Measurability of derivative of Lebesgue integral function

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.

$\Phi(x)$ is an absolutely continuous function of $x$ on $[a,b]$ so it is Borel-measurable.
• Thank you so much! Why is $\Phi'$ measurable too? I heartily thank you again! – Self-teaching worker Oct 28 '14 at 17:41
• Every absolutely continuous function on $[a,b]$ is of bounded variation on $[a,b]$ so it has (measurable) derivative a.e. on $[a,b]$. – Mustafa Said Oct 28 '14 at 18:11
• 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!!! – Self-teaching worker Oct 30 '14 at 15:36