## Completions of Measures

We’ve shown that we can uniquely extend a -finite measure on a ring to a unique -finite measure on the -ring . But, of course, we actually found that we could restrict the outer measure to the -ring of -measurable sets, which may be larger than . Luckily, we can get this extra ground without having to go through the outer measure.

What’s the essential difference? What do we know about that we don’t know about ? The measure on is complete. The smaller -ring may not contain all negligible sets.

So, let’s throw them in; if is a measure on a -ring , define to be the class of all sets , where , and is a negligible with respect to the measure , or “-negligible”. This collection is a -algebra, and the set function defined on by is a complete measure called the “completion” of .

First, given sets and with , we have the two equations

These tell us that any set that can be written as the symmetric difference of a set in and a measurable set can also be written as the union of two other such sets, and vice versa. That is, we can also characterize as the class of sets of the form instead of .

This characterization makes it clear that is closed under countable unions. Indeed, just write each set in a sequence as a union of a set from and a -negligible set . The countable union of the is still in , and the countable union of the negligible sets is still negligible. Thus is a -ring.

Now, if we have two ways of writing a set in , say , then we also have the equation . And, therefore, . Therefore, , and the above definition of is unambiguous.

Using the characterization of in terms of unions, it’s easy to verify that is a measure. We only need to check countable additivity, which is perfectly straightforward; the union of a pairwise disjoint sequence of sets is formed by taking the union of the and the . The measure is countably additive on the , and the union of the is still negligible.

Finally, we must show that is complete. But if we have a set with , then and with . Thus any subset is also a subset of , with . Writing it as , we find , and so is complete.

There’s just one fly in the ointment: we don’t really know that this complete measure is the same as the one we get by restricting from to -measurable sets. It turns out that if is -finite on , then the completion of the extension of to *is* the same as this restriction.

Since we’ve been using for the completion, let’s write for the class of -measurable sets. Since restricted to is complete, it follows that , and that and coincide on . We just need to show that .

In light of the fact that is also -finite on , we just need to show that if has finite outer measure, then . But in this case has a measurable cover with . Since these are finite, we find that . But *also* has a measurable cover , with . And so we can write , showing that .

## 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 .