Image of continuously differentiable curve $\gamma:[0,1]\rightarrow \mathbb{R}^n$ has volume zero Let $n>1$. Then the  image of the $C^1$ curve  $\gamma = (\gamma_1, ... , \gamma_n)$,  has $n$-dim volume (area if $n=2$) equal $0$. Hence the image of $\gamma$ can not have interior points; in particular, it would be far from a Peano curve!!
[$C^1$ means that all $\gamma_j$ are continuously differentiable; it is actually sufficient that they are Lipschitz-continuous. 
Volume $0$ means that for every $\epsilon>0$ there is a countable (here finite!) covering by balls $B(a_j,r_j)$ s.t. $\sum r_j^n < \epsilon$.]
Question: How to prove that statement?
Show first that a $C^1$ curve is Lipschitz. Note that Lipschitz is a strong ("quantitative") form of uniform continuity. Then find a corresponding covering!
I have shown $C^1$ curve is Lipschitz. But I am unable to find that covering.
If someone can tell me about this covering I will be grateful for it.
 A: You have shown that $\gamma$ is Lipschitz, then :
$$\exists L >0,\|\gamma(x)-\gamma(y)\| \leq L |x-y|.$$
Let $m \in \mathbb{N}$, $i \in \lbrace{ 0, \dots, m-1 \rbrace}$ and define : 
$$B_i=B(\gamma(i/m),r_m)$$
With $r_m=\dfrac {L+1}{m}$.
Then : $$\gamma([0,1]) \subset\bigcup_{i=0}^{m} B_i.$$
Indeed, let $y=\gamma(x)$. $\exists i \in \lbrace{ 0, \dots, m-1 \rbrace}$ such that $i/m \leq x \leq (i+1)/m$, then:
$$\|y-\gamma(i/m)\| \le L|x-i/m| \le L/m<\dfrac{L+1}{m}=r_m.$$
Finally, since $n>1$ :
$$\sum_{i=0}^{m-1}Vol(B_i)=m Vol\left(B\left(0,\dfrac{L+1}{m}\right)\right) = m\left(\dfrac {L+1}{m}\right)^n Vol(B(0,1)) \underset{m \to \infty} \to 0.$$
A: You use that $\gamma$ has finite arc length $L$. This means that you can cover it with $N+1$ balls of radius $L/N$, which means the area is less than $(N+1)(L/N)^2\pi = L^2\pi(1/N + 1/N^2) < L^2\pi/N$. Thus the area must be zero.
A: Let $K$ be a Lipschitz constant for all the co-ordinates. For $m\in \Bbb N,$ for $0\leq j\leq m-1,$ the image of $[j/m,(j+1)/m]$ is covered by a closed  $n$-cube centered at $\gamma (j/n)$ with volume $(2K/m)^n$. So the image of $\gamma$ is covered by  a closed set of volume $m\cdot (2K/m)^n,$ which $\to 0$ as $m\to \infty.$
