Let $G$ be a finitely generated group with symmetric generating set $S$. Then $S$ induces a distance $d$ on $G$ by letting $d(a,b) = $ the minimum $n$ such that there are generators $s_1,...,s_n$ with $as_1...s_n = b$. In cases where $G$ is of polynomial growth, it is known that the metric space $(G,d)$ satisfies the doubling property, that there is a constant $C$ such that every ball of radius $r$ can be covered by $C$ or fewer balls of radius $r/2$. However, it is also known that examples exist of metric spaces with the doubling property that do not satisfy the Besicovitch covering theorem (BCT), such as the Heisenberg group. My question is this: what additional conditions on the metric space $(G,d)$ guarantee that $(G,d)$ will satisfy the BCT? Both necessary and sufficient conditions would be helpful, so any help with either is appreciated.
Here is one formulation of BCT: there is a constant $n$ dependent only on $G$ and $d$ such that the following holds. Let $F$ be a family of balls in $(G,d)$, that is a family of pairs $(g,r)$ representing the ball centered at $g$ of radius $r$, where $g \in G$ and $r$ is a natural number. Assume that every element of a finite set $E$ appears as the $g$ in some pair $(g,r)$ in $F$. Then there is a subfamily $B$ of $F$ such that $B$ covers $E$ and every element $x$ of $E$ is contained in at most $n$ elements of $B$.