In my last post I asked the following question:
Given a metric space $X$, let $B_r(x):=\{y\in X\mid d(x,y)\leq r\}$ be the closed balls. Fix a real number $p>0$. My question now is: Is there a Borel-measure $\mu$ on $X$ such that \begin{align*} \big(\operatorname{diam} B_r(x)\big)^p=\mu(B_r(x)),\qquad r>0,x\in X? \end{align*}
The question had been answered with "No (in general)", and a counterexample can easily be found. Since I do only need inequality instead of equality in the above statement, my new question reads as follows:
Under the above circumstances, is there a Borel-measure $\mu$ on $X$ such that \begin{align*} \big(\operatorname{diam} B_r(x)\big)^p\leq\mu(B_r(x)),\qquad r>0,x\in X? \end{align*}
Thanks in advance.
Sure, you can let $\mu(E)=\infty$ for every nonempty set $E$.
But if you want $\mu(B_r(x))$ to be finite, then no. Consider $\mathbb{R}$ with the metric $d(x,y)=\min(|x-y|,1)$. Then $B_1(0)$ contains $B_{1/3}(n)$ for every $n\in\mathbb{Z}$, and all these balls are disjoint, so $\mu(B_1(0))=\infty$.
The property you stated is related to what is known as "lower Ahlfors regularity" in the literature; specifically, $X$ is lower Ahlfors regular with exponent $p$ if there is $c>0$ such that $\mathcal H^p(B_r(x))\ge Cr^p$ for all $x$ and $r$.