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Let $(X,d)$ be an $\mathcal{H}^n$-rectifiable metric space, i.e. there exits a collection of Lipschitz maps from measurable subsets of $\mathbb{R}^n$ to $X$ such that $ \mathcal{H}^n(X \backslash \cup_i f_i(A_i)) = 0 $.

Is it true that for any subset $A \subset X$, $$ \mathcal{H^n}(A) = \mathcal{H}^n_\infty (A) \ .$$

The claim is true on $X = \mathbb{R}^n$. Note that the same equality fails horribly if we consider $\mathcal{H}^k$ for $k<n$ -- think of an infinitely long curve inside a bounded set.

If this helps: I am interested in small scales, so, you might consider asymptotic behavior as $\text{diam} (A) \to 0$.

Thanks for your consideration.

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    $\begingroup$ What if X is a circle in the plane with the metric inherited from the plane? To handle the asymptotic version consider a union of countably many circles with shrinking radii. $\endgroup$ – Yuval Peres Oct 29 '19 at 17:08
  • $\begingroup$ What do you mean when you write $\mathcal{H}^n_{\infty}$? $\endgroup$ – Amir Sagiv Oct 29 '19 at 17:09
  • $\begingroup$ Amir, for the definition of Hausdorff content see e.g. math.stonybrook.edu/~bishop/all2.pdf $\endgroup$ – Yuval Peres Oct 29 '19 at 17:14
  • $\begingroup$ Thanks @YuvalPeres I in fact like your answer better because the problem is not just at a single point in your example, unlike the accepted answer. $\endgroup$ – Behnam Esmayli Oct 31 '19 at 3:34
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In general, no. For example, $X$ may be a countably infinite collection of lines through the origin in $\mathbb{R}^2$. Then $X$ is $1$-rectifiable.

For any ball $B$ centered at the origin, $B\cap X$ has finite Hausdorff $1$-content but infinite Hausdorff $1$-measure.

If you have some kind of Ahlfors regularity of the measure, then you can get comparability of Hausdorff measure and Hausdorff content.

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  • $\begingroup$ Thank you lots! $\endgroup$ – Behnam Esmayli Oct 31 '19 at 3:35

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