Let $1 \le k \le n-1$. We say $M \subset \mathbb{R}^n$ is countably $\mathcal{H}^k$-rectifiable if $M$ is $\mathcal{H}^k$-measurable and there exist countably many Lipschitz maps $f_h : \mathbb{R}^k \rightarrow \mathbb{R}^n$ such that $$ \mathcal{H}^k(M \setminus \bigcup_{h=1}^\infty f_h(\mathbb{R}^k)) = 0. $$

In the book "Sets of Finite Perimeter and Geometric Variational Problems" on page 97 it is claimed that $M$ is countably $\mathcal{H}^k$-rectifiable if and only if there exist a Borel set $M_0$, countably many Lipschitz maps $f_h : \mathbb{R}^k \rightarrow \mathbb{R}^n$ and Borel sets $F_h \subset \mathbb{R}^k$ such that $$ M = M_0 \cup \bigcup_{h=1}^\infty f_h(F_h), \quad \mathcal{H}^k(M_0) = 0. $$

Assuming that $M$ is a Borel set, it is easy to prove the decomposition. However, I do not see how we can pick $M_0$ and $F_h$ to be Borel sets when $M$ is merely $\mathcal{H}^k$-measurable. Can someone help?

  • 1
    $\begingroup$ I haven't looked at your reference, but possibly the requirement that $M_0$ be a Borel set is an oversight by the author (or you misquoted from the book), and all that is required is that $M_0$ be an $\mathcal{H}^k$ measure zero set. Perhaps the mathoverflow question Why is there a $\mathcal{H}^d$-null set in the definition of d-rectifiable set? will be of help. $\endgroup$ Sep 18 at 18:16
  • $\begingroup$ I don't have a copy of this book (or access to it), but I found an errata list here. There is nothing for p. 97, but since you are presumably reading through the book (or at least using it for something), I thought this errata list would be useful. $\endgroup$ Sep 18 at 18:23
  • $\begingroup$ @DaveL.Renfro Thanks for the comment! I read the claim carefully one more time and $M_0$ is in fact claimed to be a Borel set. I also believe that this was an oversight. But in the following pages there is a similar statement with $M_0$ taken to be Borel again. I was curious if I was missing something simple. I am aware of the errata but thank you nevertheless! $\endgroup$ Sep 18 at 19:19


You must log in to answer this question.