It's known that for a metric space with doubling measure $(X,\mu)$, the Lebesgue differentiation theorem holds , i.e. If $f:X\to \mathbb{R}$ is a locally integrable function, then $\mu$-a.e. points are Lebesgue points.

Can we relax the doubling condition to local doubling or local uniformly doubling?

  • $\begingroup$ I would guess so, since Lebesgue differentiation theorem is essentially local in nature. $\endgroup$ Sep 17, 2015 at 8:56

1 Answer 1


Yes, one can relax the doubling condition to

$$\limsup_{r \to 0} \frac{\mu(B(x,2r))}{\mu(B(x,r))}<\infty \ \text{for}\ \mu-\text{a.e.}\ x \in X$$

This is done in Section 3.4 of the book ``Sobolev Spaces on Metric Measure Spaces" by Heinonen, Koskela, Shanmugalingam, and Tyson.

The proof is essentially the same as in the doubling case (i.e., it still boils down to the Vitali covering property), but somewhat more techincal.


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