# Doubling metrics, doubling measures, Lebesgue density

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. 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.)

• I don't think I agree with the second paragraph. Let $X = [0,1]$ with its usual metric, let $\nu$ be Lebesgue measure (which is doubling) and let $\mu$ be the probability measure with cdf $F(x) = e^{1-1/x}$. Then $\mu \ll \nu$ but $\mu$ is not doubling. – Nate Eldredge Jul 19 '16 at 21:08
• Ack, I see my mistake. Do you agree that the 2nd paragraph holds if I additionally assume $\nu\ll\mu$? – Aryeh Kontorovich Jul 19 '16 at 21:15
• I believe my counterexample still applies. Note that the density $d\mu/d\nu$ is $F'(x)$ which is a.e. strictly positive. – Nate Eldredge Jul 19 '16 at 21:16
• Oh I see. I had mistakenly assumed that $F'(x)$ blows up at $0$, but now I see this isn't the case. I'll edit the question to remove that paragraph -- but the main question remains for now. – Aryeh Kontorovich Jul 19 '16 at 21:26
• Right. It's true that $d\nu/d\mu = 1/F'$ blows up near 0 but that's not a problem, it's still in $L^1(\mu)$. – Nate Eldredge Jul 19 '16 at 21:33