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 or , 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 is a finite signed measure and if is any signed measure such that , then to every there is a so that for every measurable with .
So, let’s say that the conclusion fails, and there’s some for which we can find a sequence of measurable with , and yet for each . Then we can define , and find
for each , and thus . But we also find, since is finite,
But this contradicts the assertion that .
Let’s make the connection even further by proving the following proposition: if is a signed measure and if is integrable with respect to then we can integrate with respect to and define
for every measurable . It’s easy to see that is a finite signed measure, and I say that . Indeed, if then . Thus we see that
and so 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 is reflexive — — and transitive — and together imply — and so it forms a preorder on the collection of signed measures. Two measures and so that and are said to be equivalent, and we write .
For example, we can verify that . Indeed, , and we know this implies that . Similarly, , which implies that . This is useful because it allows us to show that for a measurable set if and only if for all measurable subsets . If , then since is a measure, and then since . Conversely, if for all measurable subsets , then in particular , and thus since .
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 defined on the measurable sets of a measure space is absolutely continuous if for every there is a so that implies that .
But now I want to change this definition. Given a measurable space and two signed measures and defined on we say that is absolutely continuous with respect to — and write — if for every measurable set for which . It still essentially says that is small whenever is small, but here we describe “smallness” of by itself, while we describe “smallness” of by its total variation .
This situation is apparently asymmetric, but only apparently; If and are signed measures, then the conditions
are equivalent. Indeed, if is a Hahn decomposition with respect to then whenever we have both
Thus if the first condition holds we find
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 can be equivalently expressed in terms of its total variation does not mean that smallness in can be equivalently expressed in terms of the signed measure itself. Indeed, consider the following two functions on the unit interval with Lebesgue measure :
and define to be the indefinite integral of . We can tell that the total variation is the Lebegue measure itself, since . Thus if then we can easily calculate
and so . However, it is not true that for every measurable with . Indeed, , and yet we calculate
By the way: it’s tempting to say that this integral is actually equal to , 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 and are any two measures, then . Indeed, if then by the positivity of measures we must have both and , the latter of which shows the absolute continuity we’re after.