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I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа that if $f:[a,b]\to\mathbb{R}$ is a Lebesgue-summable function on its domain then the derivative $\Phi'$ of the integral function, defined for any $x\in[a,b]$ by$$\Phi(x)=\int_{[a,x]}f d\mu$$where the Lebesgue integral is intented to be calculated according to the "canonical" linear Lebesgue measure, is measurable.

I should specify that I cannot use the fact that $\Phi'(x)=f(x)$ because that equality is proven by Kolmogorov-Fomin's by using the very fact that $\Phi'$ is measurable. I thank anybody for any clarification.

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$\Phi(x)$ is an absolutely continuous function of $x$ on $[a,b]$ so it is Borel-measurable.

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  • $\begingroup$ Thank you so much! Why is $\Phi'$ measurable too? I heartily thank you again! $\endgroup$ Oct 28, 2014 at 17:41
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    $\begingroup$ Every absolutely continuous function on $[a,b]$ is of bounded variation on $[a,b]$ so it has (measurable) derivative a.e. on $[a,b]$. $\endgroup$ Oct 28, 2014 at 18:11
  • $\begingroup$ I've reached the point where Kolmogorov-Fomin's deals with absolute continuity and, after studying several preliminary lemmas, I've realised that the text does show this property without using the fact that $\Phi'(x)=f(x)$. Thank you so much again!!! $\endgroup$ Oct 30, 2014 at 15:36
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    $\begingroup$ @DavideZena you are very welcome. $\endgroup$ Oct 30, 2014 at 18:56

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