If we start with a measure on a ring , we can extend it to an outer measure on the hereditary -ring . And then we can restrict this outer measure to get an actual measure on the -ring of -measurable sets. And so we ask: how does the measure relate to the measure .
First, we will show that any set is -measurable. Then, since is a -ring containing it will also contain the smallest such -ring .
So, if and , and if , then the definition of the induced outer measure tells us that there exists a sequence so that is contained in their union and
Since this holds for all , we conclude that , and thus that is -measurable.
Now is defined for every set in . And since it’s a measure on each of the rings and , we can induce outer measures from each! But since these are -rings, life is a little easier. For one thing, the hereditary -ring each one generates is just again; for another, the induced outer measure of a set is just the infimum of the measure of any set containing . And, as it turns out, both of the induced outer measures are exactly ! That is, for any we find:
Indeed, by definition we have
since everything in the first set is also in the second, and everything in the third set is also in the fourth. The equality in the middle holds because every sequence in can be replaced by a disjoint sequence, and the sum of the measures can then be replaced by the measure of the disjoint union, and so we only ever need to use one set in .
But since for every , we must have . Thus all the inequalities above are equalities, as we claimed.
Yesterday, we showed that — given an outer measure on a hereditary -ring — the collection of -measurable sets forms a ring. In fact, it forms a -ring. That is, given a countably infinite sequence of -measurable sets, their union is also -measurable. Even better if the are pairwise disjoint, then
To see this, we start with a disjoint sequence and an equation we came up with yesterday:
We can keep going like this, adding in more and more of the :
for every natural number . This finite union is -measurable. This, along the fact that , tells us that
This is true for every , so we may pass to the limit and use the countable subadditivity of
But this is enough to show that is -measurable, and so is closed under countable disjoint unions. And this shows that is closed under countable unions in general, by our trick of replacing a sequence by a disjoint sequence with the same partial unions.
Since the previous inequalities must then actually be equalities, we see that we must have
It’s tempting to simply subtract from both sides, but this might be an infinite quantity. Instead, we’ll simply replace with , which has the same effect of giving us
as we claimed.
If we replace by in this equation, we find that — when restricted to the -ring — is actually countably additive. That is, if we define for , then is actually a measure. Even better, it’s a complete measure.
Indeed, if and , then as well. We must show that is actually -measurable, and so exists and equals zero. But we can easily see that for any
and this is enough to show that is -measurable, and thus that is complete.
An outer measure on a hereditary -ring is nice and all, but it’s not what we really want, which is a measure. In particular, it’s subadditive rather than additive. We want to fix this by restricting to a nice collection of sets within .
Every set splits every other set into two pieces: the part that’s in and the part that’s not. What we want to focus on are the sets that split every other set additively. That is, sets so that for every set we have
We call such sets “-measurable”. Actually, to show that is -measurable we just need to show that for every , because the opposite inequality follows from the subadditivity of .
This condition seems sort of contrived at first, and there’s really not much to justify it at first besides the foregoing handwaving. But we will soon see that this definition turns out to be useful. For one thing, the collection of -measurable sets is a ring!
The proof of this fact is straightforward, but it feels like pulling a rabbit out of a hat, so follow closely. Given sets -measurable sets and , we need to show that their union and difference are both -measurable as well. Saying that is -measurable means that for every we have
Saying that is -measurable means that for every we have
We can take each of these and plug them into the first equation to find the key equation
Now this key equation works for as well as . We know that and , but . So, sticking into the key equation we find
But the three terms on the right are the first three terms in the key equation. And so we can replace them and write
which establishes that is -measurable! Behold, the rabbit!
Let’s see if we can do it again. This time, we take and stick it into the key equation. We find
Again we can find the three terms on the right of this equation on the right side of the key equation as well. Replacing them in the key equation, we find
which establishes that is -measurable as well!
That is, can be covered by a countable collection of sets in . For every such cover, sum up the -measures of all the sets in the cover, and define to be the greatest lower bound of such sums. Then is an outer measure, which extends to all of . Further, if is -finite, then will be too. We call the outer measure “induced by” .
First off, if itself, then we can cover it with itself and an infinite sequence of empty sets. That is, . Thus we must have . On the other hand, if is contained in the union of a sequence , then monotonicity tells us that , and thus . That is, must be equal to for sets ; as a set function, indeed extends . In particular, we find that .
If and are sets in with and is a sequence covering , then it must cover as well. Thus can be at most , and may be even smaller. This establishes that is monotonic.
We must show that is countably subadditive. Let and be sets so that is contained in the union of the . Let be an arbitrarily small positive number, and for each choose some sequence that covers such that
This is possible because the definition of tells us that we can find a covering sequence whose measure-sum exceeds by an arbitrarily small amount. Then the collection of all the constitute a countable collection of sets in which together cover . Thus we conclude that
Since was arbitrary, we conclude that
and so is countably subadditive.
Finally, if , we can pick a cover . If is -finite, we can cover each of these sets by a sequence so that . The collection of all the is then a countable cover of by sets of finite measure; the extension is thus -finite as well.
We’re going to want a modification of the notion of a measure. But before we introduce it, we have (of course) a few definitions.
First of all, a collection of sets is called “hereditary” if it includes all the subsets of each of its sets. That is, if and , then as well. It’s not very useful to combine this with the definition of an algebra, because an algebra must contain itself; the only hereditary algebra is itself. Instead, we define a “ring” of sets (or a -ring) to be closed under union (countable unions for -rings) and difference operations, but without the requirement that it contain ; complements and intersections are also not guaranteed, since we built these from differences using itself. Pretty much everything we’ve done so far with algebras can be done with rings, and hereditary -rings will be interesting objects of study.
Just like we found for algebras and monotone classes, the intersection of two hereditary collection is again hereditary. We can thus construct the “smallest” hereditary -ring containing a given collection , and we’ll call it . In fact, it’s not hard to see that this is the collection of all sets which can be covered by a countable union of sets in ; any -ring containing must contain all such countable unions, and a hereditary collection must then contain all the subsets.
Now, an extended real-valued set function on a collection is called “subadditive” whenever , , and their union are in , we have the inequality
It’s called “finitely subadditive” if for every finite collection whose union is also contained in we have the inequality
and “countably subadditive” if for every sequence of sets in whose union is also in , we have
Note that these differ from additivity conditions in two ways: we only ask for an inequality to hold, and we don’t require the unions to be disjoint.
Finally, we can define an “outer measure” to be an extended real-valued, non-negative, monotone, and countably subadditive set function , defined on a hereditary -ring , and such that . Just as for a measure, we say that is “finite” or “-finite” if every set has finite or -finite outer measure.
First, we need to define the “symmetric difference” of two sets. This is the collection of points in one or the other set, but not in both:
So now we define an extended real-valued function by
I say that this has almost all the properties of a metric function. First of all, it’s clearly symmetric and nonnegative, so that’s two of the four right there. It also satisfies the triangle inequality. That is, for any three sets , , and in , we have the inequality
Indeed, points in are either in or not. If not, then they’re in , while if they are they’re in . Similarly, points in are either in or . That is, the symmetric difference is contained in the union of the symmetric difference and the symmetric difference . And so monotonicity tells us that
establishing the triangle inequality.
What’s missing is the assertion that if and only if . But there may be plenty of sets with measure zero, and any one of them could arise as a symmetric difference; as written, our function is not a metric. But we can fix this by changing the domain.
Let’s define a relation: if and only if . This is clearly reflexive and symmetric, and the triangle inequality above shows that it’s transitive. Thus is an equivalence relation, and we can pass to the collection of equivalence classes. That is, we consider two sets and to be “the same” if .
This trick will handle the obstruction to being a metric, but only if we can show that gives a well-defined function on these equivalence classes. That is, if and , then . But means , and similarly for . Thus we find
and so the two are equal. We define the distance between two -equivalence classes by picking a representative of each one and calculating between them.
This relation turns out to be extremely useful. That is, as we go forward we will often find things simpler if we consider two sets to be “the same” if they differ by a set of measure zero, or by a subset of such a set. We will call subsets of sets of measure zero “negligible”, since we can neglect things that only happen on such a set.
Again we start with definitions. An extended real-valued set function on a collection of sets is “continuous from below” at a set if for every increasing sequence of sets — that is, with each — for which — remember that this limit can be construed as the infinite union of the sets in the sequence — we have . Similarly, is “continuous from above” at if for every decreasing sequence for which and which has for at least one set in the sequence we have . Of course, as usual we say that is continuous from above (below) if it is continuous from above (below) at each set in its domain.
Now I say that a measure is continuous from above and below.
First, if is an increasing sequence whose limit is also in , then . Let’s define and calculate
where we’ve used countable (and finite) additivity to turn the disjoint union into a sum and back.
Next, if is a decreasing sequence whose limit is also in , and if at least one of the has finite measure, then . Indeed, if has finite measure then by monotonicity, and thus the limit must have finite measure as well. Now is an increasing sequence, and we calculate
And thus a measure is continuous from above and from below.
On the other hand we have this partial converse: Let be a finite, non-negative, additive set function on an algebra . Then if either is continuous from below at every or is continuous from above at , then is a measure. That is, either one of these continuity properties is enough to guarantee countable additivity.
Since is defined on an algebra, which is closed under finite unions, we can bootstrap from additivity to finite additivity. So let be a countably infinite sequence of pairwise disjoint sets in whose (disjoint) union is also in , and define the two sequences in :
If is continuous from below, is an increasing sequence converging to . We find
On the other hand, if is continuous from above at , then is a decreasing sequence converging to . We find
First, we make a couple of definitions. A set function on a collection of sets is “monotone” if whenever we have and in with , then . We say that is “subtractive” if whenever further and , then .
Now I say that any measure on an algebra is both monotone and subtractive. Indeed, since is an algebra, then is guaranteed to be in as well, and it’s disjoint from . And so we see that
Since is non-negative, we must have , and so is monotone. And if is finite, then we can subtract it from both sides of this equation to show that is subtractive as well.
Next, say is a measure on an algebra . If we have a set and a finite or (countably) infinite sequence of sets so that , then we have the inequality . To show this, we’ll invoke a very useful trick: if is a sequence of sets in an algebra , then there is a pairwise disjoint sequence of sets so that each and . Indeed, we define
That is, is the same as , and after that point each is everything in that hasn’t been covered already.
So we start with , and come up with a new sequence . The (disjoint) union of the is, like the union of the . Then, by the countable additivity of , we have . But the monotonicity says that , and also that , and so .
On the other hand, if is a pairwise disjoint and we have , then . Indeed, if is a finite sequence, then the union is in . Monotonicity and finite additivity then tell us that
However if is an infinite sequence, then what we just said applies to any finite subsequence. Then the sum is the limit of the sums . Each of these is less than or equal to , and so their limit must be as well.
From this point in, I will define a “set function” as a function whose domain is some collection of subsets . It’s important to note here that is not defined on points of the set , but on subsets of . For some reason, a lot of people find that confusing at first.
We’re primarily concerned with set functions which take their values in the “extended real numbers” . That is, the value of is either a real number, or , or , with the latter two being greater than all real numbers and less than all real numbers, respectively.
We say that such a set function is “additive” if whenever we have disjoint sets and in with disjoint union also in , then we have
Similarly, we say that is finitely additive if for every finite, pairwise disjoint collection whose union is also in we have
And we say that is countably additive of for every pairwise-disjoint sequence of sets in whose union is also in , we have
Now we can define a “measure” as an extended real-valued, non-negative, countably additive set function defined on an algebra , and satisfying . With this last assumption, we can show that a measure is also finitely additive. Indeed, given a collection , just define for to get a sequence. Then we find
If is a measure on , we say a set has finite measure if . We say that has “-finite” measure if there is a sequence of sets of finite measure () so that . If every set in has finite (or -finite) measure, we say that is finite (or -finite) on .
Finally, we say that a measure is “complete” if for every set of measure zero, also contains all subsets of . That is, if , , and , then . At first, this might seem to be more a condition on the algebra than on the measure , but it really isn’t. It says that to be complete, a measure can only assign to a set if all of its subsets are also in .
Now we want to move from algebras of sets closer to -algebras by defining a new structure: a “monotone class”. This is a collection so that if we have an increasing sequence of sets
with , then the “limit” of this sequence is in . Similarly, if we have a decreasing sequence of sets
with , then the “limit” of this sequence is in . In each case, we can construe this limit more simply. In the case of an increasing sequence, each term contains all the terms before it, and is thus the same as the union of the whole sequence up to that point. Thus we can define
Similarly, for a decreasing sequence we can define
Now, if is an algebra of subsets of , and is a monotone class containing , then also contains , the smallest -algebra containing .
As an aside, what do we mean by “smallest”? Well, it’s not hard to see that the intersection of a collection of algebras is itself a -algebra, just like we did way back when we showed that we had lower bounds in the lattice of ideals. So take all the -algebras containing — there is at least one, because itself is one — and define to be the intersection of all of them. This is a -algebra contained in all the others, thus “smallest”.
Similarly, there is a smallest monotone class containing ; we will assume that is this one without loss of generality, since if it contains then all the others do as well. We will show that this is actually a -algebra, and then it must contain !
Given , define to be the collection of so that . This is itself a monotone class, and it contains , and so it must contain . But it’s defined as being contained in , and so we must have . Thus as long as and .
Now take a and define to be the collection of so that . What we just showed is that , and is a monotone class. And so , and whenever both and are in .
We can repeat a similar argument to the previous two paragraphs to show that for . Start by defining as the collection of so that both and are in .
So is both a monotone class and an algebra. We need to show that it’s a -algebra by showing that it’s closed under countable unions. But if then is a monotone increasing sequence of set in , since they’re all finite unions. The countable union is the limit of this sequence, which contains by virtue of being a monotone class.
And so is a -algebra containing , and so it contains .