The full problem reads:

Prove that if $f$ is absolutely continuous on $[0,1]$ and $g$ is continuous on $[0,1]$ such that $f'=g$ a.e., then $f$ is differentiable on $[0,1]$ and $f'=g$.

My analysis skills are very rusty and I'm having a hard time seeing how to prove this. Thanks in advance for any advice!


1 Answer 1


By Lebesgue's fundamental theorem of calculus, $$f(x)=f(0)+\int_0^xf'. $$ By hypothesis, $$f(x)=f(0)+\int_0^x g.$$ By the standard fundamental theorem of calculus, $f'(x)=g(x)$ for all $x$.

  • $\begingroup$ That's correct, but you should fill in a few details to make it useful for the OP and other users. $\endgroup$
    – user436658
    Oct 9, 2017 at 18:29
  • 1
    $\begingroup$ @ProfessorVector What details? $\endgroup$
    – Aloizio Macedo
    Oct 9, 2017 at 18:31
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    $\begingroup$ The link Aloizio attaches gives any details needed I believe $\endgroup$ Oct 9, 2017 at 21:15
  • $\begingroup$ I think the proof is OK without any further details needed. Since $g$ is continuous its Lebesgue integral can be treated as a Riemann integral and then $f'=g$ everywhere in $[0, 1] $. $\endgroup$
    – Paramanand Singh
    Oct 10, 2017 at 8:42

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