TITLE: When is Hausdorff measure a Frostman measure?
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: 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)