Consider a probability space and a subset S of . The projection is the set of such that there exists a with . We can ask whether there exists a map
such that . From the definition of the projection, values of satisfying this exist for each individual . By invoking the axiom of choice, then, we see that functions with the required property do exist. However, to be of use for probability theory, it is important that should be measurable. Whether or not there are measurable functions with the required properties is a much more difficult problem, and is answered affirmatively by the measurable selection theorem. For the question to have any hope of having a positive answer, we require S to be measurable, so that it lies in the product sigma-algebra , with denoting the Borel sigma-algebra on . Also, less obviously, the underlying probability space should be complete. Throughout this post, will be assumed to be a complete probability space.
It is convenient to extend to the whole of by setting for outside of . Then, is a map to the extended nonnegative reals for which precisely when is in . Next, the graph of , denoted by , is defined to be the set of with . The property that whenever is expressed succinctly by the inclusion . With this notation, the measurable selection theorem is as follows.
Theorem 1 (Measurable Selection) For any , there exists a measurable such that and
As noted above, if it wasn’t for the measurability requirement then this theorem would just be a simple application of the axiom of choice. Requiring to be measurable, on the other hand, makes the theorem much more difficult to prove. For instance, it would not hold if the underlying probability space was not required to be complete. Note also that, stated as above, measurable selection implies that the projection of S is equal to a measurable set , so the measurable projection theorem is an immediate corollary. I will leave the proof of Theorem 1 for a later post, together with the proofs of the section theorems stated below.
A closely related problem is the following. Given a measurable space and a measurable function, , does there exist a measurable right-inverse on the image of ? This is asking for a measurable function, , from to such that . In the case where is the Borel space , Theorem 1 says that it does exist. If S is the graph then will be the required right-inverse. In fact, as all uncountable Polish spaces are Borel-isomorphic to each other and, hence, to , this result applies whenever is a Polish space together with its Borel sigma-algebra.
Continuous-time stochastic calculus is generally applied under the setting of a filtered probability space which, as usual, I am assuming to be complete. It can be important to choose the random time given by Theorem 1 to be a stopping time. For this to be possible, the stronger measurability criterion that S is in the optional sigma-algebra is required.
Theorem 2 (Optional Section) For any and , there exists a stopping time such that and
In the statement of the optional section theorem, the equality (1) has been replaced by an inequality (2). The condition ensures that is a subset of , and the measure of the difference of these events is
In fact, it is not generally possible to choose the stopping time to satisfy (1). To see this, consider the following example. Let be a random variable taking values in and such that for each positive t. For example, take to be uniformly distributed on . Let be the (completed) natural filtration of , so that is a stopping time, and set . It can be seen that is generated by the sets over and, hence, is trivial when restricted to . So, every -measurable random variable is almost-surely constant on the event . Now, any stopping time can be seen to be deterministic on the event . So, if , we have
almost surely, for some fixed positive s. Then,
Theorem 3 (Predictable Section) For any and , there exists a predictable stopping time such that and
Again, it is not generally possible to choose to satisfy (1). With the stopping time and filtration as in the example above, consider taking . Let be a predictable stopping time with . If is a sequence of stopping times announcing then, without loss of generality, we can set whenever . Using the argument above, there are deterministic times such that (a.s.) whenever . Taking the limit as n goes to infinity, there exists a deterministic time s with
almost surely. Then, again,
A straightforward consequence of the measurable selection and section theorems is that, to test whether two processes are equal, it is enough to compare them at each random time, up to zero probability sets. Throughout, I am considering processes to be equal if they are equal up to evanescence.
Theorem 4 Let X and Y be stochastic processes.
- Measurable Selection: If X,Y are jointly measurable and (a.s.) for each -measurable random time , then .
- Optional Section: If X,Y are optional and (a.s.) for each stopping time , then .
- Predictable Section: If X,Y are predictable and (a.s.) for each predictable stopping time , then .
Proof: Let . That is, S is the set of for which , and lies in the sigma-algebra generated by X and Y. Any random time with graph contained in S satisfies whenever . I will argue by contradiction, so suppose that . Then, S is not evanescent, hence .
If X,Y are jointly measurable then so is S and, by Theorem 1, there exists a random time with graph contained in S and,
contradicting the condition of the first statement.
Next, if X,Y are optional then so is S and, by Theorem 2, there exists a stopping time with graph contained in S and,
Taking small enough that the right hand side is non-negative contradicts the condition of the second statement.
Finally, if X,Y are predictable then so is S. Theorem 3 states that there exists a predictable stopping time with graph contained in S, and such that (3) holds. This contradicts the condition of the third statement. ⬜
We previously showed that a map is measurable if and only if is jointly measurable, and is a stopping time if is progressive or, equivalently, is optional. In the previous post, it was also mentioned that is a predictable stopping time if and only if is a predictable set. However, advanced techniques are required to prove this statement, so no proof was given at the time. With the aid of the predictable section theorem, I will now give a proof.
Lemma 5 A map is a predictable stopping time if and only if .
Proof: By predictable section, Theorem 3, there are predictable stopping times such that and
Compare this to the classification of predictable stopping times previously proven in these notes. There, a stopping time was shown to be predictable if and only if its graph is predictable. The proof used there relied on some advanced properties of the stochastic integral in order to show that implies that is fair. This required constructing the paths of stochastic integrals of with respect to bounded martingales. Although it requires predictable section, the proof given in this post is more direct, and is also much stronger, since it does not require to already be known to be a stopping time.