Relationship between doubling constant of a metric space and of a metric measure space

Question

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", 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.

Answer 1

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 $C_d\leq C_m^4$.