question on equivalent ideas of absolute continuity of measures The measure $\nu$ is absolute continuous with respect to $\mu$ if for each $A$, $\mu(A)=0$ implies $\nu(A)=0$ (indicated by $\nu \ll \mu$).
There is an $\epsilon$-$\delta$ idea related to this definition:  

If $\nu$ is finite, then:
  $\nu \ll \mu$  $\iff$ for every $\epsilon$ there exist a $\delta$ satisfying $\nu(A)<\epsilon$, if $\mu(A)<\delta$  

What is the necessity of finiteness of $\nu$?
What happens if $\nu$ is not finite?
Is there any example to show that $\epsilon$-$\delta$ definition does not hold if $\nu$ is infinite, even if $\nu\ll\mu$?
 A: Here is a counter-example:
Let $$\nu(E) = \int_E \frac{1}{x^2}dm(x)$$ where $m$ denotes the Lebesgue measure. The measure $\nu$ is not finite.
Then $\nu \ll m$ but 
$$\exists \epsilon=1>0 \quad \forall \delta>0 \quad \exists E=]0,\delta/2] \text{ measurable set such that } m(E)<\delta$$ but $\nu(A)≥\epsilon$, because $\nu(E) = \left[ \frac{-1}{x} \right]_0^{\delta/2} = \infty$.
A: One direction of the equivalence always holds, that is, if the $\epsilon$-$\delta$ condition holds, then $\nu<<\mu$, because if $\mu(A)=0$, then for all $\epsilon>0$ and its corresponding $\delta>0$ we have $\mu(A)<\delta\implies \nu(A)<\epsilon$, and hence $\nu(A)=0$.
To see why the finiteness of $\nu$ is important, we therefore must examine the other direction. We can trivially contradict it if we assume $\mu$ to be arbitrarily small, such as the Lebesgue measure, because then we can consistently define a $\mu$-absolutely-continuous measure by
$$\nu(A) = \begin{cases}0 & \mu(A)=0\\ \infty & \mu(A)>0\end{cases}$$
However, to gain more insight, we may examine the proof of the other direction when $\nu$ is finite, as presented, e.g., in this answer based on Folland's Real Analysis. The proof hinges on the continuity from above of finite measures, and unsurprisingly, the measure $\nu$ we defined above is a good counterexample to that property.
