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I wanted to check my understanding on the Cantor set having derivative $0$ almost everywhere but not being differentiable everywhere.

I know that the Cantor Ternary set has measure zero.

Definition of Almost everywhere: A property P holds almost everywhere if there exists a Null set $N$ (in the sigma algebra) such that the property is true for all $x \in N^{c} $ i.e. the complement of $N.$

If we denote the Cantor function as $\psi: [0, 1] \rightarrow [0, 1]$ where $ \psi \subseteq [0, 1]$ Let $x \in [0, 1]$ with ternary expansion $0.a_{1}a_{2}...$ Let $N$ be the first $n \in \mathbb{N}$ such that $a_{n} = 1.$ If for all $n \in \mathbb{N}, a_{n} \in [0, 2],$ let $N = \infty.$

Define $b_{n} = \frac{a_{n}}{2},$ for all $n < \mathbb{N}$ and $b_{N} = 1.$ Then $$ \psi(x) = \sum_{n=1}^{N} \frac{b_{n}}{2^{n}}$$

Then for all points outside the Cantor function the derivative is $0$ (since they are just numbers).

Therefore it is differentiable almost everywhere.

But i'm not sure how to prove that if $x$ belongs to the Cantor set then $\psi$ is not differentiable at $x$.

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  • $\begingroup$ Have you tried using the definition of derivative, showing that the limit doesn't exist (i.e. isn't a finite, real number)? $\endgroup$ – Arthur Jan 11 '18 at 11:16
  • $\begingroup$ I worked it out. Thanks! $\endgroup$ – VBACODER Jan 13 '18 at 13:09

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