Equivalent ideas of absolute continuity of measures

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 easy way is just to note the following:

if $\nu(A)=0$ and $\mu <<\nu $, then setting $\epsilon =1/n$, we can find a $\delta_n>0$ s.t.$\ \nu(A)<\delta_n\Rightarrow \mu(A)<1/n$.

But this condition is vacuously satisfied for all $\delta_n$, because $\nu(A)=0$ so $\mu(A)<1/n$ for all $n\in \mathbb N$ and this implies that $\mu(A)=0$.