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

$$f(x)=\begin{cases}0, &x=0,\\ \\ \sum\limits_{r_n<x } 2^{-n}, & 0 \lt x \le 1, \end{cases} $$

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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The same function is used in an exercise on p.13 of Carothers' Real analysis. The reader is asked in this exercise to show that "$f$ is everywhere discontinuous on $[0,1]$ but that $f$ is everywhere continuous when considered as a function on only $[0,1]\setminus\mathbb Q$." –  Martin Sleziak Nov 18 '11 at 18:45
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And the same function is used in Understanding analysis by Stephen Abbott in an exercise on p.169, where the author claims that it is continuous at every irrational point. –  Martin Sleziak Nov 18 '11 at 18:50
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@Martin: It’s not hard to show that $f$ is continuous at every irrational. However, it need not be differentiable at every irrational, since if $a$ is any irrational in $(0,1)$, you can use a Hilbert’s hotel trick to construct the enumeration of $\mathbb{Q}\cap(0,1)$ in such a way that $f$ is not differentiable at $a$. Thus, the naive approach can’t work. –  Brian M. Scott Nov 18 '11 at 19:35
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It seems this has been asked and answered at mathoverflow mathoverflow.net/questions/81411/… –  David Mitra Nov 20 '11 at 21:15
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@MartinSleziak: So it appears the statement in Carothers is incorrect, since the function is continuous at the irrationals. –  Nate Eldredge Nov 21 '11 at 19:27

2 Answers 2

up vote 3 down vote accepted

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. Thanks for copying the answer across (although I don't think it needs to be community wiki). –  George Lowther Nov 21 '11 at 17:23

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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Note, though, that the example in part $3$ of that paper shows that a function can be discontinuous on a dense set and differentiable almost everywhere even though the set of points of differentiability must then be of first category. –  Brian M. Scott Nov 19 '11 at 18:38
    
Ack; so, I'm completely off here. –  David Mitra Nov 19 '11 at 18:54

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