Absolute Continuity I

We’ve shown that indefinite integrals are absolutely continuous, but today we’re going to revise and extend this notion. But first, to review: we’ve said that a set function $\nu$ defined on the measurable sets of a measure space $(X,\mathcal{S},\mu)$ is absolutely continuous if for every $\epsilon>0$ there is a $\delta$ so that $\mu(E)<\delta$ implies that $\lvert\nu(E)\rvert<\epsilon$ .

But now I want to change this definition. Given a measurable space $(X,\mathcal{S})$ and two signed measures $\mu$ and $\nu$ defined on $\mathcal{S}$ we say that $\nu$ is absolutely continuous with respect to $\mu$ — and write $\nu\ll\mu$ — if $\nu(E)=0$ for every measurable set $E$ for which $\lvert\mu\rvert(E)=0$ . It still essentially says that $\nu$ is small whenever $\mu$ is small, but here we describe “smallness” of $\nu$ by $\nu$ itself, while we describe “smallness” of $\mu$ by its total variation $\lvert\mu\rvert$ .

This situation is apparently asymmetric, but only apparently; If $\mu$ and $\nu$ are signed measures, then the conditions

$\displaystyle\begin{aligned}&\nu\ll\mu\\&\nu^+\ll\mu\quad\mathrm{and}\quad\nu^-\ll\mu\\&\lvert\nu\rvert\ll\mu\end{aligned}$

are equivalent. Indeed, if $X=A\uplus B$ is a Hahn decomposition with respect to $\nu$ then whenever $\lvert\mu\rvert(E)=0$ we have both

$\displaystyle\begin{aligned}0\leq\lvert\mu\rvert(E\cap A)&\leq\lvert\mu\rvert(E)=0\\{0}\leq\lvert\mu\rvert(E\cap B)&\leq\lvert\mu\rvert(E)=0\end{aligned}$

Thus if the first condition holds we find

$\displaystyle\begin{aligned}\nu^+(E)=\nu(E\cap A)&=0\\\nu^-(E)=-\nu(E\cap B)&=0\end{aligned}$

and the second condition must hold as well. If the second condition holds we use the definition

$\displaystyle\lvert\nu\rvert(E)=\nu^+(E)+\nu^-(E)$

to show that the third must hold. And if the third holds, then we use the inequality

$\displaystyle0\leq\lvert\nu(E)\rvert\leq\lvert\nu\rvert(E)$

to show that the first must hold.

Now, just because smallness in $\nu$ can be equivalently expressed in terms of its total variation does not mean that smallness in $\mu$ can be equivalently expressed in terms of the signed measure itself. Indeed, consider the following two functions on the unit interval $[0,1]\subseteq\mathbb{R}$ with Lebesgue measure $\mu$ :

$\displaystyle\begin{aligned}f_1(x)&=\chi_{\left[0,\frac{1}{2}\right]}-\chi_{\left(\frac{1}{2},1\right]}\\f_2(x)&=x\end{aligned}$

and define $\nu_i$ to be the indefinite integral of $f_i$ . We can tell that the total variation $\lvert\nu_1\rvert$ is the Lebegue measure $\mu$ itself, since $\lvert f_1\rvert=1$ . Thus if $\lvert\nu\rvert(E)=0$ then we can easily calculate

$\displaystyle\nu_2(E)=\int\limits_Ex\,d\mu=\int x\chi_E\,d\mu=0$

and so $\nu_2\ll\nu_1$ . However, it is not true that $\nu_2(E)=0$ for every measurable $E$ with $\nu_1(E)=0$ . Indeed, $\nu_1(\left[0,1\right])=0$ , and yet we calculate

$\displaystyle\nu_2(\left[0,1\right])=\int x\,d\mu\geq\int\frac{1}{2}\chi_{\left[\frac{1}{2},1\right]}\,d\mu=\frac{1}{4}$

By the way: it’s tempting to say that this integral is actually equal to $\frac{1}{2}$ , but remember that we only really know how to calculate integrals by taking limits of integrals of simple functions, and that’s a bit more cumbersome than we really want to get into right now.

One first quick result about absolute continuity: if $\mu$ and $\nu$ are any two measures, then $\nu\ll\mu+\nu$ . Indeed, if $\mu(E)+\nu(E)=0$ then by the positivity of measures we must have both $\mu(E)=0$ and $\nu(E)=0$ , the latter of which shows the absolute continuity we’re after.

July 1, 2010 - Posted by John Armstrong | Analysis, Measure Theory

8 Comments »

[…] Continuity II Now that we’ve redefined absolute continuity, we should tie it back to the original one. That definition makes precise the […]

Pingback by Absolute Continuity II « The Unapologetic Mathematician | July 2, 2010 | Reply
[…] Another relation between signed measures besides absolute continuity — indeed, in a sense the opposite of absolute continuity — is singularity. We say that […]

Pingback by Singularity « The Unapologetic Mathematician | July 5, 2010 | Reply
[…] Before the main business, a preliminary lemma: if and are totally finite measures so that is absolutely continuous with respect to , and is not identically zero, then there is a positive number and a measurable […]

Pingback by The Radon-Nikodym Theorem (Statement) « The Unapologetic Mathematician | July 6, 2010 | Reply
[…] to the general case, we know that the absolute continuity is equivalent to the conjunction of and , and so we can reduce to the case where is a finite measure, not just a […]

Pingback by The Radon-Nikodym Theorem (Proof) « The Unapologetic Mathematician | July 7, 2010 | Reply
[…] , while on . For the first case, let be a set for which . Since , we must have , and so . Then by absolute continuity, we conclude that , and thus on . The proof that on is […]

Pingback by The Radon-Nikodym Theorem for Signed Measures « The Unapologetic Mathematician | July 8, 2010 | Reply
[…] As we’ve said before, singularity and absolute continuity are diametrically opposed. And so it’s not entirely surprising that if we have two totally […]

Pingback by Lebesgue Decomposition « The Unapologetic Mathematician | July 14, 2010 | Reply
[…] from a measure space to a totally -finite measure space so that the pushed-forward measure is absolutely continuous with respect to . Then we can select a non-negative measurable […]

Pingback by Pulling Back and Pushing Forward Structure « The Unapologetic Mathematician | August 2, 2010 | Reply
[…] If is the metric space associated to a measure space , and if is a finite signed measure that is absolutely continuous with respect to . Then defines a continuous function on […]

Pingback by Associated Metric Spaces and Absolutely Continuous Measures I « The Unapologetic Mathematician | August 16, 2010 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects

Subjects
Archives

July 2010

M T W T F S S

1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17 18

19 20 21 22 23 24 25

26 27 28 29 30 31

« Jun Aug »
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider