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

$f(x)=\begin{cases}0, &x=0,\\ \\ \sum\limits_{r_n

where $\{r_n \}_{n\in \mathbb N} =\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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    @MartinSleziak: So it appears the statement in Carothers is incorrect, since the function is continuous at the irrationals.2011-11-21

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The following is the elementary answer that Pietro Majer gave to this question on MO. I copy the answer here so we can consider this question answered.

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 $f'(x)=0\ .$

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    +1. Tha$n$ks for copying the answer across (although I don't think it needs to be community wiki).2011-11-21
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You won't get anywhere if you try to prove that $f$ is differentiable (with 0 derivative) at every irrational point. See here, whose result implies that there is a subset of the irrational numbers, dense on the interval, over which $f$ is not differentiable. (This question, however, neatly illustrates the difference between small in the sense of Baire category and small in the sense of measure.)

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    Ack; so, I'm completely off here.2011-11-19