As stated in this question,
[Lebesgue differentiation theorem holds on locally doubling space?](https://mathoverflow.net/questions/218457/lebesgue-differentiation-theorem-holds-on-locally-doubling-space)
and proved here,
<http://www.math.uiuc.edu/~tyson/595f15lecture2.pdf>
the Lebesgue differentiation theorem (LDT) holds in doubling measure spaces.
It is also known that every complete doubling metric supports a doubling measure:
<http://www.ams.org/journals/proc/1998-126-02/S0002-9939-98-04201-4/S0002-9939-98-04201-4.pdf>


[From this I conclude that if $\mu$ is a probability measure on a complete doubling metric space $(X,d)$ and $\mu\ll\nu$, where $\nu$ is a doubling measure (whose existence is guaranteed), then $\mu$ is itself a doubling measure and hence satisfies the LDT. **EDIT**: this is false, ignore.]


Does the LDT hold for probability measures on complete doubling metric spaces? (Can additionally assume $(X,d)$ is compact if needed.)



Doubling metrics, doubling measures, Lebesgue density
A version of Lebesgue differentiation for non-doubling measures on doubling metric space appears in section 3 of:


T. Hytonen, A FRAMEWORK FOR NON-HOMOGENEOUS ANALYSIS ON METRIC SPACES, AND THE RBMO SPACE OF TOLSA
<http://www.raco.cat/index.php/PublicacionsMatematiques/article/download/191394/387513>


I leave it to your judgment whether this is of use to you.


