Let $\mu$ be a finite, compactly supported, non-zero measure on $\mathbb{R}^d$ for an integer $d$. Let $B(x,r)$ denote the ball of radius $r>0$ centered at $x \in \mathbb{R}^d$. For $\delta \in [0,d]$, we define two conditions: $\mu$ satisfies (C1)-$\delta$ if
$$ \mu(B(x,r)) \leq r^\delta $$
for all $x \in \mathbb{R}^d$ and $r >0$; and $\mu$ satisfies (C2)-$\delta$ if
$$ \mu(B(x,r)) \geq r^\delta$$
for all $r \in (0,1]$ and $x \in \text{supp}(\mu)$, the closed support of $\mu$.
Frostman's Lemma characterizes the sets which support measures satisfying (C1)-$\delta$.
Frostman's Lemma. A Borel set $A \subset \mathbb{R}^d$ supports a non-zero measure satisfying (C1)-$\delta$ if and only if $A$ has positive $\delta$-dimensional Hausdorff measure.
I am interested in a complimentary result concerning measures which satisfy (C2)-$\delta$. Observe that any non-empty set supports a point measure, which satisfies (C2)-$0$ and hence satisfies (C2)-$\delta$ for any $\delta > 0$. So any non-empty set trivially supports a measure satisfying (C2)-$\delta$. The question is rather when a given set is equal to the closed support of such a measure. In this case we only need consider closed sets. We will also restrict to $A$ being compact.
It can be shown using a covering argument that if $A$ is equal to the support of a (C2)-$\delta$ measure, then its Hausdorff dimension is at most $\delta$. This gives a necessary condition. The problem is then as follows.
Problem. Give a sufficient (or necessary and sufficient) condition (e.g. in terms of Hausdorff measure) on a compact set $A \subset \mathbb{R}^d$ which implies the existence of a finite measure $\mu$ satisfying (C2)-$\delta$ such that the closed support of $\mu$ equals $A$.
