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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.

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    $\begingroup$ $F$ is not only in $L^1$, its bounded. $\endgroup$ – Arctic Char Oct 29 '20 at 21:34
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    $\begingroup$ Indeed, thanks to the dominated convergence theorem applied to the function, your function $F$ is continuous on $\Omega$ (which is bounded) so that this is $L^1$ ! $\endgroup$ – L-E Oct 29 '20 at 21:57
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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 $\int_0^{1}\int_0^{1}\chi_{(0,x)}(s) |f(s)|dsdx= \int_0^{1} (1-s)|f(s)|ds \leq \int_0^{1} |f(s)| ds <\infty$.

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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 $3).$

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