Let $(X,d,m)$ be a metric measure space. We say that it is *doubling in the sense of metric spaces* if for every:
$x\in X$ and every $r>0$ there exists some **(metric) doubling constant** $C\_d\geq 0$ such that
$$
Ball(x,r) \mbox{ can be covered by at-most $C\_d$ balls of radius $r/2$}.
$$


There is a different, [related and in some spirit "equivalent"](https://www.ams.org/journals/proc/1998-126-02/S0002-9939-98-04201-4/S0002-9939-98-04201-4.pdf), notion of *doubling in the sense of metric measure spaces*, which states that there is a constant $C\_m\geq 0$ such that: for every $x\in X$ and each $r>0$
$$
m(Ball(x,r)) \leq C\_m m(Ball(x,r/2)).
$$


If $(X,d,m)$ is doubling in the sense of metric measure spaces, with constant $C\_m$, then is it doubling in the sense of metric spaces? And if so, can we $C\_m$ to deduce an upper-bound for $C\_d$?


Note, I'm most interested in the case where $m$ is an $s$-dimensional Hausdorff measure.



Relationship between doubling constant of a metric space and of a metric measure space
Apart from the obvious counterexample of the measure being $0$, if $(X,d,m)$ is doubling in the sense of metric measure spaces it will be doubling in the sense of metric spaces.


Consider a ball $B(x,r)$. If for some $n$, $B(x,r)$ cannot be covered by $n$ balls of radius $\frac{r}{2}$, then we can obtain by recursion a sequence of points $x\_1,\dots,x\_n$ in $B(x,r)$ which are pairwise at distance $\geq\frac{r}{2}$. Thus the balls $B(x\_i,\frac{r}{4})$ are disjoint, and they are all contained in $B(x,2r)$. Suppose $m(B(x\_1,\frac{r}{4}))$ is the smallest of all the $m(B(x\_i,\frac{r}{4}))$.


Then $m(B(x\_1,4r))\geq m(B(x,2r))\geq\sum\_{i=1}^nm(B(x\_i,\frac{r}{4}))\geq nB(x\_1,\frac{r}{4})$, so either $n\leq C\_m^4$ or $m(B(x\_1,4r))=0$. If you only consider finite distances the second option implies that $m(X)=0$.


So an upper bound would be $C\_d\leq C\_m^4$.


