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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.

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  • $\begingroup$ 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? $\endgroup$ – Anthony Quas Jan 22 '17 at 17:11
  • $\begingroup$ 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? $\endgroup$ – Longyearbyen Jan 22 '17 at 17:45
  • $\begingroup$ $\mathcal H^1_\delta$-measurable? $\endgroup$ – Anthony Quas Jan 22 '17 at 18:24
  • $\begingroup$ But why should this set be Borel? $\endgroup$ – Longyearbyen Jan 22 '17 at 18:28
  • $\begingroup$ Sorry - I didn't read your question properly. $\endgroup$ – Anthony Quas Jan 22 '17 at 18:37
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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 $0 \leq m < n$.

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  • $\begingroup$ Exercise 2.4.5 in the book by Ambrosio and Tilli has a hint so you might want to look into that book if you are interested how to prove it. $\endgroup$ – Piotr Hajlasz Mar 23 '18 at 18:37

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