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The Cantor function has weak derivative equal to $0$ a.e.

Its distributional derivative should be the $\log_3 2$-Hausdorff measure restricted to the Cantor set, but I'm having troubles doing the computation because of the peculiar definition of the Cantor function.

How can we compute the distributional derivative of Cantor function?

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  • $\begingroup$ It will be a measure supported on the Cantor set. $\endgroup$ – Lord Shark the Unknown Apr 6 at 1:40
  • $\begingroup$ @LordSharktheUnknown Yes, it should be the $\log_3 2$ Hausdorff measure on the Cantor set. But how can we do the computation? $\endgroup$ – Riku Apr 6 at 1:43
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Use the self-similarity.

You want to show that the characteristic function of [0,a) integrated against the derivative is equal to the Hausdorff measure of $[0,a)\cap C$ (and equal to the Cantor function at a). That is true for a=1 (by suitable normalization), for a=1/3 by self-similarity of the Cantor function and then true for $1/3<a<2/3$ trivially, since the function is constant on that interval. Expand a in base 3 and use the self-similarity to show the general result.

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  • $\begingroup$ Could you add some details? Even starting from the beginning "You want to show that the characteristic function of [0,a) integrated against the derivative is equal to the Hausdorff measure..." Your answer is not quite clear to me. $\endgroup$ – Riku Apr 9 at 11:37
  • $\begingroup$ Let $\phi$ be the Cantor function and h the Hausdorff measure.Then $\phi$ is increasing and bounded and hence in BV and its derivative is a Radon measure (see Rudin, Chapter 3.5 for example), let us call it $m$. You want to show that m([a,b))= h([a,b)) for all a<=b (or (a,b], I don't quite remember). By additivity it suffices to consider a=0. Now the measure is defined such that m([a,b))=\phi(b)-\phi(a). Proceed from there as in the first answer. $\endgroup$ – user494137 Apr 9 at 19:50

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