[proofplan]
The proof is a direct application of Markov's inequality to the $L^p$-difference quotient. Given that $u$ is $L^p$-differentiable at $x$ with derivative $L$, we set up the level set
\begin{align*}
E_\varepsilon(r) := \left\{ y \in B(x, r) : \frac{|u(y) - u(x) - L(y - x)|}{|y - x|} > \varepsilon \right\}
\end{align*}
and bound its relative measure by the $p$-th power of the difference-quotient average. Since the average tends to zero by hypothesis, the relative measure of $E_\varepsilon(r)$ tends to zero, which is precisely the definition of approximate differentiability with derivative $L$.
[/proofplan]
[step:Recall the hypothesis and the goal]
The hypothesis is that $u: \Omega \to \mathbb{R}$ is $L^p$-differentiable at $x \in \Omega$ with derivative $L \in (\mathbb{R}^n)^*$ (a linear functional on $\mathbb{R}^n$, written here as $L(y-x) = \ell \cdot (y-x)$ for some $\ell \in \mathbb{R}^n$): there exists $r_0 > 0$ with $\overline{B}(x, r_0) \subseteq \Omega$ such that
\begin{align*}
\Phi_p(x, r) := \fint_{B(x, r)} \left( \frac{|u(y) - u(x) - L(y-x)|}{|y - x|} \right)^p \, d\mathcal{L}^n(y) \to 0 \qquad (r \to 0^+).
\end{align*}
The integrand is well-defined a.e. on $B(x, r)$ because $\mathcal{L}^n(\{x\}) = 0$.
The goal is to show that $u$ is approximately differentiable at $x$ with derivative $L$, namely:
\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*}
where
\begin{align*}
E_\varepsilon(r) := \left\{ y \in B(x, r) \setminus \{x\} : \frac{|u(y) - u(x) - L(y - x)|}{|y - x|} > \varepsilon \right\}.
\end{align*}
The set $E_\varepsilon(r)$ is measurable since $u$ is measurable and the function $y \mapsto |u(y) - u(x) - L(y-x)|/|y-x|$ is measurable on $B(x, r) \setminus \{x\}$.
[/step]
[step:Apply Markov's inequality with exponent $p$]
Define the auxiliary function
\begin{align*}
\Psi: B(x, r) \setminus \{x\} &\to [0, \infty) \\
y &\mapsto \frac{|u(y) - u(x) - L(y - x)|}{|y - x|}.
\end{align*}
Then $E_\varepsilon(r) = \{y \in B(x, r) \setminus \{x\} : \Psi(y) > \varepsilon\}$.
Markov's inequality (the elementary bound: for any non-negative measurable function $\Psi$, any $\lambda > 0$, and any $q \ge 1$, $\mu(\{\Psi > \lambda\}) \le \lambda^{-q} \int \Psi^q \, d\mu$, valid because $\mathbb{1}_{\{\Psi > \lambda\}} \le \Psi^q / \lambda^q$ pointwise on $\{\Psi > \lambda\}$ and the inequality $0 \le 0$ holds on its complement) applied with $q = p$, $\lambda = \varepsilon$, $\mu = \mathcal{L}^n \lfloor B(x, r)$:
\begin{align*}
\mathcal{L}^n(E_\varepsilon(r)) \le \frac{1}{\varepsilon^p} \int_{B(x, r) \setminus \{x\}} \Psi(y)^p \, d\mathcal{L}^n(y) = \frac{1}{\varepsilon^p} \int_{B(x, r)} \frac{|u(y) - u(x) - L(y-x)|^p}{|y - x|^p} \, d\mathcal{L}^n(y),
\end{align*}
where the last equality holds because $\mathcal{L}^n(\{x\}) = 0$.
Dividing both sides by $\mathcal{L}^n(B(x, r))$:
\begin{align*}
\frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x, r))} \le \frac{1}{\varepsilon^p} \cdot \fint_{B(x, r)} \frac{|u(y) - u(x) - L(y-x)|^p}{|y - x|^p} \, d\mathcal{L}^n(y) = \frac{\Phi_p(x, r)}{\varepsilon^p}.
\end{align*}
[/step]
[step:Pass to the limit $r \to 0^+$]
By the $L^p$-differentiability hypothesis, $\Phi_p(x, r) \to 0$ as $r \to 0^+$. Hence for every fixed $\varepsilon > 0$,
\begin{align*}
\lim_{r \to 0^+} \frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x, r))} \le \lim_{r \to 0^+} \frac{\Phi_p(x, r)}{\varepsilon^p} = 0.
\end{align*}
Since the relative measure is non-negative, the limit equals zero.
This is the definition of approximate differentiability of $u$ at $x$ with approximate derivative $L$. The proof is complete.
[/step]
