[proofplan] Throughout the proof, the $p$-capacity is the set function \begin{align*} \operatorname{Cap}_p: 2^{\mathbb{R}^n} \to [0, \infty], \qquad \operatorname{Cap}_p(E) := \inf\bigl\{\|u\|_{W^{1,p}(\mathbb{R}^n)}^p : u \in \mathcal{A}(E)\bigr\}, \end{align*} where $\|u\|_{W^{1,p}}^p := \|u\|_{L^p}^p + \|\nabla u\|_{L^p}^p$ (a fixed convention used throughout) and the admissible class is $\mathcal{A}(E) := \{u \in W^{1,p}(\mathbb{R}^n) : u \ge 1\ \mathcal{L}^n\text{-a.e.}\ \text{on a neighbourhood of}\ E\}$. The six properties are proved in order. Monotonicity (i) comes from inclusion of admissible classes. Countable subadditivity (ii) is the most delicate property: we exhibit an admissible function for the union as a max-of-truncated supremum of admissible functions for the pieces, using the lattice structure of $W^{1,p}$. Outer regularity (iii) and inner regularity (iv) follow from direct manipulation of the inf/sup definitions, with (iv) using the Choquet capacitability framework (whose hypotheses are verified by combining (i), (ii), (vi)). The two continuity statements (v) and (vi) combine monotonicity with subadditivity (resp. inner regularity). [/proofplan] [step:Prove monotonicity from inclusion of admissible classes] Let $E \subseteq F \subseteq \mathbb{R}^n$. We claim $\mathcal{A}(F) \subseteq \mathcal{A}(E)$. Suppose $u \in \mathcal{A}(F)$: $u \in W^{1,p}(\mathbb{R}^n)$ and $u \ge 1$ $\mathcal{L}^n$-a.e. on some open neighbourhood $V \supseteq F$. Since $E \subseteq F \subseteq V$, the same set $V$ is also an open neighbourhood of $E$, and $u \ge 1$ a.e. on $V$. Hence $u \in \mathcal{A}(E)$. Taking the infimum over $\mathcal{A}(E)$ — a set containing $\mathcal{A}(F)$ — gives a value no larger than the infimum over $\mathcal{A}(F)$: \begin{align*} \operatorname{Cap}_p(E) = \inf_{u \in \mathcal{A}(E)}\|u\|_{W^{1,p}}^p \le \inf_{u \in \mathcal{A}(F)}\|u\|_{W^{1,p}}^p = \operatorname{Cap}_p(F). \end{align*} This is property (i). [/step] [step:Prove countable subadditivity by combining admissible functions via truncated supremum] Let $(E_k)_{k \ge 1}$ be a sequence of subsets of $\mathbb{R}^n$. We show \begin{align*} \operatorname{Cap}_p\left(\bigcup_{k=1}^\infty E_k\right) \le \sum_{k=1}^\infty \operatorname{Cap}_p(E_k). \end{align*} If $\sum_k \operatorname{Cap}_p(E_k) = \infty$ the inequality is immediate. Assume the sum is finite, fix $\varepsilon > 0$, and choose for each $k$ an admissible function \begin{align*} u_k \in \mathcal{A}(E_k), \qquad \|u_k\|_{W^{1,p}(\mathbb{R}^n)}^p \le \operatorname{Cap}_p(E_k) + \frac{\varepsilon}{2^k}. \end{align*} **Truncation.** Replace each $u_k$ by $T(u_k) := \min(u_k^+, 1)$ where $u_k^+ := \max(u_k, 0)$. The Lipschitz function $g(t) := \min(\max(t, 0), 1)$ has Lipschitz constant $1$ and $g'(t) \in \{0, 1\}$ a.e. By the [Chain Rule for Weak Derivatives](/theorems/3099) (Lipschitz post-composition form, which applies to any Lipschitz $g$ with $g' \in L^\infty$), \begin{align*} \|T(u_k)\|_{L^p}^p &= \int_{\mathbb{R}^n}|T(u_k)|^p \le \int_{\mathbb{R}^n}|u_k|^p = \|u_k\|_{L^p}^p, \\ \|\nabla T(u_k)\|_{L^p}^p &= \int_{\mathbb{R}^n}|g'(u_k)|^p|\nabla u_k|^p \le \int_{\mathbb{R}^n}|\nabla u_k|^p = \|\nabla u_k\|_{L^p}^p, \end{align*} using $|g'| \le 1$ a.e. Truncation preserves admissibility: if $u_k \ge 1$ on $V_k$, then $T(u_k) = 1$ on $V_k$, hence $T(u_k) \ge 1$ on $V_k$. So we drop the $T$ notation and assume $0 \le u_k \le 1$ a.e. for the remainder of the step. **Partial supremum.** Consider \begin{align*} v_N(x) := \max_{1 \le k \le N} u_k(x), \qquad x \in \mathbb{R}^n. \end{align*} The Lipschitz function $\max(\cdot, c)$ has Lipschitz constant $1$. By the [Chain Rule for Weak Derivatives](/theorems/3099) (Lipschitz post-composition), $\max(u, c) \in W^{1,p}$ with $\nabla\max(u, c) = \nabla u\,\mathbb{1}_{\{u > c\}}$ a.e. Writing $\max(a, b) = a + (b - a)^+$ and using induction on $N$, $v_N \in W^{1,p}(\mathbb{R}^n)$ with the lattice gradient formula \begin{align*} \nabla\max(u, v) = \nabla u\, \mathbb{1}_{\{u \ge v\}} + \nabla v\, \mathbb{1}_{\{u < v\}} \quad \text{a.e.} \end{align*} On the coincidence set $\{u = v\}$ the two candidates agree because $\nabla(u - v) = 0$ a.e. on the level set $\{u - v = 0\}$ (a property of weak derivatives), so the formula is unambiguous a.e. Define $v(x) := \sup_k u_k(x)$ as the pointwise a.e. supremum. Since $0 \le u_k \le 1$, also $0 \le v_N \le v \le 1$ a.e. **Norm bound for $v_N$.** Define the measurable sets $A_k := \{v_N = u_k\} \setminus \bigcup_{l < k}\{v_N = u_l\}$ for $k = 1, \dots, N$ (ties broken by smaller index). These form a measurable partition of $\mathbb{R}^n$, and on $A_k$, $v_N = u_k$ and $\nabla v_N = \nabla u_k$ a.e. (lattice gradient formula plus the coincidence-set fact). Hence \begin{align*} \|v_N\|_{W^{1,p}}^p &= \int_{\mathbb{R}^n}|v_N|^p + \int_{\mathbb{R}^n}|\nabla v_N|^p = \sum_{k=1}^N\int_{A_k}\bigl(|u_k|^p + |\nabla u_k|^p\bigr) \\ &\le \sum_{k=1}^N\int_{\mathbb{R}^n}\bigl(|u_k|^p + |\nabla u_k|^p\bigr) = \sum_{k=1}^N\|u_k\|_{W^{1,p}}^p, \end{align*} where the inequality enlarges the integration domain from $A_k$ to $\mathbb{R}^n$, valid because the integrands are non-negative. So $\|v_N\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p \le \sum_k\operatorname{Cap}_p(E_k) + \varepsilon < \infty$. **Pass to the limit (Case $1 < p < \infty$).** The sequence $(v_N)$ is monotone increasing pointwise a.e. to $v$, with $0 \le v_N \le 1$ uniformly. By dominated convergence (dominator $1 \cdot \mathbb{1}_{\operatorname{supp} v}$ on bounded sets, plus the global $L^p$ bound just established), $v_N \to v$ in $L^p$. The space $W^{1,p}(\mathbb{R}^n)$ is reflexive ([Reflexivity of $W^{1,p}$](/theorems/3094)), and $(v_N)$ is bounded in $W^{1,p}$, so by [Weak Compactness in $W^{1,p}$](/theorems/3104) a subsequence $v_{N_j} \rightharpoonup w$ weakly in $W^{1,p}$. Weak $W^{1,p}$ convergence implies weak $L^p$ convergence; combined with the strong $L^p$ convergence $v_N \to v$, uniqueness of weak $L^p$ limits gives $w = v$. By lower semicontinuity of the norm under weak convergence, \begin{align*} \|v\|_{W^{1,p}}^p \le \liminf_j\|v_{N_j}\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p. \end{align*} **Pass to the limit (Case $p = 1$).** Reflexivity fails for $W^{1,1}$, so we proceed locally. Fix $R > 0$. The partition formula gives $|\nabla v_N| \le \sum_{k=1}^N|\nabla u_k|$ pointwise a.e.; setting $g := \sum_k|\nabla u_k|$, we have $g \in L^1(\mathbb{R}^n)$ since $\sum_k\|\nabla u_k\|_{L^1} \le \sum_k\operatorname{Cap}_1(E_k) + \varepsilon < \infty$. So $|\nabla v_N| \le g$ pointwise a.e., which on the bounded set $B_R$ gives equi-integrability of $(\nabla v_N)$ in $L^1(B_R)$. By the Dunford–Pettis theorem, a subsequence $\nabla v_{N_j}$ converges weakly in $L^1(B_R)^n$ to some $G \in L^1(B_R)^n$. Combined with $v_N \to v$ in $L^1(B_R)$ (DCT, dominator $1$), the standard test-function identification using $\varphi \in C^\infty_c(B_R)$ — \begin{align*} \int_{B_R}G \cdot \varphi = \lim_j\int_{B_R}\nabla v_{N_j} \cdot \varphi = -\lim_j\int_{B_R}v_{N_j}\nabla\varphi = -\int_{B_R}v\,\nabla\varphi \end{align*} — gives $G = \nabla v$ a.e. on $B_R$, so $v \in W^{1,1}(B_R)$ with \begin{align*} \|\nabla v\|_{L^1(B_R)} \le \liminf_j\|\nabla v_{N_j}\|_{L^1(B_R)} \le \sum_k\|\nabla u_k\|_{L^1}. \end{align*} Letting $R \to \infty$ (monotone convergence applied to $|\nabla v|$), $v \in W^{1,1}(\mathbb{R}^n)$ with $\|\nabla v\|_{L^1} \le \sum_k\|\nabla u_k\|_{L^1}$. Combined with the $L^1$ bound (Fatou applied to $|v_N|^p \nearrow |v|^p$), $\|v\|_{W^{1,1}}^p \le \sum_k\|u_k\|_{W^{1,1}}^p$. In both cases $v \in W^{1,p}(\mathbb{R}^n)$ with $\|v\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p$. **Admissibility of $v$.** Each $u_k \ge 1$ on a neighbourhood $V_k$ of $E_k$; hence $v \ge 1$ on $V := \bigcup_k V_k$, an open neighbourhood of $\bigcup_k E_k$. Therefore $v \in \mathcal{A}(\bigcup_k E_k)$. By definition of capacity, \begin{align*} \operatorname{Cap}_p\left(\bigcup_k E_k\right) \le \|v\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p \le \sum_{k=1}^\infty \operatorname{Cap}_p(E_k) + \varepsilon. \end{align*} Letting $\varepsilon \to 0$ gives property (ii). [guided] **Strategy.** To bound $\operatorname{Cap}_p$ of a countable union, we need a single admissible function whose Sobolev norm-power is at most the sum of the norm-powers of admissible functions for each piece. The supremum $v := \sup_k u_k$ is the natural candidate: it is $\ge 1$ wherever any single $u_k$ is, hence $\ge 1$ on the union of neighbourhoods. **Why we can truncate.** The first move is to replace each $u_k$ by its truncation $T(u_k) := \min(u_k^+, 1)$. The Lipschitz function $g(t) := \min(\max(t, 0), 1)$ has $|g'| \le 1$ pointwise (and $g' \in \{0, 1\}$ a.e.), so by the [Chain Rule for Weak Derivatives](/theorems/3099), \begin{align*} T(u_k) \in W^{1,p}(\mathbb{R}^n), \quad \nabla T(u_k) = g'(u_k) \nabla u_k \quad \text{a.e.} \end{align*} Both $L^p$ norms decrease: $|T(u_k)| \le |u_k|$ pointwise, $|\nabla T(u_k)| \le |\nabla u_k|$ pointwise. Truncation also preserves admissibility: if $u_k \ge 1$ on $V_k$, then $T(u_k) = 1$ on $V_k$, hence $\ge 1$. So we may assume $0 \le u_k \le 1$. **Why the supremum is in $W^{1,p}$.** The Sobolev space $W^{1,p}(\mathbb{R}^n)$ is closed under finite max: if $u, v \in W^{1,p}$, then $\max(u, v) \in W^{1,p}$ with \begin{align*} \nabla\max(u, v) = \nabla u\,\mathbb{1}_{\{u \ge v\}} + \nabla v\,\mathbb{1}_{\{u < v\}}\quad \text{a.e.} \end{align*} This is a consequence of the Chain Rule applied to the Lipschitz function $\max(t, c)$. By induction, $v_N := \max(u_1, \dots, u_N) \in W^{1,p}$. **Norm bound for $v_N$.** Decompose $\mathbb{R}^n$ into disjoint measurable sets $A_1, \dots, A_N$ where $A_k$ is the set on which $u_k$ achieves the max (with ties broken by smaller index). On $A_k$, $v_N = u_k$ and $\nabla v_N = \nabla u_k$. Then \begin{align*} \|v_N\|_{W^{1,p}(\mathbb{R}^n)}^p = \sum_{k=1}^N \int_{A_k}\bigl(|u_k|^p + |\nabla u_k|^p\bigr)\, d\mathcal{L}^n \le \sum_{k=1}^N \int_{\mathbb{R}^n}\bigl(|u_k|^p + |\nabla u_k|^p\bigr) = \sum_{k=1}^N \|u_k\|_{W^{1,p}}^p, \end{align*} where we expanded the integration domain from $A_k$ to $\mathbb{R}^n$ in each summand (the integrand is non-negative). So $\|v_N\|_{W^{1,p}}^p$ is uniformly bounded by $\sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p$, which is finite by our choice of $u_k$'s with $\|u_k\|^p \le \operatorname{Cap}_p(E_k) + \varepsilon/2^k$. **Pass to the limit.** The sequence $(v_N)$ is pointwise non-decreasing (each new $u_{N+1}$ in the max can only increase the value), bounded above by $1$, hence converges $\mathcal{L}^n$-a.e. to $v(x) := \sup_k u_k(x) \in [0, 1]$. The dominated convergence theorem (with dominator $1 \cdot \mathbb{1}_{\operatorname{supp} v}$ for local convergence, and the global $L^p$ bound from Fatou) gives $v_N \to v$ in $L^p$ for $p \in [1, \infty)$. For the gradient we split the argument into two cases. *Case $1 < p < \infty$.* The sequence $(v_N)$ is bounded in $W^{1,p}$, and $W^{1,p}$ is reflexive ([3094](/theorems/3094)), so a subsequence converges weakly to some $w \in W^{1,p}$ ([3104](/theorems/3104)). By uniqueness of $L^p$-weak limits (the weak $W^{1,p}$ limit projects to a weak $L^p$ limit, which must coincide with the strong $L^p$ limit $v$), $w = v$. Lower semicontinuity of the norm under weak convergence gives \begin{align*} \|v\|_{W^{1,p}}^p \le \liminf_N\|v_N\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p. \end{align*} *Case $p = 1$.* Reflexivity fails for $W^{1,1}$. Instead, work locally on each ball $B_R$. By the partition formula, $|\nabla v_N| \le \sum_{k=1}^N|\nabla u_k| \le g := \sum_k|\nabla u_k| \in L^1(\mathbb{R}^n)$. So $(\nabla v_N)$ is dominated by $g$ pointwise a.e., hence equi-integrable on $B_R$. By the Dunford–Pettis theorem, a subsequence $\nabla v_{N_j} \rightharpoonup G$ weakly in $L^1(B_R)^n$. Combined with $v_N \to v$ in $L^1(B_R)$ (DCT), the test-function identification $\int G \cdot \varphi = -\int v\,\nabla\varphi$ for $\varphi \in C^\infty_c(B_R)$ gives $G = \nabla v$ a.e. on $B_R$, so $v \in W^{1,1}(B_R)$ with $\|\nabla v\|_{L^1(B_R)} \le \liminf_j\|\nabla v_{N_j}\|_{L^1(B_R)} \le \sum_k\|\nabla u_k\|_{L^1}$. Letting $R \to \infty$, $v \in W^{1,1}(\mathbb{R}^n)$ with $\|\nabla v\|_{L^1} \le \sum_k\|\nabla u_k\|_{L^1}$. (Alternative for $p = 1$: identify $\operatorname{Cap}_1$ with the variational $1$-capacity defined via $BV$ functions. By the Fleming–Rishel coarea formula, $\operatorname{Cap}_1(E) = \inf\{P(F) : F \supset E$ on a neighbourhood$\}$ where $P$ is the $BV$ perimeter. Subadditivity then reduces to subadditivity of the perimeter on disjoint pieces, a classical $BV$ fact derived from the chain rule for $BV$ functions.) **Admissibility of $v$.** The supremum $v = \sup_k u_k \ge u_k$ pointwise, so on the open neighbourhood $V_k$ where $u_k \ge 1$, we have $v \ge 1$. Hence $v \ge 1$ a.e. on $V := \bigcup_k V_k$, an open neighbourhood of $\bigcup_k E_k$. So $v \in \mathcal{A}(\bigcup_k E_k)$. **Final estimate and the $\varepsilon$ trick.** Since $v$ is admissible for $\bigcup_k E_k$, \begin{align*} \operatorname{Cap}_p\left(\bigcup_{k=1}^\infty E_k\right) \le \|v\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\|u_k\|_{W^{1,p}}^p \le \sum_{k=1}^\infty\bigl(\operatorname{Cap}_p(E_k) + \varepsilon/2^k\bigr) = \sum_{k=1}^\infty\operatorname{Cap}_p(E_k) + \varepsilon. \end{align*} Sending $\varepsilon \to 0$ gives the claimed subadditivity. [/guided] [/step] [step:Prove outer regularity on compact sets] Let $K \subset \mathbb{R}^n$ be compact. We show \begin{align*} \operatorname{Cap}_p(K) = \inf\{\operatorname{Cap}_p(U) : U \supset K\ \text{open}\}. \end{align*} **Inequality $\le$.** For every open $U \supset K$, monotonicity (Step 1) gives $\operatorname{Cap}_p(K) \le \operatorname{Cap}_p(U)$. Taking the infimum over such $U$ preserves the inequality: $\operatorname{Cap}_p(K) \le \inf_U \operatorname{Cap}_p(U)$. **Inequality $\ge$.** Fix $\varepsilon > 0$. By the definition of $\operatorname{Cap}_p(K)$, there exists $u \in \mathcal{A}(K)$ with $\|u\|_{W^{1,p}}^p < \operatorname{Cap}_p(K) + \varepsilon$. By definition of admissibility, there is an open neighbourhood $V$ of $K$ on which $u \ge 1$ a.e. Set $U := V$ — an open neighbourhood of $K$. Then $u \in \mathcal{A}(U)$ as well: $U \subseteq V$ (in fact $U = V$), and $u \ge 1$ a.e. on $V \supseteq U$. Therefore \begin{align*} \operatorname{Cap}_p(U) \le \|u\|_{W^{1,p}}^p < \operatorname{Cap}_p(K) + \varepsilon. \end{align*} This produces, for every $\varepsilon > 0$, an open $U \supset K$ with $\operatorname{Cap}_p(U) < \operatorname{Cap}_p(K) + \varepsilon$. Hence \begin{align*} \inf_U \operatorname{Cap}_p(U) \le \operatorname{Cap}_p(K) + \varepsilon \quad \text{for all } \varepsilon > 0, \end{align*} and taking $\varepsilon \to 0$, $\inf_U \operatorname{Cap}_p(U) \le \operatorname{Cap}_p(K)$. Combining both inequalities, $\operatorname{Cap}_p(K) = \inf_U \operatorname{Cap}_p(U)$. This is property (iii). [/step] [step:Prove inner regularity on open sets via Choquet capacitability] Let $U \subseteq \mathbb{R}^n$ be open. We show \begin{align*} \operatorname{Cap}_p(U) = \sup\{\operatorname{Cap}_p(K) : K \subset U\ \text{compact}\}. \end{align*} **Inequality $\ge$.** For every compact $K \subset U$, monotonicity (Step 1) gives $\operatorname{Cap}_p(K) \le \operatorname{Cap}_p(U)$, so $\sup_K\operatorname{Cap}_p(K) \le \operatorname{Cap}_p(U)$. **Inequality $\le$.** This is the substantive direction, and the cleanest route is to invoke Choquet's capacitability theorem as a black box. The theorem applies to any set function $C : 2^{\mathbb{R}^n} \to [0, \infty]$ that is a *Choquet capacity*, meaning it satisfies: (C1) **Monotonicity.** $E \subseteq F \implies C(E) \le C(F)$. (C2) **Continuity along increasing sequences.** $E_j \nearrow E$ (arbitrary sets) $\implies C(E_j) \nearrow C(E)$. (C3) **Continuity along decreasing sequences of compact sets.** $L_j \searrow L$ with each $L_j$ compact $\implies C(L_j) \to C(L)$. **Choquet's capacitability theorem** (Choquet, *Theory of Capacities*, 1953/54, Theorem 1; Dellacherie–Meyer, *Probabilities and Potential*, Chapter III) then asserts that every analytic set $E \subseteq \mathbb{R}^n$ — in particular, every Borel set, and *a fortiori* every open set — is *capacitable*: \begin{align*} C(E) = \sup\{C(K) : K \subseteq E,\ K\ \text{compact}\}. \end{align*} We do not reprove this theorem; we invoke it. **The Sobolev $p$-capacity is a Choquet capacity.** That $\operatorname{Cap}_p$ satisfies (C1), (C2), (C3) is a standard fact in nonlinear potential theory (see Heinonen–Kilpeläinen–Martio, *Nonlinear Potential Theory of Degenerate Elliptic Equations*, Theorem 2.2; Adams–Hedberg, *Function Spaces and Potential Theory*, Section 2.3; Mazya, *Sobolev Spaces*, Section 11.4 for the $p = 1$ case). Of these: - (C1) is property (i), proved in Step 1. - (C3) is property (vi), proved in Step 6 below (the proof there uses only outer regularity on compacts (Step 3) and the topological compactness pinch, both available independently of the present step). The hypotheses of Step 6 are met for any decreasing sequence of compacts in $\mathbb{R}^n$: the argument there is purely local (it uses sequential compactness inside the fixed compact $L_1$, which is automatic for any starting compact set), and applies verbatim to compacts in $\mathbb{R}^n$. - (C2) is the standard increasing-set continuity of the Sobolev $p$-capacity, established in the references above. Its proof for arbitrary increasing sequences uses the lattice structure of $W^{1,p}$ in the same spirit as Step 2, but requires either strong subadditivity of $\operatorname{Cap}_p$ or an approximation through outer regularity; we cite it rather than reproducing the argument. With (C1), (C2), (C3) in hand, Choquet's theorem applies and gives, for every Borel set $B \subseteq \mathbb{R}^n$, \begin{align*} \operatorname{Cap}_p(B) = \sup\{\operatorname{Cap}_p(K) : K \subseteq B,\ K\ \text{compact}\}. \end{align*} Specialising to $B = U$ open yields $\operatorname{Cap}_p(U) \le \sup_K\operatorname{Cap}_p(K)$, the desired inequality. Combining with the easy direction, $\operatorname{Cap}_p(U) = \sup\{\operatorname{Cap}_p(K) : K \subset U\ \text{compact}\}$. This is property (iv). **Recovering an exhausting sequence.** As a corollary, define the standard exhaustion \begin{align*} K_m := \left\{x \in \mathbb{R}^n : |x| \le m,\ \operatorname{dist}(x, U^c) \ge \tfrac{1}{m}\right\}, \quad m = 1, 2, \dots, \end{align*} where $U^c := \mathbb{R}^n \setminus U$ is closed and $\operatorname{dist}(\cdot, U^c)$ is $1$-Lipschitz. Each $K_m$ is closed (intersection of two sets defined by Lipschitz inequalities) and bounded (subset of $\overline{B(0, m)}$), hence compact; $K_m \subseteq K_{m+1}$; and $\bigcup_m K_m = U$, since any $x \in U$ has $\operatorname{dist}(x, U^c) > 0$ and so lies in $K_m$ for $m \ge \max(|x|, 1/\operatorname{dist}(x, U^c))$. Every compact $K \subset U$ is contained in $K_m$ for $m$ large: the open sets $V_m := \{|x| < m+1,\ \operatorname{dist}(x, U^c) > 1/(m+1)\} \subseteq K_{m+1}$ form an open cover of $K$ (since $V_m \nearrow U$ and $K \subset U$), and compactness of $K$ extracts a single $V_{m_0}$ containing $K$, hence $K \subseteq K_{m_0+1}$. Therefore by monotonicity \begin{align*} \sup\{\operatorname{Cap}_p(K) : K \subset U\ \text{compact}\} = \lim_{m \to \infty}\operatorname{Cap}_p(K_m), \end{align*} and inner regularity is realised concretely along the sequence $K_m$. [guided] **Strategy.** Inner regularity for an open set $U$ asserts that $\operatorname{Cap}_p(U)$ is approximated from below by the capacity of compact subsets. The forward inequality is monotonicity (Step 1); the reverse direction is the substantive content. **Why a naive cutoff fails.** Given a near-minimiser $u \in \mathcal{A}(U)$ with $\|u\|^p < \operatorname{Cap}_p(U) + \varepsilon$, one might hope to multiply $u$ by a smooth cutoff $\eta_m$ supported on a slight thickening of an exhausting compact $K_m$ to produce $u_m := u\eta_m$ admissible for $K_m$. The product rule gives $\nabla u_m = \eta_m\nabla u + u\nabla\eta_m$, and the term $u\nabla\eta_m$ has $|\nabla\eta_m| \sim m$ on a transition shell of width $\sim 1/m$ near $\partial U$. Its $L^p$ contribution scales as $m^p \cdot \mathcal{L}^n(\text{shell})^{1/p}$, which need not vanish unless $|u|$ is small near $\partial U$ — and $u \ge 1$ on a neighbourhood of $U$ is precisely the opposite. So the cutoff approach gives only the *upper* bound $\operatorname{Cap}_p(K_m) \le \operatorname{Cap}_p(U) + O(1)$, not the matching lower bound. A more powerful tool is required. **The Choquet capacitability framework.** That tool is **Choquet's capacitability theorem**. Choquet (1953/54) defined a *capacity* on $\mathbb{R}^n$ as a set function $C : 2^{\mathbb{R}^n} \to [0, \infty]$ satisfying (C1) monotonicity: $E \subseteq F \implies C(E) \le C(F)$; (C2) continuity along increasing sequences of arbitrary sets; (C3) continuity along decreasing sequences of compact sets; and proved that for any such $C$, every analytic set $E$ (in particular, every Borel set, *a fortiori* every open set) is *capacitable*: \begin{align*} C(E) = \sup\{C(K) : K \subseteq E\ \text{compact}\} = \inf\{C(O) : E \subseteq O\ \text{open}\}. \end{align*} The first equality, applied to an open $E = U$, is exactly inner regularity (iv). This theorem is genuinely deep — its proof uses the Souslin operation and analytic-set machinery from descriptive set theory — and is not something we reprove here. Standard references include Choquet's original *Theory of Capacities* (Annales de l'Institut Fourier, 1953/54), Doob's *Classical Potential Theory and its Probabilistic Counterpart* (Chapter 1), and Dellacherie–Meyer's *Probabilities and Potential* (Chapter III). **Sobolev $p$-capacity is a Choquet capacity.** What we *do* need to know is that our $\operatorname{Cap}_p$ falls within the framework — i.e., satisfies (C1), (C2), (C3). This is itself a standard fact in nonlinear potential theory, proved (with some labour) in: - Heinonen–Kilpeläinen–Martio, *Nonlinear Potential Theory of Degenerate Elliptic Equations*, Theorem 2.2 (for $1 < p < \infty$); - Adams–Hedberg, *Function Spaces and Potential Theory*, Theorem 2.3.10; - Mazya, *Sobolev Spaces*, Section 11.4 (covering $p \ge 1$). Of the three axioms: - (C1) is our property (i), proved cleanly in Step 1. - (C3) is our property (vi), proved in Step 6 below — and that proof depends only on Step 1 (monotonicity) and Step 3 (outer regularity), so we may cite Step 6 here without circularity. Step 6's argument applies to any decreasing sequence of compacts in $\mathbb{R}^n$ (the sequential compactness used there sits inside the starting compact $L_1$, which is supplied by the sequence itself). - (C2) is the increasing-set continuity of $\operatorname{Cap}_p$ on *arbitrary* sets — a stronger statement than property (v) of this theorem, which concerns increasing sequences of *open* sets only. The proof of (C2) for general sequences requires more than countable subadditivity (Step 2): one shows that $\operatorname{Cap}_p$ is *strongly subadditive* — $\operatorname{Cap}_p(A \cup B) + \operatorname{Cap}_p(A \cap B) \le \operatorname{Cap}_p(A) + \operatorname{Cap}_p(B)$ — by combining the truncation and lattice partition argument from Step 2 applied to two near-minimisers $u_A \in \mathcal{A}(A)$ and $u_B \in \mathcal{A}(B)$ via $\max(u_A, u_B) \in \mathcal{A}(A \cup B)$ and $\min(u_A, u_B) \in \mathcal{A}(A \cap B)$. Strong subadditivity plus monotonicity plus countable subadditivity yields (C2). Rather than reproducing this here, we cite the standard references above. **Why we do not have a logical loop.** The reader may worry that property (v) of this very theorem (continuity along increasing *open* sequences, Step 5) is invoked somewhere in the chain, creating circularity. It is not: Step 5 *uses* (iv), but (iv) here is established via Choquet's theorem, whose hypotheses (C1)/(C2)/(C3) are imported as a single black-box fact about $\operatorname{Cap}_p$ from external references. (C2) is *not* the same as (v): (v) restricts to open sequences, whereas (C2) is the much stronger statement on arbitrary sequences. So the dependency runs Steps 1, 3, 6 + external-fact $\implies$ (iv), then (iv) + Step 1 $\implies$ Step 5 (v); no circle. **Applying Choquet's theorem.** With (C1), (C2), (C3) verified (Steps 1, 6 + cited (C2)), Choquet's theorem gives, for every Borel $B$, \begin{align*} \operatorname{Cap}_p(B) = \sup\{\operatorname{Cap}_p(K) : K \subseteq B\ \text{compact}\}. \end{align*} Specialised to the open set $U$ (which is Borel, in particular analytic), this is property (iv). **Concrete realisation along an exhaustion.** The supremum is realised by any sequence of compacts that *exhausts* $U$ from within. The standard choice is \begin{align*} K_m := \left\{x \in \mathbb{R}^n : |x| \le m,\ \operatorname{dist}(x, U^c) \ge 1/m\right\}, \end{align*} which is compact (closed and bounded), increasing, with $\bigcup_m K_m = U$. Every compact $K \subset U$ lies in $K_m$ for some $m$ (take $m \ge \max(\sup_{x \in K}|x|, 1/\inf_{x \in K}\operatorname{dist}(x, U^c))$, both finite by compactness of $K$ and continuity of the relevant functions). Therefore $\sup_K \operatorname{Cap}_p(K) = \lim_m \operatorname{Cap}_p(K_m)$, and inner regularity reads \begin{align*} \operatorname{Cap}_p(U) = \lim_{m \to \infty}\operatorname{Cap}_p(K_m). \end{align*} **Case $p = 1$.** All the pieces (Steps 1, 3, 6) are valid for $p = 1$ as written, and Mazya's text supplies (C2) for $\operatorname{Cap}_1$. Choquet's theorem then yields inner regularity for the $1$-capacity without modification. [/guided] [/step] [step:Prove continuity along increasing sequences of open sets] Let $E_1 \subseteq E_2 \subseteq \cdots$ be open subsets of $\mathbb{R}^n$, and set $E := \bigcup_k E_k$, which is open. We show \begin{align*} \operatorname{Cap}_p(E) = \lim_{k \to \infty}\operatorname{Cap}_p(E_k). \end{align*} **Inequality $\ge$.** Monotonicity gives $\operatorname{Cap}_p(E_k) \le \operatorname{Cap}_p(E)$ for each $k$, hence $\lim_k \operatorname{Cap}_p(E_k) \le \operatorname{Cap}_p(E)$. The limit exists because the sequence is non-decreasing (monotonicity in $k$). **Inequality $\le$.** By inner regularity (Step 4), for any $\varepsilon > 0$ there exists a compact $K \subset E$ with $\operatorname{Cap}_p(K) > \operatorname{Cap}_p(E) - \varepsilon$. The compact set $K$ is covered by the open increasing family $(E_k)$: $K \subset E = \bigcup_k E_k$. By compactness, there exists $k_0$ such that $K \subseteq E_{k_0}$ (since the cover is increasing, a single $E_{k_0}$ suffices once $k_0$ is large enough — pick a finite subcover and take $k_0$ above all its indices). Then by monotonicity, \begin{align*} \operatorname{Cap}_p(E_{k_0}) \ge \operatorname{Cap}_p(K) > \operatorname{Cap}_p(E) - \varepsilon. \end{align*} Since the sequence $(\operatorname{Cap}_p(E_k))$ is non-decreasing in $k$, $\lim_k \operatorname{Cap}_p(E_k) \ge \operatorname{Cap}_p(E_{k_0}) > \operatorname{Cap}_p(E) - \varepsilon$. Letting $\varepsilon \to 0$, $\lim_k \operatorname{Cap}_p(E_k) \ge \operatorname{Cap}_p(E)$. Combining both inequalities, $\lim_k \operatorname{Cap}_p(E_k) = \operatorname{Cap}_p(E)$. This is property (v). [/step] [step:Prove continuity along decreasing sequences of compact sets] Let $K_1 \supseteq K_2 \supseteq \cdots$ be compact subsets of $\mathbb{R}^n$, and set $K := \bigcap_k K_k$, which is compact (closed subset of the compact $K_1$). We show \begin{align*} \operatorname{Cap}_p(K) = \lim_{k \to \infty}\operatorname{Cap}_p(K_k). \end{align*} **Inequality $\le$.** Monotonicity gives $\operatorname{Cap}_p(K) \le \operatorname{Cap}_p(K_k)$ for each $k$, hence $\operatorname{Cap}_p(K) \le \lim_k\operatorname{Cap}_p(K_k)$. The limit exists as the limit of the non-increasing sequence. **Inequality $\ge$.** By outer regularity (Step 3), for any $\varepsilon > 0$ there exists an open $U \supset K$ with $\operatorname{Cap}_p(U) < \operatorname{Cap}_p(K) + \varepsilon$. We claim that for $k$ sufficiently large, $K_k \subseteq U$. Suppose for contradiction that $K_k \not\subseteq U$ for infinitely many $k$. Then for each such $k$, choose $x_k \in K_k \setminus U$. Since $(x_k)$ is a sequence in the compact set $K_1$, it has a convergent subsequence $x_{k_l} \to x^* \in K_1$. Moreover $x_{k_l} \in K_{k_l}$, and the $K_k$ are nested decreasing — so for any fixed $m$, $x_{k_l} \in K_m$ for all $l$ with $k_l \ge m$, hence $x^* \in K_m$ (since $K_m$ is closed). This holds for every $m$, so $x^* \in \bigcap_m K_m = K$. Also $x_{k_l} \in U^c$ for all $l$ — but $U^c$ is closed, so $x^* \in U^c$. Therefore $x^* \in K \cap U^c$, contradicting $K \subset U$. The contradiction proves: there is some $k_0$ with $K_k \subseteq U$ for all $k \ge k_0$. By monotonicity, for $k \ge k_0$, \begin{align*} \operatorname{Cap}_p(K_k) \le \operatorname{Cap}_p(U) < \operatorname{Cap}_p(K) + \varepsilon. \end{align*} Hence $\lim_k \operatorname{Cap}_p(K_k) \le \operatorname{Cap}_p(K) + \varepsilon$. Letting $\varepsilon \to 0$, $\lim_k \operatorname{Cap}_p(K_k) \le \operatorname{Cap}_p(K)$. Combining both inequalities, $\lim_k \operatorname{Cap}_p(K_k) = \operatorname{Cap}_p(K)$. This is property (vi). [guided] **Why this direction is delicate.** The forward inequality $\operatorname{Cap}_p(K) \le \lim_k \operatorname{Cap}_p(K_k)$ is just monotonicity. The reverse inequality requires that the capacity does not "drop" abruptly at the intersection — equivalently, that approximating $K$ from the outside (which outer regularity provides) is consistent with approximating from inside via the sequence $K_k$. **Compactness pinch.** The key topological lemma is: for a decreasing sequence of compact sets $K_k$ with $K = \bigcap_k K_k$, every open set $U \supset K$ contains $K_k$ for all $k$ sufficiently large. The standard proof: suppose not, so there are infinitely many $k$ with $K_k \setminus U \ne \varnothing$. Pick $x_k \in K_k \setminus U$ for each such $k$. The sequence $(x_k)$ lies in the compact $K_1$, so by sequential compactness it has a subsequence $x_{k_l} \to x^*$. We claim $x^* \in K = \bigcap_m K_m$. Indeed, for each fixed $m$, $x_{k_l} \in K_{k_l} \subseteq K_m$ for $k_l \ge m$, and $K_m$ is closed, so the limit $x^* \in K_m$. This holds for every $m$, hence $x^* \in K$. But also $x_{k_l} \in U^c$, and $U^c$ is closed, so $x^* \in U^c$. Therefore $x^* \in K \cap U^c$ — but $K \subset U$, contradicting $x^* \in U^c$. **Applying outer regularity.** Outer regularity gives, for each $\varepsilon > 0$, an open $U \supset K$ with $\operatorname{Cap}_p(U) < \operatorname{Cap}_p(K) + \varepsilon$. The compactness pinch then gives a $k_0$ such that $K_k \subseteq U$ for all $k \ge k_0$. Monotonicity then gives $\operatorname{Cap}_p(K_k) \le \operatorname{Cap}_p(U) < \operatorname{Cap}_p(K) + \varepsilon$ for $k \ge k_0$. The sequence $\operatorname{Cap}_p(K_k)$ is non-increasing in $k$ (by monotonicity, each $K_k \supseteq K_{k+1}$ gives $\operatorname{Cap}_p(K_k) \ge \operatorname{Cap}_p(K_{k+1})$), so it has a limit $L = \lim_k\operatorname{Cap}_p(K_k) \in [\operatorname{Cap}_p(K), \infty)$. We have shown $L \le \operatorname{Cap}_p(K) + \varepsilon$ for every $\varepsilon > 0$, hence $L \le \operatorname{Cap}_p(K)$. Combined with the easy direction $L \ge \operatorname{Cap}_p(K)$, we get $L = \operatorname{Cap}_p(K)$. **Why a similar argument fails for arbitrary decreasing sequences.** This argument relies crucially on $K$ being **compact** — used in two places: first, $K_1$ is compact, so the sequence $(x_k) \subset K_1$ has a convergent subsequence; second, the compactness pinch (every open neighbourhood of a compact set contains all but finitely many of an inwardly-decreasing sequence of compacts) is a topological fact that fails for general decreasing sequences. For decreasing sequences of, say, open sets or general Borel sets, the analogous continuity statement is generally false — a reflection of the fact that capacity, like outer measures, is well-behaved on compact sets but not on arbitrary sets. [/guided] [/step]