In this post I will give a proof of the measurable projection theorem. Recall that this states that for a complete probability space and a set S in the product sigma-algebra , the projection, , of S onto , is in . The previous post on analytic sets made some progress towards this result. Indeed, using the definitions and results given there, it follows quickly that is -analytic. To complete the proof of measurable projection, it is necessary to show that analytic sets are measurable. This is a consequence of Choquet’s capacitability theorem, which I will prove in this post. Measurable projection follows as a simple consequence.
The condition that the underlying probability space is complete is necessary and, if this condition was dropped, then the result would no longer hold. Recall that, if is a probability space, then the completion, , of with respect to consists of the sets such that there exists with and . The probability space is complete if . More generally, can be uniquely extended to a measure on the sigma-algebra by setting , where B and C are as above. Then is the completion of .
In measurable projection, then, it needs to be shown that if is the projection of a set in , then A is in the completion of . That is, we need to find sets with with . In fact, it is always possible to find a in which minimises , and its measure is referred to as the outer measure of A. For any probability measure , we can define an outer measure on the subsets of , by approximating from above,
Similarly, we can define an inner measure by approximating A from below,
It can be shown that A is -measurable if and only if . We will be concerned primarily with the outer measure , and will show that that if A is the projection of some , then A can be approximated from below in the following sense: there exists in for which . From this, it will follow that A is in the completion of .
It is convenient to prove the capacitability theorem in slightly greater generality than just for the outer measure . The only properties of that are required is that it is a capacity, which we now define. Recall that a paving on a set X is simply any collection of subsets of X, and we refer to the pair as a paved space.
Definition 1 Let be a paved space. Then, an -capacity is a map which is increasing, continuous along increasing sequences, and continuous along decreasing sequences in . That is,
- if then .
- if is increasing in n then as .
- if is decreasing in n then as .
As was claimed above, the outer measure defined by (1) is indeed a capacity.
Lemma 2 Let be a probability space. Then,
- for all .
- For all , there exists a with and .
- is an -capacity.
Proof: The first statement is clear — the definition (1) gives that . For any we can take to get to obtain the reverse inequality.
We show that satisfies the three properties required of an -capacity. That it is increasing is immediate from (1). For the second condition, consider a sequence increasing to the limit . By definition, there exists with and . Replacing by , if necessary, we can suppose that is an increasing sequence. Denote its limit by . Then and,
So . Next, consider a sequence decreasing to the limit . Then, using the first statement of the lemma and monotone convergence of the measure ,
So satisfies the defining properties of an -capacity.
Finally, for any , , the second statement of the lemma follows from (2) above (we can take ). ⬜
Next, recall that our definition of analytic sets made use of an auxiliary compact paved space. Before proceeding further, we will require some basic facts concerning compact pavings.
One way to construct a compact paving is to take any collection of closed subsets of a compact topological space. In fact, every compact paving arises in this way. Given a paved space , we can always consider the topology on K whose closed sets are generated by . The collection of closed sets under this topology is the closure of under taking finite unions and arbitrary intersections, which we will denote by . The Alexander subbase theorem states that if the original paving is compact, then the topology that it generates is also compact. Alexander’s theorem is usually stated in terms of open sets rather than closed ones, in which case the collection of open sets generating the topology is known as a subbasis. The statement in terms of closed sets is equivalent, simply by taking complements.
To be precise, when I refer to `finite unions’ this includes the empty union, which is just the empty set. Similarly, `finite intersections’ and `arbitrary intersections’ of a paving on a set X is taken to include the empty intersection, which is the whole of X.
Theorem 3 (Alexander Subbase Theorem) Let be a compact paved space, and be the closure of under finite unions and arbitrary intersections.
Then, is a compact paving.
Proof: Before proceeding with the proof, I note that we will make use of Zorn’s Lemma, which is equivalent to the Axiom of Choice. In fact, uncountable choice is not really required for our applications of analytic sets. For one thing, in the definition of analytic sets we could have restricted to countable compact pavings, for which the application of Zorn’s Lemma can be removed from the argument below. For another thing, we could have restricted from the start to compact paved spaces consisting of the closed subsets of a compact topological space, and the subbase theorem would not have been required. So, while this theorem is used in the approach I give here, it is not a fundamental part of the theory.
We need to show that if is a subcollection of with empty intersection then there is a finite subset of with empty intersection. We use proof by contradiction, so suppose that every finite subset of has nonempty intersection. The idea is to look at and make use of compactness of . Although it is immediate that finite subsets of also have nonempty intersection, it is not necessarily the case that the intersection of is empty.
Instead we suppose that has been chosen maximal subject to the properties it has empty intersection and that all finite subsets of have nonempty intersection. The existence of such a maximal is a consequence of Zorn’s lemma. It then just needs to be shown that also has empty intersection, and compactness of will give the contradiction.
Consider a finite union where each is in . We can show that at least one of the is also in . If not then, by maximality, there exists a finite subset of with empty intersection. As this must necessarily contain , it can be written as . Then, is a finite subset of such that . So, is a finite subset of with empty intersection, contradicting the choice of . Hence, at least one of the must be in .
Now, consider any . As , it can be written as an intersection where each is a finite union of elements of . For each i, the inclusion implies that the intersection of any finite subset of contains the intersection of and, hence, is nonempty. By maximality, this shows that . As is a finite union of elements of , at least one of which must be in (by what we showed above), there is a with . So,
As this holds for all , it follows that is empty, as required. ⬜
Recall from the discussion in the previous post that the difficulty in giving an elementary proof of measurable projection is the following: if is a sequence of subsets of a product space , decreasing to a limit S, then the projections need not decrease to . That is, projection is not continuous along decreasing sequences. However, we do obtain continuity so long as we restrict to sets arising from the product of a compact paving on K. This is the third statement of the lemma below, and is the point where compactness enters the theory in a fundamental way. Recall that we use and to represent, respectively, the collections of countable unions and countable intersections of .
Lemma 4 Let and be paved spaces such that is closed under finite unions and pairwise intersections, and is compact. Let be the closure of the paving under finite unions and pairwise intersections. Then,
- for all .
- for all .
- if decreases to a limit S, then decreases to .
- If I is an -capacity then,
is an -capacity.
Proof: Note that in the statement of the lemma, is only required to be closed under `pairwise intersections’ rather than `finite intersections’. The difference is simply that we are not requiring to contain the whole set X (i.e., the empty intersection). This is just a very small matter, and we state the lemma in this way as it will be a slight convenience when applying the result.
Without loss of generality, by theorem 3, we may suppose that is closed under finite unions and arbitrary intersections. It follows that the intersection of a pair of sets in is again in ,
So, consists of finite unions of .
Clearly, if is in , then the projection is equal to E if L is nonempty, and is the empty set otherwise. In either case, . Now, any is of the form for so, as is closed under finite unions, is in , proving the first statement of the lemma.
Next, consider a sequence decreasing to a limit S. Then, is also decreasing. It just needs to be shown that
This is the point where compactness enters the theory in a fundamental way. For any , consider the slices
The condition is equivalent to being nonempty. Furthermore, for any n, if we write as then we have,
So, is a decreasing sequence of nonempty sets in . By compactness, the intersection is nonempty, so as required. This proves (3), giving the third statement of the lemma.
By definition, any is the intersection of a sequence . Without loss of generality, we can suppose that is decreasing (replace by if required). By the third statement of the lemma which was proved above,
proving the second statement.
Finally, let be an -capacity and consider . If increases to a limit S, then, increases to . So, as I is a capacity, increases to . Next, if decreases to a limit S then, by the first and third statements of the lemma, is a sequence in decreasing to . The fact that I is an -capacity means that decreases to . Hence, is an -capacity as required. ⬜
Given a paved space and an -capacity, I, a set is called I-capacitable if it can be approximated from below in an appropriate sense. There are a couple of equivalent ways of stating this, given by the following lemma.
Lemma 5 Let be a paved space closed under finite unions and I be an -capacity. For any , the following are equivalent.
- There exists such that and .
- For all , there exists with and,
Proof: If the first statement holds then, by definition of , we can write for a sequence . Using the fact that the capacity I is continuous along increasing sequences,
So, for , we can take for sufficiently large n, and (4) will be satisfied. Furthermore, since is closed under finite unions, B will be in as required.
For the converse, suppose that, the second statement holds. So, there is a sequence in with . Then, is in and , as claimed. ⬜
We now prove the capacitability theorem, which states that analytic sets are capacitable. This was first stated in the generality given here by Sion (On capacitability and measurability, 1963).
Theorem 6 (Choquet’s Capacitability Theorem) Let be a paving closed under finite unions and pairwise intersections, and I be an -capacity. Then, for any -analytic set A, there exists such that and .
Proof: We start by showing that any can be approximated from below by a as in the statement of the theorem, before extending to all analytic sets.
By definition, for some sequence . Then, for some . The idea is that, by truncating the union, we can approximate from below by members of , so that A is approximated from below by an element of satisfying (4).
For any finite sequence of positive integers , write
These are in , as it is closed under finite unions and pairwise intersections. Fixing any , we can inductively define an infinite sequence such that
for all r. First, for r equal to 1, note that increases to A as . Hence, as the capacity I is continuous along increasing sequences, (5) will be satisfied for large enough . Next, suppose that have been chosen satisfying (5) for . Letting increase to infinity and using continuity of I along increasing sequences,
We now set,
By construction, and, using the continuity of I along decreasing sequences in ,
So, (4) is satisfied, as required.
Now, consider any -analytic set A. By definition, there is a compact paved space and an such that A is the projection . Letting be the closure of under finite unions and pairwise intersections, Lemma 4 states that is an -capacity. So, by what we have shown above, for any , there exists an such that and,
An immediate consequence of the capacitability theorem is the measurability of analytic sets.
Corollary 7 Let be a complete probability space. Then, every -analytic set is in . That is,
So, since is a sigma-algebra, both B and C are in . As and , this shows that A is in the completion of . By the completeness assumption, . ⬜
Measurable projection follows by combining the capacitability theorem with the properties of analytic sets established in the previous post.
Theorem 8 (Measurable Projection) Let be a complete probability space. Then, for all .
Proof: Let be the collection of compact intervals in , which is a compact paving generating the sigma-algebra . As the complement of any compact interval is a countable union of compact intervals, it follows from corollary 4 of the post on analytic sets that the sigma-algebra generated by consists of -analytic sets,
Finally, we note that the reference to a probability measure can be removed from the statement of the projection theorem. Rather than referring to a complete probability space, a universally complete sigma-algebra can be used. On a measurable space , a set is said to be universally measurable if it is in the completion of with respect to every probability measure on . To remove ambiguity, we will also say that A is universally -measurable. The universal completion of is the collection of universally -measurable sets,
The intersection here is taken over all probability measures on . Then, and are referred to as universally complete if every universally measurable set is in . That is, if . In particular, if is a complete probability space then it is immediate that every universally -measurable set is in . Corollary 7 above can be generalised using universal measurability.
Corollary 9 Let be a measurable space. Then, every -analytic set is universally -measurable. That is,
Proof: By corollary 7, every -analytic set is in for any probability measure on and, hence, is universally -measurable. ⬜
Similarly, measurable projection can be generalized.
Theorem 10 (Universally Measurable Projection) Let be a measurable space. Then, is universally -measurable, for all .
Proof: If is a probability measure on then, as , the set S is measurable. By theorem 8, . This holds for all such probability measures , so is in . ⬜