Suppose $(E, \mathcal{A})$ is a measurable space. Let $\mu$ and $\gamma$ be two distinct measures of this space. Now we say that $\gamma$ is absolutely continuous with respect to $\mu$ if for every $A \in \mathcal{A} , \mu(A) = 0 \Rightarrow \gamma(A) = 0 $.
I am unable to understand the essence of this definition of "absolutely continuous". I mean what is so absolute and continuous about these measures?
For simplicity, assume that $\gamma$ is a finite measure. One can show that this condition is equivalent to the following: For each $\varepsilon > 0$ there exists some $\delta > 0$ such that $\mu(A)<\delta$ implies $\gamma(A) < \varepsilon$. Hence, $\gamma$ can be regarded continuous with respect to $\mu$. In this case, there exists a measurable function $f$ such that $\gamma(A) = \int_A f\,d\mu$. If the measure space is $\mathbb R$ and $\mu$ is the Lebesgue measure, then $\gamma((-\infty,x]) = \int_{-\infty}^xf(t)\,dt$ and this function (in $x$) is absolutely continuous.