For $0< s < n$, denote by $\mathscr{H}^s$ the $s$-dimensional Hausdorff measure. Let $E\subseteq \mathbb{R}^n$ be $\mathscr{H}^s$-measurable, and $f:\mathbb{R}^n \to \mathbb{R}^n$ a bi-Lipschitz mapping, this is, there exist two constants $C_1, C_2 >0$ such that $$C_1 |x-y| \le |f(x) - f(y)| \le C_2|x-y|,\;\forall x,y \in \mathbb{R}^n.$$
Question: Can we conclude that $f(E)$ is $\mathscr{H}^s$-measurable?
I have known
- For any Borel subset $B \subseteq \mathbb{R}^n$, the set $f(B)$ is Borel;
- If $E\subseteq \mathbb{R}^n$ with $\mathscr{H}^s(E)=0$, then $\mathscr{H}^s(f(E))=0$.
As we all know, a Lebesgue measurable set can be written as the union of a Borel set and a Lebesgue null set. Thus, if $E$ is Lebesgue measurable, then $f(E)$ is also Lebesgue measurable. However we don't know the $\sigma$-algebra of $\mathscr{H}^s$-measurable sets.