Let $f:I\rightarrow\mathbb{R}$ be a continuous function which derivative exists a.e. and $\int f'$ exists in I. Also, suppose $f$ satisfies the Luzin N property. Show that $f$ is absolutely continuous.

So I don't seem to find any way to tackle this problem. This seems to be covered everywhere when the derivative is defined everywhere, but no such case when it's only a.e.

What I know is that using bounded variation it's possible to get absolute continuity, however I can't find how to prove the bounded variation condition.

Is there a reference or a way to solve this problem?


1 Answer 1


See 2. here. The theorem that's being proven assumes bv, but that condition is only used to establish that $f$ has a derivative a.e. that is integrable, which are exactly the conditions you have.

  • $\begingroup$ Although the claim of looking for BV is true. However, your next claim is false, consider Cantor Lebesgue function, which is not of BV. $\endgroup$
    – Rono
    Dec 13, 2016 at 23:31
  • $\begingroup$ I did have a false claim which is that my expression for the variation assumes that we already have absolute continuity. The Cantor Lebesgue function is bv (it's monotone) but it's not Luzin (it maps the Cantor set to $[0,1]$). $\endgroup$
    – user293794
    Dec 14, 2016 at 0:14

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