# Bilipschitzian maps and densities

Let $A \subseteq \mathbf{R}^{m}$ and suppose that $\mathbf{R}^{m} \setminus A$ has $m$ dimensional density equals $0$ at a point $a \in A$. Let $B \subseteq \mathbf{R}^{m}$ and let $f : A \rightarrow B$ be a bilipschitzian map onto $B$. Is it true that $\mathbf{R}^{m} \setminus B$ has $m$ dimensional density equals $0$ at $f(a)$?

I suspect it should be false. Anyone knows some counterexample?

First extend $f$ to a Lipschitz map $F : \mathbb R^m \to \mathbb R^m$ using for example Kirszbraun’s extension theorem (or the McShane-Whitney extension theorem in each coordinate separately). Since $a$ is a density point of $A$ we can find for any $\epsilon \in (0,1)$ an arbitrary small radius $r > 0$ such that $$\mathcal H^{m-1}(A \cap \partial B(a,r)) \geq (1-\epsilon)\mathcal H^{m-1}(\partial B(a,r)).$$ If we choose $\epsilon$ small enough, then $F(\partial B(a,r))$ doesn't contain $f(a)$ and hence the degree $\deg(B(a,r),F,f(a))$ is well defined. Indeed $\operatorname{dist}(f(a),\partial B(a,r)) \geq c r$, where $c$ depends on the Lipschitz constants of $f,f^{-1}$ and $F$ (assuming $\epsilon$ is small enough). Since $f$ is bi-Lipschitz on $A$ and $F$ is Lipschitz we see that $$\lim_{r\to 0+} \frac{\mathcal H^{m}(F(B(a,r)\setminus A))}{\mathcal H^{m}(F(B(a,c\operatorname{Lip}(F)^{-1}r)\cap A))} = 0.$$ Hence for small $r$ there is a positive measure set of points $p \in B(f(a),cr)\cap F(B(a,r) \cap A)$ for which the preimage $B(a,r) \cap F^{-1}(p)$ consists of a single point $q \in A$ at which $F$ is differentiable. Therefore $$\deg(B(a,r),F,f(a)) = \operatorname{sign}\operatorname{det}DF_p = \pm 1.$$ Note that $DF$ is almost everywhere nondegenerate on $A$ which is implied by the chain rule and the fact that $f$ is bi-Lipschitz on $A$. The formula above for the mapping degree of a Lipschitz map can be found for example in Corollary 4.1.26 of Federer's geometric measure theory book. From a property of the mapping degree we now obtain that $B(f(a),cr) \subset F(B(a,r))$. As we have used more or less already above \begin{equation} \lim_{r\to 0+}\frac{\mathcal H^{m}(B(f(a),cr)\setminus F(A))}{r^m} \leq \lim_{r\to 0+}\frac{\mathcal H^{m}(F(B(a,r)\setminus A))}{r^m} = 0. \end{equation} Hence $f(a)$ is a density point of $F(A)=B$.