Let $(X,d)$ be a metric space and let $\mathcal{H}^s$ be the $s$-dimensional Hausdorff measure on $X$.

For a measure $\mu$ on $X$, we say that $\mu$ is a Frostman measure (sometimes referred as upper Ahlfors regular measure) if for some $\alpha >0$, $\mu(B(x,r))\leq C r^\alpha$ for every $x \in X$ and $r>0$.

Under what conditions on $s$ and $d$ is $\mathcal{H}^s$ a Frostman measure?

Some (silly) partial answers. For $X=\mathbb{R}^d$, $\mathcal{H}^d$ is clearly a Frostman measure, as it is a scalar multiple of Lebesgue measure. For $s>d$, $\mathcal{H}^s$ satisfies, trivially, $0=\mathcal{H}^s(B(x,r)) \leq C r^s$, so the answer is positive also in this case. For $s<d$, we have $\mathcal{H}^s(B(x,r)) = \infty$ so the inequality can't hold.

Obviously, if the metric space equipped with the measure $\mathcal{H}^s$ is $s$-Ahlfors regular, the measure is Frostman.

I've tried to think about the more general case, but I fail to have any intuition on a metric space $X$, so apologies if the answer turns out to be easy.

  • $\begingroup$ Related: mathoverflow.net/questions/191828/… $\endgroup$ Dec 22, 2015 at 11:40
  • $\begingroup$ @JohannesHahn Thanks! I've seen that question (it's the first Google result googling "Hausdorff measure and Frostman measure") and it's very interesting, even though not too related. $\endgroup$ Dec 22, 2015 at 11:44

1 Answer 1


K. Falconer, Fractal Geometry: Mathematical Foundations and Applications (3rd ed, 2014, Wiley). page 77

Corollary 4.12 Let $F$ be a Borel subset of $\mathbb R^n$ with $0 < \mathcal H^s(F) \le \infty$. Then there is a compact set $E \subset F$ such that $0 < \mathcal H^s(E) < \infty$ and a constant $b$ such that $$ \mathcal H^s\big(E \cap B(x,r)\big) \le br^s $$ for all $x \in \mathbb R^n$ and $r>0$.

$ $

Corollary 4.12, which may be regarded as a converse of the Mass distribution principle 4.2, is often called 'Frostman's lemma'.

The notes suggest for a complete proof: P. Mattila Geometry of Sets and Measures in Euclidean Space (1999, Cambridge Univ Press)

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    $\begingroup$ 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. $\endgroup$ Dec 22, 2015 at 13:50
  • $\begingroup$ 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. $\endgroup$ Dec 22, 2015 at 15:13
  • $\begingroup$ @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. $\endgroup$ Dec 22, 2015 at 15:49
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    $\begingroup$ @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. $\endgroup$ Dec 22, 2015 at 18:24

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