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Define $f:[0,1]\to [0,1]$ by $f(0)=0$, and $$f(x)=\sum\limits_{r_n\le x} 2^{ -n }$$ with $0\lt x\le 1$ where $[r_n]_{n\in \mathbb{Z^+} } = \mathbb{ Q} \cap (0,1) $.

How to show that the derivative $f'(x)=0$ a.e.?

I can show this function is increasing and discontinuous at every rational, and how to word on?

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closed as too localized by Gjergji Zaimi, Alain Valette, George Lowther, S. Carnahan Nov 20 '11 at 16:25

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Why do you think $f'(x)=0$ a.e.? – GH from MO Nov 20 '11 at 11:49
It would have been nice to mention that you asked this same question yesterday on math.SE: – Theo Buehler Nov 20 '11 at 12:21
In principle, the question seems too elementary for this site, or at least borderline. But, since I gave an answer, I vote to reopen, not to loose the information. In general, note that the eventuality of a unsuited question with a good answer (I'm not claiming this is the case, of course) has been contemplated in this site; consider e.g. the badge "Reversal". – Pietro Majer Nov 21 '11 at 10:38
@Pietro: you could always post your answer to the stackexchange site too. – George Lowther Nov 21 '11 at 13:22
thank you, good idea – Pietro Majer Nov 21 '11 at 18:34
up vote 4 down vote accepted

The function $f(x)$ is $\mu([0,x])$ where $\mu$ is the radon measure $\sum_{n\in\mathbb{Z} _ +} 2^{-n}\delta _ {q_n}$, and $\mu$ is singular the Lebesgue measure $\lambda$ (in fact, $\operatorname{supp}(\mu)=(0,1)\cap\mathbb{Q}$). So the absolutely continuous part $\mu ^ a$ $\lambda$ is zero, and the Radon-Nikodym derivative $d\mu ^ a/d \lambda$ is also zero; but this coincides a.e. with the derivative of $f$. Note that (depending on the particular chosen enumeration of $(0,1)\cap\mathbb{Q}$) there might be infinitely many irrational points $x$ where $f$ is continuous and derivable with any value of $f'(x)$; a Lebesgue null set though.

edit. A more elementary argument. Consider the nested family of open nbd's of $(0,1)\cap\mathbb{Q},$

$$A_\epsilon:=\cup_{n\in\mathbb{Z} _ + } (r_n- \epsilon 2^{-n/3},r_n+ \epsilon 2^{-n/3}), \qquad \epsilon > 0.$$ So $|A _\epsilon|=O(\epsilon)$ and $A:=\cap _ {\epsilon > 0} A _ \epsilon$ has measure zero. Let $x \in (0,1) \setminus A$: There exists $\epsilon > 0$ such that for any $n\in\mathbb{Z} _ +$ there holds $ \epsilon 2^{-n/3}\le |x-r _ n|$. Thus, for any $y\in (0,1)$

\begin{align*} |f(x)-f(y)| &\le \sum_{|x- r _ n|\le|x- y| } 2^{-n}\\ &= \frac{1}{\epsilon^2}\sum_{|x- r _ n|\le|x- y| } 2^{-n/3}(\epsilon 2^{-n/3})^2\\ &\le \frac{1}{\epsilon^2}\bigg(\sum_{n=1}^\infty 2^{-n/3}\bigg)|x-y|^2\\ &= \frac{|x-y|^2}{\epsilon^2(2^{1/3}-1))} \end{align*} showing that $f'(x)=0.$

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I like the elementary argument. – Gerald Edgar Nov 20 '11 at 17:58
@Pietro: hope you don't mind, I copied the second part of your answer to the Math.SE variant of this question. – Willie Wong Nov 21 '11 at 17:07
good idea, thank you – Pietro Majer Nov 21 '11 at 18:32

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