I am stuck on what I think may be the very last line of the proof I am seeking.
Let $(X, \mathcal{B})$ be a measurable space which has associated with it the finite measures $\mu$ and $\nu$ s.t. $\nu \ll \mu$. I aim to show that $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\forall A \in \mathcal{B}$,
$\mu(A) < \delta \implies \nu(A) < \epsilon$
Fix $\epsilon > 0$.
For all $n \in \mathbb{N}$, let $\delta_n = \frac{1}{n^2}$.
For all $n \in \mathbb{N}$, let $A_n \in \mathcal{B}$ s.t. if $\exists E \in \mathcal{B}$ s.t. $\mu(E) < \delta_n$ and $\nu(E) > \epsilon$, then set $A_n = E$. Otherwise, set $A_n = \emptyset \in \mathcal{B}$.
Suppose for sake of contradiction that $|\{A_n\}| = \infty$, so that no matter the $\delta > 0$, we could find a $\delta_n = \frac{1}{n^2} < \delta$ which has associated with it a measurable $A_n \ne \emptyset$ with $\mu(A_n) < \delta_n < \delta$ and $\nu(A_n) > \epsilon$.
Now if we let $\underset{n \rightarrow \infty}{\text{limsup}}$ $A_n = S$, we have (from a prior problem) that $\mu(S) = 0$ since $\mu$ is a finite measure and $\sum_{n=1}^\infty \mu(A_n) \le \sum_{n=1}^\infty \delta_n = \sum_{n=1}^\infty \frac{1}{n^2} < \infty$. Since $\nu \ll \mu$ we therefore have $\nu(S) = 0$ as well.
Yet $\nu(S) = \nu(\bigcap_{n=1}^\infty \bigcup_{n=m}^\infty A_m) \ge \epsilon > 0$ since...
and it's here where I'm stuck in the proof.