Fix a metrix space $(X,d)$ and consider the Hausdorff (outer) measures $\mathcal{H}^s$ on $X$.

A Frostman measure on $X$ is a finite Borel measure $\mu$ such that there exists $C,t,r_0>0$ with $\mu(B_r(x)) \leq C r^t$ for all $x\in X$ and all $0<r\leq r_0$. Let's call the supremum of such $t$ the Frostman-Exponent $FrostExp(\mu)$.

It is well known that any such measure satisfies $$\mu(A) \leq C \mathcal{H}^t(A)$$

In particular: The Hausdorff dimension satisfies $$\dim_\mathcal{H}(A)\geq \sup\lbrace FrostExp(\mu) \mid \mu \text{ Frostman measure with }\mu(A)>0\rbrace$$ Frostman himself proved that in fact this is essentially an equality: For any Borel set $A$ with $\mathcal{H}^t(A)>0$ there exists a Frostman measure of exponent $\geq t$ with $\mu(A)>0$.

Now my question is: Can we also say something about the precise value of the Hausdorff measure in terms of Frostman measures?

A naive conjecture might be that the above inequality is sharp in the sense that if $\infty>\mathcal{H}^t(A)>0$ then $$\sup\lbrace \mu(A) \mid \mu \text{ Borel measure with } \exists r_0 \forall 0<r<r_0: \sup_{x\in X}\mu(B_r(x))r^{-t} \leq 1\rbrace = \mathcal{H}^t(A)$$ I suppose that this is not true. Mostly because it seems to me that this would be so nice that it would be explicitely stated somewhere in textbooks that prove Frostman's lemma and I have yet to see this claim in the literature.

I have fiddled around with some of the constructions (like the one using the max-flow-min-cut theorem on an infinite tree) that come up in the proof of Frostman's lemma to see if they produce measures that come close to something like this. I haven't had any luck so far.

I would be glad if someone could tell me if a similar equality actually holds. Feel free to add additional assumptions on $X$ or $A$, I do not need the most general case although I am of course interested in it.

  • $\begingroup$ Just a half-backed thought: one could try to prove that the Hausdorff measure itself could be optimal, but this would not work in a hyperbolic manifold (the Hausdorff measure is the usual volume up to normalization, and hyperbolic balls have volume strictly greater than their Euclidean counterparts). So, maybe it is worth trying the case of a hyperbolic manifold, trying to prove that any measures satisfying the required inequality on balls must be bounded above by $f\,\mathrm{dvol}$ with $f<1-\varepsilon$, where $\varepsilon$ would depend on the injectivity radius of the manifold. $\endgroup$ Dec 31, 2014 at 11:00
  • $\begingroup$ @BenoîtKloeckner Does that really work?? All riemannian manifolds can be isometrically embedded into some euclidean space and isometries respect Hausdorff (and Frostman) measures so that there shouldn't be any difference between (hyperbolic) manifolds and subsets of euclidean spaces. $\endgroup$ Jan 6, 2015 at 18:24
  • $\begingroup$ I do not know whether this really work, it is half-backed after all. But is there a reason why things should be much better for $X$ a subset of Euclidean space? $\endgroup$ Jan 6, 2015 at 19:50
  • $\begingroup$ No, there isn't. I just wanted to point out that there is no particular reason to focus on "complicated" objects because anything that does happen already happens inside some $\mathbb{R}^n$. $\endgroup$ Jan 6, 2015 at 21:11
  • $\begingroup$ My remark is meant to help find possible counter-examples: trying all subset of $\mathbb{R}^n$ is hopeless, while hyperbolic (or negatively curved) manifold have that special property that make them plausible counter-examples. $\endgroup$ Jan 7, 2015 at 12:10

1 Answer 1


I think I found an answer for $X\subseteq\mathbb{R}^n$ (which is good enough for me). In Mattila's book there is a theorem about densities stating that the upper density of the $t$-dimensional Hausdorff measure on a $\mathcal{H}^t$-finite set $A\subseteq\mathbb{R}^n$ is $\mathcal{H}^t$-almost everywhere bounded by 1, i.e.

$$ \limsup_{\delta \downarrow 0} \mathcal{H}^t(A\cap B_\delta(x))/(2\delta)^t \leq 1$$

for $\mathcal{H}^t$-almost all $x\in A$. Moreover there is a comment (of the form "one can show" without proof) that more generally $$\lim_{\delta\downarrow 0} \sup \left\lbrace \mathcal{H}^t(A\cap E)/diam(E)^t \mid x\in E, diam(E)<\delta\right\rbrace = 1$$ holds for $\mathcal{H}^t$-almost all $x\in A$. This would means that the Hausdorff measure itself is a Frostman-like measure up to a null set: There is a set $A_0\subseteq A$ of full measure such that the equality holds and thus for any $\epsilon>0$ there is a $r_0$ with $\mathcal{H}^t(A\cap E) \leq (1+\epsilon) diam(E)^t$ for all $r<r_0$ and all sufficiently small $E\subseteq A_0$. Thus $\mu(C):=(1+\epsilon)^{-1}\mathcal{H}^t(A_0\cap C)$ is a finite Borel measure supported on $A$ satisfying $\mu(E) \leq diam(E)^t$ and $\mu(A)=(1+\epsilon)^{-1}\mathcal{H}^t(A)$ so that one finds the equation $$\mathcal{H}^t(A) = \sup \left\lbrace \mu(A) \mid \mu \text{ Borel measure s.t. }\exists\delta>0:\sup\limits_{diam(E)<\delta} \mu(E)diam(E)^{-t}\leq 1\right\rbrace$$

The presence of the factor $2^t$ in the first inequality makes it implausible to me that my conjectural equation with the supremum over all Frostman measures really holds. Instead a similar density theorem for the spherical Hausdorff measure proves that the analogue of my conjecture holds for the spherical Hausdorff measure. (In particular: The idea with the hyperbolic manifolds will not work because for $t=n$ the usual and the spherical Hausdorff measure coincide)

I'm very satisfied with this result although I wouldn't mind any additions regarding more general situations.


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