6
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$ Oct 29, 2019 at 17:08
  • $\begingroup$ What do you mean when you write $\mathcal{H}^n_{\infty}$? $\endgroup$
    – Amir Sagiv
    Oct 29, 2019 at 17:09
  • $\begingroup$ Amir, for the definition of Hausdorff content see e.g. math.stonybrook.edu/~bishop/all2.pdf $\endgroup$ Oct 29, 2019 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. $\endgroup$ Oct 31, 2019 at 3:34

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you lots! $\endgroup$ Oct 31, 2019 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.