[proofplan] The strategy reduces classical differentiability of $u \in W^{1,p}_{\mathrm{loc}}(\Omega)$ for $p > n$ to Rademacher's theorem applied to a Lipschitz truncation $v_\lambda$ of $u$. The construction uses the centred $L^p$-Hardy-Littlewood maximal function $M_p(|\nabla u|) := M(|\nabla u|^p)^{1/p}$, whose sublevel set $E_\lambda := \{x \in U : M_p(|\nabla u|)(x) \le \lambda\}$ is closed (by lower semicontinuity) and exhausts $U$ in measure as $\lambda \to \infty$ (by the weak-$(p,p)$ maximal inequality, which uses $p > 1$). On $E_\lambda \cap \overline{Q}$, $u$ is Lipschitz with constant $L_\lambda := \max(C_n \lambda, C_0)$, where $C_n \lambda$ controls pairs at small separation via a chaining argument and $C_0$ depends on $\|u\|_{W^{1,p}(U)}$ and the geometry. The McShane extension $v_\lambda: \mathbb{R}^n \to \mathbb{R}$ is $L_\lambda$-Lipschitz and agrees with $u$ on $E_\lambda \cap \overline{Q}$. By [Rademacher's Theorem](/theorems/3069), $v_\lambda$ is classically differentiable a.e. The transfer from $v_\lambda$ to $u$ for *all* $h \to 0$ (not merely on a density-1 set) rests on the **pointwise discrepancy estimate** $|u(y) - v_\lambda(y)| \le C_n \cdot L_\lambda \cdot \operatorname{dist}(y, E_\lambda \cap \overline{Q})$ — a consequence of Sobolev–Morrey continuity of $u$ on small balls centred at $E_\lambda$-points and the McShane Lipschitz bound. Combined with the distance estimate $\operatorname{dist}(x_0 + h, E_\lambda) = o(|h|)$ at a Lebesgue density-$1$ point of $E_\lambda$, this gives $|u(x_0 + h) - v_\lambda(x_0 + h)| = o(|h|)$ unconditionally over all $h \to 0$, transferring the Rademacher expansion of $v_\lambda$ to a classical expansion of $u$. Identification of $\nabla v_\lambda(x_0)$ with $\nabla u(x_0)$ at Lebesgue points of $\nabla u$ uses a directional sequence in $E_\lambda$ together with the affine chaining argument applied to $\widetilde u(y) := u(y) - \nabla u(x_0) \cdot (y - x_0)$. The hypothesis $p > n$ enters via [Morrey's Inequality](/theorems/62), which gives a continuous representative of $u$ and the Sobolev oscillation bound on small balls. [/proofplan] [step:Setup: localise via cubes and pass to the precise representative] Cover $\Omega$ by a countable family of open cubes $\{Q_k\}_{k \ge 1}$ with $Q_k \subset\subset \Omega$. Since the conclusion is local — a property holding $\mathcal{L}^n$-a.e. on each $Q_k$ extends to a property holding $\mathcal{L}^n$-a.e. on $\Omega$ — it suffices to prove classical differentiability $\mathcal{L}^n$-a.e. on each $Q_k$. Fix one such $Q := Q_k$ and choose an open set $U$ with $Q \subset\subset U \subset\subset \Omega$. Then $u \in W^{1,p}(U)$, and we set \begin{align*} \delta := \operatorname{dist}(Q, \partial U) > 0. \end{align*} By [Morrey's Inequality](/theorems/62) on $U$ (a bounded Lipschitz domain), $u$ has a continuous representative $u^* \in C(\overline{U})$ with Hölder seminorm controlled by $\|\nabla u\|_{L^p(U)}$. We claim further that $u^*$ obeys the $L^\infty$ bound \begin{align*} \|u^*\|_{L^\infty(U)} \le C_M \|u\|_{W^{1,p}(U)}, \end{align*} where $C_M$ depends only on $n, p$, and the geometry of $U$ (in particular $\operatorname{diam}(U)$ and $\mathcal{L}^n(U)$). To justify this, fix any $x \in U$. Morrey's Hölder bound yields, for all $y \in U$, \begin{align*} |u^*(x) - u^*(y)| \le C_M^{(1)} |x-y|^{1 - n/p} \|\nabla u\|_{L^p(U)} \le C_M^{(1)} \operatorname{diam}(U)^{1 - n/p} \|\nabla u\|_{L^p(U)}, \end{align*} with $C_M^{(1)} = C_M^{(1)}(n,p,U)$ the Morrey constant. Averaging over $y \in U$ and applying Hölder's inequality on the bounded domain $U$, \begin{align*} |u^*(x)| \le \fint_U |u^*(y)| \, d\mathcal{L}^n(y) + C_M^{(1)} \operatorname{diam}(U)^{1 - n/p} \|\nabla u\|_{L^p(U)} \le \mathcal{L}^n(U)^{-1/p} \|u\|_{L^p(U)} + C_M^{(1)} \operatorname{diam}(U)^{1 - n/p} \|\nabla u\|_{L^p(U)}, \end{align*} which is bounded by $C_M \|u\|_{W^{1,p}(U)}$ for $C_M := \max(\mathcal{L}^n(U)^{-1/p}, C_M^{(1)} \operatorname{diam}(U)^{1 - n/p})$. The hypothesis $p > n$ ensures the embedding $W^{1,p}(U) \hookrightarrow C(\overline{U})$. Henceforth replace $u$ by $u^*$ and write $u(x)$ for the pointwise value; for the continuous representative, every point of $U$ is a Lebesgue point of $u$. By the [Lebesgue Differentiation Theorem](/theorems/3068) applied componentwise to $|\nabla u| \in L^p(U) \subseteq L^1(U)$, the precise representative \begin{align*} (\nabla u)^*: U &\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*} satisfies $(\nabla u)^*(x) = \nabla u(x)$ for $\mathcal{L}^n$-a.e. $x \in U$. Henceforth $\nabla u$ denotes this precise representative. [/step] [step:Define the maximal function level sets $E_\lambda$ and verify their measure-exhaustion] Define the centred $L^p$-Hardy-Littlewood maximal function of $|\nabla u|$ extended by zero outside $U$: \begin{align*} M_p(|\nabla u|): \mathbb{R}^n &\to [0, \infty] \\ x &\mapsto \sup_{r > 0} \Big( \fint_{B(x, r)} |\nabla u(y)|^p \mathbb{1}_U(y) \, d\mathcal{L}^n(y) \Big)^{1/p}. \end{align*} Equivalently $M_p(|\nabla u|)(x) = M(|\nabla u|^p)(x)^{1/p}$, where $M$ is the standard Hardy-Littlewood maximal function. The function $M_p(|\nabla u|)$ is lower semicontinuous on $\mathbb{R}^n$: each averaging functional $x \mapsto \big( \fint_{B(x,r)} |\nabla u|^p \mathbb{1}_U \, d\mathcal{L}^n \big)^{1/p}$ is continuous, hence the supremum is lower semicontinuous. Lower semicontinuity implies the sublevel set $\{M_p(|\nabla u|) \le \lambda\}$ is closed in $\mathbb{R}^n$. For $\lambda > 0$, define the level set \begin{align*} E_\lambda := \{x \in U : M_p(|\nabla u|)(x) \le \lambda\}. \end{align*} This is the intersection of the closed set $\{M_p(|\nabla u|) \le \lambda\}$ with $U$, hence Borel. By the weak-$(p,p)$ [Hardy-Littlewood Maximal Inequality](/theorems/3070) (which is equivalent to the strong $(p,p)$ for $p > 1$ by Marcinkiewicz interpolation), \begin{align*} \mathcal{L}^n(U \setminus E_\lambda) \le \mathcal{L}^n\big( \{ x \in \mathbb{R}^n : M_p(|\nabla u|)(x) > \lambda \} \big) = \mathcal{L}^n\big( \{ x \in \mathbb{R}^n : M(|\nabla u|^p)(x) > \lambda^p \} \big) \le C(n, p) \cdot \lambda^{-p} \cdot \big\| |\nabla u|^p \mathbb{1}_U \big\|_{L^1(\mathbb{R}^n)} = C(n, p) \cdot \lambda^{-p} \cdot \big\| |\nabla u| \big\|_{L^p(U)}^p, \end{align*} for a constant $C(n, p)$ depending only on $n$ and $p$. In particular, $M_p(|\nabla u|) < \infty$ $\mathcal{L}^n$-a.e. on $\mathbb{R}^n$, and $\mathcal{L}^n(U \setminus E_\lambda) \to 0$ as $\lambda \to \infty$. Since $E_\lambda \subseteq E_{\lambda'}$ for $\lambda \le \lambda'$, the union has full measure: \begin{align*} \mathcal{L}^n\Big( U \setminus \bigcup_{\lambda > 0} E_\lambda \Big) = 0. \end{align*} Note that for $x \in E_\lambda$ and any ball $B(x, r) \subseteq \mathbb{R}^n$, Jensen's inequality gives \begin{align*} \fint_{B(x, r)} |\nabla u| \cdot \mathbb{1}_U \, d\mathcal{L}^n \le \Big( \fint_{B(x, r)} |\nabla u|^p \mathbb{1}_U \, d\mathcal{L}^n \Big)^{1/p} \le M_p(|\nabla u|)(x) \le \lambda, \end{align*} so the standard Hardy-Littlewood maximal function is also bounded: $M(|\nabla u|)(x) \le M_p(|\nabla u|)(x) \le \lambda$ for $x \in E_\lambda$. [/step] [step:Lipschitz estimate on $E_\lambda \cap Q$ with constant $L_\lambda := \max(C_n \lambda, C_0)$] We separate the analysis into a small-separation regime, where the maximal-function bound gives a $C_n \lambda$ Lipschitz constant, and a large-separation regime, where the Morrey $L^\infty$ bound gives a $\lambda$-independent Lipschitz constant. [claim:Lipschitz estimate on $E_\lambda \cap Q$] There exist a dimensional constant $C_n > 0$ (depending on $n, p$) and a constant $C_0 = C_0(n, p, \delta, \|u\|_{W^{1,p}(U)})$, both independent of $\lambda$, such that with \begin{align*} L_\lambda := \max\big(C_n \lambda, \, C_0\big), \end{align*} for every $\lambda > 0$ and every $x_1, x_2 \in E_\lambda \cap \overline{Q}$, \begin{align*} |u(x_1) - u(x_2)| \le L_\lambda \cdot |x_1 - x_2|. \end{align*} [/claim] *Proof of claim.* Fix $x_1, x_2 \in E_\lambda \cap \overline{Q}$ with $x_1 \ne x_2$, and set $r := |x_1 - x_2|$. **Case A: $r < \delta/4$ (small-separation regime).** In this case $B(x_1, 2 r) \cup B(x_2, 2 r) \subseteq U$, since $\operatorname{dist}(x_i, \partial U) \ge \delta$ and $2 r < \delta/2$. Let $z := (x_1 + x_2)/2$ be the midpoint, and write $u_{B(z, r)} := \fint_{B(z, r)} u \, d\mathcal{L}^n$. The triangle inequality gives \begin{align*} |u(x_1) - u(x_2)| \le |u(x_1) - u_{B(z, r)}| + |u(x_2) - u_{B(z, r)}|. \end{align*} We bound each term by chaining centred at $x_1$ (the second is symmetric). Set $B_k := B(x_1, 2^{-k+1} r)$ for $k = 0, 1, 2, \dots$, so $B_0 = B(x_1, 2 r) \supseteq B(z, r)$ (since $|z - x_1| = r/2$, so $B(z, r) \subseteq B(x_1, 3 r/2) \subseteq B_0$). For each $k$, $B_{k+1} \subseteq B_k$ with volume ratio $2^{-n}$. By Jensen's inequality, \begin{align*} |u_{B_{k+1}} - u_{B_k}| \le \fint_{B_{k+1}} |u - u_{B_k}| \, d\mathcal{L}^n \le 2^n \fint_{B_k} |u - u_{B_k}| \, d\mathcal{L}^n. \end{align*} By the [Poincaré Inequality](/theorems/3071) on $B_k$ ($W^{1,1}$ form, valid because $u \in W^{1,p}(U) \hookrightarrow W^{1,1}(B_k)$ since $B_k \subset U$ is bounded; $C_P$ denotes the dimensional Poincaré constant), \begin{align*} \fint_{B_k} |u - u_{B_k}| \, d\mathcal{L}^n \le C_P \cdot 2^{-k+1} r \cdot \fint_{B_k} |\nabla u| \, d\mathcal{L}^n \le C_P \cdot 2^{-k+1} r \cdot M(|\nabla u|)(x_1) \le C_P \cdot 2^{-k+1} r \cdot \lambda, \end{align*} using $M(|\nabla u|)(x_1) \le M_p(|\nabla u|)(x_1) \le \lambda$ since $x_1 \in E_\lambda$. Since $x_1$ is a Lebesgue point of the continuous representative $u$, $u_{B_k} \to u(x_1)$ as $k \to \infty$, and the telescoping series gives \begin{align*} |u(x_1) - u_{B_0}| \le \sum_{k = 0}^\infty |u_{B_{k+1}} - u_{B_k}| \le \sum_{k=0}^\infty 2^n C_P \cdot 2^{-k+1} r \lambda = 2^{n+2} C_P \cdot r \lambda, \end{align*} using $\sum_{k \ge 0} 2^{-k+1} = 4$. Transferring from $B_0$ to $B(z, r)$ via $B(z, r) \subseteq B_0$ with volume ratio $2^n$, \begin{align*} |u_{B(z, r)} - u_{B_0}| \le 2^n \fint_{B_0} |u - u_{B_0}| \, d\mathcal{L}^n \le 2^n \cdot C_P \cdot 2 r \lambda = 2^{n+1} C_P r \lambda. \end{align*} Combining, \begin{align*} |u(x_1) - u_{B(z, r)}| \le \big( 2^{n+2} + 2^{n+1} \big) C_P r \lambda = 3 \cdot 2^{n+1} C_P r \lambda. \end{align*} The symmetric estimate at $x_2$ adds another $3 \cdot 2^{n+1} C_P r \lambda$. Setting the dimensional constant \begin{align*} C_n := 2 \cdot 3 \cdot 2^{n+1} C_P = 3 \cdot 2^{n+2} C_P, \end{align*} we obtain $|u(x_1) - u(x_2)| \le C_n \cdot r \lambda = C_n \lambda \cdot |x_1 - x_2|$. **Case B: $r \ge \delta/4$ (large-separation regime).** From the Morrey $L^\infty$ bound, \begin{align*} |u(x_1) - u(x_2)| \le 2 \|u\|_{L^\infty(U)} \le 2 C_M \|u\|_{W^{1,p}(U)} \le \frac{8 C_M \|u\|_{W^{1,p}(U)}}{\delta} \cdot |x_1 - x_2|, \end{align*} using $|x_1 - x_2| = r \ge \delta/4$. Set \begin{align*} C_0 := \frac{8 C_M \|u\|_{W^{1,p}(U)}}{\delta}, \end{align*} which depends on $n, p, \delta, \|u\|_{W^{1,p}(U)}$ but is independent of $\lambda$. Combining the two cases, for every $x_1, x_2 \in E_\lambda \cap \overline{Q}$, \begin{align*} |u(x_1) - u(x_2)| \le \max\big(C_n \lambda, \, C_0\big) \cdot |x_1 - x_2| = L_\lambda \cdot |x_1 - x_2|. \end{align*} $\square$ The Lipschitz constant $L_\lambda$ is finite for every $\lambda > 0$. As $\lambda \to \infty$, $L_\lambda = C_n \lambda$ eventually (once $C_n \lambda \ge C_0$). [/step] [step:Extend to a Lipschitz function $v_\lambda$ on $\mathbb{R}^n$ via McShane] If $E_\lambda \cap \overline{Q} = \varnothing$, define $v_\lambda \equiv 0$ on $\mathbb{R}^n$; the next step's claim is then vacuous. Otherwise, by [McShane's Extension Theorem](/theorems/3067), the $L_\lambda$-Lipschitz function $u|_{E_\lambda \cap \overline{Q}}: E_\lambda \cap \overline{Q} \to \mathbb{R}$ extends to \begin{align*} v_\lambda: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \inf_{w \in E_\lambda \cap \overline{Q}} \big( u(w) + L_\lambda |x - w| \big), \end{align*} satisfying $\operatorname{Lip}(v_\lambda) \le L_\lambda$ and $v_\lambda|_{E_\lambda \cap \overline{Q}} = u|_{E_\lambda \cap \overline{Q}}$. By [Rademacher's Theorem](/theorems/3069), $v_\lambda: \mathbb{R}^n \to \mathbb{R}$ is classically differentiable at $\mathcal{L}^n$-a.e. $x \in \mathbb{R}^n$. Let $N_\lambda \subseteq \mathbb{R}^n$ be the set where $v_\lambda$ fails to be classically differentiable; $\mathcal{L}^n(N_\lambda) = 0$. [/step] [step:Set up the good set $G_\lambda$] Define \begin{align*} D_\lambda &:= \{x \in U : x \text{ is a Lebesgue density-1 point of } E_\lambda\}, \\ L_{\nabla u}^{(p)} &:= \big\{x \in U : \fint_{B(x, r)} |\nabla u(y) - \nabla u(x)|^p \, d\mathcal{L}^n(y) \to 0 \text{ as } r \to 0^+ \big\}, \\ G_\lambda &:= \big( (E_\lambda \cap Q) \setminus N_\lambda \big) \cap D_\lambda \cap L_{\nabla u}^{(p)}. \end{align*} By the [Lebesgue Density Theorem](/theorems/3072), $\mathcal{L}^n((E_\lambda \cap Q) \setminus D_\lambda) = 0$. By the [Lebesgue Differentiation Theorem](/theorems/3068) applied to $|\nabla u(\cdot) - c|^p \in L^1_{\mathrm{loc}}(U)$ for each rational $c \in \mathbb{Q}^n$ (a countable union of full-measure sets remains full-measure), $\mathcal{L}^n(U \setminus L_{\nabla u}^{(p)}) = 0$. Combined with $\mathcal{L}^n(N_\lambda) = 0$, \begin{align*} \mathcal{L}^n((E_\lambda \cap Q) \setminus G_\lambda) = 0. \end{align*} [/step] [step:Pointwise discrepancy bound $|u - v_\lambda| \le C_n L_\lambda \operatorname{dist}(\cdot, E_\lambda)$] The transfer of differentiability from $v_\lambda$ to $u$ for *all* $h \to 0$ rests on a uniform pointwise bound on the discrepancy $|u - v_\lambda|$ in terms of the nearest-point distance to $E_\lambda$. Fix $x_0 \in G_\lambda$ and choose \begin{align*} \rho_0 := \min\big( \delta/16, \, \operatorname{dist}(x_0, \partial Q)/4 \big) > 0, \end{align*} so that $B(x_0, 4 \rho_0) \subseteq Q$ and $\rho_0 < \delta/16$. Set \begin{align*} V := \overline{B(x_0, 2 \rho_0)} \subseteq Q. \end{align*} [claim:Pointwise discrepancy estimate] There exists a dimensional constant $C_n^* > 0$ (depending only on $n, p$) such that for every $y \in V$, \begin{align*} |u(y) - v_\lambda(y)| \le C_n^* \cdot L_\lambda \cdot \operatorname{dist}(y, E_\lambda \cap \overline{Q}). \end{align*} [/claim] *Proof of claim.* Fix $y \in V$ and set $s := \operatorname{dist}(y, E_\lambda \cap \overline{Q})$. The set $E_\lambda \cap \overline{Q}$ is closed and non-empty (it contains $x_0$), so the infimum is attained: pick $z \in E_\lambda \cap \overline{Q}$ with $|y - z| = s$. If $s = 0$, then $y \in E_\lambda \cap \overline{Q}$, so $v_\lambda(y) = u(y)$ and the bound holds with both sides equal to zero. Assume $s > 0$, so $y \notin E_\lambda$. Since $y \in V$ and $x_0 \in E_\lambda \cap \overline{Q}$, $s \le |y - x_0| \le 2 \rho_0$, hence $4 s \le 8 \rho_0 < \delta/2$. Both $y$ and $z$ lie in $\overline{Q}$ and $|y - z| = s$, with $\operatorname{dist}(z, \partial U) \ge \delta$, so $B(z, 4 s) \subseteq U$. **Step (i): McShane bound on $|v_\lambda(y) - v_\lambda(z)|$.** Since $v_\lambda$ is $L_\lambda$-Lipschitz and $v_\lambda(z) = u(z)$, \begin{align*} |v_\lambda(y) - u(z)| = |v_\lambda(y) - v_\lambda(z)| \le L_\lambda \cdot s. \end{align*} **Step (ii): Sobolev–Morrey oscillation bound on $|u(y) - u(z)|$.** Since $z \in E_\lambda$, the average of $|\nabla u|^p$ over $B(z, 4 s) \subseteq U$ satisfies \begin{align*} \fint_{B(z, 4 s)} |\nabla u|^p \, d\mathcal{L}^n \le M_p(|\nabla u|)(z)^p \le \lambda^p, \end{align*} hence \begin{align*} \big\| |\nabla u| \big\|_{L^p(B(z, 4 s))} = \Big( \int_{B(z, 4 s)} |\nabla u|^p \, d\mathcal{L}^n \Big)^{1/p} \le \omega_n^{1/p} \cdot (4 s)^{n/p} \cdot \lambda, \end{align*} where $\omega_n$ denotes the volume of the unit ball in $\mathbb{R}^n$. We now invoke [Morrey's Inequality](/theorems/62) in its scaled form on the ball $B(z, 4s)$. On the unit ball $B(0,1)$, Morrey's inequality gives the Hölder seminorm bound \begin{align*} [u]_{C^{0, \alpha}(B(0, 1))} := \sup_{\xi \ne \eta} \frac{|u(\xi) - u(\eta)|}{|\xi - \eta|^\alpha} \le C_M^{(2)}(n, p) \cdot \big\| |\nabla u| \big\|_{L^p(B(0, 1))}, \qquad \alpha := 1 - n/p, \end{align*} with $C_M^{(2)}(n, p)$ the dimensional Morrey constant on the unit ball. By the standard rescaling argument — substitute $u_r(\xi) := u(z + r \xi)$ on $B(0, 1)$, observe $\nabla u_r(\xi) = r \nabla u(z + r \xi)$ and $\|\nabla u_r\|_{L^p(B(0,1))} = r^{1 - n/p} \|\nabla u\|_{L^p(B(z, r))}$, and the Hölder seminorm transforms as $[u_r]_{C^{0,\alpha}(B(0,1))} = r^\alpha [u]_{C^{0,\alpha}(B(z,r))}$ — the inequality applied to $u_r$ rearranges to \begin{align*} [u]_{C^{0, \alpha}(B(z, r))} \le C_M^{(2)}(n, p) \cdot r^{1 - \alpha - n/p} \cdot \big\| |\nabla u| \big\|_{L^p(B(z, r))} = C_M^{(2)}(n, p) \cdot \big\| |\nabla u| \big\|_{L^p(B(z, r))}, \end{align*} since $1 - \alpha - n/p = 1 - (1 - n/p) - n/p = 0$. The radius cancels: the constant $C_M^{(2)}(n, p)$ is independent of $r$. Applied with $r = 4 s$, for any $\xi, \eta \in B(z, 4 s)$, \begin{align*} |u(\xi) - u(\eta)| \le C_M^{(2)}(n, p) \cdot |\xi - \eta|^\alpha \cdot \big\| |\nabla u| \big\|_{L^p(B(z, 4 s))}. \end{align*} Specialising to $\xi = y$ and $\eta = z$ (both in $B(z, 4 s)$ since $|y - z| = s < 4 s$) and substituting the $L^p$-gradient bound from above, \begin{align*} |u(y) - u(z)| \le C_M^{(2)} \cdot s^{\alpha} \cdot \omega_n^{1/p} (4 s)^{n/p} \lambda = C_M^{(2)} \cdot \omega_n^{1/p} \cdot 4^{n/p} \cdot s^{\alpha + n/p} \cdot \lambda = C_n'' \cdot s \cdot \lambda, \end{align*} since $\alpha + n/p = (1 - n/p) + n/p = 1$, with the dimensional constant $C_n'' := C_M^{(2)} \omega_n^{1/p} 4^{n/p}$ depending only on $n, p$. **Step (iii): Combine.** By the triangle inequality, \begin{align*} |u(y) - v_\lambda(y)| \le |u(y) - u(z)| + |u(z) - v_\lambda(y)| \le C_n'' s \lambda + L_\lambda s. \end{align*} Since $L_\lambda \ge C_n \lambda$ (from step 3), $\lambda \le L_\lambda / C_n$, so $C_n'' \lambda \le (C_n''/C_n) L_\lambda$. Setting \begin{align*} C_n^* := \frac{C_n''}{C_n} + 1, \end{align*} a dimensional constant depending only on $n, p$, we obtain \begin{align*} |u(y) - v_\lambda(y)| \le C_n^* \cdot L_\lambda \cdot s = C_n^* \cdot L_\lambda \cdot \operatorname{dist}(y, E_\lambda \cap \overline{Q}). \end{align*} $\square$ The estimate holds at every point of $V$ (no measure-zero exception set), since $u$ is the continuous representative and $v_\lambda$ is continuous (being Lipschitz). [/step] [step:Classical differentiability of $u$ at $x_0$ for *every* $h \to 0$] [claim:Distance-to-$E_\lambda$ vanishes faster than $|h|$] For $x_0 \in G_\lambda \subseteq D_\lambda$, \begin{align*} \operatorname{dist}(x_0 + h, E_\lambda \cap \overline{Q}) = o(|h|) \qquad \text{as } |h| \to 0. \end{align*} [/claim] *Proof of claim.* Suppose not: there exist $\eta_0 > 0$ and a sequence $h_m \to 0$ with $\operatorname{dist}(x_0 + h_m, E_\lambda \cap \overline{Q}) \ge \eta_0 |h_m|$. Then $B(x_0 + h_m, \eta_0 |h_m|/2) \cap E_\lambda = \varnothing$ for $|h_m|$ small enough that $B(x_0 + h_m, \eta_0 |h_m|/2) \subseteq Q$. The ball lies in $B(x_0, |h_m|(1 + \eta_0/2))$ and has volume $\omega_n (\eta_0/2)^n |h_m|^n$, so the density of $U \setminus E_\lambda$ in $B(x_0, |h_m|(1 + \eta_0/2))$ is at least \begin{align*} \frac{\omega_n (\eta_0/2)^n |h_m|^n}{\omega_n (1 + \eta_0/2)^n |h_m|^n} = \Big( \frac{\eta_0/2}{1 + \eta_0/2} \Big)^n > 0, \end{align*} contradicting density $1$ of $E_\lambda$ at $x_0$. $\square$ [claim:Classical differentiability of $u$ at every $x_0 \in G_\lambda$] For every $x_0 \in G_\lambda$, with $\ell_\lambda := \nabla v_\lambda(x_0)$, \begin{align*} u(x_0 + h) - u(x_0) = \ell_\lambda \cdot h + o(|h|) \qquad \text{as } |h| \to 0, \end{align*} where the $o(|h|)$ holds for *every* $h \to 0$ (no density-1 restriction). [/claim] *Proof of claim.* Since $x_0 \in E_\lambda \cap \overline{Q}$, $u(x_0) = v_\lambda(x_0)$. Since $x_0 \notin N_\lambda$, $v_\lambda$ is classically differentiable at $x_0$ with gradient $\ell_\lambda$: \begin{align*} v_\lambda(x_0 + h) - v_\lambda(x_0) = \ell_\lambda \cdot h + o(|h|) \qquad \text{as } |h| \to 0, \end{align*} holding *unconditionally* over all $h \to 0$ — Rademacher's expansion is unconditional, not restricted to a density-1 set. For $|h| < \rho_0$, $x_0 + h \in V$, so by the pointwise discrepancy estimate from the previous step, \begin{align*} |u(x_0 + h) - v_\lambda(x_0 + h)| \le C_n^* \cdot L_\lambda \cdot \operatorname{dist}(x_0 + h, E_\lambda \cap \overline{Q}). \end{align*} By the distance claim, $\operatorname{dist}(x_0 + h, E_\lambda \cap \overline{Q}) = o(|h|)$ as $|h| \to 0$ unconditionally. Hence \begin{align*} |u(x_0 + h) - v_\lambda(x_0 + h)| \le C_n^* L_\lambda \cdot o(|h|) = o(|h|), \end{align*} unconditionally over all $h \to 0$. By the triangle inequality and $u(x_0) = v_\lambda(x_0)$, \begin{align*} u(x_0 + h) - u(x_0) &= \big( u(x_0 + h) - v_\lambda(x_0 + h) \big) + \big( v_\lambda(x_0 + h) - v_\lambda(x_0) \big) \\ &= o(|h|) + \big( \ell_\lambda \cdot h + o(|h|) \big) = \ell_\lambda \cdot h + o(|h|), \end{align*} holding unconditionally over all $h \to 0$. $\square$ This is classical differentiability of $u$ at $x_0$ with gradient $\ell_\lambda$, for *all* $h \to 0$. [/step] [step:Identify $\ell_\lambda = \nabla u(x_0)$ via a directional sequence in $E_\lambda$] It remains to identify $\ell_\lambda = \nabla v_\lambda(x_0)$ with the precise representative $\nabla u(x_0)$. [claim:Identification $\ell_\lambda = \nabla u(x_0)$] For every $x_0 \in G_\lambda$, $\ell_\lambda = \nabla u(x_0)$. [/claim] *Proof of claim.* Fix any unit vector $e \in S^{n-1}$. We will show $(\ell_\lambda - \nabla u(x_0)) \cdot e = 0$. Since $e$ is arbitrary, this forces $\ell_\lambda = \nabla u(x_0)$. *Construction of a sequence $h_m \in E_\lambda - x_0$ with $h_m / |h_m| \to e$.* By density 1 of $E_\lambda$ at $x_0$, for every cone \begin{align*} C_\theta(e) := \{ h \in \mathbb{R}^n \setminus \{0\} : h \cdot e \ge (1 - \theta) |h| \} \end{align*} of half-aperture parameter $\theta \in (0, 1)$, \begin{align*} \frac{\mathcal{L}^n(\{ h \in C_\theta(e) \cap B(0, r) : x_0 + h \in E_\lambda \})}{\mathcal{L}^n(C_\theta(e) \cap B(0, r))} \to 1 \qquad \text{as } r \to 0^+. \end{align*} In particular, the density-1 set $\{h : x_0 + h \in E_\lambda\}$ intersects $C_\theta(e) \cap B(0, r)$ for every $\theta > 0$ and every $r$ small enough. Pick $r_m \to 0^+$ and $\theta_m \to 0^+$, and choose $h_m$ with $x_0 + h_m \in E_\lambda$, $h_m \in C_{\theta_m}(e)$, and $|h_m| \le r_m$. Then $h_m \to 0$ and $h_m / |h_m| \to e$. *Apply the previous step's expansion at $h_m$:* \begin{align*} u(x_0 + h_m) - u(x_0) = \ell_\lambda \cdot h_m + o(|h_m|). \tag{$*$} \end{align*} *Affine Lebesgue-point expansion at $h_m$.* Define the affine remainder \begin{align*} \widetilde u(y) := u(y) - \nabla u(x_0) \cdot (y - x_0). \end{align*} Then $\nabla \widetilde u(y) = \nabla u(y) - \nabla u(x_0)$ in the weak sense, and $\widetilde u \in W^{1, p}(U)$. Set $w_m := x_0 + h_m$ and $r_m := |h_m|$. We bound $|\widetilde u(w_m) - \widetilde u(x_0)|$ via the Sobolev–Morrey oscillation estimate on the ball $B(x_0, 3 r_m) \subseteq U$ (which contains both $x_0$ and $w_m$, since $|w_m - x_0| = r_m < 3 r_m$ and $3 r_m < 3 \rho_0 < \delta$). The strategy combines [Morrey's Inequality](/theorems/62) — converting an $L^p$ bound on $\nabla \widetilde u$ to a pointwise Hölder bound — with the *$L^p$-Lebesgue-point* property of $\nabla u$ at $x_0$, which provides the $L^p$ decay on shrinking balls. By [Morrey's Inequality](/theorems/62) applied to $\widetilde u \in W^{1, p}(B(x_0, 3 r_m))$ in the same scaled form as in step 6 (the dimensional Morrey constant $C_M^{(2)}(n, p)$ is independent of the radius), for any $\xi, \eta \in B(x_0, 3 r_m)$, \begin{align*} |\widetilde u(\xi) - \widetilde u(\eta)| \le C_M^{(2)} |\xi - \eta|^{1 - n/p} \big\| |\nabla \widetilde u| \big\|_{L^p(B(x_0, 3 r_m))}. \end{align*} Apply with $\xi = w_m$ and $\eta = x_0$: since $|w_m - x_0| = r_m$ and both points lie in $B(x_0, 3 r_m)$, \begin{align*} |\widetilde u(w_m) - \widetilde u(x_0)| \le C_M^{(2)} \cdot r_m^{1 - n/p} \cdot \big\| |\nabla \widetilde u| \big\|_{L^p(B(x_0, 3 r_m))}. \end{align*} Since $x_0 \in L_{\nabla u}^{(p)}$ (by the good-set definition in step 5), \begin{align*} \fint_{B(x_0, r)} |\nabla u(y) - \nabla u(x_0)|^p \, d\mathcal{L}^n(y) \to 0 \qquad \text{as } r \to 0^+. \end{align*} Define \begin{align*} \omega_{x_0}^{(p)}(r) := \sup_{0 < t \le 4 r} \Big( \fint_{B(x_0, t)} |\nabla u(y) - \nabla u(x_0)|^p \, d\mathcal{L}^n(y) \Big)^{1/p}; \end{align*} then $\omega_{x_0}^{(p)}(r) \to 0$ as $r \to 0^+$. Substituting, \begin{align*} \big\| |\nabla \widetilde u| \big\|_{L^p(B(x_0, 3 r_m))}^p = \int_{B(x_0, 3 r_m)} |\nabla u - \nabla u(x_0)|^p \, d\mathcal{L}^n = \omega_n (3 r_m)^n \cdot \fint_{B(x_0, 3 r_m)} |\nabla u - \nabla u(x_0)|^p \, d\mathcal{L}^n, \end{align*} hence \begin{align*} \big\| |\nabla \widetilde u| \big\|_{L^p(B(x_0, 3 r_m))} \le \omega_n^{1/p} (3 r_m)^{n/p} \cdot \omega_{x_0}^{(p)}(r_m). \end{align*} Plugging into the Morrey bound, \begin{align*} |\widetilde u(w_m) - \widetilde u(x_0)| \le C_M^{(2)} \cdot r_m^{1 - n/p} \cdot \omega_n^{1/p} (3 r_m)^{n/p} \cdot \omega_{x_0}^{(p)}(r_m) = C_n''' \cdot r_m \cdot \omega_{x_0}^{(p)}(r_m), \end{align*} combining the powers $r_m^{1 - n/p} \cdot r_m^{n/p} = r_m$ and absorbing $C_M^{(2)} \omega_n^{1/p} 3^{n/p}$ into the dimensional constant $C_n'''$. Since $\omega_{x_0}^{(p)}(r_m) \to 0$, \begin{align*} |\widetilde u(w_m) - \widetilde u(x_0)| = o(r_m) = o(|h_m|). \end{align*} Translating back via $\widetilde u(w_m) - \widetilde u(x_0) = u(x_0 + h_m) - u(x_0) - \nabla u(x_0) \cdot h_m$, \begin{align*} u(x_0 + h_m) - u(x_0) = \nabla u(x_0) \cdot h_m + o(|h_m|). \tag{$**$} \end{align*} *Subtract.* From $(*)$ and $(**)$, \begin{align*} \ell_\lambda \cdot h_m + o(|h_m|) = \nabla u(x_0) \cdot h_m + o(|h_m|), \end{align*} so $(\ell_\lambda - \nabla u(x_0)) \cdot h_m / |h_m| \to 0$ as $m \to \infty$. Since $h_m / |h_m| \to e$, \begin{align*} (\ell_\lambda - \nabla u(x_0)) \cdot e = 0. \end{align*} Since $e \in S^{n-1}$ was arbitrary, $\ell_\lambda = \nabla u(x_0)$. $\square$ Combining with the differentiability claim of the previous step, for every $x_0 \in G_\lambda$, \begin{align*} u(x_0 + h) - u(x_0) - \nabla u(x_0) \cdot h = o(|h|) \qquad \text{as } |h| \to 0, \end{align*} unconditionally over all $h \to 0$. This is classical differentiability of $u$ at $x_0$ with classical gradient equal to the precise representative $\nabla u(x_0)$. [/step] [step:Assemble the full-measure set of classical differentiability] By the previous steps, for each $\lambda > 0$ there is a set $G_\lambda \subseteq E_\lambda \cap Q$ of full measure in $E_\lambda \cap Q$ on which $u$ is classically differentiable for all $h \to 0$ with classical gradient $\nabla u(x_0)$. Define \begin{align*} G := \bigcup_{m = 1}^\infty G_m \subseteq Q. \end{align*} Since $\bigcup_m E_m \cap Q$ has full measure in $Q$ (by the level-set step) and $G_m$ has full measure in $E_m \cap Q$, the set $G$ has full measure in $Q$: \begin{align*} \mathcal{L}^n(Q \setminus G) \le \mathcal{L}^n\Big(Q \setminus \bigcup_m (E_m \cap Q)\Big) + \sum_m \mathcal{L}^n((E_m \cap Q) \setminus G_m) = 0 + 0 = 0. \end{align*} At every $x_0 \in G$, $u$ is classically differentiable for all $h \to 0$ with classical gradient $\nabla u(x_0)$. Repeating this argument on each cube $Q_k$ in the countable cover of $\Omega$ and taking the union of the resulting full-measure subsets produces a full-measure subset of $\Omega$ at every point of which $u$ is classically differentiable with classical gradient equal to the precise representative of the weak gradient. The proof is complete. [/step]