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.