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?

If $P$ is not a finite measure, then the direction "$\implies$" does not work. Let $\mu$ be the Lebesgue-measure and $P$ the measure with Lebesgue-density $|x|$. Then $\mu([a, a + \delta]) = \delta$ but $P([a, a + \delta]) = a\delta + \frac{\delta^2}{2}$ for any $a \ge 0$.
To see that the other direction holds, let $(A_i)_{i \in \mathbb{N}}$ be a Partition of the measurable space so that $P$ and $\lambda$ are both finite on every set $A_i$ and let $M$ be a $\mu$-nullset. Then $P|_{A_i} \ll\mu|_{A_i}$ by the assumption and this yields $P(M) = \sum \limits_{i \in \mathbb{N}} P|_{A_i}(M) = 0$, since $M$ is a $\mu|_{A_i}$-nullset.
