As stated in this question, 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.)