# Constructing a strictly increasing function with zero derivatives

I'm trying to construct a fuction described as follows: $f:[0,1]\rightarrow R$ such that $f'(x)=0$ almost everywhere,f has to be continuous and strictly increasing. (I'd also conlude that this functions is not absolutely continuous)
The part in bracket is easy to prove.
I'm in troubles with constructing the function:
I thought about considering the function as an infinite sum of sucession of increasing and conitnuous function, I considered the Cantor-Vitali functions which is defined on $[0,1]$ and is continuous and incresing (not strictly).
So $f(x)=\sum_{k=0}^{\infty} 2^{-k}\phi(3^{-k}x)$ where $2^{-k}$ is there to makes the sum converging and $\phi$ is the Cantor-Vitali function.
The sum convergethe function in continuous (as sum of) and is defined as asked.But I'm in trouble while proving that is strictly increasing.Honestly it seems to be but I don't know how to prove it.
I know that $0\leq x \leq y\leq 1$ always exist a k such that $\phi(3^{-k}x)\leq \phi(3^{-k}y)$ but then I stucked.
I'm looking for help in proving this part.

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See here for other ideas. – David Mitra Dec 4 '12 at 11:30
For an annotated bibliography of 82 items on this topic, see Bibliography for Singular Functions. – Dave L. Renfro May 25 at 14:03

By $\phi$ we denote Cantor-Vitali function. Let $\{(a_n,b_n):n\in\mathbb{N}\}$ be the set of all intervals in $[0,1]$ with rational endpoints. Define $$f_n(x)=2^{-n}\phi\left(\frac{x-a_n}{b_n-a_n}\right)\qquad\qquad f(x)=\sum\limits_{n=1}^{\infty}f_n(x)$$ I think you can show that it is continuous and have zero derivative almost everywhere. As for strict monotonicity consider $0\leq x_1<x_2\leq 1$ and find interval $(a_n,b_n)$ such that $(a_n,b_n)\subset(x_1,x_2)$, then $$f(x_2)-f(x_1)\geq f(b_n)-f(a_n)\geq f_n(b_n)-f(a_n)=2^{-n}>0$$ So $f$ is strictly monotone.
Did you mean $n^{-2}$ or $2^{-n}$ I think could work in both ways..or not? – Laura Dec 4 '12 at 11:38