Absolute continuity of measure and $\epsilon-\delta$ condtion

Question

I am reading Folland's Real Analysis p. 89

Theorem:

Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X,M) $. Then $\nu << \mu$ iff for every $\epsilon >0$ there exists $\delta>0$ such tant $|\nu(E)|<\epsilon$ whenever $\mu(E)<\delta$.

In the proof, it says that "clearly" the $\epsilon-\delta$ condition implies that $\nu << \mu$.

I have trouble on seeing why this holds.
How does this imply $\mu(E) = 0 \Rightarrow \nu(E) = 0 $ with $E\in M$?

Thanks!

Answer 1

Let $\mu(E)=0$. Then for all $\varepsilon>0$ we have $|\nu(E)|<\varepsilon$. So then $|\nu(E)|=...$