Can anyone exhibit a finite-dimensional metric space (preferably, $R^d$) equipped with a measure that does not satisfy the conclusions of the Lebesgue Density Theorem? Such examples exist in infinite-dimensional spaces (e.g., Hilbert space with the Gaussian measure) but what about a finite-dimensional one?
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asked Aug 29, 2016 at 23:27
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1 Answer
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It is a theorem of Besicovitch that measures on $\mathbb R^d$ do satisfy the density theorem.
Fremlin, Measure Theory, Chap. 47
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Besicovitch, around 1930, extended his density properties of sets to those of finite Hausdorff measure.
source
next: D. G. Larman, "A new theory of dimension", Proc. London. Math. Soc. 17 (1967) 178-192
Def: a metric space is finite-dimensional in the sense of Larman iff there is a constant $K$ such that every ball of radius $2R$ can be covered by at most $K$ balls of radius $R$.
Larman proves that such finite-dimensional spaces have a Vitali-type property, which of course implies the density theorem for all measures. (Lebesgue's and Besicovitch's proofs used Vitali coverings.)
My student Manav Das investigated metric spaces with various Vitali-type properties. For example:
Nonlinear Anal. 46 (2001) 457-463
Real Analysis Exchange 27 (2001/02) 7-15
answered Aug 29, 2016 at 23:48
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$\begingroup$ Thanks, @Gerald! What about a more general (but still finite-dimensional) metric space? $\endgroup$ Commented Aug 30, 2016 at 7:15
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$\begingroup$ Gerald, metric spaces with finite Larman dimension are what's called today "doubling metric spaces" (i.e., those with a finite doubling dimension), right? I know that any such space equipped with a doubling measure satisfies LDT, but Larman shows something stronger -- that it holds for all measures, even non-doubling ones -- correct? $\endgroup$ Commented Aug 30, 2016 at 15:42