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Feb 28, 2022 at 13:36 comment added Iosif Pinelis @A_K : Yes, except that I think you meant $r^D$ in place of $n^D$.
Feb 28, 2022 at 9:14 comment added ABIM So if insead only required $\mu(b(x,r))\lesssim n^D$ (and not the lower-inequality) then this would only imply that $\mu$ has a density with respect to $\lambda$ but not that this density is bounded away from $0$ and only bounded away from $\infty$?
Feb 24, 2022 at 15:26 comment added Iosif Pinelis If $d\ne D$, then the inequality $cr^d\le Cr^D$ cannot hold for all $r>0$. (My previous comment was incorrect.)
Feb 24, 2022 at 15:19 comment added ABIM Right, and my definition were on a Riemannian manifold, then the same (argument...but patched locally) would work to conclude that $\mu$ must have a density with respect to the Riemannian Volume measure; no?
Feb 24, 2022 at 14:21 vote accept ABIM
Feb 24, 2022 at 14:17 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 24, 2022 at 14:11 history answered Iosif Pinelis CC BY-SA 4.0