Set of measure zero and $C^{1}$ functions

Question

Does a $C^{1}$ function map a set of measure zero into a set of measure zero?

Answer 1

It maps a set of measure zero to a set of measure zero. To see that notice it is enough to prove this for measure zero sets contained in closed intervals. Because then we can take a countable union of the image, and a countable union of measurable zero sets are measure zero.

Now let us assume $K\subset [a,b]$. The function $f\in C^{1}[a,b]$, so in particular $f'$ is continuous on $[a,b]$ and has bounded maximum and minimum. For convenience let $|f'|\le C$ on $[a,b]$. Now if any open interval $(a_1, b_1)$ covers a subset of $K$, its image has measure at most $C(b_1-a_1)$. So by taking the outer-measure for $K$ overall possible open coverings we conclude $f(K)$ must have measure zero as well. The argument should generalize without difficulty to $\mathbb{R}^{n}$.