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Let $(M,d)$ be a metric space (separable, complete, better?) and let $\mu$ be a ($\sigma$-additive, positive, locally finite, regular?) Borel measure on $M$. For $x\in M$ consider the volume growth function $f_x:[0,+\infty)\to[0,+\infty]$ given by $$ f_x(r)=\mu(B(x,r)),\quad B(x,r)=\{y\in M\,|\quad d(x,y)<r\}. $$

Question: What are some standard sufficient conditions on $(M,d,\mu)$ such that $f_x$ is absolutely continuous? I need references.

This will require certain homogeneity of $\mu$ and $(M,d)$, and even then may not be true for all $x\in M$ but only some. Seems like a classical problem but I cannot find it addressed in any of the many treatments of metric spaces. Let me stress again that I am looking for a reference where this problem is considered.

Thank you.

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  • $\begingroup$ If we start from a measure $\nu$ on $(X,d)$ with convenient boundedness assumptions, a measure $\mu$ like that should be $$\mu(E):=\int_0^\infty \nu(E\cap \{d(x,a)=t\})dt.$$ It is not completely clear if the resulting measure $\mu$ is independent of $a$, nor if any $\mu$ can be represented this way by disintegration, since $\nu$ will not be sigma additive on $X$. $\endgroup$
    – Pietro Majer
    Commented Jul 30, 2023 at 12:21
  • $\begingroup$ That $\mu$ depends on $a$ is not a problem. I don't need $f_a$ to be a.c. for all $a$. But $\mu$ indeed needs to be $\sigma$-additive. This is, however, not really an answer to the question, because checking whether a given $\mu$ is of this form may be as hard as verifying a.c. directly. $\endgroup$
    – Bedovlat
    Commented Jul 30, 2023 at 18:55
  • $\begingroup$ Sorry, above I meant "$\nu$ will not be sigma finite ". $\endgroup$
    – Pietro Majer
    Commented Jul 30, 2023 at 19:36

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