Problem 3

Exercise 4.3 (Lipschitz functions preserve measure zero sets)   Prove that a Lipschitz function $f : \mathbb{R} \to \mathbb{R}$ maps sets of Lebesgue measure zero to sets of Lebesgue measure zero. For which values of $n$ and $m$ does the same statement hold for Lipschitz functions $f : \mathbb{R}^n \to \mathbb{R}^m$?

Proof. If $n<m$, then measure can spring up from nothing, as in the following example. Consider a line segment in $\mathbb{R}^2$, having zero measure. The projection map is Lipschitz, and sends the line segment to a subset of full measure in $[0,1]$.

If $n=m$, identify a Lipschitz coefficient $M$ and proceed by covering the image of a measure zero set $A$

$$A \subseteq \bigcup_{k=1}^\infty B_{\delta_k}(x_k) \quad \sum_{k=1}^\infty \mu(B_{\delta_k}) < \epsilon/M^n$$ (4.7)

Then the image is contained also in balls with expanded radii

$$f(A) \subseteq \bigcup_{k=1}^\infty B_{M\delta_k}(f(x_k))$$ (4.8)

A dilation by $M$ introduces a factor $M^n$, so that

$$\sum_{k=1}^\infty \mu(B_{M\delta_k}) = \sum_{k=1}^\infty M^n\mu(B_{\delta_k}) = M^n \sum_{k=1}^\infty \mu(B_{\delta_k}) < M^n\epsilon/M^n=\epsilon$$ (4.9)

If $n>m$, then zero measure sets remain zero measure by realizing that cubes in $\mathbb{R}^m$ have zero measure in $\mathbb{R}^n$ by the construction of the product measure. From this it follows that balls also have measure zero when included into higher dimensional spaces. $\qedsymbol$