Una curva no rectificable

Proponemos un ejemplo de curva no rectificable.

Enunciado
Se considera la curva del plano $$\Gamma:\left \{ \begin{matrix} \displaystyle\begin{aligned} & x=t\quad\text{si}\quad0\le t\le1\\& y=\left \{ \begin{matrix} \displaystyle\begin{aligned} & t\cos \frac{1}{t}\quad \text{si}\quad0<t\le 1\\& 0\quad \text{si}\quad t=0.\end{aligned}\end{matrix}\right. \end{aligned}\end{matrix}\right.$$ Demostrar que no es rectificable.
Sugerencia: considerar particiones de $[0,1]$ de la forma $$P:\;0,\;\frac{1}{(n-1)\pi},\;\frac{1}{(n-2)\pi},\;\ldots,\;\frac{1}{2\pi},\;\frac{1}{\pi},\;1.$$

Solución
Los puntos de la poligonal que corresponden a la partición $P$ son $$M_0=(0,0),\;M_1=\left(\frac{1}{(n-1)\pi}, \frac{1}{(n-1)\pi}\cos (n-1)\pi\right),$$ $$M_2=\left(\frac{1}{(n-2)\pi}, \frac{1}{(n-2)\pi}\cos (n-2)\pi\right), $$ $$\ldots$$ $$M_{n-2}=\left(\frac{1}{2\pi}, \frac{1}{2\pi}\cos 2\pi\right),M_{n-1}=\left(\frac{1}{\pi}, \frac{1}{\pi}\cos \pi\right),\;M_n=(1,\cos 1).$$ La suma de las distancias de los segmentos de la poligonal es $$s(P)=\left|M_0M_1\right|+\left|M_1M_2\right|+\cdots+\left|M_{n-2}M_{n-1}\right|+\left|M_{n-1}M_n\right|$$ $$=\left|\left(\frac{1}{(n-1)\pi}, \frac{1}{(n-1)\pi}\cos (n-1)\pi\right)\right|$$ $$+\left|\left(\frac{1}{(n-2)\pi}-\frac{1}{(n-1)\pi},\frac{1}{(n-2)\pi}\cos (n-2)\pi-\frac{1}{(n-1)\pi}\cos (n-1)\pi\right)\right|$$ $$+\cdots +\left|\left(1-\frac{1}{\pi},\cos 1-\frac{1}{\pi}\cos\pi\right)\right|.$$ Eliminando el primer y último término de la suma anterior, $$s(P)\ge \sum_{k=1}^{n-2}\left|\left(\frac{1}{k\pi}-\frac{1}{(k+1)\pi},\frac{1}{k\pi}\cos k\pi-\frac{1}{(k+1)\pi}\cos (k+1)\pi\right)\right|$$ $$\ge \sum_{k=1}^{n-2}\left|\frac{1}{k\pi}\cos k\pi-\frac{1}{(k+1)\pi}\cos (k+1)\pi\right|$$ $$\ge \sum_{k=1}^{n-2}\left|\frac{(-1)^k}{k\pi}-\frac{(-1)^{k+1}}{(k+1)\pi}\right|=\sum_{k=1}^{n-2}\left|\frac{1}{k\pi}+\frac{1}{(k+1)\pi}\right|\ge \frac{2}{\pi}\sum_{k=1}^{n-2}\frac{1}{k+1}.$$ Entonces,$$\lim_{n\to +\infty}s(P)\ge \lim_{n\to +\infty}\frac{2}{\pi}\sum_{k=1}^{n-2}\frac{1}{k+1}=\frac{2}{\pi}\sum_{k=1}^{+\infty}\frac{1}{k+1}=+\infty, $$ lo cual implica que $\Gamma $ no es rectificable.