Image of Lipschitz map measure zero

Question

Let $f\colon\mathbb{R}^{n-1}\to\mathbb{R}^n$ be Lipschitz, i.e., $|f(x)-f(y)|\leq C|x-y|$ for some $C>0$.

How do we show that the image $f(\mathbb{R}^{n-1})$ has Lebesgue measure zero?

I can see that $\mathbb{R}^{n-1}$ itself has measure zero in $\mathbb{R}^n$, for instance $\mathbb{R}$ is the line which is measure zero in the plane $\mathbb{R}^2$.

However I am not sure how to prove the above rigorously.

Thanks!

Answer 1

Below are two proofs of this fact. Neither is self contained; the first is more classical, but I added the second as I find it very interesting.

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Let $\mu$ denote the $n$-dimensional Lebesgue measure, and let $\mathcal{H}^n$ be the $n$-dimensional Hausdorff measure. Note that

In $\mathbb{R}^n$, $\mu=c\mathcal{H}^n$, for some positive constant $c$.

Depending on the normalization of $\mathcal{H}^n$, the constant can be made $1$. The proof of this result is not trivial; I can tell you more about it, if you are interested.

Recall that, for $E \subset \mathbb{R}^n$, $\delta>0$ $$ \mathcal{H}_{\delta}^n(E)=\inf \left\{\sum_{i=1}^{\infty} \operatorname{diam}(E_i)^n \,\middle|\, \bigcup_{i=1}^{\infty}E_i \supset E, \ \operatorname{diam}(E_i)\leq \delta\right\} $$ and $$ \mathcal{H}^n(E)=\sup_{\delta>0}\mathcal{H}_{\delta}^n(E)=\lim_{\delta \to 0} \mathcal{H}_{\delta}^n(E). $$

Moreover, the infimum can be taken over open balls $B_i$.

We have, since $f$ is Lipschitz,
$$ \mathcal{H}_{\delta}^n(f(\mathbb{R}^{n-1})) \leq \sum_{i=1}^{\infty} \operatorname{diam}(f(B_i))^n \leq C^n \operatorname{diam} (B_i)^n. $$ By taking the infimum over all coverings with balls of radius less or equal to $\delta$, and then taking the limit for $\delta \to 0$, we get $$ \mathcal{H}^n(f(\mathbb{R}^{n-1})) \leq C^n \mathcal{H}^n(\mathbb{R}^{n-1})=0 $$ and so we are done.

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Let $A=f(\mathbb{R}^{n-1}) \subseteq \mathbb{R}^n$.

If we can prove that in a set of full measure $A$ has Lebesgue density smaller than $1$, then it follows that $\mu(A)=0$, as by Lebesgue density theorem, we know that for almost every point in $A$, $A$ has density $1$.

By Rademacher's theorem, $f$ is differentiable at almost every (with respect to $n-1$ dimensional Lebesgue measure) point of $\mathbb{R}^{n-1}$. The images of such points have full measure in the image, and at those point, $A$ has a tangent plane.

Let $y=f(x)$ and let $v$ be a unit vector normal to the tangent plane to $A$ at $y$. Let $r>0$ be small enough. Note that $B\left(y+\frac{r}{2}v, \frac{r}{4}\right) \cap A = \emptyset$ (a simple picture will clarify this statement).

We have that $\mu(B(y,r)\cap A) \leq c \mu(B(y,r))$, where $c<1$ does not depend on $r$. In fact, $$ B(y,r)\cap A \subseteq B(y,r) \smallsetminus B\left(y+\frac{r}{2}v, \frac{r}{4}\right), $$ and the latter ball has measure a fixed proportion of $B(y,r)$.

Then, $$ \lim_{r \to 0} \frac{\mu(B(y,r)\cap A)}{\mu(B(y,r)}=c <1. $$