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?



Bilipschitzian maps and densities
This is actually true since a density point of A is mapped to a density point of B as shown for example in:


[Z Buczolich, Density points and bi-Lipschitz functions in ^m, Proceedings of the American Mathematical Society 116 (1), 53-59](http://www.ams.org/journals/proc/1992-116-01/S0002-9939-1992-1100645-5/)


A only slightly different proof than the one offered in the paper above could probably be worked out along these lines:


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 $F(A)=B$.


