How to show that $\lim_{h\to 0}\int_0^h|f(x)|dx=0.$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a locally integrable function. How can we see that $$\lim_{h\to 0}\int_0^h|f(x)|dx=0.$$ If $f$ is bounded, then we have the result. But what about $f$ only locally integrable ?

Answer 1

$f\in L^1_{loc}$, hence $\int_0^1 |f|<+\infty$. Let $f_h:=\chi_{[0,h]}f$, where $\chi_E(x)=1$ is $x\in E$, $0$ otherwise.

For any sequence $h_n\to 0$, $f_{h_n}\to 0$ a.e., and $|f_{h_n}|\leq |f_{|[0,1]}|$ for any $h$, hence dominated convergence gives $$\int_0^1 |f_{h_n}| dx= \int_0^{h_n} |f| dx\to 0.$$ The arbitrariness of $(h_n)$ concludes the proof.