The first property, $\nu(A)=0\implies \mu(A)=0$, is the fundamental definition. The $\epsilon$-$\delta$ property is equivalent to it when $\mu$ is finite, but not in general. For example, the measure $\mu(E) = \int_E |x| \,dx$ on real line is absolutely continuous with respect to the Lebesgue measure. But we can find arbitrarily short intervals $I$ with $\mu(I)\ge 1$.
To prove the equivalence of two properties for a finite measure $\mu$, you don't need the Radon-Nikodym theorem. Suppose that the $\epsilon$-$\delta$ condition fails. Then there exists $\epsilon$ such that for all $n=1,2,3\dots$ there is a measurable set $E_n$ with $\nu(E_n)<2^{-n}$ and $\mu(E_n)\ge \epsilon$. Let $F_k = \bigcup_{n=k}^\infty E_n$ and $F = \bigcap_{k=1}^\infty F_k$.
Since $\nu(F_k) <2^{1-k}$, we have $\nu(F)=0$. On the other hand, $\mu(F_k)\ge \epsilon$ for every $k$, which (because $\mu$ is finite) implies $\mu(F)\ge \epsilon$.
The proof is taken from Folland's ''Real Analysis'', which is generally considered a better way to learn measure theory than reading Wikipedia[citation needed].
The proof is incorrect. In the fist part of the proof, the $\delta$ you picked does not guarantee that for all measurable sets $A$, $\mu(A) < \delta$ implies $|\int_A f_n\, d\mu| < \epsilon$. The second part of the proof is almost correct, but the condition $\lim\limits_{M\to \infty} \mu(E_M) = 0$ follows from the assumption $\sup_n \int |f_n|\, d\mu < \infty$, not that $\mu$ is finite. Indeed, if $\alpha := \sup_n \int |f_n|\, d\mu < \infty$, then $\mu(E_M) \le \frac{\alpha}{M}$ for all $M > 0$; as a consequence, $\lim\limits_{M\to \infty} \mu(E_M) = 0$. If you make this fix, then the second part of the proof will be correct.
To prove the forward direction, fix $\epsilon > 0$ and choose $M > 0$ such that
$$\sup_n \int_{E_M} |f_n| \, d\mu < \frac{\epsilon}{2}.$$
Then
$$\int |f_n|\, d\mu = \int_{E_M} |f_n|\, d\mu + \int_{E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(X) < \infty$$
for all $n\in N$.
Set $\delta = \frac{\epsilon}{2M}$. For all measurable sets $A$, $\mu(A) < \delta$ implies
$$\left|\int_A f_n\, d\mu\right| \le \int_A |f_n|\, d\mu = \int_{A\cap E_M} |f_n|\, d\mu + \int_{A\cap E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(A) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Since $\epsilon$ was arbitrary, $\{f_n\}$ is uniformly absolutely continuous.
Best Answer
Let $f$ be finite almost everywhere, say, in $E$. Then if $A_n = \{ x \mid \lvert f(x) \rvert \leq M \}$, we have $\bigcup^\infty_{n = 0} A_M = E$.
Since it is an increasing sequence of sets $A_M \uparrow E$ and the measure is finite, for any $\delta > 0$ we have $M$ such that $\mu(A_M) > \mu(E) - \delta$, so that $\mu(A_M^c) < \delta$. Now, if we choose $\delta$ so that $\mu(B) < \delta \implies \int_B \lvert f \rvert < 1$, we have
$$\int \lvert f \rvert = \int_E \lvert f \rvert = \int_{A_M} \lvert f \rvert + \int_{A_M^c} \lvert f \rvert \leq M \mu(A_M) + 1 < \infty.$$