# The differentiablity of monotonic functions

## Problem

Suppose that $f:(a,b)\to\Bbb R$ is monotonic, and $D=\left\{\,x\;\big|\;f\textrm{ is not differentiable at }x\,\right\}$. Try to prove that for each $\eta>0$, $D$ could be covered by a collection (at most countable) of open intervals $\{O_n\}_{n=1}^\infty$ whose total length is smaller than $\eta$.

## Motivation

I heard that it's a well-known theorem in measure theory. I wonder whether we could work without measure theory, where our tools are so elementary, just as in calculus course.

## Thoughts

Let $$\varphi(x)=\limsup_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}-\liminf_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$, then $D=\left\{\,x\;\big|\;\varphi(x)>0\,\right\}$. Let $D_\varepsilon=\left\{\,x\;\big|\;\varphi(x)\ge\varepsilon\,\right\}$, we have $D=\bigcup_{n=1}^\infty D_{1/n}$, therefore it suffices to prove that $D_\varepsilon$ could be covered by a collection (at most countable) of open intervals whose total length is less than $\eta$, for each $\varepsilon,\eta>0$.

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It seems hopeless to me if tools like Vitali covering theorem or Lebesgue's decomposition theorem are not allowed. By the way, you cannot expect that $D_\epsilon$ or even $D$ would be covered by finitely many open intervals whose total length is arbitrarily small. – 23rd Dec 15 '12 at 16:45
@richard I agree with richard ; indeed, one can construct an increasing $f$ which is not differentiable at any point of $\mathbb Q$. Such a counterexample makes a “completely elementary” solution unlikely. – Ewan Delanoy Dec 15 '12 at 20:21
@richard Thanks. – Frank Science Dec 16 '12 at 6:58
@FrankScience: You are welcome! – 23rd Dec 16 '12 at 13:09