 # Antiderivative of lebesgue integrable function Let f : [ 0 , b ] &#x2192;< ntaraxq 2022-07-04 Answered
Antiderivative of lebesgue integrable function
Let $f:\left[0,b\right]\to \mathbb{R}$ be Lebesgue integrable. We define
$g\left(x\right)={\int }_{x}^{b}\frac{f\left(t\right)}{t}dt,\phantom{\rule{1em}{0ex}}0
Show that g(x) is Lebesgue integrable in [0,b]. Also show that
${\int }_{0}^{b}g\left(x\right)dx={\int }_{0}^{b}f\left(t\right)dt.$
If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that,
${\int }_{0}^{b}\left[{g}^{\prime }\left(x\right)-\frac{f\left(x\right)}{x}\right]dx=0$, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.
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Explanation:
Let $\mathrm{\Delta }=\left\{\left(x,t\right)\mid 0\le x\le t\le b\right\}$. Define h(x,t) on $\left[0,b\right]×\left[0,b\right]$ by $h\left(x,t\right)={\chi }_{\mathrm{\Delta }}\left(x,t\right)\frac{f\left(t\right)}{t}$. Apply Fubini's theorem to h to integrate it in two different ways. (There will be an initial step involving |h| to prove integrability of h.)
Edit: $\mathrm{\Delta }$ would more accurately be changed in order to be a subset of $\left(0,b\right]×\left(0,b\right]$

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