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 .