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$\def\R{\mathbf{R}}$ $\def\Q{\mathbf{Q}}$ Background: Using properties of uniform convergence and the definition of the function, in an exercise I proved that the Cantor Function is increasing, is differentiable except on the Cantor set, where it is just continuous. It was a very weird property that it somehow manages to increase from $0$ to $1$ while having a derivative of $0$ wherever it is defined. In some intuitive way I felt like the Cantor set had the ability to push the function from $0$ to $1$, since the function only strictly increased near cantor set points. My question is, which other sets are able to do this? Or rather, what sort of properties must the set satisfy to be able to construct such a function like this?

The Question. Say we are given a function $f:\R \to \R $, and a subset $A \subseteq \R$, with the following properties:

  • $f$ is increasing
  • $f$ is continuous
  • The derivative $f'(x) = 0$ for all $x \notin A$, and $f$ is not differentiable for any $x\in A$.
  • $f$ is not constant.

Then what can we say about the set $A$? (Open ended)

We know that $A$ must be uncountable.
Proof. $f$ is not constant, hence the range has at least two distinct values, say $a$ and $b$. (assume $a<b$ WLOG).
Since $f$ is continuous, it has the intermediate value property and hence the range contains $[a,b]$.
$f$ is increasing, but constant everywhere except on $A$. So to "bring" $f$ from $a$ to $b$, the elements of $A$ must map to every number in $[a,b]$
Since $[a,b]$ is uncountable, it follows that $A$ must be as well.
Q.E.D.

Are there any other characteristics that we can deduce of $A$? I wonder if $A$ can be the irrationals.

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  • $\begingroup$ I'm too busy to say anything all that specific, but my (closed, so might not be viewable to everyone) "question" and answer Bibliography for Singular Functions will point you to a lot of literature about this topic. $\endgroup$ Aug 24, 2022 at 17:28

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Lebesgue's theorem for the differentiability of monotone function:

Let $f:\Bbb{R}\to\Bbb{R}$ be an increasing function then $f$ is differentiable almost everywhere.

$A=\{x\in\Bbb{R} : f \text{ is not differentiable at } x\}$

Then $m(A) =0$

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  • $\begingroup$ If I remember correctly, we also have: “$f:\Bbb R\to\Bbb R$ measurable” implies: “$m(A)=0$” also $\endgroup$
    – FShrike
    Aug 24, 2022 at 16:52

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