The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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


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:


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 | Analysis, Measure Theory | 8 Comments