# Image of a closed interval by a lipschitz continuous function is a null set.

Let $f: [a,b] \rightarrow ℝ^2$ be lipschitz continuous. Show that $\{f(t): t\in[a,b]\}$ is a null set.

I've already shown that the image of every null set for lipschitz continuous functions is also a null set. Can I extend this for an interval $[a,b]$? Can anyone give me a hint?

Partition $[a, b]$ into $n$ closed intervals $U_1, U_2, \ldots, U_n$ of equal measure in the obvious way. Since $f$ is Lipschitz, it maps each of the $U_k$ into an open ball $B_k$ of radius $(a-b)K/n$ in $\mathbb R^2$, where $K$ is a Lipschitz constant of $f$. The total measure of $B_k$ is
$$n \cdot \pi \cdot \frac{(a-b)^2 K^2}{n^2} = \pi \cdot \frac{(a-b)^2 K^2}{n}$$
which is an upper bound for the measure of the image of $f$, since the $B_k$ are an open cover of $\textrm{Im} f$. Since $n \in \mathbb N$ was arbitrary, let $n \to \infty$ to obtain the result.