Let f, g be measurable functions from a probability space to measurable spaces and respectively. Consider the following three conditions:
(1) (in other words, each element in is equivalent (up to null sets) to some element in . The notation is used when an author wants to make explicit the practice of treating sub-sigma algebras ignoring null sets.)
(2) There is a measurable with a.e.
(3) There is a measure preserving map with a.e. where are the pushforward measures obtained from .
(2) implies (3) trivially. (3) implies (2) trivially (even if the measure preserving map is only a.e. defined.)
(2) implies (1) trivially. The Doob-Dynkin lemma for probability space is that (1) implies (2) if is a standard Borel space. In this post, we present some proofs of this lemma and some corollaries and others. Compare with the Doob-Dynkin lemma for measurable spaces (the previous article). As we’ll see, Doob-Dynkin lemma for probability space is a special case of that for measurable spaces. The one for probability spaces suffices for most applications in probability theory and ergodic theory.
1. Proof 1
In this proof, we simply show that the Doob-Dynkin lemma for probability space follows from that for measurable spaces. It is enough to show that the condition can be improved to after discarding all points in some null set from .
For each , there is such that the symmetric difference is a -null set. It would be nice to be able to discard all points in the union but is this a null set? Perhaps not.
Exercise 10. Show that it is enough to discard all points in the countable union where is a sequence that generates (Existence of such a sequence is guaranteed by the assumption that is a standard Borel space).
2. Proof 2
In this proof, we adapt an argument in the previous article to our setting of ignoring null sets. We divide into two cases.
Case I: when the image of g countable
In this case, there is such that a.e. for each i.
Notice that form a countable partition modulo , in other words, the intersection of and is a null set when and their union over all i has measure 1, or equivalently that for a.e. , there exists unique i for which .
Exercise 20. Show that form a countable partition mod . (the general idea is that the measure algebra homomorphism behaves like an inclusion map)
Let H be any (everywhere-defined) measurable function from X to Y such that for -a.e. x and for each i we have . It is easy to define such an H.
Now it only remains to show a.e. For a.e. , we have:
there is i for which is in ;
for that i, (by property of H);
but that (by how chosen);
so on this .
Case II: when the image of g is not countable.
WLOG assume is the unit interval.
Let . There is a measurable with a.e..
Notice that holds everywhere. It would be nice to take but we only know converges at least on for some subset of measure 1, and we don’t know if this image is measurable.
A work around is that we let H be any (everywhere-defined) measurable such that for all x for which the limit exists. Such H exists and with this H we can proceed to show a.e..
Another work around is to notice that the subset (alternatively also ) is measurable and so one can show that its measure is 1. Then we set H to be a -a.e. limit of . Then is a P-a.e. limit of . So we can proceed with this H too.
3. Proof 3
In this proof, we show that (1) implies (3) using the fact that is a Lebesgue space (which follows because is a standard Borel space) and the fact that a Lebesgue space can be approximated by a sequence of partitions. This proof is probably equivalent to the previous one, but let’s do it for illustration.
The idea in this proof is that points in X (resp. Y) more or less should correspond to appropriate nested sequence of elements in (resp. ), so in order to build H, we only need establish an appropriate procedure to transform an appropriate nested sequence of elements in to that of , but that is done for free by observing that with abuse of notation we may safely pretend that (if we identify with ). We’ll proceed without abuse of notation.
Notice that WLOG we may replace and with any other almost-isomorphic probability spaces at the minor cost of losing everywhere definedness of f, g.
WLOG there is a non-decreasing sequence of countable partitions such that for any non-increasing choice the intersection is a single point and that generates . This is possible because we may assume that the probability space Y is a disjoint union of the Cantor space and atoms, which follows from the result that any Lebesgue space is almost-isomorphic to a disjoint union of an interval and atoms).
Now we want to pick a good sequence such that for all n we have . We can assume that for each n all elements of (and hence also those of too) have positive measure. So we can ensure that all elements of have positive measure. For each n, is a countable partition modulo and the sequence is non-decreasing modulo . After discarding only countably many -null sets from X, we can ensure that form a non-decreasing sequence of countable partitions. The reason we ensure elements of and only have elements of positive measure is because we want to ensure the natural bijection between and .
For each , let be the unique element in containing x, and let be the unique element in which corresponds to in the sense that . Define H by which is a well defined map from X to Y (which is because is non-decreasing (for a fixed x) because its counterpart in is non-decreasing up to null sets because its counterpart in Y is.).
(Measurability of H) For each , it is easy to show that is measurable, moreover, equal to (where is the element corresponding to ).
(On a.e.) Notice that H is measurable and everywhere defined, so is an almost-everywhere defined measurable function. So we are comparing two a.e. defined measurable functions from to . Inverse images of under the two functions are equal up to null sets because of . Now notice that the set is a subset of where the terms in the latter are null sets because of the previous sentence. We have shown a.e..
(On measure preserving) It is easy to show that H is measure preserving from the fact that a.e.. Alternatively, it also follows from .
4. Uniqueness of H
If H and H’ satisfy (2) , then holds -a.e.. This follows from the following two observations:
- holds -a.e.
- is measurable (because is a standard Borel space).
Exercise 25. Let f, g be measurable functions from a probability space to measurable spaces and respectively. Show that if and only if there are with and with and a bi-measurable bijection such that a.e..
Exercise 30. Given two measure-theoretic dynamical systems and (measure preserving transformations on Lebesgue spaces), show that the latter is a factor of the former if and only if there is a joining (a -invariant probability measure on ) such that mod (with abuse of notation).