We first utilize the Extension Theorem to transition the domain from the local set $U$ to the global manifold $\mathbb{R}^n$, yielding a sequence with compact support that remains globally bounded in $W^{1,p}(\mathbb{R}^n)$. By convolving this extended sequence with a standard mollifier with respect to the $n$-dimensional Lebesgue measure $\mathcal{L}^n$, we generate smooth approximations that possess uniform Lipschitz continuity globally. Applying the Arzelà-Ascoli Theorem to these smooth approximations guarantees uniform convergence on compact sets for any strictly fixed mollification scale.

To eliminate the scale dependency, we employ Cantor's diagonal argument across a strictly decreasing sequence of discrete scales, extracting a singular diagonal subsequence that forms a Cauchy sequence strictly within $L^1(U)$. Finally, we apply Lebesgue interpolation between the $L^1(U)$ norm and the strictly bounded $L^{p^*}(U)$ norm provided by the continuous Gagliardo-Nirenberg-Sobolev embedding. This bounded interpolation forces the $L^q(U)$ norm of the sequence differences to strictly vanish, completely confirming strong convergence in the target space $L^q(U)$.

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Let $U \subset \mathbb{R}^n$ be a strictly bounded proper open subset with a boundary $\partial U$ of regularity $C^1$. Let $p \in \mathbb{R}$ with $1 \le p < n$, and define the strict Sobolev conjugate exponent $p^* = \frac{np}{n-p}$. Let $q \in \mathbb{R}$ be a target exponent satisfying $1 \le q < p^*$.

Consider a sequence of functions indexed by $m \in \mathbb{N}$: \begin{align*} u_m: U &\to \mathbb{R} \\ x &\mapsto u_m(x) \end{align*} Assume the sequence $\{u_m\}_{m=1}^\infty$ is uniformly bounded within the Sobolev space $W^{1,p}(U)$, meaning there exists a strict uniform constant $M > 0$ such that $\|u_m\|_{W^{1,p}(U)} \le M$ for all $m \in \mathbb{N}$.

**Step 1** [claim] The sequence extends to a uniformly bounded sequence in $W^{1,p}(\mathbb{R}^n)$ with compact support inside a strictly larger bounded domain. [/claim] [proof] Because $\partial U$ possesses $C^1$ regularity, the Extension Theorem guarantees the existence of a bounded linear extension operator $E$ mapping $W^{1,p}(U)$ into $W^{1,p}(\mathbb{R}^n)$. We define the extended functions: \begin{align*} E u_m: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto (Eu_m)(x) \end{align*} The bounded linearity strictly enforces that $\|E u_m\|_{W^{1,p}(\mathbb{R}^n)} \le C_E \|u_m\|_{W^{1,p}(U)} \le C_E M$ for a fixed operator constant $C_E > 0$.

Let $V \subset \mathbb{R}^n$ be a bounded proper open set strictly containing $U$ such that the geometric distance between $\partial U$ and $\partial V$ strictly exceeds $2$.

We define a smooth topological cutoff function $\zeta \in C_c^\infty(\mathbb{R}^n)$ via the explicit mapping: \begin{align*} \zeta: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \zeta(x) \end{align*} We require $\zeta(x) = 1$ for all $x \in U$, and we strictly enforce the support condition $\text{supp}(\zeta) \subseteq V$. We define the truncated extension sequence $\overline{u}_m = \zeta \cdot E u_m$. The product rule for weak derivatives confirms $\overline{u}_m \in W^{1,p}(\mathbb{R}^n)$. Furthermore, the sequence $\{\overline{u}_m\}_{m=1}^\infty$ is globally bounded in $W^{1,p}(\mathbb{R}^n)$ by a composite constant $M' > 0$ and satisfies the rigid support constraint $\text{supp}(\overline{u}_m) \subseteq V$ for all $m \in \mathbb{N}$. [/proof]

**Step 2** [claim] For any fixed scale $\varepsilon > 0$, the standard mollification of the sequence $\{\overline{u}_m\}_{m=1}^\infty$ admits a uniformly convergent subsequence on the compact closure $\overline{V}$. [/claim] [proof] Let $\eta \in C_c^\infty(\mathbb{R}^n)$ be the standard strictly positive mollifier supported exclusively on the unit ball $B(0,1)$, normalized such that $\int_{\mathbb{R}^n} \eta(y) \, d\mathcal{L}^n(y) = 1$. For any fixed scale $\varepsilon > 0$, we define the rescaled mollifier $\eta_\varepsilon(x) = \varepsilon^{-n} \eta(x/\varepsilon)$. We define the spatial convolution mapping for the extended sequence: \begin{align*} v_{\varepsilon, m}: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto (\eta_\varepsilon * \overline{u}_m)(x) = \int_{\mathbb{R}^n} \eta_\varepsilon(x-y) \overline{u}_m(y) \, d\mathcal{L}^n(y) \end{align*} We establish uniform pointwise boundedness strictly for a fixed $\varepsilon \in (0, 1)$. By applying Hölder's inequality over the $n$-dimensional Lebesgue measure $\mathcal{L}^n$, we obtain: \begin{align*} |v_{\varepsilon, m}(x)| \le \int_{V} |\eta_\varepsilon(x-y)| |\overline{u}_m(y)| \, d\mathcal{L}^n(y) \le \|\eta_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \|\overline{u}_m\|_{L^1(V)} \end{align*} Because the domain $V$ possesses finite Lebesgue measure, Hölder's inequality strictly bounds the $L^1$ norm by the $L^p$ norm, yielding $\|\overline{u}_m\|_{L^1(V)} \le \mathcal{L}^n(V)^{1-1/p} M'$. Consequently, the sequence $\{v_{\varepsilon, m}\}_{m=1}^\infty$ is uniformly bounded in $L^\infty(\overline{V})$.

We subsequently bound the continuous gradient mapping to secure equicontinuity: \begin{align*} |\nabla v_{\varepsilon, m}(x)| \le \int_{V} |\nabla \eta_\varepsilon(x-y)| |\overline{u}_m(y)| \, d\mathcal{L}^n(y) \le \|\nabla \eta_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \|\overline{u}_m\|_{L^1(V)} \le \varepsilon^{-n-1} \|\nabla \eta\|_{L^\infty(\mathbb{R}^n)} \mathcal{L}^n(V)^{1-1/p} M' \end{align*} This uniform global bound on the spatial gradient enforces strict Lipschitz continuity uniformly across the index $m$. By the Arzelà-Ascoli Theorem, the sequence $\{v_{\varepsilon, m}\}_{m=1}^\infty$ possesses a nested subsequence that converges uniformly on the compact topological set $\overline{V}$. Because uniform convergence on a set of finite Lebesgue measure strictly implies $L^1$ convergence, this subsequence intrinsically converges in the complete space $L^1(V)$. [/proof]

**Step 3** [claim] There exists a constant $C > 0$ such that the mollification error is globally bounded by $\|v_{\varepsilon, m} - \overline{u}_m\|_{L^1(V)} \le \varepsilon C M'$ uniformly for all $m \in \mathbb{N}$. [/claim] [proof] We express the difference pointwise using the exact integral definition of the convolution over the unit ball $B(0,1)$: \begin{align*} v_{\varepsilon, m}(x) - \overline{u}_m(x) = \int_{B(0,1)} \eta(y) \left( \overline{u}_m(x - \varepsilon y) - \overline{u}_m(x) \right) \, d\mathcal{L}^n(y) \end{align*} We apply the Fundamental Theorem of Calculus along the straight ray from $x$ to $x-\varepsilon y$, parameterized by $t \in [0,1]$ with respect to the one-dimensional Lebesgue measure $\mathcal{L}^1$: \begin{align*} \overline{u}_m(x - \varepsilon y) - \overline{u}_m(x) = \int_0^1 \frac{\partial}{\partial t} \overline{u}_m(x - t\varepsilon y) \, d\mathcal{L}^1(t) = -\varepsilon \int_0^1 \nabla \overline{u}_m(x - t\varepsilon y) \cdot y \, d\mathcal{L}^1(t) \end{align*} Taking the absolute value, we strictly integrate over the domain $V$ with respect to $\mathcal{L}^n(x)$: \begin{align*} \int_V |v_{\varepsilon, m}(x) - \overline{u}_m(x)| \, d\mathcal{L}^n(x) \le \int_V \int_{B(0,1)} \eta(y) \varepsilon |y| \left( \int_0^1 |\nabla \overline{u}_m(x - t\varepsilon y)| \, d\mathcal{L}^1(t) \right) \, d\mathcal{L}^n(y) \, d\mathcal{L}^n(x) \end{align*} We explicitly invoke Fubini's Theorem to exchange the integration order, strictly evaluating the integral over $x$ first. The translation invariance of the Lebesgue measure over the global manifold dictates $\int_V |\nabla \overline{u}_m(x - t\varepsilon y)| \, d\mathcal{L}^n(x) \le \|\nabla \overline{u}_m\|_{L^1(\mathbb{R}^n)}$: \begin{align*} \|v_{\varepsilon, m} - \overline{u}_m\|_{L^1(V)} \le \varepsilon \left( \int_{B(0,1)} \eta(y) |y| \, d\mathcal{L}^n(y) \right) \left( \int_0^1 1 \, d\mathcal{L}^1(t) \right) \|\nabla \overline{u}_m\|_{L^1(\mathbb{R}^n)} \end{align*} The integral over the sphere strictly evaluates to a constant less than $1$. We apply Hölder's inequality to bound the gradient term strictly by the Sobolev norm on the finite-measure support $V$, establishing $\|v_{\varepsilon, m} - \overline{u}_m\|_{L^1(V)} \le \varepsilon \mathcal{L}^n(V)^{1-1/p} M'$. Defining $C = \mathcal{L}^n(V)^{1-1/p}$ concludes the proof. [/proof]

**Step 4** [claim] The sequence $\{\overline{u}_m\}_{m=1}^\infty$ admits a subsequence $\{w_m\}_{m=1}^\infty$ that forms a strictly Cauchy sequence within the complete Banach space $L^1(V)$. [/claim] [proof] We iteratively construct a convergent subsequence strictly utilizing Cantor's diagonal argument. For the initial Stage $k=1$, we set $\varepsilon_1 = 1$. By the compactness proven in Step 2, we extract a subsequence defined by a strictly increasing mapping $\sigma_1: \mathbb{N} \to \mathbb{N}$ such that $\{v_{1, \sigma_1(m)}\}_{m=1}^\infty$ converges unconditionally in $L^1(V)$. For any subsequent Stage $k$, setting $\varepsilon_k = 1/k$, we extract a nested subsequence mapping $\sigma_k$ from the image of $\sigma_{k-1}$ such that $\{v_{1/k, \sigma_k(m)}\}_{m=1}^\infty$ converges in $L^1(V)$.

We define the diagonal sequence $w_m = \overline{u}_{\sigma_m(m)}$. Let $\delta > 0$ be an arbitrary strict tolerance threshold. We fix a scale integer $k \ge 1$ sufficiently large to ensure the uniform smoothing error satisfies $1/k < \delta / (3 C M')$. By Step 3, every sequence element satisfies: \begin{align*} \|v_{1/k, \sigma_m(m)} - w_m\|_{L^1(V)} \le \frac{C M'}{k} < \frac{\delta}{3} \end{align*} Since the $k$-th mollified diagonal sequence converges in $L^1(V)$, it is inherently Cauchy. There exists an integer $N \in \mathbb{N}$ such that for all $i, j \ge N$, we have $\|v_{1/k, \sigma_i(i)} - v_{1/k, \sigma_j(j)}\|_{L^1(V)} < \delta/3$. Iterating the triangle inequality over the differences unconditionally proves: \begin{align*} \|w_i - w_j\|_{L^1(V)} \le \|w_i - v_{1/k, \sigma_i(i)}\|_{L^1(V)} + \|v_{1/k, \sigma_i(i)} - v_{1/k, \sigma_j(j)}\|_{L^1(V)} + \|v_{1/k, \sigma_j(j)} - w_j\|_{L^1(V)} < \delta \end{align*} This strictly establishes $\{w_m\}_{m=1}^\infty$ as a Cauchy sequence in $L^1(V)$. Because $U \subset V$, the restriction of $\{w_m\}_{m=1}^\infty$ to the original domain $U$ forms a strictly Cauchy sequence in $L^1(U)$. [/proof]

**Step 5** [claim] The restricted diagonal subsequence $\{w_m\}_{m=1}^\infty$ converges strictly within the complete target Lebesgue space $L^q(U)$. [/claim] [proof] Because the exponent $q$ explicitly satisfies $1 \le q < p^*$, we strictly invoke Lebesgue interpolation theory. There exists a unique structural parameter $\theta \in (0, 1]$ satisfying the algebraic relation: \begin{align*} \frac{1}{q} = \frac{\theta}{1} + \frac{1 - \theta}{p^*} \end{align*} We apply the explicit $L^q$ interpolation inequality to the difference between any two elements of the restricted subsequence mapping over the domain $U$: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le \|w_i - w_j\|_{L^1(U)}^\theta \|w_i - w_j\|_{L^{p^*}(U)}^{1 - \theta} \end{align*} By the fundamental Gagliardo-Nirenberg-Sobolev inequality, the continuous embedding $W^{1,p}(U) \subseteq L^{p^*}(U)$ ensures the sequence $\{u_m\}_{m=1}^\infty$ satisfies a uniform global bound $\|u_m\|_{L^{p^*}(U)} \le C_* M$ for a dimensional constant $C_* > 0$. Because $\{w_m\}_{m=1}^\infty$ is a rigid subsequence of $\{u_m\}_{m=1}^\infty$, the triangle inequality enforces a strict maximum diameter: \begin{align*} \|w_i - w_j\|_{L^{p^*}(U)} \le \|w_i\|_{L^{p^*}(U)} + \|w_j\|_{L^{p^*}(U)} \le 2 C_* M \end{align*} Substituting this absolute bound back into the interpolation inequality directly couples the $L^q$ geometry to the $L^1$ geometry: \begin{align*} \|w_i - w_j\|_{L^q(U)} \le \|w_i - w_j\|_{L^1(U)}^\theta (2 C_* M)^{1 - \theta} \end{align*} From Step 4, the sequence $\{w_m\}_{m=1}^\infty$ is rigorously Cauchy in $L^1(U)$, forcing the right-hand term $\|w_i - w_j\|_{L^1(U)}$ to strictly vanish as $i, j \to \infty$. The structural parameter satisfies $\theta > 0$, confirming that the bounded exponential factor drives the entire product to strictly converge to $0$.

Consequently, $\{w_m\}_{m=1}^\infty$ strictly fulfills the definition of a Cauchy sequence with respect to the $L^q(U)$ norm. Because $L^q(U)$ is structurally a complete Banach space, the sequence unconditionally converges to a limit function $u \in L^q(U)$. We have thus extracted a subsequence from the initial bounded family in $W^{1,p}(U)$ that converges explicitly in $L^q(U)$, completing the rigorous proof of compact embedding. [/proof]