Let $\mu$ be a measure.
I'm looking for a reference to a proof showing that the following condition is enough to prove absolute continuity with respect to the Lebesgue measure:
$$\liminf\limits_{r \downarrow 0} \frac{\mu(B(x,r))}{\lambda(B(x,r))}<\infty \text{ for } \mu\text{-almost all }x \in \mathbb{R}$$
(With $\lambda$ as Lebesgue measure and B(x,r) as ball with radius r.)
Let us assume that $\mu$ is a $\sigma$-finite measure.
Then $\mu = \mu_s +\mu_a$ with $\mu_s \perp \lambda$ and $\mu_a <<\lambda$ (by Lebesgue-Radon-Nikodym).
It suffices to show that $\mu_s =0$.
Let us assume towards a contradiction that this is false.
By Rudin, Real and complex analysis, Theorem 8.9, we have $D\mu_s (x)=\infty$ for $\mu_s$-a.e. $x \in \Bbb{R}$, i.e. for all $x \in \Bbb{R}\setminus N$ with $\mu_s (N)=0$.
But this implies
$$ 0<\mu_s (\Bbb{R}) = \mu_s (\Bbb{R}\setminus N) \leq \mu (\Bbb{R}\setminus N), $$
so that by assumption (because the liminf is finite $\mu$-a.e.), there is some $x \in \Bbb{R}\setminus N$ with
$$ \infty > \liminf_r \frac{\mu(B_r (x))}{\lambda(B_r (x))} \geq \liminf \frac{\mu_s (B_r (x))}{\lambda (B_r (x))}=\infty, $$
a contradiction.
Observe that Theorem 8.9 is only stated for the real line. Rudin also provides a discussion of the (in)validity of the claim in higher dimensions after the theorem.