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I've tried but I cannot tell whether the following is true or not. Let $f:[0,1]\rightarrow \mathbb{R}$ be a nondecreasing and continuous function. Is it true that I can find a Lebesgue integrable function $h$ such that $$ f(x)=f(0)+\int_{0}^{x}h(x)dx $$ such that $f'=h$ almost everywhere?

Any hint on how to proceed is really appreciated it!

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@Andrew : The whole point is that we can't assume it is. We use the hypothesis to show that it is differentiable almost everywhere. –  Patrick Da Silva Jul 21 '12 at 4:56
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@Cristian: If the above formula holds, then you must have $f'(x) = h(x)$ at every Lebesgue point of $f$, which is a.e. Since $f'(x) = 0$ a.e., we have that $h$ is essentially zero. But the Cantor function is not constant. Absolute continuity is required. –  copper.hat Jul 21 '12 at 6:13
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For example, see Rudin's "Real & Complex Analysis" Theorem 7.11 and Section 7.16 for details. –  copper.hat Jul 21 '12 at 6:18
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thanks @cooper. You should put your comment as an answer to my question. It is the kind of argument I was looking for....ohhhh, and thanks for the reference to rudin's –  Cristian Jul 21 '12 at 6:22
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@copper.hat: I'm just repeating Cristian's above suggestion to post as an answer because he misspelled your username in the ping. –  joriki Jul 21 '12 at 6:33
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1 Answer

up vote 2 down vote accepted

The Cantor function (call it $f$) is a counterexample. It is monotone, continuous, non-constant and has a zero derivative a.e.

If the above integral formula holds, then you have $f'(x) = h(x)$ at every Lebesgue point of $f$, which is a.e. $x \in [0,1]$. Since $f'(x) = 0$ a.e., we have that $h$ is essentially zero. Since $f$ is not constant, we have a contradiction.

See Rudin's "Real & Complex Analysis" Theorem 7.11 and Section 7.16 for details.

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