# Counter example for absolutely continuous measure

I need a example for the following statement: "Given a pair of finite measures $(\mu,\nu)$ on a given measurable space $(\Omega, \mathbb{A})$ is said to have property $P$ if for every $\epsilon >0$ there exists a $\delta >0$ such that for all $A \in \mathbb{A}$, $\mu(A)<\delta \rightarrow \nu(A)<\epsilon$. Show that in the case that $\nu$ is not a finite measure, $\nu$ is absolutely continuous with respect to $\mu$ does not imply that $(\mu,\nu)$ has property P.

Thanks!

On $(\mathbb R_+,\mathcal B(\mathbb R_+))$, try $\mu$ the Lebesgue measure and $\nu(\mathrm dx)=\mathrm dx/x$: if $A_n=(1/(2n),1/n)$, then $\mu(A_n)=1/(2n)$ converges to $0$ but $\nu(A_n)=\log2$ does not.