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Problem : Let $f:[0,1]\to \Bbb R$ is of bounded variation. Prove that there exists $g:[0,1]\to \Bbb R$ which absolutely continuous and $g'(x)=f'(x)$ for almost every $x \in [0,1]$.

I have tried to introduce $f$ as a difference of 2 monotonic functions but I'm getting stuck. I will be very happy is someone can help me with that question

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    $\begingroup$ This is just a restatement of Lebesgue Decomposition Theorem. $\endgroup$ Mar 13, 2021 at 11:48
  • $\begingroup$ Can you explain more to me? $\endgroup$ Mar 13, 2021 at 12:02

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Hint: Prove that $f'$ is Lebesgue integrable, and then consider its indefinite (Lebesgue) integral. For the former, your approach is useful.

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