# Doubling measure is absolutely continuous with respect to Lebesgue

Let $\mu$ be a fixed finite measure on $\mathbb R$. We say that $\mu$ is doubling if there exists a constant $C>0$, such that for any two adjacent intervals $I=[x−h,x]$ and $J=[x,x+h]$, $$C^{−1}\mu(I)≤\mu(J)≤C\mu(I).$$ Assuming that $\mu$ is doubling, show that there exist positive constants $B$ and $a$, such that for every interval $I$, $$\mu(I)≤B[length(I)]^a$$

By Radon-Nikodim, I solved but I use the fact that $\mu$ is absolutely continous with respect to Lebesgue measure, but I don't know to prove this last part, i.e., that $\mu$ is absolutely continuous with respect to Lebesgue.

• A doubling measure is not necessarily absolutely continuous (see wikipedia). The statement that you want to prove is (a special case of) Lemma 2.1 of On the $\overline{\partial}$ equation in weighted $L^2$ norms in $C^1$, Michael Christ, J. Geom. Anal. 1 (1991), no. 3, 193–230, MR1120680. – user92891 Sep 3 '13 at 13:08
• possible duplicate of Prove Property of Doubling Measure on $\mathbb{R}$ – user147263 Jul 4 '15 at 23:38
• ^ this proves the statement that's the actual goal of this post, not the false statement that the OP hoped to use for it. – user147263 Jul 4 '15 at 23:39