We will shortly give a proof of measurable projection and, also, of the section theorems. Starting with the projection theorem, recall that this states that if is a complete probability space, then the projection of any measurable subset of onto is measurable. To be precise, the condition is that *S* is in the product sigma-algebra , where denotes the Borel sets in , and is the projection . Then, . Although it looks like a very basic property of measurable sets, maybe even obvious, measurable projection is a surprisingly difficult result to prove. In fact, the requirement that the probability space is complete is necessary and, if it is dropped, then need not be measurable. Counterexamples exist for commonly used measurable spaces such as and . This suggests that there is something deeper going on here than basic manipulations of measurable sets.

The techniques which will be used to prove the projection theorem involve analytic sets, which will be introduced in this post, with the proof of measurable projection to follow in the next post. [**Note:** I have since posted a more direct proof of measurable projection and section, which does not make use of analytic sets.] These results can also be used to prove the optional and predictable section theorems which, at first appearances, seem to be quite basic statements. The section theorems are fundamental to the powerful and interesting theory of optional and predictable projection which is, consequently, generally considered to be a hard part of stochastic calculus. In fact, the projection and section theorems are really not that hard to prove, although the method given here does require stepping outside of the usual setup used in probability and involves something more like descriptive set theory.

Let us consider how one might try and approach a proof of the projection theorem. As with many statements regarding measurable sets, we could try and prove the result first for certain simple sets, and then generalise to measurable sets by use of the monotone class theorem or similar. For example, let denote the collection of all for which . It is straightforward to show that any finite union of sets of the form for and are in . If it could be shown that is closed under taking limits of increasing and decreasing sequences of sets then the result would follow from the monotone class theorem. Increasing sequences are easily handled — if is a sequence of subsets of then from the definition of the projection map,

If for each *n*, this shows that the union is again in . Unfortunately, decreasing sequences are much more problematic. If for all then we would like to use something like

(1) |

However, this identity does not hold in general. For example, consider the decreasing sequence . Then, for all *n*, but is empty, contradicting (1). There is some interesting history involved here. In a paper published in 1905, Henri Lebesgue claimed that the projection of a Borel subset of onto is itself measurable. This was based upon mistakenly applying (1). The error was spotted in around 1917 by Mikhail Suslin, who realised that the projection need not be Borel, and lead him to develop the theory of analytic sets.

Actually, there is at least one situation where (1) can be shown to hold. Suppose that for each , the slices

are compact. For each , is a decreasing sequence of compact sets, so has nonempty intersection. Then, and (1) follows. This argument is connected to the use of compactness in the definition of analytic sets given below. However, the compactness property will not be central to any of the arguments here and, in fact, all the results stated below would hold if it is dropped, but it will be a vital part in the proof of the projection theorem in the next post.

One way to try and fix the method considered above to prove the projection theorem is to put additional restrictions on the collection of subsets of . That is, for , in addition to requiring that is measurable, we include constraints that ensure that is closed under limits of increasing and decreasing sequences. One way of achieving this is to let be the collection of -analytic sets, as defined below.

Roughly, then, the outline of the proof of the measurable projection theorem the following:

- Every set in is analytic.
- The projection maps analytic sets to analytic sets.
- Every analytic subset of is in .

The first point is proven below in corollary 4, and relies on showing that countable unions and intersections of analytic sets are analytic. The second point is stated as theorem 2, and is a direct consequence of the definitions. Finally, the third point uses Choquet’s capacitability theorem, and will be left until the next post.

One issue with introducing the concept of analytic sets is to settle on which definition to use. The fact that there are many different looking, but equivalent, ways of defining them just shows how they are a fundamental and flexible idea. Often, they are introduced as specific to Polish spaces. We would prefer not to restrict to this setting here. Alternatively, they can be defined using the Suslin operation. However, I will follow Bichteler (Stochastic Integration With Jumps, 2002), which seems to be one of the most direct approaches for our purposes.

I now introduce a bit of notation. The term *paving* on a set *X* will be used to denote, simply, any collection of subsets of *X*. Then, for a paving on a set *X*, the pair will be referred to as a *paved space*. Examples of paved spaces include any topological space with the paving given by either its open or its closed subsets, and any measurable space where the paving is a sigma-algebra. For a paving , we use and to represent its closure under, respectively, countable unions and countable intersections.

It is clear that , that , and that .

We say that a paving is *compact* if every with empty intersection has a finite subset with empty intersection. For example, a Hausdorff space together with its compact subsets is a compact paved space. For two paved spaces and , we use to denote the paving on the Cartesian product consisting of the sets for and . It is straightforward to show that the product of compact pavings and is itself compact. Finally, for any two sets *X* and *Y*, will be used to denote the projection .

Definition 1Let be a paved space. Then, a set is -analytic if and only if there exists a compact paved space and such that .

This definition makes use of an auxilliary compact paved space. It does not matter at the moment what this space might be, but it is possible show that it is always possible to take, for example, and to be the collection of closed bounded intervals.

We use to denote the collection of all -analytic sets in . Clearly, every element of is itself -analytic,

(2) |

From the definition of , the set *S* in definition 1 can be expressed as

for sets and . This can be rearranged as

Consider the set *C* of such that is nonempty. Compactness of the paving ensures that *C* is closed in the product topology. The set can then be expressed purely in terms of the closed set and as,

(3) |

This shows that the precise details of the auxiliary paved space in definition 1 are not important, and all that matters is which subsets of have nonempty intersection. Note that the union in (3) need not be countable, so we cannot infer the measurability of *A*.

As was promised, we show that projections of analytic sets are analytic. This is a direct consequence of the definition.

Theorem 2Let and be paved spaces, with compact. Then, for any -analytic set , the projection is -analytic.

*Proof:* As *S* is analytic, there is a compact paved space such that *S* can be expressed be expressed as a projection, for some . As both and are compact, it follows that is compact, and for . ⬜

We move on to the main statement and proof of this post, that countable unions and intersections of analytic sets are analytic. For a metric space *X* with collection of closed sets and open sets , this can be used to show that all Borel sets are both -analytic and -analytic. Similarly, for a pair of measurable spaces and , it shows that product sigma-algebra consists of -analytic sets.

Theorem 3Let be a paved space and be -analytic. Then, and are also -analytic. Equivalently,

*Proof:* As are -analytic sets, there are compact paved spaces and sets such that . We outline two different ways to embed the pavings into a larger paved space , which, respectively, will lead to the representation of the union and the intersection of the as analytic sets.

First consider the cartesian product . This comes with projections satisfying for nonempty . Let be the collection of subsets of *K* of the form for some and .

It can be seen that is a compact paving. Consider a collection of sets as runs through some index set *I*. The intersection is

The only way that this can be empty is if there is some *n* such that is empty. By compactness of , this shows that there is a finite such that for and is empty. So, is compact.

If we let be the projection , then for all . If follows that, for then is in . Using , we have that an is in if and only if, for all *n*, there is a such that . Choosing with , this is equivalent to the existence of a such that, for all *n*, . Hence,

This expresses as the projection of a set in , so is -analytic.

Next, consider the disjoint union . This comes with inclusion maps which are one-to-one and for . Let be the collection of subsets of *K* of the form for and .

It can be seen that this definition of also gives a compact paving. Consider a collection of sets as runs through some index set *I*. If for some *j* and *k*, then giving a finite subcollection with empty intersection. On the other hand, if for some fixed *n* and all *k* then

For this to be empty, the intersection of the must be empty so, by compactness of , for a finite subset . So, is compact.

Now, the sets can, by definition, be expressed as for in . Letting be the inclusion maps , then for all . As is one-to-one, it follows that is in .

The union of the sets can be written as,

The final two equalities here hold because is one-to-one and the images of and are disjoint for . This expresses as the projection of a set in , so is -analytic ⬜

Suppose, for example, that *X* is a metric space and consist of either the collection of open sets or the collection of closed sets. The following consequence of theorem 3 then implies that Borel sets are -analytic.

Corollary 4Let be a paved space such that for all .

Then, all -measurable sets are -analytic. That is,

*Proof:* As we know from theorem 3 that is closed under countable intersections and unions, this is a standard argument. Consider the collection of sets such that is in . From the condition that the complement of every is in , we have . Given any sequence , the union will be in by theorem 3, with complement,

which is again in by theorem 3. Therefore, is closed under countable unions and under complements, so is a sigma-algebra containing . ⬜

The map on the pavings of a set *X* is a closure operator.

The first two statements are immediate from the definition. For the third statement, we start with the following simple lemma.

Lemma 5For any two paved spaces and we have

*Proof:* If and then for a compact paved space and . It follows that and is -analytic. ⬜

Finally, we show that is a closure operator.

Theorem 6Let be a paving. Then, a set is -analytic if and only if it is -analytic. That is,

*Proof:* The inclusion is immediate from the definition of analytic sets, or from (2). For the reverse inclusion, suppose that . Then, for some compact paved space and . From lemma 5, so, by theorem 3, . Finally, theorem 2 says that as required. ⬜

In Corollary 4 we have to assume that $mathcal{E}$ is not empty right? Otherwise the set $B$ would be empty as well and not a sigma-algebra.

Yes, I should have said “nonempty paved space”. Thanks for pointing this out, will fix it