## Showing the derivative of this function is equal to $0$ a.e [closed]

Define $f:[0,1]\to [0,1]$ by

$f(0)=0$ , $f(x)=\sum\limits_{r_n\le x} 2^{ -n }$ 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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Why do you think $f'(x)=0$ a.e.? – GH Nov 20 2011 at 11:49
It would have been nice to mention that you asked this same question yesterday on math.SE: math.stackexchange.com/q/83365 – Theo Buehler Nov 20 2011 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 2011 at 10:38
@Pietro: you could always post your answer to the stackexchange site too. – George Lowther Nov 21 2011 at 13:22
thank you, good idea – Pietro Majer Nov 21 2011 at 18:34

## closed as too localized by Gjergji Zaimi, Alain Valette, George Lowther, S. Carnahan♦Nov 20 2011 at 16:25

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 w.r.to the Lebesgue measure $\lambda$ (in fact, $\operatorname{supp}(\mu)=(0,1)\cap\mathbb{Q}$). So the absolutely continuous part $\mu ^ a$ w.r.to $\lambda$ 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 $y\in (0,1)$ $$|f(x)-f(y)|\le \sum_{|x- r _ n|\le|x- y| } 2^{-n}= \frac{1}{\epsilon^2}\sum_{|x- r _ n|\le|x- y| } 2^{-n/3}(\epsilon 2^{-n/3})^2\le$$$$\le \frac{1}{\epsilon^2}\bigg(\sum_{n=1}^\infty 2^{-n/3}\bigg)|x-y|^2= \frac{|x-y|^2}{\epsilon^2(2^{1/3}-1))}\ ,$$ showing that $f'(x)=0\ .$