# Showing the derivative of this function is equal to $0$

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
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
@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
It seems this has been asked and answered at mathoverflow mathoverflow.net/questions/81411/… –  David Mitra Nov 20 '11 at 21:15
@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

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 $f'(x)=0\ .$
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.)
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