[proofplan]
We show that the Lebesgue point condition at $x$ implies approximate continuity at $x$ by applying the Markov inequality to convert the integral condition into a density condition. Specifically, for any $\varepsilon > 0$, the set $E_\varepsilon = \{y \in B(x,r) : |f(y) - f(x)| \geq \varepsilon\}$ has $\mathcal{L}^n$-measure bounded by $\varepsilon^{-1}$ times the integral of $|f(\cdot) - f(x)|$ over $B(x,r)$. Since $x$ is a Lebesgue point, this integral (normalized by the ball volume) tends to zero, forcing the density of $E_\varepsilon$ at $x$ to be zero. The "in particular" statement follows from [Almost Every Point Is a Lebesgue Point](/theorems/3032).
[/proofplan]
[step:Apply the Markov inequality to bound the measure of the deviation set]
Fix $\varepsilon > 0$ and $r > 0$. Define the level set
\begin{align*}
E_\varepsilon(r) := \{y \in B(x,r) : |f(y) - f(x)| \geq \varepsilon\}.
\end{align*}
Since $|f(y) - f(x)| \geq \varepsilon$ on $E_\varepsilon(r)$, the Markov inequality gives
\begin{align*}
\varepsilon \cdot \mathcal{L}^n(E_\varepsilon(r)) \leq \int_{E_\varepsilon(r)} |f(y) - f(x)| \, d\mathcal{L}^n(y) \leq \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y),
\end{align*}
where the second inequality enlarges the domain of integration from $E_\varepsilon(r) \subset B(x,r)$ to $B(x,r)$, which is valid since the integrand $|f(y) - f(x)|$ is non-negative. Dividing both sides by $\varepsilon \cdot \mathcal{L}^n(B(x,r))$:
\begin{align*}
\frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x,r))} \leq \frac{1}{\varepsilon} \cdot \frac{1}{\mathcal{L}^n(B(x,r))} \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y).
\end{align*}
[guided]
The Lebesgue point condition controls the integral average of $|f(y) - f(x)|$, while approximate continuity requires controlling the measure of the set where $|f(y) - f(x)|$ is large. These two kinds of information are connected by the Markov inequality (also called Chebyshev's inequality in its first-moment form): for a non-negative measurable function $g$ and a threshold $\varepsilon > 0$,
\begin{align*}
\mathcal{L}^n(\{y \in B(x,r) : g(y) \geq \varepsilon\}) \leq \frac{1}{\varepsilon} \int_{B(x,r)} g(y) \, d\mathcal{L}^n(y).
\end{align*}
We apply this with $g(y) = |f(y) - f(x)|$. The set $\{y \in B(x,r) : g(y) \geq \varepsilon\}$ is precisely $E_\varepsilon(r)$, and we obtain
\begin{align*}
\mathcal{L}^n(E_\varepsilon(r)) \leq \frac{1}{\varepsilon} \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y).
\end{align*}
Dividing both sides by $\mathcal{L}^n(B(x,r)) > 0$ gives the density bound
\begin{align*}
\frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x,r))} \leq \frac{1}{\varepsilon} \cdot \frac{1}{\mathcal{L}^n(B(x,r))} \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y).
\end{align*}
The right-hand side is $\varepsilon^{-1}$ times the Lebesgue point average, which we can now send to zero.
[/guided]
[/step]
[step:Send $r \to 0^+$ using the Lebesgue point hypothesis to conclude density zero]
Since $x$ is a Lebesgue point of $f$, by definition
\begin{align*}
\lim_{r \to 0^+} \frac{1}{\mathcal{L}^n(B(x,r))} \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y) = 0.
\end{align*}
Combining with the bound from the previous step:
\begin{align*}
0 \leq \lim_{r \to 0^+} \frac{\mathcal{L}^n(E_\varepsilon(r))}{\mathcal{L}^n(B(x,r))} \leq \frac{1}{\varepsilon} \cdot \lim_{r \to 0^+} \frac{1}{\mathcal{L}^n(B(x,r))} \int_{B(x,r)} |f(y) - f(x)| \, d\mathcal{L}^n(y) = 0.
\end{align*}
Therefore, for every $\varepsilon > 0$,
\begin{align*}
\lim_{r \to 0^+} \frac{\mathcal{L}^n(\{y \in B(x,r) : |f(y) - f(x)| \geq \varepsilon\})}{\mathcal{L}^n(B(x,r))} = 0.
\end{align*}
This is precisely the condition that $f$ is approximately continuous at $x$: the set $\{y : |f(y) - f(x)| \geq \varepsilon\}$ has density zero at $x$ for every $\varepsilon > 0$.
[/step]
[step:Deduce approximate continuity at $\mathcal{L}^n$-almost every point]
By the theorem [Almost Every Point Is a Lebesgue Point](/theorems/3032), the set of non-Lebesgue points of $f$ has $\mathcal{L}^n$-measure zero. Since we have shown that every Lebesgue point of $f$ is a point of approximate continuity of $f$, it follows that $f$ is approximately continuous at $\mathcal{L}^n$-almost every point of $\mathbb{R}^n$.
[/step]
