I must show that
Let $f$ be Lebesgueable integrable, then
$$\lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x \in \mathbb{R}$.
Tentative proof:
$$|f(x) - f(y)| \geq 0 \Rightarrow \lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy \geq 0$$
We must show that
$$ \lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy \leq 0$$
I will show that for any $\epsilon >0$
$$ \lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy < \epsilon$$
If $f(x) = 0$, By absolute continuity of integration, for any $\epsilon >0$, $\exists \delta >0$ such that if
$$ m({|x-y| < h}) < \delta \Rightarrow \int_{|x -y| < h}|f(y)|dy <\frac{\epsilon}{2} $$
Now $$ \lim_{h\to 0}\sup m({|x-y| < h}) \leq m({|x-y| < h}) < \delta$$
then
$$ \lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy < \frac{\epsilon}{2} < \epsilon $$
Now for $f(x) \neq 0$,
$$\int_{|x -y| < h}|f(x) - f(y)|dy \leq \int_{|x -y| < h}|f(x)|dy + \int_{|x -y| < h}|f(y)|dy $$
$$= |f(x)|m({|x -y| < h}) + \int_{|x -y| < h}|f(y)|dy $$
and $|f(x)| < \infty$ for allmost all $x \in \mathbb{R}$, since $f$ is Lebesgue integrable.
Let $\delta^{*} < \min \{\delta,\frac{\epsilon}{2|f(x)|}\}$, now by absolute continuiuty of integration, for $\epsilon > 0$, $$m({|x -y| < h}) < \delta^{*} < \delta \Rightarrow \int_{|x -y| < h}|f(x) - f(y)|dy \leq |f(x)|m({|x -y| < h}) + \int_{|x -y| < h}|f(y)|dy < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
For almost every $x \in \mathbb{R}$. Hence,
$$ \lim_{h\to 0}\sup m({|x-y| < h}) \leq m({|x-y| < h}) < \delta^{*} < \delta \Rightarrow \lim_{h\to 0}\sup\int_{|x -y| < h}|f(x) - f(y)|dy < \epsilon $$
Therefore,
$$\lim_{h\to 0}\sup \int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x \in \mathbb{R}$.