# Hausdorff approximating measures and Borel sets

Suppose $1 \leq m \leq n$ are integers and for each $0 < \delta < \infty$ let $\mathscr{H}^{m}_{\delta}$ be the size $\delta$ approximating measure of the $m$ dimensional Hausdorff measure $\mathscr{H}^{m}$ of $\mathbf{R}^{n}$. Recall $\mathscr{H}^{m}(A) = \sup_{\delta > 0} \mathscr{H}^{m}_{\delta}(A)$ whenever $A \subseteq \mathbf{R}^{n}$.

Is it true that for every Borel (compact) subset $B \subset \mathbf{R}^{n}$ there exists $0 < \delta < \infty$ such that $B$ is $\mathscr{H}^{m}_{\delta}$ measurable?

Of course it is well known (and easy to prove) that if $m < n$ (the case $m = n$ is excluded because $\mathscr{H}^{n}_{\delta}$ is equal the $n$ dimensional Lebesgue measure for every $0 < \delta < \infty$) then for every $0 < \delta < \infty$ there exists a Borel set that is not $\mathscr{H}^{m}_{\delta}$ measurable. For example if $n = 2$ and $m= 1$ it is not difficult to see that the boundary of an open ball with radius $\delta/2$ is not $\mathscr{H}^{1}_{\delta}$ measurable.

• Doesn't a non-measurable one-dimensional subset of the real line, thought of as a subset of $\mathbb R^2$ give an immediate counterexample? – Anthony Quas Jan 22 '17 at 17:11
• Thank you for your interest in this question. I am not sure to have got your point. When you say "non measurable" what measaure do you think about? – Longyearbyen Jan 22 '17 at 17:45
• $\mathcal H^1_\delta$-measurable? – Anthony Quas Jan 22 '17 at 18:24
• But why should this set be Borel? – Longyearbyen Jan 22 '17 at 18:28
• Sorry - I didn't read your question properly. – Anthony Quas Jan 22 '17 at 18:37

There is a difference between $m = n$ and $m < n$. In case $m=n$, all the outher measures $\mathscr H^n_\delta$ are equal to the Lebesgue measure. So each Borel set in $\mathbb R^n$ is $\mathscr H^n_\delta$-measurable for all $\delta > 0$.
For $m < n$, the following is given as Exercise 2.4.5 in the book "Topics on analysis in metric spaces" by Ambrosio and Tilli. It is written there that this is due to Kirchheim:
If $\delta > 0$ and $A$ is $\mathscr H^m_\delta$-measurable, then $\mathscr H^m(A) = 0$ or $\mathscr H^m(\mathbb R^n \setminus A) = 0$.
As a particular consequence, the closed unit ball in $\mathbb R^n$ is not $\mathscr H^m_\delta$-measurable if $\delta > 0$ and $0 \leq m < n$.