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I recently learned about Sard's theorem: Let $f:[a,b]\to\mathbb{R}$ and denote $E_f=\{x\in(a,b)|f'(x)\mbox{ exists and }f'(x)=0\}$, then $m(f(E_f))=0$. Now, Cantor's step function is almost everywhere differentiable and $f'(x)=0$ for every point of differentiability. The image of this set is $[0,1]$ from continuity and the fact that $f(\mathcal{C})=[0,1]$ and therefore the measure is not null. I note that this is also the statement in Frank Jones Lebesgue Integration on Euclidean Space. (unless I am mistaken). What am I missing?

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    Not familiar with Sard's theorem, but I just looked it up on Wikipedia and it appears the theorem requires that $f$ is continuously differentiable. – 2017-01-22
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    you are right. I have seen this statement, but it is not required in the statement i know and is not used in the proof – 2017-01-22
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    It probably *is* used in the proof since the Cantor function is an obvious contradiction... It may be tucked away in the proof, but it's probably used in there somewhere. – 2017-01-22
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    The Cantor function is not differentiable everywhere. I don't think Sard theorem is applicable here. – 2017-01-22
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    @JohnMa the theorem does not require differentiability everywhere – 2017-01-22
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    The Cantor function $f$ is differentiable exactly on $[0,1]\setminus \mathcal{C}$. That is the union of countably many intervals, and on each of these intervals $f$ is constant. Hence $f(E_f)$ is a countable set. – 2017-01-22
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    @yoni Can you state your reference? – 2017-01-22
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    @JohnMa Lebesgue Integration on Euclildean Spaces by Frank Jones chapter 15.The theorems are not numbered – 2017-01-22
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    It says: Suppose $\Omega \subset \mathbb R^n$ is open and $\Phi : \Omega \to \mathbb R$. Suppose $E\subset \Phi$ and **that for every points $x\in E$, $\Phi$ is differentiable at $x$** and $J(x) = 0$. Then $\Phi(E)$ is a null set. – 2017-01-22
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    $E$ is the set $E_f$ i defined, $\Omega = [0,1]$ and $n=1$. I dont see how switching to the open interval changes anything. – 2017-01-22
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    It doesn't. But $E_f = [0,1]\setminus \mathcal{C}$, and $f(E_f)$ is countable, hence a null set, for the Cantor function $f$. – 2017-01-22
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    @DanielFischer I think i get it.$[0,1]\ \mathcal{C}$ is open and hence a union of countably many open sets (rationals). In the construction of Cantor function every in every one of countably many steps we add a finite number of image points, resulting in a dense (but countable) set in $[0,1]$. Finally, the points in cantor's set complete the image set in a continous manner. Thanks a lot – 2017-01-22

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