We use a lemma to prove Radon-Nikodym theorem.
Lemma Let $\mu$ and $\nu$ be finite positive measures on a measurable space $(X, \mathcal{A}) .$ Either $\mu \perp \nu$ or else there exists $\varepsilon>0$ and $G \in \mathcal{A}$ such that $\mu(G)>0$ and $G$ is a positive set for $\nu-\varepsilon \mu .$
However if $\mu \perp \nu$ , $\mu$ is concentrated on $E$ , $\nu$ is concentrated on $F$ and $E\cap F=\emptyset,$ is F a positive set for $\nu-\varepsilon\mu(\forall \varepsilon)$?
For all $\varepsilon$,take a subset $G$ of $F$ then $\nu(G)-\varepsilon\mu(G)=\nu(G)\geq0.$
If this is right,then in any case,we always find a $G$ with such property.
Am I right?Any help will be thanked.
$\mu$ is concentrated in $E$ and $E \cap F$ is empty. So $\mu (G) \leq \mu (F)=0$. So you cannot take $G \subset F$.