measure theory - Showing that $nu ll mu$ implies $forall epsilon > 0$, $exists delta > 0$ s.t. $mu(A) < delta implies nu(A) < epsilon$

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. $\forall A \in \mathcal{B}$,

$$\mu(A) < \delta \implies \nu(A) < \epsilon$$

1. Fix $\epsilon > 0$.




2. For all $n \in \mathbb{N}$, let $\delta_n = \frac{1}{n^2}$.




3. 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 $A_n = \emptyset \in \mathcal{B}$.




4. 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 $\nu(A_n) > \epsilon$.




5. 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.




6. 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.

Answer

You are almost there. Notice that since $\nu(S) = 0$,
$$0 = \nu(S) = \nu \left( \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} A_m \right) = \lim_{n \to \infty} \nu \left( \bigcup_{m=n}^{\infty} A_m \right).$$
Therefore, there exists $n$ such that
$$\nu(A_n) \le \nu \left( \bigcup_{m=n}^{\infty} A_m \right) < \varepsilon,$$

a contradiction of the choice of $A_n$.

I must say though that the right way to formulate this proof would be this : Suppose by sake of contradiction that the result is false, i.e. that there exists an $\varepsilon > 0$ such that for all $\delta > 0$, there is $C_{\delta} \in \mathcal B$ with
$$\mu(C_{\delta}) < \delta, \quad \nu(C_{\delta}) \ge \varepsilon.$$
Then you can choose $\delta_n = \frac 1{n^2}$ and $A_n = C_{\delta_n}$ and then continue by following our steps.

Hope that helps,