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The wikipedia article on 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 $\lambda$:

$\mu<<\lambda$ 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 $\mu<<\nu$? 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?

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I think you got it slightly wrong: δ should be allowed to depend on I (making absolute continuity a local concept). –  Harald Hanche-Olsen Dec 30 '09 at 2:34
    
Thanks for pointing this out, I corrected the post. I am still trying to prove this and Fedja's equivalence.. –  kweinert Dec 30 '09 at 7:50
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1 Answer

up vote 5 down vote accepted

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).

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