# Lebesgue differentiation theorem holds on locally doubling space?

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?

• I would guess so, since Lebesgue differentiation theorem is essentially local in nature. – leo monsaingeon Sep 17 '15 at 8:56

$$\limsup_{r \to 0} \frac{\mu(B(x,2r))}{\mu(B(x,r))}<\infty \ \text{for}\ \mu-\text{a.e.}\ x \in X$$
This is done in Section 3.4 of the book Sobolev Spaces on Metric Measure Spaces" by Heinonen, Koskela, Shanmugalingam, and Tyson.