Absolute Continuity and Upper Bound Constraints

Question

I'm reading the theorem of the following proof:

Let $\nu$ be a finite signed measure and $\mu$ be a positive measure on $(X,\mathcal{M})$. Then $\nu << \mu$ (<< means absolutely continuous) iff for all $\epsilon > 0$ there exists $\delta >0$ such that $\mu(E) < \delta \implies |\nu(E)|<\epsilon$.

The proof begins by justifying why it suffices to consider only $\nu$ positive. Then it says:

Clearly the epsilon-delta condition implies $\nu << \mu$

This is not clear to me. Consider the following reasoning: $$\mu(E) = 0 \implies \nu(E) = 0 \text{ (by def)} \\ \nu(E) > 0 \implies \mu(E) > 0 \text{ (contrapositive)}$$ In particular, absolute continuity is giving us lower bound gaurentees, not upper bound. So how does the epsilon-delta condition follow?

Answer 1

Let $\epsilon' > 0$ and $\mu(E) = 0$. Note that for all $\delta > 0$, $\mu(E) < \delta$, so the premise $\mu(E) < \delta$ in the epsilon-delta condition:

$$\forall \epsilon > 0, \exists \delta > 0: \mu(E)<\delta \implies \nu(E) < \epsilon$$

is always satisfied. This shows that $\nu(E)<\epsilon'$. Since the choice of $\epsilon'$ is arbitrary, we conclude that $\nu(E) = 0$.

Answer 2

Suppose $\mu(E)=0,\ $ and let $\epsilon>0$. We need to show that $\nu(E)=0.$

Now, we know that for $\epsilon$ there is a $\delta>0$ such that $\nu(E)<\epsilon$ whenever $\mu(E)<\delta.$

So, take $\epsilon_n=1/n$, and the $\delta_n$ that corresponds to it, according to hypothesis. Then, since $\mu(E)=0$, obviously $\mu(E)<\delta_n$ so in fact $\nu(E)<1/n$ for any integer $n$. And this implies that $\nu(E)=0.$