \infty$ when $p > 1$ (geometric ratio $2^{-(p-1)/p} < 1$); the $p = 1$ case is handled separately below.[/step]

[guided]We are estimating the capacity of the open set where $u_{k+1}$ and $u_k$ disagree by more than the threshold $\lambda_k$. The trick — sometimes called the *Sobolev–Markov inequality for capacity* — is to scale up the function so that it dominates the indicator of the level set, and use the capacity definition. **Sobolev–Markov inequality (statement and proof).** For $f \in W^{1,p}(\mathbb{R}^n)$ continuous and $\lambda > 0$, \begin{align*} \operatorname{Cap}_p(\{|f| > \lambda\}) \le \lambda^{-p}\, \|f\|_{W^{1,p}}^p. \end{align*} *Proof.* The set $E_\lambda = \{|f| > \lambda\}$ is open since $f$ is continuous. The function $g := |f|/\lambda$ lies in $W^{1,p}(\mathbb{R}^n)$ (because $|f|$ does, by the chain rule for $W^{1,p}$), and $g \ge 1$ on $E_\lambda$. By the definition of $p$-capacity (admissible functions are those in $W^{1,p}$ with $w \ge 1$ on the open set), $g$ is admissible for $E_\lambda$, so \begin{align*} \operatorname{Cap}_p(E_\lambda) \le \|g\|_{W^{1,p}}^p = \lambda^{-p}\,\|f\|_{W^{1,p}}^p. \qquad\square \end{align*} **Choice of threshold.** Apply with $f = u_{k+1} - u_k$ and threshold $\lambda_k$ to be determined. We get \begin{align*} \operatorname{Cap}_p(G_k) \le \lambda_k^{-p}\, \|u_{k+1} - u_k\|_{W^{1,p}}^p < \lambda_k^{-p}\, (\tfrac{3}{4})^p\, 2^{-kp}. \end{align*} For the bound $\operatorname{Cap}_p(G_k) < 2^{-k}$ we need $\lambda_k^{-p}\, (\tfrac{3}{4})^p\, 2^{-kp} = 2^{-k}$, i.e. $\lambda_k^p = (\tfrac{3}{4})^p\, 2^{-k(p-1)}$, giving $\lambda_k = \tfrac{3}{4}\, 2^{-k(p-1)/p}$. With this choice, \begin{align*} \operatorname{Cap}_p(G_k) < 2^{-k}. \end{align*} **Why this threshold also yields uniform Cauchy convergence (for $p > 1$).** In step 4 we will need $\sum_k \lambda_k < \infty$ so that off $\bigcup_{k \ge m} G_k$ the partial sums of $|u_{k+1} - u_k|$ form a Cauchy sequence uniformly. With $\lambda_k = \tfrac{3}{4}\, 2^{-k(p-1)/p}$ this is a geometric series with ratio $2^{-(p-1)/p}$, which is strictly less than $1$ iff $p > 1$. So for $1 < p < \infty$ the threshold simultaneously gives summable capacity and summable threshold. **The case $p = 1$.** When $p = 1$, $\lambda_k = \tfrac{3}{4}$ is constant, so the level sets $G_k$ control only differences of size $\ge 3/4$. The summable-capacity bound $\operatorname{Cap}_1(G_k) < 2^{-k}$ still holds and gives a small bad set in capacity, but uniform convergence off the bad set requires a slightly different setup. The standard fix is to apply the Sobolev–Markov inequality at $\lambda_k = 2^{-k}$ directly (giving $\operatorname{Cap}_1(G_k) < (\tfrac{3}{4})\, 2^0 = \tfrac{3}{4}$, not summable), then use the *truncation* trick: pass to a subsequence converging in $W^{1,1}$ at rate $\|u_{k_j} - u\|_{W^{1,1}} < 2^{-2j}$, so consecutive differences satisfy $\|u_{k_{j+1}} - u_{k_j}\|_{W^{1,1}} < \tfrac{3}{4}\, 2^{-2j}$, and Sobolev–Markov at threshold $\lambda_j = 2^{-j}$ gives $\operatorname{Cap}_1(G_j) < \tfrac{3}{4}\, 2^{-j}$, with $\sum \lambda_j < \infty$. This is the same kind of geometric-rate bookkeeping but with a faster sequence. The detailed bookkeeping for $p = 1$ follows the same pattern; see Heinonen-Kilpelainen-Martio §4 for the full argument. For $1 < p < \infty$ — which is the regime we work in below — the single rate from step 1 with the threshold $\lambda_k$ above suffices. **Sum.** The capacity sum is bounded by \begin{align*} \sum_{k \ge 1} \operatorname{Cap}_p(G_k) < \sum_{k \ge 1} 2^{-k} = 1 < \infty. \end{align*} This is the analytic Borel–Cantelli input we need.[/guided]

[step:Define the bad sets $H_m$ and verify their capacity tends to zero]For $m \ge 1$, set \begin{align*} H_m := \bigcup_{k \ge m} G_k. \end{align*} By the countable subadditivity of $p$-capacity (see [$p$-Capacity Properties](/theorems/3106)), \begin{align*} \operatorname{Cap}_p(H_m) \le \sum_{k \ge m} \operatorname{Cap}_p(G_k) < \sum_{k \ge m} 2^{-k} = 2^{-(m-1)} \to 0 \quad \text{as } m \to \infty. \end{align*} Each $H_m$ is open, being a union of open sets.[/step]

[guided]The level sets $G_k$ have summable capacity. The Borel–Cantelli–for-capacity heuristic says that the set of points lying in $G_k$ for arbitrarily large $k$ should have capacity zero. We make this precise by tracking the *upper envelope* $H_m = \bigcup_{k \ge m} G_k$ — the set of points where the smooth approximation is still bad at some scale $\ge m$. **Hypothesis check for countable subadditivity.** The countable subadditivity property $\operatorname{Cap}_p(\bigcup_i E_i) \le \sum_i \operatorname{Cap}_p(E_i)$ holds for any countable family of subsets of $\mathbb{R}^n$ (no measurability or topological hypotheses are needed); this is part of the standard list of capacity properties in [$p$-Capacity Properties](/theorems/3106). The family $(G_k)_{k \ge m}$ is countable, so the hypothesis is satisfied. **The capacity tail bound.** \begin{align*} \operatorname{Cap}_p(H_m) \le \sum_{k \ge m} 2^{-k} = 2^{-(m-1)} \to 0 \quad \text{as } m \to \infty. \end{align*} For any prescribed $\varepsilon > 0$ we can choose $m$ large with $\operatorname{Cap}_p(H_m) < \varepsilon$. Each $H_m$ is the union of countably many open sets $G_k$, hence is itself open. This will matter when we want to remove $H_m$ to obtain the closed "good" set on which we control the limit.[/guided]

[step:Show that $u_k$ converges uniformly on $\mathbb{R}^n \setminus H_m$ to a continuous function $\tilde{u}$]Fix $m \ge 1$ and any point $x \notin H_m$. Then for every $k \ge m$, $x \notin G_k$, so $|u_{k+1}(x) - u_k(x)| \le \lambda_k = \tfrac{3}{4}\, 2^{-k(p-1)/p}$. For $\ell > k \ge m$, \begin{align*} |u_\ell(x) - u_k(x)| \le \sum_{j = k}^{\ell - 1} |u_{j+1}(x) - u_j(x)| \le \tfrac{3}{4} \sum_{j = k}^{\ell - 1} 2^{-j(p-1)/p} \le \tfrac{3}{4}\, \frac{2^{-k(p-1)/p}}{1 - 2^{-(p-1)/p}}. \end{align*} The right-hand side is independent of $\ell$ and uniform in $x \in \mathbb{R}^n \setminus H_m$, and tends to zero as $k \to \infty$ (since $p > 1$). Hence $(u_k)_{k \ge m}$ is uniformly Cauchy on $\mathbb{R}^n \setminus H_m$; let $\tilde{u}$ denote its uniform limit on this set. Since each $u_k$ is continuous on $\mathbb{R}^n \supseteq \mathbb{R}^n \setminus H_m$ and the convergence is uniform on $\mathbb{R}^n \setminus H_m$, $\tilde{u}$ is continuous on $\mathbb{R}^n \setminus H_m$. The sets $H_m$ are nested decreasing ($H_m \supseteq H_{m+1}$), and $\operatorname{Cap}_p(H_m) \to 0$. Define \begin{align*} N := \bigcap_{m = 1}^\infty H_m. \end{align*} Then $\operatorname{Cap}_p(N) \le \operatorname{Cap}_p(H_m) \to 0$, so $\operatorname{Cap}_p(N) = 0$. On $\mathbb{R}^n \setminus N = \bigcup_m (\mathbb{R}^n \setminus H_m)$, $\tilde{u}$ is well defined as the limit. Extend $\tilde{u}$ to $\mathbb{R}^n$ by setting $\tilde{u} := 0$ on $N$ (any choice on $N$ leaves the $p$-quasicontinuity property unchanged, since $N$ has capacity zero).[/step]

[guided]We assemble the pointwise limit. For $x \notin H_m$, the sequence $(u_k(x))_{k \ge m}$ is Cauchy: bounding consecutive differences by $\lambda_k$ (because $x \notin G_k$ for any $k \ge m$) and summing, \begin{align*} |u_\ell(x) - u_k(x)| \le \sum_{j = k}^{\ell - 1} \lambda_j = \tfrac{3}{4}\sum_{j = k}^{\ell - 1} 2^{-j(p-1)/p} \le C_p\, 2^{-k(p-1)/p}, \end{align*} where $C_p := \tfrac{3}{4}/(1 - 2^{-(p-1)/p})$ depends only on $p$ (and is finite for $p > 1$). The estimate is uniform in $x \in \mathbb{R}^n \setminus H_m$, so the sequence is uniformly Cauchy. By the Cauchy criterion in $C_b(\mathbb{R}^n \setminus H_m)$ (bounded continuous functions with the sup norm; uniform limits of bounded continuous functions are bounded and continuous), there is a uniform limit $\tilde{u} \in C(\mathbb{R}^n \setminus H_m)$. **Why the limit is continuous on $\mathbb{R}^n \setminus H_m$.** Each $u_k$ is in $C^\infty(\mathbb{R}^n)$, hence continuous on $\mathbb{R}^n \setminus H_m$, and uniform convergence preserves continuity in any metric space. **Domain of $\tilde{u}$.** The pointwise limit is initially defined only on $\bigcup_m (\mathbb{R}^n \setminus H_m) = \mathbb{R}^n \setminus N$ where $N = \bigcap_m H_m$. We claim $\operatorname{Cap}_p(N) = 0$. Indeed $N \subseteq H_m$ for every $m$, so by monotonicity of capacity ([$p$-Capacity Properties](/theorems/3106)), \begin{align*} \operatorname{Cap}_p(N) \le \operatorname{Cap}_p(H_m) < 2^{-(m-1)}. \end{align*} Since this holds for all $m$, $\operatorname{Cap}_p(N) = 0$. We extend $\tilde{u}$ from $\mathbb{R}^n \setminus N$ to $\mathbb{R}^n$ by setting $\tilde{u}(x) = 0$ for $x \in N$; the choice does not affect the $p$-quasicontinuity property (since $N$ has zero capacity, any redefinition on $N$ leaves $\tilde{u}$ unchanged $p$-q.e.).[/guided]

[step:Verify $\tilde{u}$ is $p$-quasicontinuous and equals $u$ $\mathcal{L}^n$-a.e.]*Quasicontinuity.* Given $\varepsilon > 0$, choose $m$ with $\operatorname{Cap}_p(H_m) < \varepsilon$ (possible since $\operatorname{Cap}_p(H_m) \to 0$). The set $H_m$ is open, hence $\mathbb{R}^n \setminus H_m$ is closed, and $\tilde{u}|_{\mathbb{R}^n \setminus H_m}$ is continuous (step 4). By definition, $\tilde{u}$ is $p$-quasicontinuous. *$\mathcal{L}^n$-a.e. equality.* We show $u = \tilde{u}$ $\mathcal{L}^n$-a.e. via two ingredients. **(i) $L^p$ subsequence extraction with pointwise limit.** The standard "$L^p$-convergence implies a.e. convergence along a subsequence" lemma: > If $f_k \to f$ in $L^p(\mathbb{R}^n)$, $1 \le p < \infty$, then a subsequence $f_{k_j} \to f$ pointwise $\mathcal{L}^n$-a.e. *Proof.* By passing to a subsequence we may assume $\|f_k - f\|_{L^p} < 2^{-k}$. By Chebyshev, \begin{align*} \mathcal{L}^n(\{|f_k - f| > 2^{-k/(p+1)}\}) \le 2^{kp/(p+1)}\,\|f_k - f\|_{L^p}^p < 2^{kp/(p+1)}\,2^{-kp} = 2^{-kp/(p+1)}, \end{align*} which is summable in $k$ (since $p/(p+1) > 0$). By the measure-theoretic Borel–Cantelli lemma (if $\sum_k \mathcal{L}^n(E_k) < \infty$ then $\mathcal{L}^n(\limsup_k E_k) = 0$), the set of $x$ at which $|f_k(x) - f(x)| > 2^{-k/(p+1)}$ infinitely often has measure zero. Off this null set, $|f_k(x) - f(x)| \le 2^{-k/(p+1)} \to 0$, i.e., $f_k(x) \to f(x)$. $\square$ Apply with $f_k = u_k$, $f = u$. Hypotheses: $u_k \to u$ in $L^p(\mathbb{R}^n)$, since $\|u_k - u\|_{L^p} \le \|u_k - u\|_{W^{1,p}} < 2^{-k-1} \to 0$ (by step 1). Conclusion: a subsequence $u_{k_j} \to u$ pointwise $\mathcal{L}^n$-a.e. **(ii) Capacity zero implies Lebesgue measure zero, via the Sobolev embedding.** We show $\mathcal{L}^n(N) = 0$. By outer regularity of capacity (cf. [$p$-Capacity Properties](/theorems/3106): for any $E \subseteq \mathbb{R}^n$, $\operatorname{Cap}_p(E) = \inf\{\operatorname{Cap}_p(U) : U \supseteq E,\ U \text{ open}\}$), for each $\delta > 0$ there is an open $U \supseteq N$ with $\operatorname{Cap}_p(U) < \delta$. By the definition of $p$-capacity, there is an admissible $w \in W^{1,p}(\mathbb{R}^n)$ with $w \ge 1$ on $U$ and $\|w\|_{W^{1,p}} < \delta^{1/p}$. Hypothesis check for outer regularity: it holds for arbitrary subsets of $\mathbb{R}^n$ (no measurability assumption); $N \subseteq \mathbb{R}^n$ qualifies. Apply the [Sobolev embedding theorem](/theorems/61). Hypotheses: $1 \le p < \infty$ (given) and $w \in W^{1,p}(\mathbb{R}^n)$ on the whole space (constructed); both are satisfied. The conclusion depends on the regime: - *Case $1 \le p < n$:* $W^{1,p}(\mathbb{R}^n) \hookrightarrow L^{p^*}(\mathbb{R}^n)$ with $p^* = np/(n-p)$, so $\|w\|_{L^{p^*}} \le C(n,p)\,\|w\|_{W^{1,p}} < C\,\delta^{1/p}$. Since $w \ge 1$ on $U$, $\mathbf{1}_U \le w$ a.e., so $\mathcal{L}^n(U)^{1/p^*} \le \|w\|_{L^{p^*}} < C\,\delta^{1/p}$, hence $\mathcal{L}^n(U) < C^{p^*}\,\delta^{p^*/p} \to 0$ as $\delta \to 0$. - *Case $p = n$:* $W^{1,n}(\mathbb{R}^n) \hookrightarrow L^q(\mathbb{R}^n)$ for every $n \le q < \infty$. Take $q = n + 1$: $\mathcal{L}^n(U)^{1/(n+1)} \le \|w\|_{L^{n+1}} \le C\,\|w\|_{W^{1,n}} < C\,\delta^{1/n}$, so $\mathcal{L}^n(U) \le (C\,\delta^{1/n})^{n+1} \to 0$. - *Case $p > n$:* $W^{1,p}(\mathbb{R}^n) \hookrightarrow L^\infty(\mathbb{R}^n)$ with $\|w\|_{L^\infty} \le C\,\|w\|_{W^{1,p}} < C\,\delta^{1/p}$. For $\delta < C^{-p}$, $w < 1$ everywhere, contradicting $w \ge 1$ on $U$ unless $U = \emptyset$; hence $U = \emptyset$ for small $\delta$ and $\mathcal{L}^n(N) = 0$. In all three regimes $\mathcal{L}^n(N) \le \mathcal{L}^n(U) \to 0$ as $\delta \to 0$, hence $\mathcal{L}^n(N) = 0$. **(iii) Combining.** From step 4, $u_k \to \tilde{u}$ pointwise on $\mathbb{R}^n \setminus N$, and $\mathcal{L}^n(N) = 0$ by (ii); so $u_k \to \tilde{u}$ pointwise $\mathcal{L}^n$-a.e. By (i), a subsequence $u_{k_j} \to u$ pointwise $\mathcal{L}^n$-a.e. The same subsequence converges to both $u$ and $\tilde{u}$ at $\mathcal{L}^n$-a.e. point; uniqueness of pointwise limits gives $u = \tilde{u}$ $\mathcal{L}^n$-a.e.[/step]

[guided]We verify the two assertions about $\tilde{u}$ in turn. *Quasicontinuity.* The definition of $p$-quasicontinuity requires: for every $\varepsilon > 0$, there exists an open set $U_\varepsilon$ with $\operatorname{Cap}_p(U_\varepsilon) < \varepsilon$ such that $\tilde{u}|_{\mathbb{R}^n \setminus U_\varepsilon}$ is continuous. We exhibit such a $U_\varepsilon$: choose $m$ large enough that $\operatorname{Cap}_p(H_m) < \varepsilon$, possible because $\operatorname{Cap}_p(H_m) < 2^{-(m-1)} \to 0$. Set $U_\varepsilon := H_m$. We have already verified: (a) $H_m$ is open (union of open sets $G_k$), (b) $\operatorname{Cap}_p(H_m) < \varepsilon$, (c) $\tilde{u}$ restricted to $\mathbb{R}^n \setminus H_m$ is the uniform limit of the continuous functions $u_k|_{\mathbb{R}^n \setminus H_m}$, hence continuous. This certifies $p$-quasicontinuity. *Almost-everywhere equality.* We assemble the implication $\operatorname{Cap}_p(N) = 0 \implies \mathcal{L}^n(N) = 0$ from two named results, then combine with $L^p$ subsequence extraction. **Outer regularity of capacity.** The $p$-capacity is *outer regular* on $\mathbb{R}^n$: \begin{align*} \operatorname{Cap}_p(E) = \inf\{\operatorname{Cap}_p(U) : U \supseteq E,\ U \text{ open}\} \qquad\text{for every } E \subseteq \mathbb{R}^n. \end{align*} This is one of the standard properties of $p$-capacity collected in [$p$-Capacity Properties](/theorems/3106). Hypothesis check: outer regularity holds for arbitrary subsets of $\mathbb{R}^n$ (no Borel/measurability hypothesis is needed because the infimum over admissible $W^{1,p}$ functions automatically detects open neighbourhoods); $N \subseteq \mathbb{R}^n$ qualifies. Conclusion: for every $\delta > 0$ there is an open $U \supseteq N$ with $\operatorname{Cap}_p(U) < \delta$. Given such $U$, by the definition of $p$-capacity ([$p$-Capacity Properties](/theorems/3106)), \begin{align*} \operatorname{Cap}_p(U) = \inf\{\|w\|_{W^{1,p}}^p : w \in W^{1,p}(\mathbb{R}^n),\ w \ge 1 \text{ on } U\} < \delta, \end{align*} so there is an admissible $w \in W^{1,p}(\mathbb{R}^n)$ with $w \ge 1$ a.e. on $U$ and $\|w\|_{W^{1,p}}^p < \delta$, hence $\|w\|_{W^{1,p}} < \delta^{1/p}$. **Sobolev embedding.** Theorem 61 ([Sobolev embedding](/theorems/61)) states $W^{1,p}(\mathbb{R}^n) \hookrightarrow L^q(\mathbb{R}^n)$ for an exponent $q = q(n,p) \in [p, \infty]$ depending on the regime, with a continuous inclusion $\|w\|_{L^q} \le C(n,p,q)\,\|w\|_{W^{1,p}}$: - $1 \le p < n$: $q = p^* := np/(n-p)$. - $p = n$: $q$ is any finite exponent $\ge n$. - $p > n$: $q = \infty$ (the embedding actually lands in Hölder-continuous functions, but $L^\infty$ is the relevant consequence). Hypotheses: $1 \le p < \infty$ and $w \in W^{1,p}(\mathbb{R}^n)$. Both are satisfied by construction. **Conclusion of (ii).** In each regime, $\mathbf{1}_U \le w$ pointwise (a.e.) gives: - $p < n$: $\mathcal{L}^n(U)^{1/p^*} \le \|w\|_{L^{p^*}} < C\,\delta^{1/p}$, so $\mathcal{L}^n(U) < C^{p^*}\,\delta^{p^*/p}$. Since $p^*/p > 1$, $\mathcal{L}^n(U) \to 0$ faster than linearly in $\delta$. - $p = n$: with $q = n + 1$, $\mathcal{L}^n(U) \le (C\,\delta^{1/n})^{n+1} = C^{n+1}\,\delta^{(n+1)/n}$, again decaying super-linearly. - $p > n$: for $\delta$ small, $\|w\|_{L^\infty} < 1$, contradicting $w \ge 1$ on $U$ unless $U = \emptyset$; so $\mathcal{L}^n(U) = 0$ for $\delta < C^{-p}$. In every regime, $\mathcal{L}^n(N) \le \mathcal{L}^n(U) \to 0$ as $\delta \to 0$, hence $\mathcal{L}^n(N) = 0$. **$L^p$ subsequence extraction.** Lemma: > If $f_k \to f$ in $L^p(\mathbb{R}^n)$, $1 \le p < \infty$, then a subsequence $f_{k_j} \to f$ pointwise $\mathcal{L}^n$-a.e. *Proof.* By passing to a subsequence we may assume $\|f_k - f\|_{L^p} < 2^{-k}$. By Chebyshev, \begin{align*} \mathcal{L}^n(\{|f_k - f| > 2^{-k/(p+1)}\}) \le 2^{kp/(p+1)}\,\|f_k - f\|_{L^p}^p < 2^{-kp/(p+1)}, \end{align*} summable in $k$. Borel–Cantelli (measure-theoretic) gives $\mathcal{L}^n(\limsup_k\{|f_k - f| > 2^{-k/(p+1)}\}) = 0$. Off this null set, $f_k \to f$ pointwise. $\square$ Hypotheses for our application: $\|u_k - u\|_{L^p} \le \|u_k - u\|_{W^{1,p}} < 2^{-k-1} \to 0$ (step 1), so $u_k \to u$ in $L^p$. Conclusion: a subsequence $u_{k_j} \to u$ pointwise $\mathcal{L}^n$-a.e. **Combining.** Step 4 gave $u_k \to \tilde{u}$ pointwise on $\mathbb{R}^n \setminus N$, where $\mathcal{L}^n(N) = 0$ by the Sobolev embedding chain; so $u_k \to \tilde{u}$ pointwise $\mathcal{L}^n$-a.e. Subsequence extraction gives $u_{k_j} \to u$ pointwise $\mathcal{L}^n$-a.e. At every $\mathcal{L}^n$-a.e. point both pointwise limits exist and are unique, so $u = \tilde{u}$ $\mathcal{L}^n$-a.e.[/guided]

[step:Establish uniqueness up to $p$-quasieverywhere equality]Suppose $v: \mathbb{R}^n \to \mathbb{R}$ is $p$-quasicontinuous with $v = u$ $\mathcal{L}^n$-a.e. Then $v = \tilde{u}$ $\mathcal{L}^n$-a.e. We show $v = \tilde{u}$ $p$-quasieverywhere. **Strategy.** A quasicontinuous function is determined a.e. on the closed set on which it is continuous; we exploit that capacity-small sets are also Lebesgue-small (via the Sobolev embedding established in step 5(ii)), so the closed complement $F = \mathbb{R}^n \setminus W$ of a small open set $W$ fills positive Lebesgue volume in every Euclidean ball. A continuity argument then converts a single bad point into a positive-$\mathcal{L}^n$-measure violation of the a.e. equality $v = \tilde{u}$. **Reduction.** Suppose for contradiction $\operatorname{Cap}_p(\{v \ne \tilde{u}\}) > 0$. By countable subadditivity ([$p$-Capacity Properties](/theorems/3106)), $\operatorname{Cap}_p(\{v > \tilde{u}\}) > 0$ or $\operatorname{Cap}_p(\{v < \tilde{u}\}) > 0$; without loss of generality assume the former. Stratifying by gap size, \begin{align*} \{v > \tilde{u}\} = \bigcup_{k \ge 1} A_k,\qquad A_k := \{v - \tilde{u} > 1/k\}. \end{align*} By countable subadditivity, $\operatorname{Cap}_p(A_k) > 0$ for some $k$. Fix such $k$ and set $c := 1/k > 0$. We derive a contradiction by showing $\operatorname{Cap}_p(A_k) < \varepsilon$ for arbitrary $\varepsilon > 0$. **Construction of the bad open set.** Fix $\varepsilon > 0$. By quasicontinuity of $\tilde{u}$ and $v$, pick open $U, V$ with $\operatorname{Cap}_p(U), \operatorname{Cap}_p(V) < \varepsilon/2$ such that $\tilde{u}|_{\mathbb{R}^n \setminus U}$ and $v|_{\mathbb{R}^n \setminus V}$ are continuous. Set $W := U \cup V$. By countable subadditivity, $\operatorname{Cap}_p(W) \le \operatorname{Cap}_p(U) + \operatorname{Cap}_p(V) < \varepsilon$. On $\mathbb{R}^n \setminus W$, both $\tilde{u}$ and $v$ are continuous (continuity is preserved under restriction to a smaller closed set), hence so is $v - \tilde{u}$. **Lebesgue smallness of $W$.** By the [Sobolev embedding](/theorems/61) chain in step 5(ii), there is a constant $C = C(n,p)$ and an exponent $\beta = \beta(n,p) > 1$ such that for any open set $\Omega$ with $\operatorname{Cap}_p(\Omega) < \delta$, \begin{align*} \mathcal{L}^n(\Omega) \le C\,\delta^{\beta}. \end{align*} (Specifically, $\beta = p^*/p$ for $p < n$, $\beta = (n+1)/n$ for $p = n$, and $\Omega$ may be forced to be empty for $p > n$ when $\delta$ is small enough; in each regime $\beta > 1$. Hypothesis check on theorem 61: $1 \le p < \infty$ and the admissible $w \in W^{1,p}(\mathbb{R}^n)$ are supplied as in step 5(ii).) Apply with $\Omega = W$ and $\delta = \varepsilon$: \begin{align*} \mathcal{L}^n(W) \le C\,\varepsilon^{\beta}. \end{align*} **Continuity-driven contradiction.** Suppose, for contradiction with the eventual conclusion $A_k \subseteq W$, that $A_k \not\subseteq W$, i.e., there is $x_0 \in A_k \setminus W$. Then $v(x_0) - \tilde{u}(x_0) > c$, and both $\tilde{u}$ and $v$ are continuous at $x_0$ (since $x_0 \in \mathbb{R}^n \setminus W$). Continuity of $v - \tilde{u}$ at $x_0$ on $\mathbb{R}^n \setminus W$ gives an $r > 0$ such that \begin{align*} v(y) - \tilde{u}(y) > c/2 \qquad \text{for every } y \in B_r(x_0) \cap (\mathbb{R}^n \setminus W). \end{align*} The Lebesgue measure of $B_r(x_0) \cap (\mathbb{R}^n \setminus W)$ is bounded below by \begin{align*} \mathcal{L}^n(B_r(x_0) \cap (\mathbb{R}^n \setminus W)) \ge \mathcal{L}^n(B_r(x_0)) - \mathcal{L}^n(W) \ge \omega_n\, r^n - C\,\varepsilon^{\beta}, \end{align*} where $\omega_n = \mathcal{L}^n(B_1(0))$. **Choose $\varepsilon$ small enough** (depending on $r$) that $C\,\varepsilon^{\beta} < \tfrac{1}{2}\,\omega_n\, r^n$; we may do this because $\beta > 1$ and the right-hand side is fixed. Then $\mathcal{L}^n(B_r(x_0) \cap (\mathbb{R}^n \setminus W)) \ge \tfrac{1}{2}\,\omega_n\, r^n > 0$. So a set of positive $\mathcal{L}^n$-measure satisfies $v - \tilde{u} > c/2 > 0$. But $v = \tilde{u}$ $\mathcal{L}^n$-a.e. by hypothesis, so $\{v - \tilde{u} > c/2\}$ has $\mathcal{L}^n$-measure zero. **Contradiction.** Hence $A_k \subseteq W$, and \begin{align*} \operatorname{Cap}_p(A_k) \le \operatorname{Cap}_p(W) < \varepsilon. \end{align*} **Subtle point: the radius $r$ depends on $x_0$, which depends on $\varepsilon$.** To make the argument unconditional, we must show that for *some* $\varepsilon > 0$ small enough, $A_k \setminus W = \emptyset$ — i.e., we cannot have a single $x_0$ with arbitrarily small $r$. The fix: pick $\varepsilon > 0$ to be chosen at the end. Suppose $A_k \setminus W \ne \emptyset$ and pick any $x_0$ in it; the continuity radius $r = r(x_0, \varepsilon)$ is positive. The argument above produces a positive-$\mathcal{L}^n$-measure violation for any $\varepsilon$ with $C\,\varepsilon^{\beta} < \tfrac{1}{2}\, \omega_n\, r^n$. This is achievable — but $r$ depends on $x_0$, which depends on $\varepsilon$. The clean way: argue contrapositively. Fix any $r_0 > 0$ and consider only $\varepsilon$ with $C\,\varepsilon^{\beta} < \tfrac{1}{2}\, \omega_n\, r_0^n$. Then for any $x_0$ with continuity radius $r \ge r_0$, the contradiction holds. The continuity radius $r$ depends on the modulus of continuity of $v - \tilde{u}$ at $x_0$ on $\mathbb{R}^n \setminus W$ and on the gap $c$; given $c > 0$ fixed, the modulus of continuity provides a uniform $r_0 > 0$ on any compact subset of $\mathbb{R}^n \setminus W$. Since $A_k \setminus W$ may be unbounded, restrict attention to $A_k \cap K$ for an arbitrary compact $K$; the compact restriction has a uniform $r_0 = r_0(K) > 0$, and the contradiction shows $(A_k \cap K) \setminus W = \emptyset$, i.e., $A_k \cap K \subseteq W$. Since $K$ is arbitrary, $A_k \subseteq W$. **Conclusion.** $\operatorname{Cap}_p(A_k) < \varepsilon$ for arbitrary $\varepsilon > 0$, hence $\operatorname{Cap}_p(A_k) = 0$ — contradicting our standing assumption $\operatorname{Cap}_p(A_k) > 0$. The contradiction shows $\operatorname{Cap}_p(\{v > \tilde{u}\}) = 0$. By the symmetric argument $\operatorname{Cap}_p(\{v < \tilde{u}\}) = 0$, hence $\operatorname{Cap}_p(\{v \ne \tilde{u}\}) = 0$, i.e., $v = \tilde{u}$ $p$-quasieverywhere.[/step]

[guided]**Why a purely topological argument fails.** A natural attempt is: on the closed set $F := \mathbb{R}^n \setminus W$, both representatives are continuous, so $A_k \cap F$ is relatively open in $F$; if non-empty, pick a point $x_0$ and a neighbourhood $B_r(x_0) \cap F$ of $x_0$ in $F$; this neighbourhood "should" have positive $\mathcal{L}^n$-measure, contradicting $\mathcal{L}^n(A_k) = 0$. The flaw: $B_r(x_0) \cap F$ need not have positive $\mathcal{L}^n$-measure if $F$ is sufficiently sparse around $x_0$. Concretely, if $W$ is an open set of small capacity that happens to be dense in $\mathbb{R}^n$ (e.g. the union of small balls centred at the rationals — small total capacity, but dense), then $F = \mathbb{R}^n \setminus W$ has empty interior, and a relatively open subset of $F$ can be a single point with $\mathcal{L}^n$-measure zero. The fix: $W$ is small not just in capacity but also in $\mathcal{L}^n$-measure — by the Sobolev embedding chain in step 5(ii), $\mathcal{L}^n(W) \le C\,\operatorname{Cap}_p(W)^{\beta}$ with $\beta > 1$. So the $\mathcal{L}^n$-bulk of any small ball survives the removal of $W$, and the continuity argument detects a positive-measure violation of "$v = \tilde{u}$ a.e." **Reduction (recap).** Assume $\operatorname{Cap}_p(\{v \ne \tilde{u}\}) > 0$. Stratify: \begin{align*} \{v \ne \tilde{u}\} = \{v > \tilde{u}\} \cup \{v < \tilde{u}\},\qquad \{v > \tilde{u}\} = \bigcup_k A_k,\quad A_k := \{v - \tilde{u} > 1/k\}. \end{align*} By countable subadditivity ([$p$-Capacity Properties](/theorems/3106)), $\operatorname{Cap}_p(A_k) > 0$ for some $k$. Fix such $k$ and set $c := 1/k$. **Construction of $W$.** Fix $\varepsilon > 0$. By quasicontinuity, pick open $U, V$ with $\operatorname{Cap}_p(U), \operatorname{Cap}_p(V) < \varepsilon/2$ and continuity of $\tilde{u}|_{\mathbb{R}^n \setminus U}$, $v|_{\mathbb{R}^n \setminus V}$. Set $W := U \cup V$. By subadditivity, $\operatorname{Cap}_p(W) < \varepsilon$. **Lebesgue smallness of $W$.** By the Sobolev embedding chain (step 5(ii)), \begin{align*} \mathcal{L}^n(W) \le C\,\varepsilon^{\beta}, \end{align*} where $\beta = p^*/p > 1$ for $p < n$, $\beta = (n+1)/n > 1$ for $p = n$, and $W = \emptyset$ for $\delta$ small in the regime $p > n$. (Hypothesis check: outer regularity of capacity supplies an admissible $w \in W^{1,p}$ with $\|w\|_{W^{1,p}} < \varepsilon^{1/p}$; theorem 61 then gives the $L^q$-estimate; both hypotheses match step 5(ii).) **Continuity-driven contradiction.** Suppose $A_k \not\subseteq W$, i.e., there is $x_0 \in A_k \setminus W$. Both $\tilde{u}$ and $v$ are continuous at $x_0$ on $\mathbb{R}^n \setminus W$, so there is $r > 0$ with $v(y) - \tilde{u}(y) > c/2$ for $y \in B_r(x_0) \cap (\mathbb{R}^n \setminus W)$. The Lebesgue measure of this set is at least \begin{align*} \omega_n\, r^n - \mathcal{L}^n(W) \ge \omega_n\, r^n - C\,\varepsilon^{\beta}. \end{align*} Since $\beta > 1$, choosing $\varepsilon$ small (uniformly over a compact range of $x_0$, as discussed in the exact version) makes this strictly positive. Then $\{v - \tilde{u} > c/2\}$ has positive $\mathcal{L}^n$-measure, contradicting $v = \tilde{u}$ $\mathcal{L}^n$-a.e. Hence $A_k \subseteq W$, so $\operatorname{Cap}_p(A_k) \le \operatorname{Cap}_p(W) < \varepsilon$. **Letting $\varepsilon \to 0$.** $\operatorname{Cap}_p(A_k) = 0$ for each $k$, hence $\operatorname{Cap}_p(\{v > \tilde{u}\}) = 0$ by countable subadditivity. By symmetry $\operatorname{Cap}_p(\{v < \tilde{u}\}) = 0$, hence $\operatorname{Cap}_p(\{v \ne \tilde{u}\}) = 0$, i.e., $v = \tilde{u}$ $p$-quasieverywhere. The new ingredient relative to the failed topological attempt is that $W$ is small not only in capacity but also in Lebesgue measure (by the Sobolev embedding chain); this lets a single point $x_0 \in A_k \setminus W$ generate a positive-$\mathcal{L}^n$-measure violation via continuity, no matter how the residual closed set $\mathbb{R}^n \setminus W$ is distributed.[/guided]