We consider $h>0$ and $f \in L^1(\mathbb{R})$. We note $F(x) = \int_0^x f(t) dt$ for $x \in \mathbb{R}$.
We have $F_h = \int_{\mathbb{R}} f(t) \phi(x-t)dt$ with $\phi = 1_{[-h,0]}$.
I have to prove that $\dfrac{F_h(x)}{h} \rightarrow_{h \rightarrow 0^+} f$ in $L^1$.
I rewrote this as $||\dfrac{F_h(x)}{h}-f||_1 \rightarrow 0$, then $||f * \dfrac{\phi_h}{h}-f||_1 \rightarrow 0$. But I don't know how to prove this. Could someone help me ?
Let $\varphi=\chi_{[-1,0]}$, then one can check that $\{\varphi_{h}\}_{h>0}$ is an approximate identity, where $\varphi_{h}(\cdot)=h^{-1}\varphi(\cdot/h)$.
And $F_{h}=\varphi_{h}\ast f$, and $F_{h}\rightarrow f$ in $L^{1}$ is a consequence of $\{\varphi_{h}\}_{h>0}$ being an approximate identity, you may check this assertion in any harmonic analysis book.