If $f\in L^1$, then $F(x):=\int_0^x{f(s)ds}$ is also in $L^1$

Question

Let $\Omega=]0,1[$ and $f\in L^1(\Omega)$, show that
$$F(x):=\int_0^x{f(s)ds}, \forall x \in\Omega$$ is also in $L^1(\Omega).$

I was working with this exercise and would appreciate some ideas/ways to solve this.

My idea:

Notice: $$\int_0^1{\lvert f\rvert} < \infty$$ Then $\forall x\in\Omega$:$$\int_0^x{\lvert f(s)\rvert ds} < \infty$$

We can apply Fubini-Tonelli for: $$\int_0^1 \left( \int_0^x{\lvert f(s)\rvert ds}\right) dx $$

which is where I got stuck. Thanks.

Answer 1

To complete your argument note that (by Tonnelli's Theorem) $\int_0^{1}\int_0^{1}\chi_{(0,x)}(s) |f(s)|dsdx=\int_0^{1}\int_0^{1}\chi_{(s,1)}(x) |f(s)|dxds$ because $\chi_{(0,x)}(s)=\chi_{(s,1)}(x)$ for all $x$ and $s$. Hence $\int_0^{1}\int_0^{1}\chi_{(0,x)}(s) |f(s)|dsdx= \int_0^{1} (1-s)|f(s)|ds \leq \int_0^{1} |f(s)| ds <\infty$.

Answer 2

Hints: show $F$ is continuous by noting that

$1).\ \mathbb 1_{\{|f| > \lambda\}}|f| \le |f|$ and so $\underset{\lambda \to\infty}\lim\int_{\{|f| > \lambda\}}|f|\ d\mu = 0$

$2).\ \int_{x_0}^{x_0+h}|f|\ d\mu = \int_{[x_0,x_0+h] \cap \{|f| > \lambda\}}|f|\ d\mu + \int_{[x_0,x_0+h] \cap \{|f| \le \lambda\}}|f|\ d\mu$

$3).\ $ With $\delta <\frac{\epsilon}{2\lambda}$, choose $h$ so that $m([x_0,x_0+h])<\delta$

$4).\ $ Combine $2).$ and $3).$