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Rudin asked:

If $f$ is a real function on $[0,1]$ and $\gamma(t)=t+if(t)$, the length of the graph of $f$ is, by definition, the total variation of $\gamma$ on $[0,1]$. Show that this length is finite if and only if $f\in BV$. Suppose $f(0)=0$, $f$ is continuous and nondecreasing, and $f_{s}$ is the singular part of $f$. Prove that the length of the graph of $f$ is $$f_{s}(1)+\int^{1}_{0}\sqrt{1+|f'(t)|^{2}}dt$$

How does this formula if $f\in BV$ but $f$ is not necessarily monotonic? How long is the graph of the cantor function?

I do not know what is a good strategy to prove this formula, precisely because of the existence of singular functions. If we define a differentiable function's graph's length via the classical formula, then the above formula makes a lot of sense. However it is not clear to me when there is a singular function involved. By Lesbegue decomposition theorem we have $$f(t)=f_{s}(x)+\int^{x}_{-\infty}f'(t)dt$$

It is clear the same formula would not make sense because it ignored $f_{s}$ part completely. While it is not difficult to show Cantor's function's graph has length 1, this probably does not hold in general for singular functions. So my questions are as follows:

  1. How to compute the length of a non-decreasing singular function (of bounded variation)?
  2. How to prove the formula given above rigorously?
  3. What changes substantially when $f$ not monotone?
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There is a paper that can be helpfull to you: For the first question, the answer is Corollary 2.10: the length of a singular function f between a and b is |f(a)-f(b)|+|b-a|.

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