A piecewise $C^1$ curve has Jordan measure zero. $\newcommand{\Reals}{\mathbb{R}}\gamma:[0,1]\to \Reals^2$ is an injective parametrization of a curve $\Gamma$, which is piecewise $C^1$ and the length of the curve is $L(\Gamma_k)<\infty$. 
1.1.: Show that for every $n\in N$ there is a decomposition $0=t_0<t_1< \dots <t_n=1$ of $[0,1]$ so that $L(\Gamma_k) = \frac{L}{n}$ for $k=1,\dots ,n$ and $\Gamma_k=\operatorname{Im}(\gamma_k)$, $\gamma_k=\gamma|_{[t_{k-1},t_k]}$. 
1.2.: Show that you can cover $\Gamma$ with a system of spheres of radius $\frac{L}{n}$. 
1.3.: Use 1.2. and 1.3. to show that $\Gamma$ has Jordan measure zero. 
I wasn't there when they discussed when a function has Jordan measure zero  and how those things correlate with this exercise so if someone could give me some hints that would be great. 
 A: Here's a detailed sketch that (I hope) doesn't spoil all the fun:
Define the arc length function by
$$
s(t) = \int_{0}^{t} \|\gamma'(\tau)\|\, d\tau.
$$
Because $\gamma$ is an injective parametrization and piecewise $C^{1}$, the function $s:[0, 1] \to [0, L]$ is strictly increasing and continuously differentiable. (This probably looks obvious to a physicist, but a mathematician requires proof. The idea is, the speed $\|\gamma'\|$ is continuous, non-negative, and does not vanish identically on any open interval, so its integral is strictly increasing.)
1.1. follows at once. (Subdivide the image into equal-length pieces and pick $t_{k}$ accordingly.)
1.2. Note that if $0 \leq a < b \leq 1$, then 
$$
\|\gamma(b) - \gamma(a)\| \leq s(b) - s(a);
\tag{1}
$$
in words, the straight-line distance between two points on $\Gamma$ does not exceed the arc length (along $\Gamma$) between the points.
(Again, this probably looks obvious to a physicist. Here's an argument to satisfy a mathematician: If $\gamma(a) = \gamma(b)$ there's nothing to show. Otherwise, let
$$
u = \frac{\gamma(b) - \gamma(a)}{\|\gamma(b) - \gamma(a)\|}
$$
be the unit vector parallel to the displacement from $\gamma(a)$ to $\gamma(b)$, and observe that if $\theta$ is the angle between $\gamma'(\tau)$ and $u$, then
$$
\gamma'(\tau) \cdot u = \|\gamma'(\tau)\| \cos\theta \leq \|\gamma'(\tau)\|.
$$
Integrating the left-hand side from $a$ to $b$ gives the straight-line distance between the endpoints; integrating the right-hand side gives the arc length.)
Now, for each $k = 1, \dots, n$, let $\gamma(\bar{t}_{k})$ denote the point "halfway along $\Gamma$ between $\gamma(t_{k-1})$ and $\gamma(t_{k})$". That is, let $\bar{t}_{k}$ be the unique point such that $s(\bar{t}_{k}) = (k - \frac{1}{2}) \frac{L}{n}$.
It suffices to show that if $I = [t_{k-1}, t_{k}]$, then the image $\Gamma(I)$ is contained in the disk of radius $\frac{L}{n}$ about $\gamma(\bar{t}_{k})$. This is a special case of (1). (And, of course, $n$ such disks cover $\Gamma$.)
