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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.2012-07-21
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    I just realised this, thanks.2012-07-21
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    Hi, what I do know is that f is differentiable almost everywhere on [0,1] because it is a nondecreasing function. I could in principle let h be equal to f' a.e. but I don't know if I can draw the conclusion that f can be written as an integral of this function...2012-07-21
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    Look at the Cantor function. Wiki has a nice picture http://en.wikipedia.org/wiki/Cantor_function.2012-07-21
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    If I could answer real quick, I would say "yes". But proving this is a 5-page proof of multiple theorems about functions with bounded variation. It is definitely not straight-forward ; Jordan's theorem, Vitali's Lemma and some argument involving Dini's derivatives is the shortest proof I have seen. Cantor's function IS differentiable almost everywhere and satisfies the above, by the way.2012-07-21
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    @cooper.hat so if I were to add absolutely continuity to f then the answer to my question would be yes by using the derivative of f. But I am concerned about the existence of some function (not necessarily the derivative of f) such that the integral representation holds...2012-07-21
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    @Patrick I am not quite sure I got your comment. What you say is that, yes, there exists some function h such that the above holds true? I got confused because of your final line: ' Cantor's function IS differentiable almost everywhere and satisfies the above, by the way'...the above ??? Thank you!2012-07-21
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    @PatrickDaSilva: The Cantor function is monotone, continuous, and has a zero derivative a.e., so clearly it cannot satisfy the above. It is the standard example of a singular function.2012-07-21
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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.2012-07-21
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    For example, see Rudin's "Real & Complex Analysis" Theorem 7.11 and Section 7.16 for details.2012-07-21
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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's2012-07-21
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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.2012-07-21
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    @joriki: Thanks!2012-07-21
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    @Cristian : Sorry if my answer was misleading. I think I read some hypothesis wrong in my measure theory notes. I understand after giving some though to copper.hat's comment that I am not right.2012-07-21
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    @PatrickDaSilva no harm done Patrick. Thanks for help!2012-07-22

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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.