The [wikipedia article on absolute continuity](https://en.wikipedia.org/wiki/Absolute_continuity) gives a delta-epsilon definition for a measure $\mu$ defined on the Borel $\sigma$-algebra on the real line, with respect to the Lebesgue measure $\lambda$:


$\mu\ll\lambda$ if and only if for every $\epsilon>0$ and for every bounded real interval $I$ there is a $\delta>0$ such that for every (finite or infinite) sequence of pairwise disjoint sub-intervals [$x\_i,y\_i$] of $I$ with


$\sum\_{i} |y\_i - x\_i| < \delta$


it follows that


$\sum\_{i} |\mu((-\infty, y\_i])-\mu((-\infty, x\_i])| < \epsilon$.


My questions are: Does this connection generalise? What would be a topological reformulation of $\mu\ll\nu$? If the notion does not generalise for arbitrary measures, does it generalise for the Lebesgue measure on $R^n$? How would a sketch of the proof look like?



What is the "continuity" in "absolute continuity", in general?
For every $\varepsilon>0$, there exists $\delta>0$ such that every measurable set of $\nu$-measure less than $\delta$ has $\mu$-measure less than $\varepsilon$. There are some technical assumptions to be made to have this equivalent to $\mu\ll\nu$ (say, that both measures are finite) but otherwise it is as simple as that. In many decent measure spaces, it suffices to check the inequality just for some nice sets $E$ (the definition you quoted is just this imequality for finite unions of half-open intervals).


