As stated in this question,
http://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.

Does the LDT still hold if I drop the assumption $\mu\ll\nu$ -- i.e., only assume that $\mu$ is a probability measure on a complete doublign metric space? (Can additionally assume $(X,d)$ is compact if needed.)