Timeline for When is Hausdorff measure a Frostman measure?
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| Dec 22, 2015 at 18:24 | comment | added | Pablo Shmerkin | @SilviaGhinassi . Frostman's Lemma for analytic subsets of complete separable metric spaces is due to Howroyd: [Howroyd, J. D. On dimension and on the existence of sets of finite positive Hausdorff measure. Proc. London Math. Soc. (3) 70 (1995), no. 3, 581--604]. See also Chapter 8 of Mattila's book for the proof in the compact case. | |
| Dec 22, 2015 at 15:49 | comment | added | Silvia Ghinassi | @PabloShmerkin I've never heard of any version of Frostman lemma outside $\mathbb R ^d$. I would love a reference for that. Also the condition requires passing to a subset, so should I gather from that that we can't get the job done with only restrictions on $d$ and $s$ (like for instance in $\mathbb R^d$ for $s \geq d$)? Thanks. | |
| Dec 22, 2015 at 15:13 | comment | added | Pablo Shmerkin | The point of Gerald's answer is that the Frostman measure can be taken as the restriction of Hausdorff measure to a compact subset. I think this version of Frostman's Lemma should work in a separable metric space, showing that there is a wild abundance of metric spaces with the desired property. | |
| Dec 22, 2015 at 13:50 | comment | added | Silvia Ghinassi | Thanks. I am familiar with Frostman lemma, and this result from Falconer (I looked at both books you mention before asking), but it doesn't answer the main question, which is about general metric spaces. Also, Frostman lemma doesn't really apply here, as it guarantees the existence of a Frostman measure, but it doesn't guarantee that such a measure is the (positive and finite) Hausdorff measure. | |
| Dec 22, 2015 at 13:47 | history | answered | Gerald Edgar | CC BY-SA 3.0 |