Prove that image of a continuously differential path is a null set I got by the following exercise: prove that the image of a continuously differential path $\gamma:[0,1] \rightarrow R^2$ is a null set.
I know that I need to find some cover of intervals $I_j=[a_j,b_j]$ of the image that satisfy  $$\sum_{j=1}^\infty |I_j| < \epsilon$$ for any $\epsilon>0$.
However, I am not really sure how to do it. 
Any help would be appreciated.

Let $\gamma: [0,1]\to\Bbb{R}^n$ for $n\geq 2$ be continuously differentiable. Since it is continuously differentiable, it is Lipschitz continuous with Lipschitz constant $C$, i.e. there exists a $C\geq 0$ such that for all $t_1,t_2\in[0,1]$, $$\| x_2-x_1\| \leq C|t_2-t_1|$$ where $x_i = \gamma(t_i)$. This implies that $$\gamma([t_1,t_2])\subset 
B(x_1,C|t_2-t_1|),$$ a closed ball centered at $x_1$ with radius $C|t_2-t_1|$.
Take $k+1$ equally spaced points along $[0,1]$ with $$0=t_0 < t_1 < \dots < t_{k-1} < t_k = 1$$ Then, for each $i$, $$\gamma([t_i,t_{i+1}])\subset B(x_i,C|t_{i+1}-t_i|) = B\left(x_i,\frac{C}{k}\right) := B_i$$ so that the curve $\gamma([0,1]) \subset \bigcup\limits_{i}B_i$.
The volume of a ball with radius $r$ in $\mathbb{R}^n$ is proportional to $r^n$, so the volume of each $B_i$ is proportional to $\frac{1}{k^n}$. Then, the sum of the volumes is an upper bound for the volume of the union, and it is proportional to $$k\cdot \frac{1}{k^n} = \frac{1}{k^{n-1}}$$ Then, as $k\to\infty$, the upper bound for the volume goes to $0$ as long as $n\geq 2$.
Thus, we can create closed ball cover of the curve with arbitrarily small volume, so the curve must be a null set.
