A Problem on $m$ dimensional Lebesgue Measure I am currently solving my assignment on Lebesgue measures on $\mathbb{R}^n$ and I got stuck in one problem.
Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a $C^1$ function i.e. all first order partial derivatives of the components of $f$ is continuous. If  $m>n$, then show that the image of $f$ a zero $m$- dimensional Lebesgue measure.
I have no idea how to start. I only know that any line in $\mathbb{R}^2$ has  $2$-dimensional Lebesgue measure zero.
Thank you in advance for help.
 A: I will answer your question from the comments here.
Assuming that you have already proven that $f$ is Lipschitz-continuous on every compact subset of $\mathbb{R}^n$, let $K \subset \mathbb{R}^n$ be compact and let $M > 0$ such that $$\forall x,y \in K: ||f(x) - f(y)||_{\mathbb{R}^m} \leq M \cdot ||x-y||_{\mathbb{R}^n}. $$ We want to show that $f(K)$ has Lebesgue measure zero. To this end, we show that for every $\varepsilon > 0$, there is a collection $\{B_n \}_{n \in \mathbb{N}}$ of open balls in $\mathbb{R}^m$ such that $$f(K) \subset \bigcup_{n=1}^\infty B_n \text{ and } \sum_{n=1}^\infty \mathrm{vol}(B_n) < \varepsilon. $$ There is, however, an issue: the volume in $\mathbb{R}^n$ is different than the volume in $\mathbb{R}^m$ when $n \neq m$. A quick get-away is the following idea: define $$\pi: \mathbb{R}^m \simeq \mathbb{R}^n \times \mathbb{R}^{m-n} \to \mathbb{R}^m, \pi(x,y) = f(x). $$ Observe that $\pi$ is Lipschitz-continuous on every compact subset of $\mathbb{R}^m$ and it is also $C^1$. We also have that $\pi(K \times \{0\}^{m-n}) = f(K)$ and it is very easy to show that $K \times \{0\}^{m-n}$ has Lebesgue measure zero in $\mathbb{R}^m$, as $m > n$ (this is because you can cover $K$ with balls in $\mathbb{R}^n$, which have zero volume when regarded in $\mathbb{R}^m$ - alternatively, you can use open rectangles instead of open balls). Now, you only need to show that a Lipschitz-continuous function from $\mathbb{R}^m$ to $\mathbb{R}^m$ maps measure-zero sets to measure-zero sets. A solution can be found here: Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?
Proving the stronger statement (that $f(K)$ has Jordan content zero) follows immediately using the same idea as above: $K \times \{0\}^{m-n}$ even has Jordan content zero in $\mathbb{R}^m$, as it is compact (so it can be covered by a finite number of open balls).
