The smallest -ring containing is also the smallest one containing the collection of semiclosed intervals. As it turns out, it’s also a -algebra. Indeed, we can write the whole real line as the countable disjoint union of elements of .
and so itself must be in . We call the -algebra of “Borel sets” of the real line.
Our measure — defined on elements of by — is not just -finite, but actually finite on . And thus its extension to will still be -finite. The above decomposition of into a countable collection of sets of finite -measure shows us that the extended measure is, in fact, totally -finite.
But our measure might not be complete. As the smallest -algebra containing , might not contain all subsets of sets of -measure zero. And thus we form the completions of our -algebra and of our measure. We call the -algebra of “Lebesgue measurable sets”, and is “Lebesgue measure” (remember, it’s pronounced “luh-BAYG”). In fact, the incomplete measure on Borel sets is also often called Lebesgue measure.
At last we can show that the set function we defined on semiclosed intervals is a measure. It’s clearly real-valued and non-negative. We already showed that it’s monotonic, and this will come in handy as we show that it’s countably additive.
So, if is a countable disjoint sequence of semiclosed intervals whose union is also a semiclosed interval , then our first monotonicity property shows that for any finite we have
and so in the limit we must still have
But the sequence covers , and so our other monotonicity property shows that
which gives us the equality we want.
But this still isn’t quite a measure. Why not? It’s only defined on the collection of semiclosed intervals, and not on the ring of finite disjoint unions. But we’re in luck: there is a unique finite measure on extending on . That is, if , then .
Every set in is a finite disjoint union of semiclosed intervals, but not necessarily uniquely. Let’s say we have both
Then for each we have
which represents as a finite disjoint union of other sets in . Since is finitely additive, we must have
But since these sums are finite we can switch their order with no trouble. Thus we can unambiguously define
which doesn’t depend on how we represent as a finite disjoin union of semiclosed intervals.
This function clearly extends , since if we can just use itself as our finite disjoint union. It’s also easily seen to be finitely additive, and that there’s not really any other way to define a finitely additive set function to extend . But we still need to show countable additivity.
So, let be a disjoint sequence of sets in whose union is also in . Then for each we have
If happens to be in , then the collection of all the is countable and disjoint, and we can use the countable additivity of we proved above to show
In general, though, is a finite disjoint union
and we can apply the previous result to each of the :
From here on out, we’ll just write instead of for this measure on .
First off, let be a finite, disjoint collection of semiclosed intervals, all of which are contained in another semiclosed interval . Then we have the inequality
Indeed, we can write , , and without loss of generality assume that . Then our hypotheses tell us that
On the other hand, if is a closed interval contained in the union of a finite number of bounded open intervals , then we have the strict inequality
We can rearrange the open intervals by picking to contain . Then if we have and we can discard all the other sets since they only increase the right hand side of the inequality. But if , we can pick some containing . Now we repeat, asking whether is greater or less than . Eventually we’ll have a finite collection of satisfying , , and . It follows that
What does this have to do with semiclosed intervals? Well, if is a countable sequence of semiclosed intervals that cover another semiclosed interval , then we have the inequality
If , then this is trivially true, so we’ll assume it isn’t, and let be a positive number with . Then we have the closed set . We can also pick any positive number and define .
Now is smaller than , and each is larger than the corresponding , and so we find that is a closed interval covered by the open intervals . But the Heine-Borel theorem says that is compact, and so we can find a finite collection of the which cover . Renumbering the open intervals, we have
and our above result tells us that
Since we can pick and to be arbitrarily small, the desired inequality follows.
Before we go any further, let’s work towards an actual example of a measure. This one, in the long run, will be useful to us.
The underlying space we’re interested in is the real line. We need to start with a class of sets we’re interested in measuring. Specifically, we’re going to take to be the class of finite intervals, open on the right and closed on the left. That is, given finite real numbers we consider the interval
Such a bounded interval we’ll call “semiclosed”. We’ll also throw into and let this count as a degenerate sort of semiclosed interval.
Now, given two semiclosed intervals, their intersection is again a semiclosed interval. One possibility is that one interval contains the other, in which case the intersection is the smaller interval. Another possibility is that the intervals are disjoint, in which case their intersection is empty. The last possibility is that they overlap: we consider and with . Then their intersection is , which is a semiclosed interval.
The difference of two semiclosed intervals may or may not be a semiclosed interval. If intervals overlap, as above, then , and . If the intervals are disjoint, then the difference is just the original interval. But if contains , then the difference is . This isn’t a semiclosed interval, but it’s a finite disjoint union of semiclosed intervals.
But we know that these properties are exactly what we need to show that the collection of finite disjoint unions of intervals in is a ring. We could have started with open intervals or closed intervals, but then we wouldn’t have such a nice ring pop out.
We will define a finite set function . For an interval , we define . For the empty set, we define . This is the function that will be developed into our measure.
If and are disjoint sets in , then
Here, we take to be a measurable cover of and to be a measurable kernel of . The difference must be contained in , and so
On the other hand, we can take to be a measurable kernel of and to be a measurable cover of . Now the difference is contained in , and we find
Now, if , then for every whatsoever, we have
We can take and and stick them into the previous result to find
But since , we know that , and this establishes our result.
Interestingly, we can use this method of inner measures as an alternative approach to our extension theorems. If is a -finite measure on a ring , and if is the induced outer measure on , then for every set of finite measure and every we have
Then if and are two sets in such that , then we find
and so we can use this formula as the definition of the inner measure . Then we can define a set with to be -measurable if the inner and outer measures match: . And from here, the rest of the theory is as before.
There are a bunch of useful facts that we can prove with the help of measurable covers and measurable kernels. These will allow us to move statements we want to show about outer and inner measures to the realm of proper measures, where we can use nice things like additivity.
For example, given a pairwise disjoint sequence , we can show that
Just pick a measurable kernel for each . Then we can use countable additivity to show
For another, given a set and a disjoint sequence whose union is , then
This time, let be a measurable kernel of , so that
On the other hand, is the union of the , and so we can use the previous result to get the opposite inequality.
Next: if , then clearly . But conversely, if , then . To see this, let be a measurable kernel and be a measurable cover of . Then we calculate
But , so (by the completeness of ), and thus . Thus sets of finite measure in are exactly those in for which the outer and inner measures coincide.
Interestingly, notice how the last step of this proof echoes our earlier result that a set is Jordan measurable if and only if the Jordan content of its boundary is zero.
A measurable kernel is the flip side of a measurable cover. Specifically, given , a measurable kernel of is a set such that , and if for every with we have . And, as it happens, every set has a measurable kernel.
To find it, let be a measurable cover of . Then let be a measurable cover of , and set . Since contains , we find
If , then
Since was picked to be a measurable cover of , we conclude that , as we hoped.
Now if is a measurable kernel of , then . Indeed, since , we have . If this inequality is strict then , and there must be some with and . But , while , contradicting the fact that was chosen to be a measurable kernel of .
The symmetric difference of any two measurable kernels is negligible. Given two measurable kernels and , we know that . This implies that , and thus . Similarly, , and thus .
A quick one for today.
In analogy with the outer measure induced on the hereditary -ring by the measure on the -ring , we now define the “inner measure” that induces on the same hereditary -ring . We’ve seen that the outer measure is
Accordingly, the inner measure is a set function defined by
In a way, the properties of are “dual” to those of . The easy ones are the same: it’s non-negative, monotone, and .
We could also define in terms of the completed measure. Since , it’s clear
On the other hand, the definition of the completion says that for every there is a with and , and so this is actually an equality.
So, we’ve got a -finite measure on a ring , and we extend it to a measure on the -ring . But often it’s a lot more convenient to work with itself than the whole of . So, to what extent can we do this efficiently?
As it turns out, if has finite measure and , then we can find a set so that .
Any set can be covered by a sequence of sets in , and we know that
That is, we can find such a cover satisfying
But since is continuous, we see that
The sequence of numbers increases until it’s within of its limit. That is, there is some so that if we define to be the union of the first sets in the sequence, we have
But now we can find
And thus .
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.
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 .