The Unapologetic Mathematician

Mathematics for the interested outsider

Absolute Continuity II

Now that we’ve redefined absolute continuity, we should tie it back to the original one. That definition makes precise the idea of “smallness” as being bounded in size below some \epsilon or \delta, but the new one draws a sharp condition that “small” sets are those of measure zero. As it turns out, in the presence of a finiteness condition the two are the same: if \nu is a finite signed measure and if \mu is any signed measure such that \nu\ll\mu, then to every \epsilon>0 there is a \delta so that \lvert\nu\rvert(E)<\epsilon for every measurable E with \lvert\mu\rvert(E)<\delta.

So, let’s say that the conclusion fails, and there’s some \epsilon for which we can find a sequence of measurable E_n with \lvert\mu\rvert(E_n)<\frac{1}{2^n}, and yet \lvert\nu\rvert(E_n)>\epsilon for each n. Then we can define E=\limsup\limits_{n\to\infty}E_n, and find


for each n, and thus \lvert\mu\rvert(E)=0. But we also find, since \nu is finite,

\displaystyle\lvert\nu\rvert(E)=\lim\limits_{n\to\infty}\lvert\nu\rvert\left(\bigcup\limits_{i=n}^\infty E_i\right)\geq\limsup\limits_{n\to\infty}\lvert\nu\rvert(E_n)\geq\epsilon

But this contradicts the assertion that \nu\ll\mu.

Let’s make the connection even further by proving the following proposition: if \mu is a signed measure and if f is integrable with respect to \lvert\mu\rvert then we can integrate f with respect to \mu and define


for every measurable E. It’s easy to see that \nu is a finite signed measure, and I say that \nu\ll\mu. Indeed, if \lvert\mu\rvert(E)=0 then \mu^+(E)=0=\mu^-(E). Thus we see that


and so \nu\ll\mu as asserted. This extends our old result that indefinite integrals with respect to measures are absolutely continuous in our old sense.

It’s easy to verify that the relation \ll is reflexive — \mu\ll\mu — and transitive — \mu_1\ll\mu_2 and \mu_2\ll\mu_3 together imply \mu_1\ll\mu_3 — and so it forms a preorder on the collection of signed measures. Two measures \mu and \nu so that \nu\ll\mu and \mu\ll\nu are said to be equivalent, and we write \mu\equiv\nu.

For example, we can verify that \mu\equiv\lvert\mu\rvert. Indeed, \mu\ll\mu, and we know this implies that \lvert\mu\rvert\ll\mu. Similarly, \lvert\mu\rvert\ll\lvert\mu\rvert, which implies that \mu\ll\lvert\mu\rvert. This is useful because it allows us to show that \lvert\mu\rvert(E)=0 for a measurable set E if and only if \mu(F)=0 for all measurable subsets F\subseteq E. If \lvert\mu\rvert(E)=0, then \lvert\mu\rvert(F)=0 since \lvert\mu\rvert is a measure, and then \mu(F)=0 since \mu\ll\lvert\mu\rvert. Conversely, if \mu(F)=0 for all measurable subsets F\subseteq E, then in particular \mu(E)=0, and thus \lvert\mu\rvert(E)=0 since \lvert\mu\rvert\ll\mu.


July 2, 2010 Posted by | Analysis, Measure Theory | 1 Comment

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