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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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    @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.