simple integration question 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!

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.
