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.