Prove the image of $\gamma\colon\left[0,1\right]\to\mathbb{R}^{2}$ has measure zero

Question

Let $\gamma\colon\left[0,1\right]\to\mathbb{R}^{2}$ be a curve such that $\gamma\in\mathcal{C}^{1}$ (Continuously Differentiable). I need to show that $\gamma\left(\left[0,1\right]\right)$ has mesure zero, that is to show there are generalized rectangles $R_{1},R_{2},\ldots$ such that $\gamma\left(\left[0,1\right]\right)\subseteq\bigcup_{i=1}^{\infty}R_{i}$ and $\sum_{i=1}^{\infty}V\left(R_{i}\right)<\varepsilon$ where $V$ is the volume function.

Now I know how to show that the graph of a continuous function $f\colon\mathbb{R}^{n}\to\mathbb{R}$ has measure zero by using the uniform continuity, but that argument doesn't work here.

Any help?

Answer 1

Hint: A $\mathcal{C}^1$ curve $\gamma : [0, 1] \to \mathbb{R}^2$ is Lipschitz with some constant $L$. For each $n \in \mathbb{N}$ and $1 \leqslant j \leqslant n$ let $I_j = \left[ \frac{j-1}{n}, \frac{j}{n} \right]$ so that $I_j$ are of length $\frac{1}{n}$ and cover $[0, 1]$. Now let $m_j$ be the center of $I_j$ so that for each $x \in I_j$ we have that $|x - m_j| \leqslant \frac{1}{2n}$.

Can you cover $\gamma( [0, 1] )$ with a union of $n$ balls of a small radius?

Answer 2

As $\gamma$ is $C^1$ on a compact set, it is Lipschitz. Let $K$ be a Lipschitz constant for $\gamma$. Then, for each $N\in \Bbb N$, $\gamma([0,1])$ can be covered by $N$ squares, each of sidelength $2K/N$.