Counterexample to a proposition on absolute continuous measures Let $P$ be a probability and let $\mu$ be a $\sigma$-finite measure. It holds that the two following properties are equivalent:
• $P \ll \mu$
• $ \forall \epsilon > 0, \exists \delta > 0 , \mu (A) < \delta \Rightarrow P(A) < \epsilon$
My question is: if $ \mu $, $\nu$ are two $\sigma$-finite measures are still the two conditions equivalent? 
I think that the answer is no, but I can't find a counterexample. Actually I think that $( \Leftarrow )$ holds also for $\sigma$-finite measures while the other not.
Any suggestions?

Let $\nu$ be the counting measure on $\mathbb{N}$, and $\mu$ the measure with
$$\mu(A) = \sum_{n\in A} 2^{-n}.$$
Then $\nu \ll \mu$ - both measures have no nontrivial null sets, but for every $\delta > 0$ there is an $A \subset \mathbb{N}$ with $\mu(A) < \delta$, and $\nu(A) = +\infty$.
The implication in the other direction holds, whether we consider $\sigma$-finite measures or completely arbitrary (positive) measures [on the same $\sigma$-algebra], since $\mu(A) = 0$ then implies $\nu(A) < \varepsilon$ for all $\varepsilon > 0$, and that implies $\nu(A) = 0$.
