The length of a regular curve is different from its measure? (for rectifiable non absolute continuous path) Consider a path $\pi : [a,b]\to C$ where $C$ is a curve. By definition, then length of the path is $$\ell(\pi)=\sup \sum_{i=0}^n\|\pi(t_i)-\pi(t_{i+1})\|,$$
where $\{t_0,...,t_{n+1}\}$ is a partition of $[a,b]$, and the sup is taken over all partition.
I know that if $\pi$ is rectifiable but not absolutely continuous, then 
$$\ell(\pi)\geq \int_a^b \|\pi '(t)\|dt= m(C),$$
where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...
 A: This is indeed the case. The Cantor function $c$:


*

*is continuous everywhere but has zero derivative almost everywhere. Hence $\displaystyle \int_0^1 \sqrt{1+\left(\dfrac{dc}{dt}\right)^2} \ dt$ is equal to $1$.

*However its arc length is greater or equal than $2$. See Arc length of the Cantor function.

A: An example is the graph $\mathcal{G}(f)$ of the Cantor function $f$. 
The Cantor function is a continuous, increasing function from $f:[0,1]\to[0,1]$ with $f(0)=0$, $f(1)=1$. Let 
$$\gamma(t)=(t,f(t)),\qquad t\in [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $\gamma$ is larger than $\sqrt{2}$.
For any partition $0=t_0<t_1<\dots <t_n=1$ we have
\begin{align*}\ell(\mathcal{G}(f))\geq\sum_{i=0}^{n-1}\|\gamma(t_{i+1})-\gamma(t_i)\|&=\sum_{i=0}^{n-1}\sqrt{|t_{i+1}-t_i|^2+|f(t_{i+1})-f(t_i)|^2} \geq \\ 
\text(Cauchy-Schwarz)\quad &\geq \frac{1}{\sqrt{2}}\left[
\sum_{i=0}^{n-1}|t_{i+1}-t_i|+\sum_{i=0}^{n-1}\left|f(t_{i+1})-f(t_i)\right|\right]=\\
\text{(}f\textit{ is increasing)}\quad &=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\sum_{i=0}^{n-1}\left(f(t_{i+1})-f(t_i)\right)
=\frac{1}{\sqrt{2}}\left[1+f(t_n)-f(t_0)\right]=\\ 
(f(0)=0,\;f(1)=1)\quad&=\frac{1}{\sqrt{2}}\left[1+f(1)+f(0)\right] =\sqrt{2}
\end{align*}
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $\sqrt{2}$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality 
\begin{align*}\sum_{i=0}^{n-1}\|\gamma(t_{i+1})-\gamma(t_i)\|&=\sum_{i=0}^{n-1}\sqrt{|t_{i+1}-t_i|^2+|f(t_{i+1})-f(t_i)|^2} \leq \\ 
&\leq
\sum_{i=0}^{n-1}|t_{i+1}-t_i|+\sum_{i=0}^{n-1}\left|f(t_{i+1})-f(t_i)\right|=2\end{align*}
and taking the $\sup$ over all partitions of $[0,1]$, we get that $\mathcal{G}(f)$ is, indeed, a rectifiable curve, with $\ell(\mathcal{G}(f))\leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence 
$$\ell (\mathcal{G}(f))\geq \sqrt{2} > 1 =\int_0^1dt=\int_0^1\sqrt{1^2+|f'(t)|^2}dt=\int_0^1\|\gamma'(t)\|dt$$
