I read online that the classical Hardy-Littlewood maximal function with respect to Lebesgue measure can be generalized to the so-called doubling measures on metric spaces. I am not sure how to show that this generalization is measurable. To be more specific, let $\mu$ be a doubling Borel measure on a metric space $X$, let $\lambda$ be a complex Borel measure on $X$, and let $$ (M\lambda)(x)=\sup_{0<r<\infty} \frac{\lambda(B(x,r))}{\mu(B(x,r))}. $$ Why is $M\lambda$ Borel measurable?

For the classical Euclidean case where $\mu$ is Lebesgue measure on $R^{k}$, $M\lambda$ is lower semicontinuous, and the proof I read (Section 7.2 of Rudin's RCA) seems to rely crucially on a "homogeneity" property of Lebesgue measure: for every pair of points $x$ and $y$ in $R^{k}$ and for every pair of positive real numbers $r$ and $\alpha$, $$ \mu(B(x,\alpha r))=\alpha^{k}\mu(B(y,r)).\tag{$*$} $$ This condition seems stronger than the doubling condition, and I'd come to suspect that $M\lambda$ need not be lower semicontinuous in the general case, but I'm unsure of how to show this or whether this is true.

I'm also curious if there are other examples of measures that satisfy ($*$) other than Lebesgue measure on Euclidean space or the circle, or the counting measure on the integers. I played around with density functions to no avail...

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    $\begingroup$ What's a doubling measure? $\endgroup$ – zhw. Jun 24 '17 at 22:32
  • $\begingroup$ @zhw. a Borel measure $\mu$ on a metric space $X$ is doubling if $$0<\mu(B(x,2r))\leq C\mu(B(x,r))<\infty$$ for every point $x$ in $X$ and positive number $r$, where $C$ is a fixed positive constant $\endgroup$ – user363464 Jun 24 '17 at 22:36

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