## 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 .

## Regular Outer Measures

As usual, let be a ring of sets, be the smallest -algebra containing , and be the smallest hereditary -algebra containing . We’ve asked about the relation between a measure on , the outer measure it induces on , and the measure we get by restricting to . But for now, let’s consider what happens when we *start* with an outer measure on .

Okay, so we’ve got an outer measure on a hereditary -ring — like . We can define the -ring of -measurable sets and restrict to a measure on . And then we can turn around and induce an outer measure on the hereditary -ring .

Now, in general there’s no reason that these two should be related. But we *have* seen that if came from a measure (as described at the top of this post), then , and the measure induced by is just back again!

When this happens, we say that is a “regular” outer measure. And so we’ve seen that any outer measure induced from a measure on a ring is regular. The converse is true as well: if we have a regular outer measure , then it is induced from the measure on . Induced and regular outer measures are the same.

Doesn’t this start to look a bit like a Galois connection?

## Measurable Covers

Again, let be a ring, let be the smallest -ring containing , and let be the smallest hereditary -ring containing . Given a measure on , it induces an outer measure on , which restricts to a measure on .

The sets in are easily described — they’re everything that can be countably covered by sets in — but we don’t have a measure on this collection. Instead, we need good approximations of these sets by sets in , where we *do* have a measure. And so we say that a set is a “measurable cover” of if , and if for every with we have . That is, may be larger than , but only negligibly.

If has -finite outer measure, then it has a measurable cover with . Indeed, our hypothesis is that can be written as the countable disjoint union of sets of finite outer measure, so if we can show this for sets of finite measure the -finite case will follow.

By the comparisons we showed last time, for every there is some set so that and . Taking to be the intersection of this sequence, we must have

Since is arbitrary, we must have . If fits into , then . And so:

Since we can subtract it off and find .

In fact, any measurable cover must have . Further, any two of them have a negligible difference.

Indeed, if we have two measurable covers, and , then , which tells us that . But since is a measurable cover, we conclude that , and similarly that . But is the union of these two differences, and so .

If , then any measurable cover must have as well. On the other hand, if is finite, then there exists at least one measurable cover with . Then any other measurable cover differs negligibly, and so has the same measure.

And so if the measure on is -finite, then so are the measures on and on . We’ve already seen that must be -finite in this situation, and so any -measurable set can be covered by a countable sequence of -finite sets. Applying the above results to each of the sets in this sequence gives our result.