Extensions of Measures
Oops, forgot to post this this earlier…
We can put together what we’ve been doing recently to state the following theorem:
If is a -finite measure on the ring , then there is a unique measure on the -ring extending . That is, if , then . Further, the extended measure is also -finite.
The existence is straightforward. We can induce an outer measure, and then restrict it to get . It’s straightforward to verify from the definitions that . And we know that since is -finite, so is , and thus .
What we need to show is that is unique. To this end, let and be two measures on that both extend . Let be the class of sets on which and agree; this obviously contains .
Now, if one of these two measures — say — is finite, and if is a monotone sequence of sets on which and agree, then the limit of this sequence is again in . Indeed, since measures are continuous, we must have
and similarly for . Since is finite, and agrees with on , we have a sequence of finite measures . The limits of these sequences must then agree, and so as well. Thus is a monotone class. Since it contains , it must contain , and thus .
On the other hand, neither measure may be finite. In this case, let be some fixed set of finite measure. Now — the collection of intersections of sets in with — is again a ring, and is the smallest -ring containing it. Restricting and to gives finite measures, and we can use the argument above.
Now every set can be covered by a countable, pairwise disjoint collection of sets . For each one, we have , and so we must find . From here, countable additivity finishes the theorem.
In light of the uniqueness of this extension, we will just call the extended measure again, rather than .