<s>[INST] Elementary ways to calculate the arc length of the Cantor function (and singular function in general) Cantor's function: http://en.wikipedia.org/wiki/Cantor_function
There is an elementary way to prove that the arc length of the Cantor function is 2?
In this article (http://www.math.helsinki.fi/analysis/seminar/esitelmat/stat0312.pdf) they use the following result:

If $f:[a,b] \rightarrow \mathbb{R}$ is a continuous monotone function, then $f$ is singular if and only if
  $$L_a^b = |f(a)-f(b)|+|a-b|$$

But, there is a way for calculate the arc length of singular function without using this property? like using the arc length definition

If $X$ is a metric space with metric $d$, then we can define the ''length'' of a curve $\!\,\gamma : [a, b] \rightarrow X$ by $$\text{length} (\gamma)=\sup \left\{ \sum_{i=1}^n d(\gamma(t_i),\gamma(t_{i-1})) : n \in \mathbb{N} \text{ and } a = t_0 < t_1 < \cdots < t_n = b \right\}. $$

where the sup is over all $n$ and all partitions $t_0 < t_1 < \cdots < t_n$ of $[a, b]$.
 [/INST] Here is an elementary way to derive the arc length of the Cantor Function:
We know that the following sequence of functions converges uniformly to the Cantor Function. Letting $c_0(x) = x$ and $n\ge 0$, we define
$$c_{n+1}(x)= \begin{cases} 
      \frac{1}{2}c_n(3x) & 0 \le x \le \frac{1}{3},\\
      \frac{1}{2} & \frac{1}{3} \le x \le \frac{2}{3},\\
      \frac{1}{2}c_n(3x-2)+\frac{1}{2} & \frac{2}{3}\le x \le 1.
   \end{cases}
$$
We also know that once the sequence of functions goes constant on an interval, it stays constant on that interval. Then from the definition, we see that at $c_n$, $n\ge1$, we add $2^{n-1}$ constant portions of length $3^{-n}$ each. So for the totally horizontal (i.e. constant) portions of $c_n$, we have that they have total length $\displaystyle \sum_{i=0}^n \frac{2^k}{3^{k+1}}$. To find the lengths of the positively sloped portions for a given $c_n$, we note that from the definition, there are $2^n$ sloped portions, which increase the function $2^{-n}$ over a length of $3^{-n}$. Then we have that the total length of the positively sloped portions is
$$2^n\sqrt{(\frac{1}{2^n})^2 + \frac{1}{3^n})^2} = \sqrt{\frac{(2^n)^2((2^n)^2 + (3^n)^2)}{(2^n)^2(3^n)^2}} = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}.$$
So then $L(c_n) = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$, and so
$$\lim_{n \to \infty}L(c_n) = \lim_{n \to \infty} \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$$
$$ = \sqrt{\lim_{n\to \infty}\frac{2^{2n}+3^{2n}}{3^{2n}}} + \lim_{n \to \infty} \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k} = \sqrt{\lim_{n \to \infty}(\frac{2}{3})^n +1} + \frac{1}{3}\left(\frac{1}{1-\frac{2}{3}}\right)$$
$$= 1 + \frac{1}{3}\left(\frac{1}{\frac{1}{3}}\right) = 2.$$
Letting $c$ be the Cantor function, we have $L(c) \ge 2$, since $L(c) \ge L(c_n)$ for all $n$. To see this, note that on any constant portion of $c_n$, $c$ agrees with $c_n$. But on the diagonal portions of $c_n$, the graph of $c$ is broken up into more and more horizontal and diagonal portions. In essence, if on any $(a,b)$ on which $c_n$ is diagonal, $c$ is less diagonal. Then since the shortest distance between any two points is given by the straight line between them, we have that the length of $c$ is greater than the length of $c_n$ on that portion. Then we have that $L(c) \ge L(c_n)$ for all $n$, so $L(c) \ge 2$. To see that $L(c) \le 2$, note that we have $L(c) = \displaystyle \sup_{P} \sum_{i=1}^n \sqrt{(\triangle x_i)^2 + (\triangle c(x_i))^2} \le \sup_{P}\sum_{i=1}^n |\Delta x_i| + |\Delta c(x_i)| =  \sup_{P}\sum_{i=1}^n \Delta x_i + \Delta c(x_i) =2$, since the total length of the horizontal segments is $1$, and since $c$ is monotonically increasing the total length of the vertical growth is $1$. Thus $L(c) =2$.
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