Can anyone exhibit a finitedimensional metric space (preferably, $R^d$) equipped with a measure that does not satisfy the conclusions of the Lebesgue Density Theorem? Such examples exist in infinitedimensional spaces (e.g., Hilbert space with the Gaussian measure) but what about a finitedimensional one?
1 Answer
It is a theorem of Besicovitch that measures on $\mathbb R^d$ do satisfy the density theorem.
Fremlin, Measure Theory, Chap. 47
added
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) 178192
Def: a metric space is finitedimensional 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 finitedimensional spaces have a Vitalitype 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 Vitalitype properties. For example:
Nonlinear Anal. 46 (2001) 457463
Real Analysis Exchange 27 (2001/02) 715

$\begingroup$ Thanks, @Gerald! What about a more general (but still finitedimensional) metric space? $\endgroup$ Aug 30, 2016 at 7:15

$\begingroup$ 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 nondoubling ones  correct? $\endgroup$ Aug 30, 2016 at 15:42