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We're all familiar with the standard way to prove that the arc length of the Cantor function $C:[0,1] \rightarrow [0,1]$ is $2$. We also know that if a function $f:[a,b] \rightarrow \mathbb{R}$ is continuous and has bounded variation than the length of its graph is

$$ L = V(f_{s}, [a, b]) + \int_{a}^{b}\sqrt{1 + (f'(x))^2}\,dx, $$

where $f_{s}$ is the singular part of $f$ and $V(f_{s}, [a, b])$ is the total variation of the singular part on $[a,b]$. How can I compute the arc length of the Cantor function $C$ using the formula defined above? Mainly, how do I compute the total variation part? I've only managed to find one similar question How to compute the arclength of a singular function? but the answer is not satisfactory.

LucasS
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  • Please include why you find the answer not satisfactory. What remaining question to you have, not "satisfied", at that link, that leads to your judgment? It would help users here to hone their answers. – amWhy Jan 30 '23 at 11:56
  • Is $f'(x)$ meaningful if it is undefined or infinite on an uncountable subset of $[a,b]$? Intuitively, if you exclude that subset (zero-measure here) from the integral then $f'(x)=0$ for the rest so you get $b-a$ for integral; meanwhile the singular part is where $f$ increases so the variation is $f(b)-f(a)$. Add these together to get the known result – Henry Jan 30 '23 at 11:57
  • Could you tell me if my reasoning is correct? There is a theorem, which states that if $f \in NBV$ is nondecreasing, then there exists a nondecreasing function (the singular part of $f$) $f_{s} \in NBV$, such that $f'{s} = 0$ a.e. and $f(x) = f_s(x) + \int{-\infty}^{x}f'(t),dt$ for every $x \in \mathbb{R}$. In our case, we can put $C(x) = 0$ for $x < 0$. Thus $C \in NBV$. Since the singular part $C_{s}$ is nondecreasing, we have $V(C_{s}, [a, b]) = C_{s}(1) - C_{s}(0) = 1$. For the integral part, we know that $C'(x) = 0$ a.e, therefore the integral is equal to $1$ and we have the result. – LucasS Jan 30 '23 at 13:54
  • Does this answer your question? Elementary ways to calculate the arc length of the Cantor function (and singular function in general). See David C. Ulrich's solution. – Mittens Jan 30 '23 at 14:02

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