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Let f, g be functions from a non-empty set to measurable spaces and respectively. Consider the following two conditions:
(2) There is a measurable function with
By , I mean the sigma-algebra generated by , so .
The Doob-Dynkin Lemma is that the two conditions (1) and (2) are equivalent under the assumption that the measurable spaces are reasonable ones. The intuition for the lemma is this: when and are two sigma-algebras on a common underlying set such that , then intuitively it means that the information specified by is contained in the information specified by . So the condition (1) should mean that the information as to what is the value of (given some unknown point ) contains the information as to what is the value of , in other words, determines , that is, is a function of . The condition (2) says is a measurable function of .
Applications and corollaries of this lemma include
- characterization of the property of a function being measurable with respect to a sub-sigma-algebra of a given sigma-algebra
- writing a conditional expectation of a random variable given another random variable as a Borel-measurable function of the latter random variable
- characterization of the property of two measure theoretical dynamical systems being isomorphic by existence of some special kind of joining of the two.
Notice that (2) implies (1) trivially. We’ll prove that (1) implies (2) if is a standard Borel space. But before that, let’s investigate how unique H is.
1. uniqueness of H
If and are two measurable functions satisfying (2), then on . But notice that we don’t know if the image is X. We don’t even know if the image is measurable.
If is a standard Borel space, then it is known that the set is measurable. In that case, we can say that holds almost everywhere in the following sense. For the push forward measure (on ) of any probability measure on under , we have that holds -a.e. because because .
If both and are trivial sigma-algebras, i.e., if and , then it is easy to find an example where (1) holds but (2) does not. (Exercise 10. find such an example)
With sigma-algebras that separate points, I don’t know any counter-examples but we can at least say that in this case (1) implies g being a (possible non-measurable) function of f.
Exercise 20. Under the assumption that separates points of Y (in the sense that for any two distinct points there is such that and ), show first that and then that there is a function such that .
3. proof when g has a countable range, with nice Y
Let’s first prove the Doob-Dynkin lemma for the case when the image is a countable subset in a standard Borel space .
Suppose (finitely many or countably infinite) are all points in the image . For each i, the singleton is measurable. This follows from the possibly overkill assumption that is a standard Borel space and this happens to be the only time we use that assumption in this proof.
Since is measurable, (1) then implies that there is such that
Notice that form a countable partition. We do not know if form a partition of . We do not know if is X or if it is measurable. But at least we can conclude that form a countable partition modulo , by which I mean that form a partition of . In other words, for each x in the image , there exists unique i such that . (I haven’t checked if we can assume WLOG that f is onto.)
Let H be a measurable function from X to Y such that for each and for each i we have . It is easy to define such H. (Exercise 30. Define such H. Beware that may not be measurable.)
Now it only remains to show which we can check point-wise. For each , we have:
there is i for which is in ;
for that i, (by the property we specified for H);
but that (by how is chosen);
so on this .
4. proof when Y is the Cantor space
There is a slick proof of the fact that (1) implies (2) when is the Cantor space along with its Borel sigma-algebra, or equivalently the countably infinite product of the discrete measurable space (where is just the discrete sigma-algebra for .)
Let be the projection to the i’th coordinate. This is a measurable function. It is easy to see that (1) implies that for some measurable function from X to . Now define , then H is a measurable function and we have .
5. proof when Y is the unit interval
Now let’s show that (1) implies (2) when is the unit interval along with its Borel sigma-algebra. You may know that the unit interval is isomorphic to the Cantor space (as measurable spaces) (and also to the real line), so showing this is redundant, but let’s do this anyway for illustration purposes.
Let . These are approximations to . Since we have already proved Doob-Dynkin’s lemma for countable range cases, there is a measurable function with .
Notice that . You might want to define to be , but there is a little problem that we don’t know if the limit exists everywhere. At least we know that converges on . You might now want to define H to take the limit value just on the image , and to take 0 as its value outside the image, but here another problem is we don’t know if this H is measurable: The image may not be measurable.
There is a way to work around that problem. Just let H be any measurable function from X to Y such that holds for all x for which the limit exists. There are at least two ways of taking such an H:
- Let H be where the limit exits, and 0 where the limit does not exist; or
- Let H be everywhere.
Now it only remains to show . For each , we have:
(Another work around: Define a measurable H such that holds on by letting everywhere. Then proceed with this H.)
6. proof for standard Borel space case
If is a standard Borel space, then it is known that this measurable space is either discrete or isomorphic to the unit interval, but we have proved Doob-Dynkin’s lemma for both cases already.