In an attempt to solve this question, I have encountered a hint that appeals to Lebesgue’s differentiation theorem. I'm trying to fill in the detail of the proof sketch given in a lecture note.
Lebesgue’s Differentiation Let $\Omega$ be an open subset of $\mathbb R^n$ and $f \in L_1(\Omega)$. Then for a.e. $x \in \Omega$, we have $$ f(x) = \lim _{r \searrow 0} \frac{1}{|B(x, r)|} \int_{B(x, r)} f(y) \mathrm d y. $$ Here $B(x, r)$ is the open ball of radius $r$ centred at $x$, and $|E|$ denotes the $n$-dimensional Lebesgue measure of $E \subset \mathbb R^n$.
Could you have a check on my attempt?
My attempt: We define the Hardy–Littlewood maximal function $Mf: \Omega \to [0, \infty]$ of $f$ by $$ (M f)(x) :=\sup _{r>0} \frac{1}{|B(x, r)|} \int_{B(x, r)} |f(y)| \mathrm d y \quad \forall x \in \Omega. $$
Then $Mf$ is Borel measurable.
Lemma: Let $A_t := \left\{x \in \Omega \mid (M f)(x)>t\right\}$. Then $$ \left|A_t\right| \leq \frac{3^n}{t} \int_{\Omega} |f| \quad \forall t>0. $$
Proof: For each $x \in A_t$, there is $r_x>0$ such that $$ t < \frac{1}{|B(x, r_x)|} \int_{B(x, r_x)} |f| \quad \text{or} \quad |B(x, r_x)| < \frac{1}{t} \int_{B(x, r_x)} |f|. $$ Let $K$ be a compact subset of $A_t$. Clearly, $\{B(x, r_{x})\}_{x \in A_t}$ covers $K$. So there is a finite subcover. By Vitali covering lemma, there is a disjoint finite subfamily $\left\{B\left(x_i, r_i\right)\right\}_{i=1}^k$ such that $K \subset \bigcup_{i=1}^k B\left(x_i, 3 r_i\right)$. Hence $$ |K| \le \sum_{i=1}^k |B\left(x_i, 3 r_i\right)| \le 3^n\sum_{i=1}^k |B\left(x_i, r_i\right)| \le \frac{3^n}{t} \sum_{i=1}^k \int_{B(x_i, r_i)} |f| \le \frac{3^n}{t} \int_\Omega |f|. $$
Because Lebesgue measure is tight, we have $$ |A_t| \le \frac{3^n}{t} \int_\Omega |f|. $$
It suffices to prove that for a.e. $x \in \Omega$, we have $$ \limsup_{r \searrow 0} \frac{1}{|B(x, r)|} \int_{B(x, r)} |f - f(x)| = 0. $$
Now we fix $g \in C_b(\Omega)$. Then $$ \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |g - g(x)| = 0. $$
As such, $$ \begin{align} & \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f - f(x)| \\ \le & \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |g - g(x)| + \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f - g| + |f(x) - g(x)| \\ = & \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f - g| + |f(x) - g(x)| \\ \le & M (|f-g|) (x) + |f(x) - g(x)|. \end{align} $$
Now we fix $\varepsilon>0$. Then $$ \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f - f(x)| > \varepsilon \implies \left [ M (|f-g|) (x) > \frac{\varepsilon}{2} \quad \text{or} \quad |f(x) - g(x)| > \frac{\varepsilon}{2} \right ]. $$
By Tshebyshev's inequality, $$ \left | \left \{ x \in \Omega \,\middle\vert\, |f(x) - g(x)| > \frac{\varepsilon}{2} \right \}\right | \le \frac{2}{\varepsilon} \int_\Omega |f-g|. $$
By above Lemma, $$ \left | \left \{ x \in \Omega \,\middle\vert\, M (|f-g|) (x) > \frac{\varepsilon}{2} \right \}\right | \le \frac{2\cdot 3^n}{\varepsilon} \int_{\Omega} |f-g|. $$
So there is some constant $C>0$ (that depends only on $n$) such that $$ \left | \left \{ x \in \Omega \,\middle\vert\, \limsup_{r \searrow 0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f - f(x)| > \varepsilon \right \}\right | \le \frac{C}{\varepsilon} \int_{\Omega} |f-g|. $$
Because $C_b(\Omega)$ is dense in $L_1(\Omega)$, for a fixed $\varepsilon>0$ we have that $$ \frac{C}{\varepsilon} \int_{\Omega} |f-g| $$ can be arbitrarily small. This completes the proof.
