Today we start in on the representation theorem proved by Marshall Stone: every boolean ring is isomorphic (as a ring) to a ring of subsets of some set . That is, no matter what looks like, we can find some space and a ring of subsets of so that as rings.
We start by defining the “Stone space” of a Boolean ring . This is a representable functor, and the representing object is the two-point Boolean ring . That is, , the set of (boolean) ring homomorphisms from to . To be clear, consists of the two points , with the operations for addition and for multiplication, and the obvious definitions of these operations. This is a contravariant functor — if we have a homomorphism of Boolean rings we get a function between the Stone spaces , which takes a function to the function .
The Stone space isn’t just a set, though; it’s a topological space. We define the topology by giving a base of open sets. That is, we’ll give a collection of sets — closed under intersections — which we declare to be open, and we define the collection of all open sets to be given by unions of these sets. For each element , we define a basic set like so:
To see that this collection of sets is closed under intersection, consider two such sets and . I say that the intersection of these sets is the set . Indeed, if and , then
Conversely, if , then . Thus
and so , and . Similarly, . Thus we see that .
In fact, this map from to the basic sets is exactly the mapping we’re looking for! We’ve already seen that our base is closed under intersection, and that the map preserves intersections. I say that we also have . If but , then and . Then
and similarly if but . Thus . Conversely, if , then
and so either and or vice versa. Thus .
So we know that is a homomorphism of (boolean) rings. However, we don’t know yet that it’s an isomorphism. Indeed, it’s possible that has a nontrivial kernel — could be for some . We must show that given any there is some so that .
For a finite boolean ring this is easy: we pick some minimal element and define if and only if . Such a exists because there’s at least one element below — itself is one — and there can only be finitely many so we can just take their intersection. Clearly by definition, and it’s straightforward to verify that is a homomorphism of boolean rings using the fact that is an atom of .
For an infinite boolean ring, things are trickier. We define the set of all functions , not just the ring homomorphisms. This is the product of one copy of for every element of . Since each copy of is a compact Hausdorff space, Tychonoff’s theorem tells us that is a compact Hausdorff space. If is any finite subring of containing , let be the collection of those functions which are homomorphisms when restricted to and for which .
I say that the class of sets of the form has the finite intersection property. That is, if we have some finite collection of finite subrings and the finite subring they generate, then we have the relation
Indeed, is clearly contained in the generated ring . Further, if is a homomorphism on then it’s a homomorphism on each subring .
Okay, so since is a finite boolean ring, the proof given above for the finite case shows that is nonempty. Thus the intersection of any finite collection of sets is nonempty. And thus, since is compact, the intersection of all of the is nonempty.
That is, there is some function which is a homomorphism of boolean rings on any finite boolean subring containing , and with . Given any other two points and there is some finite boolean subring containing , , and , and so we must have and within this subring, and thus within the whole ring. Thus is a homomorphism of boolean rings sending to , which shows that .
Therefore, the map is a homomorphism sending the boolean ring isomorphically onto the identified base of the Stone space .
Yesterday, we saw that an absolutely continuous finite signed measure on a measure space defines a continuous function on the associated metric space , and that a sequence of such finite signed measures that converges pointwise is actually uniformly absolutely continuous with respect to .
We’re going to need to assume that is nonnegative. We’d usually do this by breaking into its positive and negative parts, but it’s not so easy to get ahold of the positive and negative parts of in this case. However, we can break each into and . Then we can take the limits and , which will still satisfy . The only difference between this decomposition and the positive and negative parts is that this pair of set functions might have some redundancy that gets cancelled off in this subtraction. And so, without loss of generality, we will assume that all the are nonnegative, and that their limit is as well.
Now, given such a sequence, define the limit function . I say that is itself a finite signed measure, and that . Indeed, is finite by assumption, and additivity is easy to check. As for absolute continuity, if , then each since , and so as the limit of the constant zero sequence.
What we need to check is continuity. We know that it suffices to show that is continuous from above at . So, let be a decreasing sequence of measurable sets whose limit is . We must show that the limit of is zero. But we know that the limit of is zero, and thus for a large enough we can make for any given . And since we know that for any there is some so that if then . Thus we can always find a large enough to guarantee that , and so the limit is zero, as asserted.
Finally, what happens if we remove the absolute continuity requirement from the ? That is: what can we say if is a sequence of finite signed measures on so that exists and is finite for each . I say that is a signed measure. What we need is to find some measure so that all the , and then we can use the above result.
Since is a finite signed measure, we can pick some upper bound . Then we define
If any , then , and so . And thus for all , as desired.
Indeed, if is any set with , then represents a point of , and defines the value of our function at this point. If is another set representing the same point, then . By absolute continuity, as well, and so . Thus our function doesn’t depend on the representative we use.
As for continuity at a point , given an , we want to find a so that if then . We calculate
Since is finite, we know that for every there is a so that if then . Using this , our assertion of continuity follows.
Now, if is a sequence of finite signed measures on that are all absolutely continuous with respect to , and if the limit exists and is finite for each , then the sequence is uniformly absolutely continuous with respect to . That is,
For any we can define the set
Since each is continuous as a function on , each of these is a closed set. Since the sequence always converges to a finite limit, it must be Cauchy for each , and so the union of all the is all of . Thus the countable union of these closed subsets has an interior point. But since is a complete metric space, it is a Baire space as well. And thus one of the must have an interior point as well.
Thus there is some , some radius , and some set so that the ball is contained in . Let be a positive number with , and so that whenever and This will suffice (by definition) for all up to . We will show that it works for higher as well. Note that if , then
so and are both inside . And so we calculate
The first term is less than by the definition of . The second and third terms are less than or equal to because and are in . Since the same works for all , the absolute continuity is uniform.
Looking over my notes from topology it seems I completely skipped over Baire sets. This was always one of those annoying topics that I never had much use for, partly because I didn’t do point-set topology or analysis. Also, even in my day the usual approach was a very classical and awkward one. Today I’m going to do a much more modern and streamlined one, and I can motivate it better from a measure-theoretic context to boot!
Basically, the idea of a Baire set is one that can’t be filled up by “negligible” sets. We’ve used that term in measure theory to denote a subset of a set of measure zero. But in topology we don’t have a “measure” to work with. Instead, we use the idea of a closed “nowhere dense” set — one for which there is no open set on which it is dense. The original motivation was a set like a boundary of a region; in the context of Jordan content we saw that such a set was negligible.
Clearly such a set has no interior — no open set completely contained inside — and any finite union of them is still nowhere dense. However, if we add up countably infinitely many we might have enough points to be dense on some open set. However, we don’t want to be able to actually fill such an open set. In the measure-theoretic context, this corresponds to the way any countable union of negligible sets is still negligible.
So, let’s be more specific: a “Baire set” is one for which the interior of every countable union of closed, nowhere dense sets is empty. Equivalently, we can characterize Baire sets in complementary terms: every countable intersection of dense open sets is dense. We can also use the contrapositive of the original definition: if a countable union of closed sets has an interior point, then one of the sets must itself have an interior point.
We’re interested in part of the famous “Baire category theorem” — the name is an artifact of the old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space is a Baire space. Let be a countable collection of open dense subsets of . We will show that their intersection is dense by showing that any nonempty open set has some point — the same point — in common with all the .
Okay, Since is dense, then is nonempty, and it contains a point . As the intersection of two open sets, it’s open, and so it contains an open neighborhood of which set can take to be an open metric ball of radius . But then is an open set, which will intersect . This process will continue, and for every we will find a point and a radius so that . We can also at each step pick .
And so we come up with a sequence of points . At each step, the ball contains the whole tail of the sequence past , and so all of these points are within of each other. Since gets arbitrarily small, this shows that is Cauchy, and since is complete, the sequence must converge to a limit . This point will be in each set , since , and it’s obviously in , as desired.
The other part of the Baire category theorem says that any locally compact Hausdorff space is a Baire space. In this case the proof proceeds very similarly, but with the finite intersection property for compact spaces standing in for completeness.
First of all, the metric on is translation-invariant. That is, if , , and are sets in with finite measure, then . Indeed, we calculate
Next we need to define an “atom” of a measure ring to be a minimal nonzero element. That is, it’s an element so that if then either or . We also define a measure ring to be “non-atomic” if it has no atoms. As a quick result: a totally -finite measure ring can have at most countably many atoms, since must contain all of them, and no two of them can have a nontrivial intersection — if there were uncountably many of them, any decomposition of could only cover countably many of them.
On the other hand, we define a metric space to be “convex” if for any two distinct elements and there is an element between them. That is, is neither nor , and it satisfies the equation . We assert that the metric space of a -finite measure ring is convex if and only if the measure ring is non-atomic.
Let and be elements of . Without loss of generality we can assume that by using the translation-invariance of . Indeed, we can replace with and with . There will be an element between and if and only if is between and .
So for any is there an element between and ? Such a will satisfy
This is only possible if , which means that , and so . But for to be between and it cannot be equal to either of them, which means that cannot be an atom for any with . Since we can decompose any into a countable union of elements of finite measure, no element of infinite measure can be an atom either.
It will be useful to have other ways of showing that a given function between boolean rings is a morphism. The definition, of course, is that and , since and here denote our addition and multiplication, respectively.
It’s also sufficient to show that and . Indeed, we can build both and from and :
So if and are preserved, then so are and .
A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if is equivalent to for a surjection from one underlying set onto the other, then is a homomorphism of Boolean rings. We first show that preserves by using its characterization as a least upper bound: , , and if and then .
So, since preserves order, we know that , and that . We can conclude from this that , and we are left to short that . But f(E)\cup f(F)=f(G)$ for some , since is surjective. Since , we must have , and similarly . But then , and so , as we wanted to show.
We thus know that preserves the operation , and we can similarly show that preserves , using the dual universal property. In order to build and from and , we need complements. But complements satisfy a universal property of their own: , and if then .
First, I want to show that by showing it is below all elements of . Indeed, by the surjectivity of , every element of is the image of some element . Thus we want to show that . But this is true because for all , and so
Now given a set and its complement , we know that . Since preserves , we must have . Let’s say is another set so that . By the surjectivity of , we have for some . Thus , and thus . This tells us that , and thus . And so .
Therefore, since preserves , , and complements, it preserves and , and thus is a morphism of Boolean rings.
We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring is complete if every set of elements of has a union.
A complete Boolean ring is clearly a Boolean -algebra. It’s an algebra, because we can get a top element by taking the union of all the elements of together, and it’s a -ring because countable unions are included under all unions. The converse is a little hairier.
First off, every totally finite measure algebra is complete. That is, if — and thus for all , then every collection of elements of will have a union.
We let be the collection of all finite unions of elements in , all of which exist since is a Boolean ring. We let be the (finite) supremum of the measures of all elements in . Since this is a finite supremum, there must be some increasing sequence so that increases to . I say that the union of the sequence — which exists because is a -ring — is the union of all of .
Indeed, we must have by the continuity of the measure . Take any set and consider the difference , which is the amount by which extends past . Our assertion is that this is always . Define the sets , which is a disjoint union since is disjoint from . Each of the is in , and the limit of the sequence is . This set has measure
but since each of the is bounded above by then so is their limit. Thus , and thus , since we consider all negligible sets to be the same.
The same is true for totally -finite measure algebras. Indeed, we can break such a measure algebra up into a countable collection of finite measure algebras by breaking into a countable number of elements so that . We define the finite measure algebra to consist of the intersections of elements of with . Then given any collection we consider its image under such intersections to get . What we said above shows that each of these collections has an intersection , and their union in is the union of the original collection .
To see this, let be a Cauchy sequence in the metric space . That is, for every there is some so that for all . Unpacking our definitions, each must be an element of the measure ring with , and thus must be (represented by) a measurable subset of finite measure. On the side of the distance function, we must have for sufficiently large and .
Let’s recast this in terms of the characteristic functions of the sets in our sequence. Indeed, we find that , and so
that is, a sequence of sets is Cauchy in if and only if its sequence of characteristic functions is mean Cauchy. Since mean convergence is complete, the sequence of characteristic functions must converge in mean to some function . But mean convergence implies convergence in measure, which is equivalent to a.e. convergence on sets of finite measure, which is what we’re dealing with.
Thus the limiting function must — like the characteristic functions in the sequence — take the value or almost everywhere. Thus it is (equivalent to) the characteristic function of some set. Since must be measurable — as the limit of a sequence of measurable functions — it’s the characteristic function of a measurable set, which must have finite measure since its measure is the limit of the Cauchy sequence . That is, , where , and is the limit of under the metric of . Thus is complete as a metric space.
Let be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that be positive — that only if — we don’t need to worry about negligible elements.
And so we write for the metric space consisting of the elements with . This has a distance function defined by . We also write for the metric associated with the measure algebra associated with the measure space . We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.
Now, if we set
then , , and itself are all uniformly continuous.
Indeed, if we take two pairs of sets , , , and , we calculate
Similarly, we find that . And thus
And so if we have control over the distance between and , and the distance between and , then we have control over the distance between and . The bounds we need on the inputs uniform, and so is uniformly continuous. The proof for proceeds similarly.
To see that is uniformly continuous, we calculate
Now if is a -finite measure space so that the -ring has a countable set of generators, then is separable. Indeed, if is a countable sequence of sets that generate , then we may assume (by -finiteness) that for all . The ring generated by the is itself countable, and so we may assume that is itself a ring. But then we know that for every and for every positive we can find some ring element so that . Thus is a countable dense set in , which is thus separable.
We’re not just interested in Boolean rings as algebraic structures, but we’re also interested in real-valued functions on them. Given a function on a Boolean ring , we say that is additive, or a measure, -finite (on -rings), and so on analogously to the same concepts for set functions. We also say that a measure on a Boolean ring is “positive” if it is zero only for the zero element of the Boolean ring.
Now, if is the Boolean -ring that comes from a measurable space , then usually a measure on is not positive under this definition, since there exist sets of measure zero. However, remember that in measure theory we usually talk about things that happen almost everywhere. That is, we consider two sets — two elements of — to be “the same” if their difference is negligible — if the value of takes the value zero on this difference. If we let be the collection of -negligible sets, it turns out that is an ideal in the Boolean -ring . Indeed, if and are negligible, then so is , so is an Abelian subgroup. Further, if and , then , so is an ideal.
So we can form the quotient ring , which consists of the equivalence classes of elements which differ by an element of measure zero. This is equivalent to our old rhetorical trick of only considering properties up to “almost everywhere”. Using this new definition of “equals zero”, any measure on a Boolean -ring gives rise to a positive measure on the quotient -ring . In particular, given a measure space , we write for the Boolean -ring it gives rise to.
We say that a “measure ring” is a Boolean -ring together with a positive measure on . For instance, if is a measure space, then is a measure ring.. If is a Boolean -algebra we say that is a measure algebra. We say that measure rings and algebras are (totally) finite or -finite the same as for measure spaces. Measure rings, of course, form a category; a morphism from one measure algebra to another is a morphism of boolean -algebras so that for all .
I say that the mapping which sends a measure space to its associated measure algebra is a contravariant functor. Indeed, let be a morphism of measure spaces. That is, is a measurable function from to , so contains the pulled-back -algebra . This pull-back defines a map . Further, since is a morphism of measure spaces it must push forward to . That is, , or in other words . And so if then , thus the ideal is sent to the ideal , and so descends to a homomorphism between the quotient rings: . As we just said, , and thus we have a morphism of measure algebras . It’s straightforward to confirm that this assignment preserves identities and compositions.