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I am looking for a reference for the fact from the title. It is stated in this page but in the given reference it seems to me like they $\textit{define}$ the length of the curve as the integral of $|\gamma'(t)|$. The definition of length of a curve I am using is the one using a supremum of sums of distances between points in a finite partition of the curve.

Edit: Sorry, as shown in this question rectifiability is not sufficient. But for my purposes it is enough with absolutely continuous curves (which are the ones mentioned in the reference from the page anyways).

Edit2: Thanks to Oliver Díaz for all his comments. As he pointed out, the result I am looking is explicitly mentioned of the book $\textit{Lebesgue Integration on Euclidean Space (Revised Edition)}$, by Frank Jones. Especifically is is proved in page 551.

Saúl RM
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  • @OliverDíaz how exactly? (At least I would appreciate a summary in a few lines of what your argument would look like. The argument cannot be extremely easy because as I say, it works for absolutely continuous curves but not for rectifiable curves) – Saúl RM Dec 16 '22 at 21:12
  • See this posting. You can commit reference to the Bochner integral in that posting since your case is simple, i.e., Euclidean space. – Mittens Dec 16 '22 at 21:14
  • Oh thanks, I will take a look – Saúl RM Dec 16 '22 at 21:14
  • @OliverDíaz But the statement from that post does not imply what I am looking for. The Cantor curve I cite in the question has bounded variation, however the integral of its derivative is $1$ and the length of the curve is $>\sqrt{2}$. Hence why we will need some extra condition like absolute continuity – Saúl RM Dec 16 '22 at 21:27
  • Your posting reads that if $\gamma$ is a.c, then it is graph is rectifiable and its length is given by certain integral. Rectifiable is the same as bounded variation and in the posting I linked there is a prove of the formula. – Mittens Dec 16 '22 at 21:30
  • @OliverDíaz I edited my question previously just so it doesn't say anything about rectifiability. I know rectifiability and bounded variation are the same, and none of them are enough to prove the question due to this counterexample, as I said in the previous comment. – Saúl RM Dec 16 '22 at 21:35
  • The Cantor function is not AC that is which the identity fails in this case. – Mittens Dec 16 '22 at 21:38
  • But it is still rectifiable/bounded variation. Also, the post you mention asks to prove that the derivatives are equal a.e, which is true for the Cantor function, but the Cantor function still doesn't satisfy what I am asking. I don't know if I am missing something. Also if you want I can take this to a discussion so that we don't fill everything with comments – Saúl RM Dec 16 '22 at 21:41
  • For the Cantor function (which is not AC, but rectifiable) see this posting – Mittens Dec 16 '22 at 21:41
  • Oh I think I see what I am missing, the answer to the question you mention actually proves something much stronger. Sorry, I assumed it just proved what the question asked for. I will read the answer in more detail – Saúl RM Dec 16 '22 at 21:44
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    FOr a more detail treatment (in the real line) see Jones, F. Lebesgue Integration in Euclidean spaces, Chapter 16, G (arc length) where functions (of bounded variation )are decomposed in their AC, Continuous and pure jump part and the arc-length is computed. A particular example discussed there is the Cantor function. – Mittens Dec 16 '22 at 21:50
  • Thanks, the book looks good. I will look at these resources and when I get an understanding of why the result I mention is true I will add it to the question – Saúl RM Dec 16 '22 at 22:03

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