Let $(X,d,m)$ be a proper metric measure space, i.e.
- closed ($d$-)balls are compact
- $m$ is a Borel atomless measure, finite and non-zero on every ball of positive radius
(If necessary, we can assume that $(X,d)$ is complete.)
Question: What are (mild) sufficient conditions such that
for every $x_0\in X$ there exists $r_0=r_0(x_0)$ such that $F_{x_0}\colon r\mapsto m B_r(x_0)$ is continuous on $[0,r_0)$
Comments:
It is clear that the assertion does not hold on arbitrary metric measure spaces (see here).
Furthermore, it is always true that for every $x_0\in X$ the function $F_{x_0}$ has up to countably many discontinuity points (see here).
Also, $F_{x_0}$ is continuous at $0$, since $m$ is assumed atomless
The property is stable under multiplication by $L^p_{\rm loc}(m)$ densities ($p>1$, but maybe not $p=1$), in the sense that if the assertion is true for $m$, then it is true for $fm$ for any density $f$ as above
It seems reasonable to expect the property to hold for $\mathsf{CD}(K,N)$ spaces, $\mathsf{MCP}(K,N)$ spaces, and maybe spaces satisfying a Bishop–Gromov-type or a Brunn–Minkowski-type inequality, but I would very much like some simpler assumptions: e.g. doubling (and, if needed, weak (1,1)-Poincaré, however I do not see why it could be needed).
5'. Bishop–Gromov seems a reasonable assumption because in that case $x\mapsto m B_r(x)$ is continuous for each fixed $r$ (see here, Lemma A.1)
A proof/reference for $\mathsf{MCP}(K,N)$ spaces would also be much appreciated.
While this seems reasonable to expect, I am not able to show the sought property for the Hausdorff measure on Ahlfors regular spaces (definition here)