TITLE: Equivalent ideas of absolute continuity of measures
QUESTION [8 upvotes]: Wikipedia says that $\mu$ is absolutely continuous with respect to $\nu$, if $\nu(A)=0 \Rightarrow \mu(A)=0$.
Okay, then I found another notion of absolute continuous measures: Let $||f||_1=1$ and $\mu(A):= \int_A |f| d \nu$, then we also have absolute continuity in the sense that if $\nu(A)  < \delta \Rightarrow \mu(A) < \varepsilon$. My question is: How are these definitions related to each other? It is clear that the second notion implies the first one, but is the converse also true? Where does this $\varepsilon$, $\delta$ comes into play? 
Edit: Afais is the converse implication the result of a special case of the Radon-Nikodym theorem and a few lines of calculation that the $\varepsilon, \delta$ stuff actually holds, right?

REPLY [5 votes]: The easy way is just to note the following: 
if $\nu(A)=0$ and $\mu <<\nu $, then setting $\epsilon =1/n$, we can find a $\delta_n>0$ s.t.$\ \nu(A)<\delta_n\Rightarrow \mu(A)<1/n$. 
But this condition is vacuously satisfied for all $\delta_n$, because $\nu(A)=0$ so $\mu(A)<1/n$ for all $n\in \mathbb N$ and this implies that $\mu(A)=0$.