Suppose $u$ and $v$ are measures on a measurable space $E$. Further suppose $u$ is finite and absolutely continuous with respect to v ($v(S)=0 \implies u(S)=0$). The problem is: Show that $\forall \epsilon>0 \;\;\;\exists \delta: v(S)<\delta \implies u(S)<\epsilon.$
I've tried the following:
Fix $\epsilon>0$.
Let $S_n=\{measurable\;\; B\in E|u(B)>\epsilon, v(B)<1/n\}$.
If $S_n$ is finite for some n we are done ($v(S)=0 \implies u(S)=0$).
Now what I am trying to show is that the other case, $S_n$ is infinite $\forall n\in\mathbb{N}$, leads to a contradiction. Any suggestions?