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

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?

-

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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: math.stackexchange.com/q/83365 – 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

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 $y\in (0,1)$
\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 $f'(x)=0.$