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 $\mu(A)>0$.



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



Calculate Hausdorff measure with Frostman measures
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.


