6
$\begingroup$

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 $\lambda$:

$\mu\ll\lambda$ 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?

$\endgroup$
2
  • $\begingroup$ I think you got it slightly wrong: δ should be allowed to depend on I (making absolute continuity a local concept). $\endgroup$ Dec 30, 2009 at 2:34
  • $\begingroup$ Thanks for pointing this out, I corrected the post. I am still trying to prove this and Fedja's equivalence.. $\endgroup$
    – kweinert
    Dec 30, 2009 at 7:50

1 Answer 1

8
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ EoM article on Luzin N-property mentions some "Banach S-property", which looks like the condition you've stated, except that it is about absolute continuity of continuous functions, not that of measures: $\lambda(f(E))$ replaces $\mu$. However, the article says that the Banach S-property is stronger than the Luzin N-property, whereas you state the equivalence. Could you clarify where the discrepancy comes from? I can't pin it down, but I guess this is because $f$ may somehow "mix" measurable sets and also map to a different space. $\endgroup$ Jan 16, 2023 at 18:17
  • $\begingroup$ @paperskilltrees They are equivalent for monotone functions but for arbitrary continuous functions there may be a discrepancy because the image folds on itself, so the measure of the image of a finite union of disjoint intervals may be much less than the sum of measures of the images of individual intervals leading to all sorts of complications (though in one direction inequality between the two is, obviously, always true giving one implication you mentioned). $\endgroup$
    – fedja
    Jan 16, 2023 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.