Real Analysis, Folland Theorem 3.5, absoulute continuity Background Information:
Suppose that $\nu$ is a signed measure and $\mu$ is a positive measure on $(X,M)$. We say that $\nu$ is absolutely continuous w.r.t. $\mu$ and write $$\nu \ll \mu$$
if $\nu(E) = 0$ for every $E\in M$ for which $\mu(E) = 0$. 
Theorem 3.5 - Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X,M)$. Then $\nu\ll\mu$ if and only if every $\epsilon > 0$ there exists a $\delta > 0$ such that $\lvert\nu(E)\rvert < \epsilon$ whenever $\mu(E) < \delta$.
Attempted proof: Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X,M)$ and $\nu\ll\mu$. Then for any measurable set $E$ $$\nu(E) = 0 \quad \text{and} \quad \mu(E) = 0.$$
I don't really understand Folland's proof for this theorem, and I want to try to do this on my own but I am stuck. Any suggestions or comments is greatly appreciated.
 A: @Wolfy , Folland's proof, in this case, follows what is possibly the short or more natural approach. However, as always Folland's proof is very terse. 
Here are the proof in details.

Theorem 3.5 - Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X,M)$. Then $\nu\ll\mu$ if and only if every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\nu(E)| < \epsilon$ whenever $\mu(E) < \delta$.

Proof: 
($\Leftarrow$)
Suppose for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\nu(E)| < \epsilon$ whenever $\mu(E) < \delta$.
It means: for every $\epsilon > 0$ there exists a $\delta_\epsilon > 0$ such that, for every measurable set $E$, if $\mu(E) < \delta_\epsilon$ then  $|\nu(E)| < \epsilon$.
Now,given any $E$ a measurable set such that $\mu(E)=0$. Then, for every   $\epsilon > 0$,  $\mu(E)=0< \delta_\epsilon$, and so $|\nu(E)| < \epsilon$. 
Since, for every   $\epsilon > 0$,  $|\nu(E)| < \epsilon$, we have $|\nu(E)| =0$, that is, $\nu(E) =0$. 
Remark: note that at this part (above) we don't need $\nu$ to be finite.  
($\Rightarrow$)
Suppose $\nu$ be a finite signed measure and $\nu\ll\mu$. It is easy to see that  $|\nu|$ be a finite (positive) measure and $|\nu| \ll\mu$.
Now, let us prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\nu|(E) < \epsilon$ whenever $\mu(E) < \delta$.
Suppose there is $\epsilon > 0$ such that for any $\delta > 0$ , there is a $E$ measurable set such that $\mu(E) < \delta$ and $|\nu|(E) \geq \epsilon$.
In particular, for each $n\in \mathbb{N}$, there is a $E_n$ measurable set such that $\mu(E_n) < \frac{1}{2^n}$ and $|\nu|(E_n) \geq \epsilon$. 
Let $F=\bigcap_k\bigcup_{n=k}^\infty E_n$. Then, we have for all $k$,
$$0\leq \mu(F)\leq \mu \left(\bigcup_{n=k}^\infty E_n \right)<\frac{1}{2^{k-1}}$$
So we have $\mu(F)=0$. So we have $|\nu|(F)=0$. 
But $\left \{\bigcup_{n=k}^\infty E_n \right \}_k$ is a monotone non-increasing sequence of measurable sets and $|\nu|$ is finite. So we have 
$$0=|\nu|(F)= |\nu|\left(\bigcap_k\bigcup_{n=k}^\infty E_n \right)=\lim_{k \to \infty} |\nu|\left(\bigcup_{n=k}^\infty E_n \right)$$
But for all $k$,
$$|\nu|\left(\bigcup_{n=k}^\infty E_n \right)\geq |\nu|(E_n)\geq \epsilon>0$$
Contradiction. 
So we have proved that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\nu|(E) < \epsilon$ whenever $\mu(E) < \delta$.
To conclude the proof, it is enough to remark that for every $E$ measurable set we have $|\nu(E)|\leq |\nu|(E)$. So we have that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|\nu(E)| < \epsilon$ whenever $\mu(E) < \delta$.
