currently I am reading the Brownian Motion Book from Peter Mörser and Yuval Peres. In section about occupation measures and Green's functions they state the following theorem.
Let $(B_s)_{s \geq 0}$ be Brownian motion and let $t>0$. Then define the following occupation measure $$ \mu_t(\cdot) = \int^{\infty}_0 \mathbf{1}_{B_s \in \cdot} ds \quad (\cdot \subset \mathbb R). $$ This definition is fine, the claim of the proposition is that $\mu_t$ is absolutly continuous w.r.t. Lebesgue measure $\lambda$.
$\textbf{Here comes the question}$: At the beginning of the proof they give the following sufficient condition: It sufficies to prove that for $\mu_t$-almost every $x \in \mathbb R$ one has $$ \liminf_{r \downarrow 0} \frac{\mu_t(B(x,r))}{\lambda(B(x,r))} < \infty $$
I have covered a course on integrations/measure theory and stochastic process, but this sufficient condition is foreign to me. Does anybody have an explanation/reference?
In Rudin's RCA (the chapter on Differentiation) it is shown that if $\mu\perp \lambda$ then $\frac{\mu(B(x,r))}{\lambda(B(x,r))} \to \infty$ as $ r\to 0$ for almost all $x$ (w.r.t Lebesgue measure). Since any positive measure is the sum of a singular and an absoluetly continuous measure it follows that this limit relation holds whenever $\mu$ is not absolutely continuous w.r.t. $\lambda$.