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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 $\eta$.


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.


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
up vote 1 down vote accepted

An elementary proof that does not employ any measure theory has been given by Riesz. An exposition of the proof can be found right at the beginning of "Functional Analysis" by Riesz and Nagy. The proof is given mainly for the case in which the function is also continuous and the extension to the full theorem only sketched.

A detailed account of how one can extend the result from the continuous to the discontinuous case is given in Rubel, Differentiability of monotonic functions 1963, Colloquium Mathematicum. The paper also references a paper (in French) by Lipinski that provides another elementary proof.

More recently, a different elementary proof has been given in Botsko, An Elementary Proof of Lebesgue's Differentiation Theorem 2003, The American Mathematical Monthly. The proof works by verifying that the set of points where the upper and lower derivative disagree has measure zero. A similiar approach is followed by Az here.

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Thanks, but I doubt whether Lemma 1 is trivial and wonder how to prove it. – Frank Science Dec 19 '12 at 1:30
@FrankScience The edit should adress your concerns. – Michael Greinecker Dec 19 '12 at 8:36

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