[proofplan] The strategy is to control the Taylor remainder $v_x(y) := u(y) - u(x) - \nabla u(x) \cdot (y - x)$ in a power-weighted average and show that for $\mathcal{L}^n$-almost every $x$, the quantity $\fint_{B(x,r)} (|v_x(y)|/|y-x|)^{p^*} \, d\mathcal{L}^n(y)$ tends to $0$ as $r \to 0$, where $p^* := np/(n-p)$. We work at points $x$ that are simultaneously Lebesgue points of $u$ and of $\nabla u$, and where the Hardy-Littlewood maximal function $M(|\nabla u - \nabla u(x)|^p)(x)$ is finite — this set is of full measure. We decompose $B(x, r)$ into dyadic shells $A_k = B(x, 2^{-k} r) \setminus B(x, 2^{-k-1} r)$, on each of which $|y - x| \asymp 2^{-k} r$, then apply Sobolev-Poincaré on the smaller ball $B(x, 2^{-k} r)$ after re-centering $v_x$ at its mean on that ball. The technical heart of the argument is a telescoping bound on the local mean $m_k(r) := (v_x)_{B(x, 2^{-k} r)}$ in terms of the gradient-deviation averages $G_j(x, r) := (\fint_{B(x, 2^{-j} r)} |\nabla u - \nabla u(x)|^p)^{1/p}$, derived by applying Sobolev-Poincaré to $\tilde u(y) := u(y) - \nabla u(x) \cdot (y - x)$ on each dyadic ball and exploiting the linear-mean-zero observation that $\tilde u_{B(x, \rho)} = u_{B(x, \rho)} - u(x) + (m_k\text{-correction})$. The shell sum is controlled by Lebesgue-Differentiation pointwise convergence, with the maximal function providing a summable dominator for the dominated convergence step. Approximate differentiability follows from $L^{p^*}$-differentiability via Markov's inequality. [/proofplan] [step:Set up: Sobolev conjugate exponent, precise representatives, and a full-measure good set] Throughout, we work under the standing hypothesis $u \in W^{1,p}_{\mathrm{loc}}(\Omega)$ with $\Omega \subseteq \mathbb{R}^n$ open and $1 \le p < n$. Define the Sobolev conjugate exponent \begin{align*} p^* := \frac{np}{n - p}, \end{align*} which is finite and satisfies $p^* > p \ge 1$ under $1 \le p < n$. We define the precise (Lebesgue) representatives of $u$ and $\nabla u$. For $u \in L^1_{\mathrm{loc}}(\Omega)$, set \begin{align*} u^*: \Omega &\to \mathbb{R} \cup \{*\} \\ x &\mapsto \begin{cases} \displaystyle \lim_{r \to 0^+} \fint_{B(x, r)} u(y) \, d\mathcal{L}^n(y) & \text{if the limit exists} \\ 0 & \text{otherwise.} \end{cases} \end{align*} For $\nabla u \in L^p_{\mathrm{loc}}(\Omega; \mathbb{R}^n) \subseteq L^1_{\mathrm{loc}}$, define analogously the componentwise precise representative \begin{align*} (\nabla u)^*: \Omega &\to \mathbb{R}^n \\ x &\mapsto \begin{cases} \displaystyle \lim_{r \to 0^+} \fint_{B(x, r)} \nabla u(y) \, d\mathcal{L}^n(y) & \text{if the (componentwise) limit exists} \\ 0 & \text{otherwise.} \end{cases} \end{align*} By the Lebesgue Differentiation Theorem, $u^*(x) = u(x)$ and $(\nabla u)^*(x) = \nabla u(x)$ for $\mathcal{L}^n$-a.e. $x \in \Omega$. We henceforth replace $u$ by $u^*$ and $\nabla u$ by $(\nabla u)^*$ and write $u(x)$, $\nabla u(x)$ for the precise pointwise values without further comment. Define the Hardy-Littlewood maximal function for a locally integrable $f: \Omega \to \mathbb{R}$ by \begin{align*} M f: \Omega &\to [0, \infty] \\ x &\mapsto \sup_{r > 0 : B(x, r) \subseteq \Omega} \fint_{B(x, r)} |f(y)| \, d\mathcal{L}^n(y). \end{align*} Fix $x_0 \in \Omega$ and $r_0 > 0$ with $\overline{B}(x_0, 2 r_0) \subseteq \Omega$. We will establish the desired conclusion at $\mathcal{L}^n$-a.e. point of $B(x_0, r_0)$; since $\Omega$ admits a countable cover by such balls, this yields the conclusion on all of $\Omega$. Set \begin{align*} \Omega_0 := \big\{ x \in B(x_0, r_0) : (i), (ii), (iii) \text{ hold} \big\}, \end{align*} where: - (i) $x$ is a Lebesgue point of $u$, so $\lim_{\rho \to 0} \fint_{B(x, \rho)} u \, d\mathcal{L}^n = u(x)$; - (ii) $x$ is a Lebesgue point of $\nabla u$ (componentwise), so $\lim_{\rho \to 0} \fint_{B(x, \rho)} |\nabla u - \nabla u(x)|^p \, d\mathcal{L}^n = 0$ (this is the stronger $L^p$-Lebesgue condition, equivalent for $|\nabla u|^p \in L^1_{\mathrm{loc}}$); - (iii) $M\big(|\nabla u - \nabla u(x)|^p \mathbb{1}_{B(x_0, 2r_0)}\big)(x) < \infty$. Each of (i), (ii), (iii) holds at $\mathcal{L}^n$-a.e. $x \in B(x_0, r_0)$: (i) and (ii) by the Lebesgue Differentiation Theorem (applied to $u \in L^1_{\mathrm{loc}}$ and to $|\nabla u - \nabla u(x)|^p \in L^1_{\mathrm{loc}}$ pointwise in $x$ — note this is the strong $L^p$ Lebesgue point condition); (iii) by the weak (1,1) inequality of the Hardy-Littlewood maximal theorem applied to $|\nabla u - \nabla u(x)|^p \mathbb{1}_{B(x_0, 2r_0)} \in L^1$ (note: the strong $(p,p)$ inequality is irrelevant here since we are in the $L^1$ class for the integrand $|\nabla u - \nabla u(x)|^p$, and the relevant statement is precisely that the maximal function of an $L^1$ function is finite a.e.). Hence $\mathcal{L}^n(B(x_0, r_0) \setminus \Omega_0) = 0$. We work with $x \in \Omega_0$ and $0 < r \le r_0$ so that $\overline{B}(x, r) \subseteq B(x_0, 2 r_0) \subseteq \Omega$. [/step] [step:Define the Taylor remainder $v_x$ and reduce $\Phi(x, r)$ to dyadic-ball averages] For $x \in \Omega_0$ and $0 < r \le r_0$, define the Taylor remainder \begin{align*} v_x: B(x, r) &\to \mathbb{R} \\ y &\mapsto u(y) - u(x) - \nabla u(x) \cdot (y - x). \end{align*} Since the affine map $y \mapsto u(x) + \nabla u(x) \cdot (y - x)$ is smooth, $v_x \in W^{1,p}(B(x, r))$ with weak gradient \begin{align*} \nabla v_x(y) = \nabla u(y) - \nabla u(x) \qquad \text{for $\mathcal{L}^n$-a.e. } y \in B(x, r). \end{align*} Write $\omega_n := \mathcal{L}^n(B(0, 1))$ so $\mathcal{L}^n(B(x, \rho)) = \omega_n \rho^n$. For a measurable set $E$ with $0 < \mathcal{L}^n(E) < \infty$ and a measurable $f$ on $E$, write $f_E := \fint_E f \, d\mathcal{L}^n$. The goal is: \begin{align*} \Phi(x, r) := \fint_{B(x, r)} \left( \frac{|v_x(y)|}{|y - x|} \right)^{p^*} d\mathcal{L}^n(y) \to 0 \qquad (r \to 0^+) \text{ for every } x \in \Omega_0. \end{align*} For $k \in \mathbb{N}_0 = \{0, 1, 2, \dots\}$, let $A_k := B(x, 2^{-k} r) \setminus B(x, 2^{-k-1} r)$ be the dyadic shell. Then $B(x, r) \setminus \{x\} = \bigsqcup_{k \ge 0} A_k$ (disjoint), and $\mathcal{L}^n(\{x\}) = 0$. For $y \in A_k$ we have $|y - x| \ge 2^{-k-1} r$, so $|y - x|^{-p^*} \le 2^{(k+1) p^*} r^{-p^*}$. Splitting $\Phi(x, r)$ over shells: \begin{align*} \Phi(x, r) &= \frac{1}{\omega_n r^n} \sum_{k = 0}^\infty \int_{A_k} \frac{|v_x(y)|^{p^*}}{|y - x|^{p^*}} \, d\mathcal{L}^n(y) \\ &\le \frac{1}{\omega_n r^n} \sum_{k = 0}^\infty \frac{2^{(k+1) p^*}}{r^{p^*}} \int_{A_k} |v_x|^{p^*} d\mathcal{L}^n. \end{align*} Since $A_k \subseteq B(x, 2^{-k} r)$ and $\mathcal{L}^n(B(x, 2^{-k} r)) = \omega_n (2^{-k} r)^n$, \begin{align*} \int_{A_k} |v_x|^{p^*} d\mathcal{L}^n \le \omega_n (2^{-k} r)^n \fint_{B(x, 2^{-k} r)} |v_x|^{p^*} d\mathcal{L}^n. \end{align*} Combining, \begin{align*} \Phi(x, r) \le \frac{2^{p^*}}{r^{p^*}} \sum_{k = 0}^\infty 2^{k(p^* - n)} \fint_{B(x, 2^{-k} r)} |v_x|^{p^*} d\mathcal{L}^n. \end{align*} This reduces the problem to bounding the dyadic-ball averages $\fint_{B(x, 2^{-k} r)} |v_x|^{p^*} \, d\mathcal{L}^n$. We will need the linear-mean-zero observation. By the change of variables $y = x + z$ on $B(x, \rho)$, and using that the Lebesgue measure on $B(0, \rho)$ is invariant under $z \mapsto -z$, \begin{align*} \fint_{B(x, \rho)} \nabla u(x) \cdot (y - x) \, d\mathcal{L}^n(y) = \nabla u(x) \cdot \fint_{B(0, \rho)} z \, d\mathcal{L}^n(z) = \nabla u(x) \cdot 0 = 0. \end{align*} Hence for the auxiliary function \begin{align*} \tilde u_x: B(x, r) &\to \mathbb{R} \\ y &\mapsto u(y) - \nabla u(x) \cdot (y - x), \end{align*} we have $\tilde u_x \in W^{1,p}(B(x, r))$ with $\nabla \tilde u_x = \nabla u - \nabla u(x)$ a.e., and the linear-mean-zero observation gives \begin{align*} (\tilde u_x)_{B(x, \rho)} = u_{B(x, \rho)} - \nabla u(x) \cdot \fint_{B(x, \rho)}(y - x) \, d\mathcal{L}^n(y) = u_{B(x, \rho)}. \end{align*} We use this auxiliary function in the next step. [/step] [step:Key lemma: telescoping bound on the local mean $m_k(r)$ in terms of $G_j(x, r)$] Define \begin{align*} m_k(r) &:= (v_x)_{B(x, 2^{-k} r)} = \fint_{B(x, 2^{-k} r)} v_x \, d\mathcal{L}^n, \\ G_j(x, r) &:= \left( \fint_{B(x, 2^{-j} r)} |\nabla u(y) - \nabla u(x)|^p \, d\mathcal{L}^n(y) \right)^{1/p}. \end{align*} By the linear-mean-zero observation of Step 2, $m_k(r) = u_{B(x, 2^{-k} r)} - u(x)$. [claim:Telescoping bound on $|m_k(r)|$] There exists a constant $C = C(n, p)$ such that for every $x \in \Omega_0$, every $0 < r \le r_0$, and every $k \in \mathbb{N}_0$, \begin{align*} |m_k(r)| \le C \cdot r \sum_{j = k}^\infty 2^{-j} G_j(x, r). \end{align*} [/claim] *Proof of claim.* Recall the Sobolev-Poincaré inequality on balls (a standard result in the theory of Sobolev functions): there is a constant $C_{\mathrm{SP}} = C_{\mathrm{SP}}(n, p)$ such that for every $w \in W^{1,p}(B(z, \rho))$ with $1 \le p < n$, \begin{align*} \left( \fint_{B(z, \rho)} |w - w_{B(z, \rho)}|^{p^*} d\mathcal{L}^n \right)^{1/p^*} \le C_{\mathrm{SP}} \cdot \rho \cdot \left( \fint_{B(z, \rho)} |\nabla w|^p \, d\mathcal{L}^n \right)^{1/p}. \end{align*} Apply this to $\tilde u_x \in W^{1,p}(B(x, r))$ on the ball $B(x, 2^{-j} r)$, using $\nabla \tilde u_x = \nabla u - \nabla u(x)$ and $(\tilde u_x)_{B(x, 2^{-j} r)} = u_{B(x, 2^{-j} r)}$: \begin{align*} \left( \fint_{B(x, 2^{-j} r)} \big| \tilde u_x(y) - u_{B(x, 2^{-j} r)} \big|^{p^*} d\mathcal{L}^n(y) \right)^{1/p^*} \le C_{\mathrm{SP}} \cdot 2^{-j} r \cdot G_j(x, r). \end{align*} For each $j \ge 0$, since $B(x, 2^{-j-1} r) \subseteq B(x, 2^{-j} r)$ with volume ratio $2^{-n}$, by Jensen's inequality (which gives $|\fint f| \le \fint |f| \le (\fint |f|^{p^*})^{1/p^*}$ for $p^* \ge 1$), \begin{align*} \big| u_{B(x, 2^{-j-1} r)} - u_{B(x, 2^{-j} r)} \big| &= \big| (\tilde u_x)_{B(x, 2^{-j-1} r)} - (\tilde u_x)_{B(x, 2^{-j} r)} \big| \\ &\le \fint_{B(x, 2^{-j-1} r)} \big| \tilde u_x - (\tilde u_x)_{B(x, 2^{-j} r)} \big| \, d\mathcal{L}^n \\ &\le 2^n \fint_{B(x, 2^{-j} r)} \big| \tilde u_x - (\tilde u_x)_{B(x, 2^{-j} r)} \big| \, d\mathcal{L}^n \\ &\le 2^n \left( \fint_{B(x, 2^{-j} r)} \big| \tilde u_x - (\tilde u_x)_{B(x, 2^{-j} r)} \big|^{p^*} d\mathcal{L}^n \right)^{1/p^*} \\ &\le 2^n C_{\mathrm{SP}} \cdot 2^{-j} r \cdot G_j(x, r). \end{align*} The first equality used $(\tilde u_x)_{B(x, \rho)} = u_{B(x, \rho)}$ for both balls. Now apply the Lebesgue Differentiation Theorem at the Lebesgue point $x$ of $u$: $u_{B(x, 2^{-j} r)} \to u(x)$ as $j \to \infty$. Hence the telescoping series \begin{align*} u(x) - u_{B(x, 2^{-k} r)} = \sum_{j = k}^\infty \big( u_{B(x, 2^{-j-1} r)} - u_{B(x, 2^{-j} r)} \big) \end{align*} converges absolutely (granted the bound just derived; we verify this in the next display), and the triangle inequality gives \begin{align*} |m_k(r)| = \big| u_{B(x, 2^{-k} r)} - u(x) \big| \le \sum_{j = k}^\infty \big| u_{B(x, 2^{-j-1} r)} - u_{B(x, 2^{-j} r)} \big| \le 2^n C_{\mathrm{SP}} \cdot r \sum_{j = k}^\infty 2^{-j} G_j(x, r). \end{align*} Absolute convergence of the right side follows from $G_j(x, r) \le M_*(x)^{1/p}$ (where $M_*(x) := M(|\nabla u - \nabla u(x)|^p \mathbb{1}_{B(x_0, 2r_0)})(x) < \infty$ by condition (iii)) and $\sum_j 2^{-j} < \infty$. The constant is $C := 2^n C_{\mathrm{SP}}$. This proves the claim. $\square$ [/step] [step:Apply Sobolev-Poincaré on each dyadic ball and combine with the telescoping bound] Apply Sobolev-Poincaré to $v_x \in W^{1,p}(B(x, r))$ on the ball $B(x, 2^{-k} r)$, using $\nabla v_x = \nabla u - \nabla u(x)$: \begin{align*} \left( \fint_{B(x, 2^{-k} r)} |v_x - m_k(r)|^{p^*} d\mathcal{L}^n \right)^{1/p^*} \le C_{\mathrm{SP}} \cdot 2^{-k} r \cdot G_k(x, r). \end{align*} By the triangle inequality in $L^{p^*}(B(x, 2^{-k} r))$ for the decomposition $v_x = (v_x - m_k(r)) + m_k(r)$, \begin{align*} \left( \fint_{B(x, 2^{-k} r)} |v_x|^{p^*} d\mathcal{L}^n \right)^{1/p^*} \le C_{\mathrm{SP}} \cdot 2^{-k} r \cdot G_k(x, r) + |m_k(r)|. \end{align*} Raising to the $p^*$-th power (a pure algebraic manipulation; not Hölder) and using $(a + b)^{p^*} \le 2^{p^*}(a^{p^*} + b^{p^*})$ for $a, b \ge 0$: \begin{align*} \fint_{B(x, 2^{-k} r)} |v_x|^{p^*} d\mathcal{L}^n \le 2^{p^*} \big( C_{\mathrm{SP}}^{p^*} (2^{-k} r)^{p^*} G_k(x, r)^{p^*} + |m_k(r)|^{p^*} \big). \end{align*} [/step] [step:Sum the dyadic shell estimates and conclude $\Phi(x, r) \to 0$ via dominated convergence] Substitute the Step 4 dyadic-ball bound into the Step 2 reduction: \begin{align*} \Phi(x, r) &\le \frac{2^{p^*}}{r^{p^*}} \sum_{k = 0}^\infty 2^{k(p^* - n)} \fint_{B(x, 2^{-k} r)} |v_x|^{p^*} d\mathcal{L}^n \\ &\le \frac{2^{p^*}}{r^{p^*}} \sum_{k = 0}^\infty 2^{k(p^* - n)} \cdot 2^{p^*} \big( C_{\mathrm{SP}}^{p^*} (2^{-k} r)^{p^*} G_k(x, r)^{p^*} + |m_k(r)|^{p^*} \big) \\ &= 2^{2 p^*} C_{\mathrm{SP}}^{p^*} \underbrace{\sum_{k = 0}^\infty 2^{-k n} G_k(x, r)^{p^*}}_{=: S_I(r)} + \underbrace{\frac{2^{2 p^*}}{r^{p^*}} \sum_{k = 0}^\infty 2^{k(p^* - n)} |m_k(r)|^{p^*}}_{=: S_{II}(r)}, \end{align*} where in the first sum we used $2^{k(p^* - n)} (2^{-k} r)^{p^*} / r^{p^*} = 2^{-k n}$. We treat $S_I(r)$ and $S_{II}(r)$ separately. Both arguments rely on the dominator $M_*(x) := M(|\nabla u - \nabla u(x)|^p \mathbb{1}_{B(x_0, 2 r_0)})(x) < \infty$, finite by condition (iii). Note: throughout this step the constants depend on $x$ via $M_*(x)$; we are proving pointwise (in $x$) convergence, not a uniform bound. **Sum I.** For each fixed $k$, condition (ii) gives $G_k(x, r)^p = \fint_{B(x, 2^{-k} r)} |\nabla u - \nabla u(x)|^p \, d\mathcal{L}^n \to 0$ as $r \to 0^+$, hence $G_k(x, r)^{p^*} \to 0$. The maximal-function dominator gives, whenever $2^{-k} r \le r_0$, \begin{align*} G_k(x, r)^p \le M_*(x), \qquad \text{so} \qquad 2^{-k n} G_k(x, r)^{p^*} \le 2^{-k n} M_*(x)^{p^*/p}. \end{align*} The bound $\sum_{k} 2^{-k n} M_*(x)^{p^*/p} = (1 - 2^{-n})^{-1} M_*(x)^{p^*/p} < \infty$ provides a summable dominator. By the Dominated Convergence Theorem applied to the counting measure on $\mathbb{N}_0$ weighted by $2^{-k n}$ with the parameter $r \to 0^+$ (the parameter version: for every sequence $r_\ell \to 0^+$, $G_k(x, r_\ell)^{p^*} \to 0$ pointwise in $k$ and is dominated by $M_*(x)^{p^*/p}$, so the weighted sum converges to $0$; this is the standard sequential reformulation of DCT for parameter limits), \begin{align*} S_I(r) \to 0 \qquad (r \to 0^+). \end{align*} **Sum II.** Apply the telescoping bound from the Step 3 claim: \begin{align*} |m_k(r)|^{p^*} \le (2^n C_{\mathrm{SP}})^{p^*} r^{p^*} \left( \sum_{j = k}^\infty 2^{-j} G_j(x, r) \right)^{p^*}. \end{align*} Using $G_j(x, r) \le M_*(x)^{1/p}$, the inner sum is bounded uniformly by $\sum_{j \ge k} 2^{-j} M_*(x)^{1/p} = 2^{1 - k} M_*(x)^{1/p}$, so \begin{align*} |m_k(r)|^{p^*} \le (2^n C_{\mathrm{SP}})^{p^*} r^{p^*} \cdot 2^{(1 - k) p^*} M_*(x)^{p^*/p}. \end{align*} Multiplying by $2^{k(p^* - n)} / r^{p^*}$ and using $2^{k(p^* - n)} \cdot 2^{(1 - k) p^*} = 2^{p^*} \cdot 2^{-k n}$: \begin{align*} 2^{k(p^* - n)} \frac{|m_k(r)|^{p^*}}{r^{p^*}} \le (2^n C_{\mathrm{SP}})^{p^*} \cdot 2^{p^*} \cdot 2^{-k n} M_*(x)^{p^*/p}. \end{align*} This is the summable dominator (in $k$) for $S_{II}(r)$, independent of $r$. For pointwise convergence in $k$ as $r \to 0^+$: for each fixed $k$ and each fixed $j \ge k$, condition (ii) gives $G_j(x, r) \to 0$ as $r \to 0^+$; the dominator $2^{-j} M_*(x)^{1/p}$ is summable in $j$. By DCT (parameter version, on the counting measure on $\{j : j \ge k\}$), \begin{align*} \sum_{j = k}^\infty 2^{-j} G_j(x, r) \to 0 \qquad (r \to 0^+) \text{ for each fixed } k. \end{align*} Hence $|m_k(r)|^{p^*} / r^{p^*} \to 0$ as $r \to 0^+$ for each fixed $k$, and consequently $2^{k(p^* - n)} |m_k(r)|^{p^*} / r^{p^*} \to 0$. By DCT once more (now on the counting measure on $\mathbb{N}_0$ with the dominator $(2^n C_{\mathrm{SP}})^{p^*} \cdot 2^{p^*} \cdot 2^{-k n} M_*(x)^{p^*/p}$), \begin{align*} S_{II}(r) \to 0 \qquad (r \to 0^+). \end{align*} **Conclusion.** $\Phi(x, r) \le 2^{2 p^*} C_{\mathrm{SP}}^{p^*} S_I(r) + S_{II}(r) \to 0$ as $r \to 0^+$ for every $x \in \Omega_0$. This is the $L^{p^*}$-differentiability statement: for $\mathcal{L}^n$-a.e. $x \in \Omega$, \begin{align*} \left( \fint_{B(x, r)} \left( \frac{|u(y) - u(x) - \nabla u(x) \cdot (y - x)|}{|y - x|} \right)^{p^*} d\mathcal{L}^n(y) \right)^{1/p^*} \to 0 \qquad (r \to 0^+). \end{align*} [/step] [step:Approximate differentiability via Markov's inequality] For $\varepsilon > 0$ and $r > 0$, define \begin{align*} E_\varepsilon(r) := \left\{ y \in B(x, r) \setminus \{x\} : \frac{|u(y) - u(x) - \nabla u(x) \cdot (y - x)|}{|y - x|} > \varepsilon \right\} \end{align*} (the singleton $\{x\}$ has measure zero, so excluding it is harmless). By Markov's inequality (the elementary bound $\mu(\{|f| > \lambda\}) \le \lambda^{-q} \int |f|^q \, d\mu$ for $q \ge 1$, $\lambda > 0$, valid because $\mathbb{1}_{\{|f| > \lambda\}} \le |f|^q / \lambda^q$ pointwise) with $q = p^*$, $\lambda = \varepsilon^{p^*}$, $\mu = \mathcal{L}^n$, applied to $y \mapsto |v_x(y)|^{p^*} / |y - x|^{p^*}$ on $B(x, r)$: \begin{align*} \mathcal{L}^n(E_\varepsilon(r)) \le \frac{1}{\varepsilon^{p^*}} \int_{B(x, r)} \frac{|v_x(y)|^{p^*}}{|y - x|^{p^*}} \, d\mathcal{L}^n(y). \end{align*} Dividing by $\mathcal{L}^n(B(x, r)) = \omega_n r^n$: \begin{align*} \frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x, r))} \le \frac{\Phi(x, r)}{\varepsilon^{p^*}}. \end{align*} By Step 5, $\Phi(x, r) \to 0$ as $r \to 0^+$ for every $x \in \Omega_0$, so \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x, r))} = 0 \qquad \text{for every } \varepsilon > 0. \end{align*} This is precisely the definition of approximate differentiability of $u$ at $x$ with approximate derivative $\nabla u(x)$. The conclusion holds for $\mathcal{L}^n$-a.e. $x \in \Omega$, completing the proof. (This is a Sobolev-function refinement of [Rademacher's Theorem](/theorems/3069), which gives classical differentiability $\mathcal{L}^n$-a.e. for Lipschitz $u$. For $u$ merely in $W^{1,p}_{\mathrm{loc}}$ with $1 \le p < n$, the gradient need not be locally bounded, and one obtains approximate (rather than classical) differentiability.) [/step]