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

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.

• I suppose it should be $m(Ball(x,r)) \leq C_m m(Ball(x,r/2))$ (instead of $=$), as in the linked paper? Mar 7, 2022 at 11:32
• What is your definition of "metric measure space"? If you allow some nonempty open subset of $X$ to have measure zero, then of course no measure doubling property can imply metric doubling. Mar 7, 2022 at 15:53

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 $$C_d\leq C_m^4$$.