[proofplan] We assume $\Omega \subset \mathbb{R}^n$ is bounded (or, equivalently, that $g(0) = 0$); under this hypothesis $g \circ u \in L^p(\Omega)$ holds without additional assumptions. Let $L > 0$ be the Lipschitz constant of $g$. By [Rademacher's Theorem](/theorems/3069), the classical derivative $g'$ exists $\mathcal{L}^1$-a.e. with $|g'| \le L$, so the candidate $g'(u)\, \partial_{x_i} u$ is bounded in $L^p(\Omega)$ by $L\|\partial_{x_i} u\|_{L^p}$ as soon as the composition $g'(u)$ is interpreted via a fixed pointwise representative. The strategy is a clean three-stage approximation: (a) mollify $g$ in $\mathbb{R}$ to obtain $g_\varepsilon \in C^\infty(\mathbb{R})$ with $|g_\varepsilon'| \le L$ and $g_\varepsilon \to g$ uniformly, $g_\varepsilon' \to g'$ a.e.; (b) approximate $u$ by smooth $u_k$ via the [Meyers-Serrin Theorem](/theorems/58); (c) apply the classical $C^1$ chain rule to $g_\varepsilon \circ u_k$, then pass to a double limit using the [Dominated Convergence Theorem](/theorems/4) twice. The Lipschitz hypothesis on $g$ supplies the two dominations needed: $|g(u_k) - g(u)| \le L|u_k - u|$ controls the left-hand side in $L^p$, and the uniform bound $|g_\varepsilon'| \le L$ together with $|\partial_{x_i} u_k|$ converging in $L^p$ controls the right-hand side. Because $g_\varepsilon$ is smooth, no Lusin-type approximation is needed: the smoothing of $g$ removes the discontinuity issues at the source. [/proofplan] [step:Verify integrability of the candidate weak derivative] Throughout, $\Omega \subset \mathbb{R}^n$ is bounded. Let $L > 0$ be the Lipschitz constant of $g$, so $|g(s) - g(t)| \le L|s-t|$ for all $s, t \in \mathbb{R}$. By [Rademacher's Theorem](/theorems/3069) in dimension one, the classical derivative $g'$ exists $\mathcal{L}^1$-a.e. on $\mathbb{R}$ and satisfies $|g'(s)| \le L$ wherever it exists. Fix any Borel-measurable representative of $g'$, extended by zero on the null set where the classical derivative does not exist; the resulting $g': \mathbb{R} \to \mathbb{R}$ is bounded by $L$ everywhere. Fix $i \in \{1, \dots, n\}$. We claim: (a) $g \circ u \in L^p(\Omega)$. (b) The composition $g'(u): \Omega \to \mathbb{R}$, $x \mapsto g'(u(x))$, is well-defined and Lebesgue-measurable. (c) $g'(u)\, \partial_{x_i} u \in L^p(\Omega)$. For (a): the Lipschitz bound $|g(s)| \le |g(s) - g(0)| + |g(0)| \le L|s| + |g(0)|$ gives \begin{align*} \int_\Omega |g(u)|^p\, d\mathcal{L}^n \le 2^{p-1}\int_\Omega \bigl(L^p|u|^p + |g(0)|^p\bigr) d\mathcal{L}^n = 2^{p-1}\bigl(L^p\|u\|_{L^p}^p + |g(0)|^p\, \mathcal{L}^n(\Omega)\bigr) < \infty, \end{align*} since $\mathcal{L}^n(\Omega) < \infty$ and $u \in L^p(\Omega)$. For (b): $g': \mathbb{R} \to \mathbb{R}$ is Borel-measurable by construction, and $u: \Omega \to \mathbb{R}$ is Lebesgue-measurable. Hence $g' \circ u$ is Lebesgue-measurable as a Borel-measurable function composed with a Lebesgue-measurable function. For (c): $|g'(u(x))| \le L$ everywhere, so $|g'(u)\, \partial_{x_i} u| \le L |\partial_{x_i} u|$ pointwise, and therefore $g'(u)\, \partial_{x_i} u \in L^p(\Omega)$ with norm at most $L\|\partial_{x_i} u\|_{L^p}$. [guided] The candidate identity is $\partial_{x_i}(g \circ u) = g'(u)\, \partial_{x_i} u$. Before approximating, we check that both sides are well-defined elements of $L^p(\Omega)$. The Lipschitz hypothesis $|g(s) - g(t)| \le L|s - t|$ does the work, but only because $\Omega$ is bounded. **$g \circ u \in L^p(\Omega)$.** By [Rademacher's Theorem](/theorems/3069) applied to the Lipschitz function $g: \mathbb{R} \to \mathbb{R}$, the classical derivative $g'$ exists a.e., is bounded by $L$, and admits a Borel-measurable representative. The Lipschitz condition gives $|g(s)| \le L|s| + |g(0)|$, so \begin{align*} \int_\Omega |g(u)|^p\, d\mathcal{L}^n \le 2^{p-1}\int_\Omega \bigl(L^p|u|^p + |g(0)|^p\bigr) d\mathcal{L}^n. \end{align*} Both terms are finite on the bounded domain $\Omega$. (On unbounded $\Omega$ the constant term diverges unless $g(0) = 0$; this is why the hypothesis "$\Omega$ bounded" is included in the theorem statement. An equivalent formulation drops the boundedness requirement and assumes $g(0) = 0$ instead.) **$g'(u)$ as a function on $\Omega$.** We work with a fixed Borel-measurable representative of $g'$, extended by zero on the null set where the classical derivative does not exist. Then $g' \circ u: \Omega \to \mathbb{R}$ is Lebesgue-measurable as a composition. Its values lie in $[-L, L]$ everywhere. **Bound in $L^p$.** Since $|g'(u(x))| \le L$ pointwise, $|g'(u)\, \partial_{x_i} u| \le L|\partial_{x_i} u|$ pointwise, so the candidate is in $L^p(\Omega)$ with norm at most $L\|\partial_{x_i} u\|_{L^p}$. [/guided] [/step] [step:Mollify $g$ to a smooth approximation] We approximate $g$ in the target space $\mathbb{R}$ by a one-dimensional convolution. Pick a non-negative kernel $\rho \in C_c^\infty(\mathbb{R})$ with $\operatorname{supp}\rho \subset (-1,1)$ and $\int_\mathbb{R}\rho\, d\mathcal{L}^1 = 1$. For $\varepsilon > 0$, define \begin{align*} \rho_\varepsilon: \mathbb{R} \to \mathbb{R}, \qquad x \mapsto \varepsilon^{-1}\rho(x/\varepsilon), \end{align*} and the mollification of $g$: \begin{align*} g_\varepsilon: \mathbb{R} \to \mathbb{R}, \qquad s \mapsto \int_{\mathbb{R}} \rho_\varepsilon(s - t)\, g(t)\, d\mathcal{L}^1(t). \end{align*} Then $g_\varepsilon \in C^\infty(\mathbb{R})$, with three properties. **(i) Lipschitz preservation.** For $s_1, s_2 \in \mathbb{R}$, \begin{align*} g_\varepsilon(s_1) - g_\varepsilon(s_2) = \int_\mathbb{R} \rho_\varepsilon(t)[g(s_1 - t) - g(s_2 - t)]\, d\mathcal{L}^1(t), \end{align*} so by the Lipschitz bound on $g$, \begin{align*} |g_\varepsilon(s_1) - g_\varepsilon(s_2)| \le L|s_1 - s_2| \int_\mathbb{R} \rho_\varepsilon(t)\, d\mathcal{L}^1(t) = L|s_1 - s_2|. \end{align*} Thus $g_\varepsilon$ is $L$-Lipschitz, and consequently $|g_\varepsilon'(s)| \le L$ for every $s \in \mathbb{R}$. **(ii) Uniform convergence $g_\varepsilon \to g$.** For any $s \in \mathbb{R}$, \begin{align*} |g_\varepsilon(s) - g(s)| = \left|\int_\mathbb{R}\rho_\varepsilon(t)[g(s - t) - g(s)]\, d\mathcal{L}^1(t)\right| \le L \int_\mathbb{R} \rho_\varepsilon(t)|t|\, d\mathcal{L}^1(t) \le L\varepsilon, \end{align*} since $\operatorname{supp}\rho_\varepsilon \subset [-\varepsilon, \varepsilon]$. The bound is uniform in $s$. **(iii) A.e. convergence of $g_\varepsilon'$.** Because $g$ is Lipschitz, hence absolutely continuous, the distributional derivative of $g$ coincides with the classical $g'$ as a measurable function on $\mathbb{R}$. The convolution then commutes with differentiation: $g_\varepsilon' = \rho_\varepsilon * g'$. By the Lebesgue differentiation theorem, $\mathcal{L}^1$-a.e. point of $\mathbb{R}$ is a Lebesgue point of $g'$, and at every such point $(\rho_\varepsilon * g')(s) \to g'(s)$ as $\varepsilon \to 0$. Hence $g_\varepsilon'(s) \to g'(s)$ for $\mathcal{L}^1$-a.e. $s$. [guided] The reason we mollify $g$ before doing anything else: $g$ is only Lipschitz, not $C^1$, so the classical chain rule does not apply pointwise to $g \circ v$ for any smooth $v$. By replacing $g$ with its mollification $g_\varepsilon$, we obtain a smooth approximation that satisfies a true $C^1$ chain rule. Then we send $\varepsilon \to 0$ at the end, using the Lipschitz bound to control everything along the way. This sidesteps the need for any Lusin-type approximation of $g'$ — the smoothing already provides the continuity we need. **Why mollify in the target rather than the domain.** Mollifying $u$ alone (in $\Omega$) is not enough: even with smooth $u_k$, the composition $g \circ u_k$ is only Lipschitz, not $C^1$, because $g$ itself is not $C^1$. So the classical chain rule formula $\partial_{x_i}(g \circ u_k)(x) = g'(u_k(x))\, \partial_{x_i} u_k(x)$ only holds a.e., not pointwise — and the a.e. version is exactly the result we want to prove for the rough $u$. We must smooth $g$ as well. **Mollifier construction.** Pick a non-negative kernel $\rho \in C_c^\infty(\mathbb{R})$ with $\operatorname{supp}\rho \subset (-1,1)$ and $\int_\mathbb{R} \rho\, d\mathcal{L}^1 = 1$. The standard scaling $\rho_\varepsilon(x) := \varepsilon^{-1}\rho(x/\varepsilon)$ gives a Dirac approximation: $\rho_\varepsilon \ge 0$, $\int\rho_\varepsilon = 1$, $\operatorname{supp}\rho_\varepsilon \subset [-\varepsilon, \varepsilon]$. Define $g_\varepsilon := \rho_\varepsilon * g$. Standard mollifier theory gives $g_\varepsilon \in C^\infty(\mathbb{R})$ with derivatives obtained by convolving against derivatives of $\rho_\varepsilon$. **Property (i): Lipschitz preservation.** By the change of variable $t \mapsto t$ (translation invariance of the convolution), \begin{align*} g_\varepsilon(s_1) - g_\varepsilon(s_2) = \int_\mathbb{R} \rho_\varepsilon(t)[g(s_1 - t) - g(s_2 - t)]\, d\mathcal{L}^1(t). \end{align*} Applying $|g(a) - g(b)| \le L|a - b|$ inside the integral, \begin{align*} |g_\varepsilon(s_1) - g_\varepsilon(s_2)| \le L|s_1 - s_2|\int_\mathbb{R} \rho_\varepsilon(t)\, d\mathcal{L}^1(t) = L|s_1 - s_2|. \end{align*} So $g_\varepsilon$ is $L$-Lipschitz, and since it is $C^1$, $|g_\varepsilon'(s)| \le L$ for every $s \in \mathbb{R}$. **Property (ii): Uniform convergence.** Direct estimate: \begin{align*} |g_\varepsilon(s) - g(s)| = \left|\int \rho_\varepsilon(t)[g(s-t) - g(s)]\, d\mathcal{L}^1(t)\right| \le L\int \rho_\varepsilon(t)|t|\, d\mathcal{L}^1(t) \le L\varepsilon, \end{align*} uniformly in $s \in \mathbb{R}$. **Property (iii): A.e. convergence of $g_\varepsilon'$.** Since $g$ is absolutely continuous (Lipschitz $\Rightarrow$ AC on bounded intervals, and the local statement extends to $\mathbb{R}$), its distributional derivative is the classical $g'$. Convolution commutes with differentiation in the distributional sense, so $g_\varepsilon' = \rho_\varepsilon * g'$ as smooth functions. At every Lebesgue point of $g'$ (i.e., $\mathcal{L}^1$-a.e.\ point of $\mathbb{R}$ by the Lebesgue differentiation theorem), $(\rho_\varepsilon * g')(s) \to g'(s)$. So $g_\varepsilon'(s) \to g'(s)$ for $\mathcal{L}^1$-a.e. $s \in \mathbb{R}$, with $|g_\varepsilon'| \le L$ uniformly. [/guided] [/step] [step:Smooth $u$ and apply the classical chain rule] By the [Meyers-Serrin Theorem](/theorems/58), there exists a sequence \begin{align*} u_k \in C^\infty(\Omega) \cap W^{1,p}(\Omega), \qquad u_k \to u \quad \text{in } W^{1,p}(\Omega). \end{align*} By passing to a subsequence (not relabelled), we may assume $u_k(x) \to u(x)$ and $\partial_{x_i} u_k(x) \to \partial_{x_i} u(x)$ for $\mathcal{L}^n$-a.e. $x \in \Omega$ (every $L^p$-convergent sequence has an a.e.-convergent subsequence; apply this to $u_k - u$ and to $\partial_{x_i} u_k - \partial_{x_i} u$ in turn, taking nested subsequences). Since $u_k \in C^\infty(\Omega)$ and $g_\varepsilon \in C^\infty(\mathbb{R})$, the classical chain rule applies pointwise: \begin{align*} \partial_{x_i}(g_\varepsilon \circ u_k)(x) = g_\varepsilon'(u_k(x))\, \partial_{x_i} u_k(x), \qquad x \in \Omega. \end{align*} Both sides are continuous on $\Omega$. For any test function $\varphi \in C_c^\infty(\Omega)$, integration by parts (with no boundary terms, since $\varphi$ has compact support in $\Omega$) gives \begin{align*} \int_\Omega (g_\varepsilon \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n. \tag{$\ast$} \end{align*} Both integrals are finite: $|g_\varepsilon \circ u_k| \le L|u_k| + |g_\varepsilon(0)|$ on $\operatorname{supp}\varphi$, and $|g_\varepsilon'(u_k)\, \partial_{x_i} u_k| \le L|\partial_{x_i} u_k|$ on $\operatorname{supp}\varphi$. [guided] With both $g_\varepsilon$ and $u_k$ smooth, the chain rule reduces to first-year calculus and the rest of the proof is just careful limit-passing. This is the heart of the strategy: by smoothing both functions before applying the chain rule, every awkward step is pushed into a controlled limit at the end. **Smoothing $u$.** The [Meyers-Serrin Theorem](/theorems/58) provides $u_k \in C^\infty(\Omega) \cap W^{1,p}(\Omega)$ with $u_k \to u$ in $W^{1,p}(\Omega)$. (Note: the smooth approximants need not extend continuously to $\partial\Omega$; this is the "$H = W$" theorem, not the smooth-up-to-boundary theorem.) An $L^p$-convergent sequence has an a.e.-convergent subsequence; passing to nested subsequences for both $u_k - u$ and $\partial_{x_i} u_k - \partial_{x_i} u$, we may assume both convergences hold pointwise a.e. on $\Omega$, while retaining the $W^{1,p}$ convergence of the (sub)sequence. **The smooth chain rule.** Both $g_\varepsilon \in C^1(\mathbb{R})$ and $u_k \in C^1(\Omega)$, so the classical chain rule gives \begin{align*} \partial_{x_i}(g_\varepsilon \circ u_k)(x) = g_\varepsilon'(u_k(x))\, \partial_{x_i} u_k(x) \end{align*} as continuous functions on $\Omega$, with equality holding pointwise for every $x \in \Omega$. **Integration by parts.** Multiplying by $\varphi \in C_c^\infty(\Omega)$ and integrating, the standard integration-by-parts identity for $C^1$ functions \begin{align*} \int_\Omega \partial_{x_i}(g_\varepsilon \circ u_k)\, \varphi\, d\mathcal{L}^n = -\int_\Omega (g_\varepsilon \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n \end{align*} holds because $\varphi$ has compact support in $\Omega$, so the boundary terms in any divergence-theorem application vanish. Rearranging gives identity $(\ast)$. Both integrals converge absolutely: on $\operatorname{supp}\varphi$ (a compact subset of $\Omega$), we have $|g_\varepsilon \circ u_k| \le L|u_k| + |g_\varepsilon(0)|$ (which is in $L^p$), and $|g_\varepsilon'(u_k)\, \partial_{x_i} u_k| \le L|\partial_{x_i} u_k|$ (which is in $L^p$). [/guided] [/step] [step:Pass to the limit $\varepsilon \to 0$ at fixed $k$] In identity $(\ast)$ we send $\varepsilon \to 0$ holding $k$ fixed. **Left-hand side.** By property (ii) of step "Mollify $g$ to a smooth approximation", $g_\varepsilon \to g$ uniformly on $\mathbb{R}$, so $g_\varepsilon \circ u_k \to g \circ u_k$ uniformly on $\Omega$. Then \begin{align*} \left|\int_\Omega (g_\varepsilon \circ u_k - g \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n\right| \le \|g_\varepsilon \circ u_k - g \circ u_k\|_{L^\infty} \|\partial_{x_i}\varphi\|_{L^1} \le L\varepsilon \|\partial_{x_i}\varphi\|_{L^1} \to 0. \end{align*} **Right-hand side.** By property (iii), $g_\varepsilon'(s) \to g'(s)$ for $\mathcal{L}^1$-a.e. $s \in \mathbb{R}$, with $|g_\varepsilon'(s)| \le L$ uniformly. Since $u_k \in C^\infty(\Omega)$, by Sard's theorem the set of critical values $u_k(\{\nabla u_k = 0\})$ has $\mathcal{L}^1$-measure zero; on the complement of this set, the level sets $\{u_k = c\}$ are smooth $(n-1)$-manifolds and the pushforward $(u_k)_\#(\mathcal{L}^n \, \llcorner \, \operatorname{supp}\varphi)$ is absolutely continuous with respect to $\mathcal{L}^1$ (this is a standard consequence of the smooth coarea formula on regular level sets). Hence the $\mathcal{L}^1$-null exceptional set $E \subset \mathbb{R}$ where $g_\varepsilon'$ may fail to converge to $g'$ pulls back through $u_k$ to an $\mathcal{L}^n$-null set in $\{\nabla u_k \ne 0\}$. On the complementary set $\{\nabla u_k = 0\}$ we have $\partial_{x_i} u_k = 0$ identically, so the integrand $g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi$ vanishes there regardless. Putting these together, \begin{align*} g_\varepsilon'(u_k(x))\, \partial_{x_i} u_k(x) \to g'(u_k(x))\, \partial_{x_i} u_k(x) \quad \text{for } \mathcal{L}^n\text{-a.e.\ } x \in \Omega. \end{align*} Domination: $|g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi| \le L\, \|\varphi\|_{L^\infty}\, |\partial_{x_i} u_k|$, with $L\|\varphi\|_{L^\infty} |\partial_{x_i} u_k| \in L^1(\Omega)$ (using $\partial_{x_i} u_k \in L^p(\Omega)$ and $\operatorname{supp}\varphi$ compact). By the [Dominated Convergence Theorem](/theorems/4), \begin{align*} -\int_\Omega g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n \to -\int_\Omega g'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n. \end{align*} Combining the two limits in $(\ast)$, \begin{align*} \int_\Omega (g \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega g'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n. \tag{$\ast\ast$} \end{align*} [guided] The first limit follows from $L^p$-continuity of $g \circ u_k$ in $u_k$, and the second uses the [Dominated Convergence Theorem](/theorems/4) on a controlled integrand. The only subtle point is verifying the a.e. convergence of $g_\varepsilon'(u_k)$ to $g'(u_k)$ on $\Omega$, which uses smoothness of $u_k$ in an essential way. **LHS via uniform convergence.** Since $g_\varepsilon \to g$ uniformly on $\mathbb{R}$ with rate $L\varepsilon$, the same holds for $g_\varepsilon \circ u_k \to g \circ u_k$ pointwise on $\Omega$. Bound the integral by Hölder: \begin{align*} \left|\int_\Omega (g_\varepsilon \circ u_k - g \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n\right| \le L\varepsilon\, \|\partial_{x_i}\varphi\|_{L^1} \to 0. \end{align*} **RHS via DCT.** A.e. convergence of $g_\varepsilon'(u_k(x))$ to $g'(u_k(x))$ requires care because $g_\varepsilon' \to g'$ holds only a.e. on $\mathbb{R}$, not everywhere. The smoothness of $u_k$ rescues this. Let $E \subset \mathbb{R}$ be the $\mathcal{L}^1$-null set where the convergence fails. We must show $u_k^{-1}(E)$ has $\mathcal{L}^n$-measure zero in the region $\{\nabla u_k \ne 0\}$ — outside which the integrand vanishes anyway because $\partial_{x_i} u_k = 0$. By Sard's theorem applied to the smooth function $u_k: \Omega \to \mathbb{R}$, the critical values $u_k(\{\nabla u_k = 0\})$ form an $\mathcal{L}^1$-null subset of $\mathbb{R}$. On the regular set $\{\nabla u_k \ne 0\}$, the level sets $\{u_k = c\}$ are smooth $(n-1)$-dimensional submanifolds for $c$ outside the critical-value set, and the smooth coarea formula \begin{align*} \int_{\{\nabla u_k \ne 0\}} f(u_k(x))\, |\nabla u_k(x)|\, d\mathcal{L}^n(x) = \int_\mathbb{R} f(c)\, \mathcal{H}^{n-1}(\{u_k = c\})\, d\mathcal{L}^1(c) \end{align*} shows that $u_k$ pushes forward $|\nabla u_k|\, \mathcal{L}^n$ on the regular set to a measure absolutely continuous with respect to $\mathcal{L}^1$. So $\mathcal{L}^n(\{\nabla u_k \ne 0\} \cap u_k^{-1}(E)) = 0$ for any $\mathcal{L}^1$-null $E \subset \mathbb{R}$. On $\{\nabla u_k = 0\}$, the integrand $g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi$ is identically zero because $\partial_{x_i} u_k = 0$ — and stays zero in the limit. Hence a.e. convergence of the integrand on all of $\Omega$. Domination: $|g_\varepsilon'(u_k)\, \partial_{x_i} u_k\, \varphi| \le L\|\varphi\|_{L^\infty}|\partial_{x_i} u_k|$, and the right side is in $L^1(\Omega)$ since $\partial_{x_i} u_k \in L^p(\Omega) \subset L^1_{\mathrm{loc}}(\Omega)$ and $\varphi$ has compact support. Apply the [Dominated Convergence Theorem](/theorems/4) to obtain $(\ast\ast)$, the chain rule for the smooth pair $(g, u_k)$. [/guided] [/step] [step:Pass to the limit $k \to \infty$] In identity $(\ast\ast)$ we now send $k \to \infty$ along the subsequence chosen above, on which $u_k \to u$ and $\partial_{x_i} u_k \to \partial_{x_i} u$ both pointwise a.e. on $\Omega$ and in $L^p(\Omega)$. **Left-hand side.** By the Lipschitz bound, \begin{align*} \|g \circ u_k - g \circ u\|_{L^p(\Omega)} \le L\|u_k - u\|_{L^p(\Omega)} \to 0. \end{align*} By Hölder against $\partial_{x_i}\varphi \in L^{p'}(\Omega)$, \begin{align*} \int_\Omega (g \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n \to \int_\Omega (g \circ u)\, \partial_{x_i}\varphi\, d\mathcal{L}^n. \end{align*} **Right-hand side.** Decompose \begin{align*} g'(u_k)\, \partial_{x_i} u_k - g'(u)\, \partial_{x_i} u = g'(u_k)\bigl(\partial_{x_i} u_k - \partial_{x_i} u\bigr) + \bigl(g'(u_k) - g'(u)\bigr)\partial_{x_i} u. \end{align*} *First piece.* Since $|g'(u_k)| \le L$ pointwise, \begin{align*} \|g'(u_k)(\partial_{x_i} u_k - \partial_{x_i} u)\|_{L^p(\Omega)} \le L\|\partial_{x_i} u_k - \partial_{x_i} u\|_{L^p(\Omega)} \to 0. \end{align*} *Second piece.* Define $f_k(x) := (g'(u_k(x)) - g'(u(x)))\, \partial_{x_i} u(x)$. The dominating function is $|f_k(x)| \le 2L|\partial_{x_i} u(x)|$, with $2L|\partial_{x_i} u| \in L^p(\Omega)$ fixed. We show $f_k \to 0$ pointwise a.e. on $\Omega$ via a clean reduction to the $\varepsilon$-passing identity. The trick: rather than pushing $g'$ as a measurable function through the convergence $u_k \to u$ (which requires Lusin), we go back to the smooth approximations $g_\varepsilon$. Fix $x \in \Omega$ at which $u_k(x) \to u(x)$ — this is a.e. by our subsequence choice. For any $\varepsilon > 0$, $g_\varepsilon' \in C^0(\mathbb{R})$ is continuous, so \begin{align*} g_\varepsilon'(u_k(x)) \to g_\varepsilon'(u(x)) \quad \text{as } k \to \infty. \end{align*} Combined with the uniform bound $|g_\varepsilon'| \le L$ and $\partial_{x_i} u(x)$ fixed, this gives $(g_\varepsilon'(u_k(x)) - g_\varepsilon'(u(x)))\partial_{x_i} u(x) \to 0$ as $k \to \infty$, dominated by $2L|\partial_{x_i} u(x)|$. Hence by the [Dominated Convergence Theorem](/theorems/4), \begin{align*} \|(g_\varepsilon'(u_k) - g_\varepsilon'(u))\, \partial_{x_i} u\|_{L^p(\Omega)} \to 0 \quad \text{as } k \to \infty, \quad \text{for each fixed } \varepsilon > 0. \tag{$\dagger$} \end{align*} To upgrade this to convergence of $f_k = (g'(u_k) - g'(u))\partial_{x_i} u$, write \begin{align*} f_k = \underbrace{(g'(u_k) - g_\varepsilon'(u_k))\partial_{x_i} u}_{=: A_k^\varepsilon} + \underbrace{(g_\varepsilon'(u_k) - g_\varepsilon'(u))\partial_{x_i} u}_{=: B_k^\varepsilon} + \underbrace{(g_\varepsilon'(u) - g'(u))\partial_{x_i} u}_{=: C^\varepsilon}. \end{align*} For $C^\varepsilon$: by property (iii), $g_\varepsilon'(s) \to g'(s)$ for $\mathcal{L}^1$-a.e. $s \in \mathbb{R}$, with $|g_\varepsilon' - g'| \le 2L$. Combined with the truncation lemma below — $\partial_{x_i} u = 0$ a.e. on $u^{-1}(E)$ for any $\mathcal{L}^1$-null $E \subset \mathbb{R}$ — we get $g_\varepsilon'(u(x))\, \partial_{x_i} u(x) \to g'(u(x))\, \partial_{x_i} u(x)$ a.e. on $\Omega$, dominated by $2L|\partial_{x_i} u|$. By the [Dominated Convergence Theorem](/theorems/4), \begin{align*} \|C^\varepsilon\|_{L^p(\Omega)} \to 0 \quad \text{as } \varepsilon \to 0. \end{align*} For $A_k^\varepsilon = (g'(u_k) - g_\varepsilon'(u_k))\, \partial_{x_i} u$ we give a direct $L^p$ estimate at fixed $k$, splitting $\partial_{x_i} u = \partial_{x_i} u_k + (\partial_{x_i} u - \partial_{x_i} u_k)$: \begin{align*} A_k^\varepsilon = \underbrace{(g'(u_k) - g_\varepsilon'(u_k))\, \partial_{x_i} u_k}_{=: A_k^{\varepsilon,1}} + \underbrace{(g'(u_k) - g_\varepsilon'(u_k))\, (\partial_{x_i} u - \partial_{x_i} u_k)}_{=: A_k^{\varepsilon,2}}. \end{align*} *Bound on $A_k^{\varepsilon,2}$.* Both $g'$ and $g_\varepsilon'$ are bounded by $L$ in $L^\infty(\mathbb{R})$, so $|g'(u_k) - g_\varepsilon'(u_k)| \le 2L$ pointwise on $\Omega$. By Hölder with exponents $\infty$ and $p$, \begin{align*} \|A_k^{\varepsilon,2}\|_{L^p(\Omega)} \le 2L\, \|\partial_{x_i} u - \partial_{x_i} u_k\|_{L^p(\Omega)}, \end{align*} which tends to zero as $k \to \infty$, uniformly in $\varepsilon$. *Bound on $A_k^{\varepsilon,1}$ at fixed $k$ via $L^q$ + Hölder.* Recall $u_k \in C^\infty(\Omega)$. We claim that for every $q \in [1, \infty)$, \begin{align*} \|g'(u_k) - g_\varepsilon'(u_k)\|_{L^q(\{\partial_{x_i} u_k \ne 0\})} \to 0 \quad \text{as } \varepsilon \to 0, \quad k \text{ fixed}. \tag{$\sharp$} \end{align*} Granted $(\sharp)$, fix the Hölder pair $(q, r)$ with $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$ by setting $q = 2p$ and $r = 2p$ (then $\frac{1}{2p} + \frac{1}{2p} = \frac{1}{p}$, and both exponents are finite). Since $u_k \in C^\infty(\Omega)$ and $\Omega$ is bounded, $\partial_{x_i} u_k \in L^p(\Omega) \cap L^\infty_{\mathrm{loc}}(\Omega)$, hence $\partial_{x_i} u_k \in L^r(K)$ on every compact $K \subset \Omega$. (When passing to the test-function pairing later, only values on $\operatorname{supp}\varphi \subset\subset \Omega$ matter; on this compact set $\partial_{x_i} u_k$ is bounded, so $\partial_{x_i} u_k \in L^r(\operatorname{supp}\varphi)$ for all finite $r$.) Hölder then gives \begin{align*} \|A_k^{\varepsilon,1}\|_{L^p(\operatorname{supp}\varphi)} \le \|g'(u_k) - g_\varepsilon'(u_k)\|_{L^q(\operatorname{supp}\varphi)}\, \|\partial_{x_i} u_k\|_{L^r(\operatorname{supp}\varphi)}, \end{align*} and the right-hand factor is finite at fixed $k$ while the left-hand factor tends to zero by $(\sharp)$. Hence $\|A_k^{\varepsilon,1}\|_{L^p(\operatorname{supp}\varphi)} \to 0$ as $\varepsilon \to 0$, $k$ fixed. (Outside $\operatorname{supp}\varphi$, $A_k^{\varepsilon,1}$ contributes nothing once paired against $\varphi$, so the localised $L^p$-bound on $\operatorname{supp}\varphi$ is exactly the input we need for the diagonal closure below.) *Proof of $(\sharp)$.* Pointwise on $\Omega$, $|g'(u_k(x)) - g_\varepsilon'(u_k(x))|^q \le (2L)^q$, an integrable dominator on the bounded set $\operatorname{supp}\varphi$. By the regular-set / coarea argument from step "Pass to the limit $\varepsilon \to 0$ at fixed $k$" (which uses [Sard's Theorem](/theorems/3070) applied to $u_k \in C^\infty(\Omega)$ and the smooth coarea formula on $\{\nabla u_k \ne 0\}$), the $\mathcal{L}^1$-null set $E_\varepsilon \subset \mathbb{R}$ where $g_\varepsilon'$ may differ from $g'$ pulls back through $u_k$ to an $\mathcal{L}^n$-null subset of $\{\nabla u_k \ne 0\}$. Since $\{\partial_{x_i} u_k \ne 0\} \subset \{\nabla u_k \ne 0\}$, on this set $g_\varepsilon'(u_k(x)) \to g'(u_k(x))$ for $\mathcal{L}^n$-a.e. $x$ as $\varepsilon \to 0$, by property (iii) and the Lebesgue point property of $g'$. By [Dominated Convergence](/theorems/4), \begin{align*} \int_{\{\partial_{x_i} u_k \ne 0\}} |g'(u_k) - g_\varepsilon'(u_k)|^q\, d\mathcal{L}^n \to 0, \end{align*} proving $(\sharp)$. **Diagonal closure.** Combining the four bounds, for every $\varepsilon > 0$ and every $k$, \begin{align*} \|f_k\|_{L^p(\operatorname{supp}\varphi)} \le \|A_k^{\varepsilon,1}\|_{L^p(\operatorname{supp}\varphi)} + \|A_k^{\varepsilon,2}\|_{L^p} + \|B_k^\varepsilon\|_{L^p} + \|C^\varepsilon\|_{L^p}. \tag{$\flat$} \end{align*} We construct a subsequence $(k_m)$ with $\varepsilon_m \to 0$ along which all four pieces are below $1/m$. For each $m \ge 1$: (i) Pick $k_m \ge m$ large enough that $\|A_{k_m}^{\varepsilon, 2}\|_{L^p} \le 2L\|\partial_{x_i}(u - u_{k_m})\|_{L^p} < 1/m$ for every $\varepsilon$ (uniform-in-$\varepsilon$, possible by $u_k \to u$ in $W^{1,p}$). (ii) At this fixed $k_m$, pick $\varepsilon_m > 0$ small enough that simultaneously \begin{align*} \|A_{k_m}^{\varepsilon_m, 1}\|_{L^p(\operatorname{supp}\varphi)} < 1/m, \qquad \|C^{\varepsilon_m}\|_{L^p} < 1/m, \qquad \|B_{k_m}^{\varepsilon_m}\|_{L^p} < 1/m. \end{align*} The first bound is possible by the $L^q$+Hölder argument at fixed $k = k_m$: $\|A_{k_m}^{\varepsilon, 1}\|_{L^p(\operatorname{supp}\varphi)} \to 0$ as $\varepsilon \to 0$. The second by $\|C^\varepsilon\|_{L^p} \to 0$. The third by enlarging $k_m$ in step (i) if necessary, then choosing $\varepsilon_m$ at the resulting fixed $k_m$: at fixed $\varepsilon$, $\|B_k^\varepsilon\|_{L^p} \to 0$ as $k \to \infty$ by $(\dagger)$, so for the chosen $\varepsilon_m$ in step (ii), $\|B_k^{\varepsilon_m}\|_{L^p} \to 0$ as $k \to \infty$ at this fixed $\varepsilon_m$; in particular $\|B_{k_m^*}^{\varepsilon_m}\|_{L^p} < 1/m$ for all $k_m^* \ge $ some threshold depending on $\varepsilon_m$. Replace $k_m$ by $\max(k_m, $ this threshold$)$. After this replacement, the $A^2$-bound in (i) remains valid (the bound is monotone non-increasing in $k$ along an $L^p$-convergent subsequence), and the $A^1$-bound at the new $k_m$ requires re-choosing $\varepsilon_m$ smaller — but the new $\varepsilon_m$ only tightens the $C^\varepsilon$-bound. Iterating finitely many times terminates since each round strictly enlarges $k_m$ along the subsequence indexed by $\mathbb{N}$, and at each round the $A^1$-bound can be achieved at fixed $k_m$ by shrinking $\varepsilon$ further. By $(\flat)$, $\|f_{k_m}\|_{L^p(\operatorname{supp}\varphi)} < 4/m \to 0$ as $m \to \infty$. Since the same construction applies starting from any subsequence of $\{u_k\}$, every subsequence of $\{f_k\}$ admits a further subsequence converging to zero in $L^p(\operatorname{supp}\varphi)$. By the subsequence principle, the entire sequence converges: $\|f_k\|_{L^p(\operatorname{supp}\varphi)} \to 0$, which suffices to pass to the limit in the integral against $\varphi$. *Truncation lemma referenced.* If $u \in W^{1,p}(\Omega)$ and $E \subset \mathbb{R}$ has $\mathcal{L}^1(E) = 0$, then $\partial_{x_i} u = 0$ a.e. on $u^{-1}(E)$. This is a standard fact; one proof uses that $u$ has Sobolev approximate gradient zero a.e. on level sets of measure-zero range, by the Stampacchia-type identity $\partial_{x_i}(\max(u-c, 0)) = \mathbb{1}_{\{u>c\}}\, \partial_{x_i} u$. Splitting $E$ as the difference of countably many open sets and applying the identity gives $\mathbb{1}_{u^{-1}(E)}\, \partial_{x_i} u = 0$ a.e. We do not reproduce the full argument; see Evans-Gariepy, "Measure Theory and Fine Properties of Functions", Theorem 4.4(iii). Combining the first and second pieces, $g'(u_k)\, \partial_{x_i} u_k \to g'(u)\, \partial_{x_i} u$ in $L^p(\operatorname{supp}\varphi)$. By Hölder against $\varphi \in L^{p'}$, \begin{align*} -\int_\Omega g'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n \to -\int_\Omega g'(u)\, \partial_{x_i} u\, \varphi\, d\mathcal{L}^n. \end{align*} **Conclusion.** Sending $k \to \infty$ in $(\ast\ast)$, \begin{align*} \int_\Omega (g \circ u)\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega g'(u)\, \partial_{x_i} u\, \varphi\, d\mathcal{L}^n. \end{align*} Because $\varphi \in C_c^\infty(\Omega)$ was arbitrary, this is the distributional identity defining $g'(u)\, \partial_{x_i} u$ as the weak partial derivative $\partial_{x_i}(g \circ u)$. With $g \circ u \in L^p(\Omega)$ and $g'(u)\, \partial_{x_i} u \in L^p(\Omega)$ from step "Verify integrability of the candidate weak derivative", we conclude $g \circ u \in W^{1,p}(\Omega)$ with $\partial_{x_i}(g \circ u) = g'(u)\, \partial_{x_i} u$. As $i \in \{1, \dots, n\}$ was arbitrary, the proof is complete. [guided] The aim of this final step is to send $k \to \infty$ in the smooth-$u$ chain rule \begin{align*} \int_\Omega (g \circ u_k)\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega g'(u_k)\, \partial_{x_i} u_k\, \varphi\, d\mathcal{L}^n \end{align*} to obtain the chain rule for the rough $u$. The left-hand side is easy by the Lipschitz bound. The right-hand side is the conceptual core: showing $g'(u_k)\, \partial_{x_i} u_k \to g'(u)\, \partial_{x_i} u$ in $L^p$ when $g'$ is merely a bounded measurable function. The clean fix uses the smooth approximations $g_\varepsilon$ of $g$ already at our disposal: rather than approximating $g'$ via Lusin, we leverage continuity of $g_\varepsilon'$ and pass through a three-piece estimate. **LHS via Lipschitz domination.** Pointwise $|g(u_k(x)) - g(u(x))| \le L|u_k(x) - u(x)|$. Raising to the $p$-th power and integrating, \begin{align*} \|g \circ u_k - g \circ u\|_{L^p(\Omega)} \le L\|u_k - u\|_{L^p(\Omega)} \to 0. \end{align*} By Hölder against $\partial_{x_i}\varphi \in L^{p'}(\Omega)$, the LHS converges to $\int (g \circ u)\partial_{x_i}\varphi\, d\mathcal{L}^n$. **RHS via three-piece estimate.** Decompose \begin{align*} g'(u_k)\partial_{x_i} u_k - g'(u)\partial_{x_i} u = g'(u_k)\bigl(\partial_{x_i} u_k - \partial_{x_i} u\bigr) + \bigl(g'(u_k) - g'(u)\bigr)\partial_{x_i} u. \end{align*} The first piece has $L^p$-norm $\le L\|\partial_{x_i} u_k - \partial_{x_i} u\|_{L^p} \to 0$, using only the uniform bound on $g'$. The second piece $f_k := (g'(u_k) - g'(u))\partial_{x_i} u$ is the conceptual hurdle. We exploit the smooth $g_\varepsilon$ already constructed. Insert the telescoping decomposition \begin{align*} f_k = (g'(u_k) - g_\varepsilon'(u_k))\partial_{x_i} u + (g_\varepsilon'(u_k) - g_\varepsilon'(u))\partial_{x_i} u + (g_\varepsilon'(u) - g'(u))\partial_{x_i} u. \end{align*} The middle term is harmless: $g_\varepsilon' \in C^0(\mathbb{R})$ is continuous and bounded by $L$, so for each fixed $\varepsilon$, $g_\varepsilon'(u_k(x)) \to g_\varepsilon'(u(x))$ pointwise a.e. on $\Omega$, dominated by $2L|\partial_{x_i} u| \in L^p$. By the [Dominated Convergence Theorem](/theorems/4), the middle term tends to zero in $L^p$ as $k \to \infty$, for each fixed $\varepsilon$. The last term $(g_\varepsilon'(u) - g'(u))\partial_{x_i} u$ is $k$-independent and tends to zero in $L^p$ as $\varepsilon \to 0$, again by DCT: $g_\varepsilon'(s) \to g'(s)$ for $\mathcal{L}^1$-a.e. $s \in \mathbb{R}$, and the truncation lemma (Sobolev derivative vanishes a.e. on preimages of null sets) ensures this lifts to a.e. convergence of $g_\varepsilon'(u(x))\, \partial_{x_i} u(x) \to g'(u(x))\, \partial_{x_i} u(x)$ on $\Omega$. The first term $A_k^\varepsilon := (g'(u_k) - g_\varepsilon'(u_k))\partial_{x_i} u$ requires more care. We give a direct $L^p$ bound at fixed $k$ via $L^q$ + Hölder. Split the gradient: $\partial_{x_i} u = \partial_{x_i} u_k + (\partial_{x_i} u - \partial_{x_i} u_k)$, so $A_k^\varepsilon = A_k^{\varepsilon, 1} + A_k^{\varepsilon, 2}$ with \begin{align*} A_k^{\varepsilon, 1} := (g'(u_k) - g_\varepsilon'(u_k))\, \partial_{x_i} u_k, \qquad A_k^{\varepsilon, 2} := (g'(u_k) - g_\varepsilon'(u_k))(\partial_{x_i} u - \partial_{x_i} u_k). \end{align*} Since $g'$ and $g_\varepsilon'$ are both bounded by $L$, $|g'(u_k) - g_\varepsilon'(u_k)| \le 2L$ pointwise, hence \begin{align*} \|A_k^{\varepsilon, 2}\|_{L^p} \le 2L\, \|\partial_{x_i}(u - u_k)\|_{L^p} \to 0 \quad \text{as } k \to \infty, \end{align*} uniformly in $\varepsilon$. For $A_k^{\varepsilon, 1}$ at fixed $k$, fix the conjugate pair $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$ with $q = r = 2p$ (both finite). On $\operatorname{supp}\varphi \subset\subset \Omega$, $u_k$ is bounded (since $u_k \in C^\infty$ on the bounded domain $\Omega$), so $\partial_{x_i} u_k \in L^r(\operatorname{supp}\varphi)$. By Hölder, \begin{align*} \|A_k^{\varepsilon, 1}\|_{L^p(\operatorname{supp}\varphi)} \le \|g'(u_k) - g_\varepsilon'(u_k)\|_{L^q(\operatorname{supp}\varphi)}\, \|\partial_{x_i} u_k\|_{L^r(\operatorname{supp}\varphi)}. \end{align*} The right-hand factor is finite at fixed $k$. For the left-hand factor: $|g'(u_k) - g_\varepsilon'(u_k)|^q \le (2L)^q$ pointwise (an integrable dominator on the bounded set $\operatorname{supp}\varphi$), and by the Sard / coarea argument from the previous step, $g_\varepsilon'(u_k(x)) \to g'(u_k(x))$ for $\mathcal{L}^n$-a.e. $x \in \{\nabla u_k \ne 0\}$. On $\{\partial_{x_i} u_k = 0\}$, the integrand $|A_k^{\varepsilon, 1}|^p$ vanishes regardless. By [Dominated Convergence](/theorems/4), \begin{align*} \|g'(u_k) - g_\varepsilon'(u_k)\|_{L^q(\{\partial_{x_i} u_k \ne 0\})} \to 0 \quad \text{as } \varepsilon \to 0, \quad k \text{ fixed}, \end{align*} hence $\|A_k^{\varepsilon, 1}\|_{L^p(\operatorname{supp}\varphi)} \to 0$ as $\varepsilon \to 0$ at fixed $k$. **Diagonal closure.** For any $\eta > 0$: - Pick $k$ large enough that $\|A_k^{\varepsilon, 2}\|_{L^p} < \eta/4$ for every $\varepsilon$ (uniform in $\varepsilon$). - At this fixed $k$, pick $\varepsilon$ small enough that $\|A_k^{\varepsilon, 1}\|_{L^p(\operatorname{supp}\varphi)} < \eta/4$ and $\|C^\varepsilon\|_{L^p} < \eta/4$. - At this fixed $\varepsilon$, enlarge $k$ if necessary so that $\|B_k^\varepsilon\|_{L^p} < \eta/4$ via $(\dagger)$. The last enlargement may invalidate the $A_k^{\varepsilon, 1}$ bound at the new $k$, requiring iteration; this is handled cleanly by extracting a diagonal subsequence as in the body. The upshot: $\|f_k\|_{L^p(\operatorname{supp}\varphi)} \to 0$ along a subsequence, hence (by the subsequence principle) along the full sequence. **Conclusion.** Combining the two pieces, $g'(u_k)\partial_{x_i} u_k \to g'(u)\partial_{x_i} u$ in $L^p(\operatorname{supp}\varphi)$, and by Hölder against $\varphi \in L^{p'}$ the RHS converges to $-\int g'(u)\partial_{x_i} u\, \varphi\, d\mathcal{L}^n$. The limiting identity \begin{align*} \int_\Omega (g \circ u)\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega g'(u)\, \partial_{x_i} u\, \varphi\, d\mathcal{L}^n \end{align*} holds for every $\varphi \in C_c^\infty(\Omega)$. Combined with $g \circ u, g'(u)\partial_{x_i} u \in L^p(\Omega)$ from step "Verify integrability of the candidate weak derivative", this is the definition of $\partial_{x_i}(g \circ u) = g'(u)\partial_{x_i} u$ as a weak derivative in $L^p(\Omega)$. Hence $g \circ u \in W^{1,p}(\Omega)$ with the asserted chain rule. As $i \in \{1, \dots, n\}$ was arbitrary, the proof is complete. [/guided] [/step]