Proofs of Doob-Dynkin Lemma

Let f, g be functions from a non-empty set \Omega to measurable spaces (X, \mathcal A) and (Y, \mathcal B) respectively. Consider the following two conditions:
(1) \sigma(f) \supset \sigma(g)
(2) There is a measurable function H: (X, \mathcal A) \to (Y, \mathcal B) with g = H \circ f

By \sigma(f), I mean the sigma-algebra generated by f, so \sigma(f) = f^{-1}(\mathcal A).

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 \mathcal C and \mathcal D are two sigma-algebras on a common underlying set such that \mathcal C \supset \mathcal D, then intuitively it means that the information specified by \mathcal D is contained in the information specified by \mathcal C. So the condition (1) should mean that the information as to what is the value of f(\omega) (given some unknown point \omega \in \Omega) contains the information as to what is the value of g(\omega), in other words, f(\omega) determines g(\omega), that is, g is a function of f. The condition (2) says g is a measurable function of f.

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 (Y, \mathcal B) is a standard Borel space. But before that, let’s investigate how unique H is.

1. uniqueness of H

If H and H' are two measurable functions satisfying (2), then H=H' on f(\Omega). But notice that we don’t know if the image f(\Omega) is X. We don’t even know if the image is measurable.

If (Y, \mathcal B) is a standard Borel space, then it is known that the set [H=H'] is measurable. In that case, we can say that H=H' holds almost everywhere in the following sense. For the push forward measure \mu (on \mathcal A) of any probability measure on \sigma(f) under f, we have that H=H' holds \mu-a.e. because \mu[H = H'] = 1 because f^{-1}[H=H'] = \Omega.

2. counterexamples

If both \mathcal A and \mathcal B are trivial sigma-algebras, i.e., if \mathcal A = \{\emptyset, X\} and \mathcal B = \{\emptyset, Y\}, 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 \mathcal B separates points of Y (in the sense that for any two distinct points y,y' there is B \in \mathcal B such that y \in B and y' \not\in B), show first that f(x) = f(x') \implies g(x) = g(x') and then that there is a function H: X \to Y such that g = H \circ f.

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 g(\Omega) is a countable subset in a standard Borel space (Y, \mathcal B).

Suppose y_1, y_2, \dots (finitely many or countably infinite) are all points in the image g(\Omega). For each i, the singleton \{y_i\} is measurable. This follows from the possibly overkill assumption that (Y, \mathcal B) is a standard Borel space and this happens to be the only time we use that assumption in this proof.

Since \{y_i\} is measurable, (1) then implies that there is A_i \in \mathcal A such that g^{-1}y_i = f^{-1}A_i

Notice that f^{-1}A_i form a countable partition. We do not know if A_i form a partition of \Omega. We do not know if f(\Omega) is X or if it is measurable. But at least we can conclude that A_i form a countable partition modulo f(\Omega), by which I mean that A_i \cap f(\Omega) form a partition of f(\Omega). In other words, for each x in the image f(\Omega), there exists unique i such that x \in A_i. (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 x \in f(\Omega) and for each i we have x \in A_i \implies H(x) = y_i. It is easy to define such H. (Exercise 30. Define such H. Beware that f(\Omega) may not be measurable.)

Now it only remains to show g = H \circ f which we can check point-wise. For each \omega \in \Omega, we have:
there is i for which f(\omega) is in A_i;
for that i, H(f(\omega)) = y_i (by the property we specified for H);
but that y_i = g(\omega) (by how A_i is chosen);
so H\circ f = g on this \omega.

4. proof when Y is the Cantor space

There is a slick proof of the fact that (1) implies (2) when (Y, \mathcal B) = (\{0,1\}^{\mathbb N}, \mathcal B) is the Cantor space along with its Borel sigma-algebra, or equivalently the countably infinite product of the discrete measurable space (\{0,1\}, \mathcal D) (where \mathcal D is just the discrete sigma-algebra for \{0,1\}.)

Let p_i: \{0,1\}^{\mathbb N} \to \{0,1\} be the projection to the i’th coordinate. This is a measurable function. It is easy to see that (1) implies that p_i \circ g = H_i \circ f for some measurable function H_i from X to \{0,1\}. Now define H(x) = (H_1(x), H_2(x), \dots), then H is a measurable function and we have g = H \circ f.

5. proof when Y is the unit interval

Now let’s show that (1) implies (2) when (Y, \mathcal B) 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 g_n := \lfloor 2^n g \rfloor / 2^n. These are approximations to g. Since we have already proved Doob-Dynkin’s lemma for countable range cases, there is a measurable function H_n: (X, \mathcal A) \to (Y, \mathcal B) with g_n = H_n \circ f.

Notice that g_n \nearrow g. You might want to define H to be \lim H_n, but there is a little problem that we don’t know if the limit exists everywhere. At least we know that H_n converges on f(\Omega). You might now want to define H to take the limit value just on the image f(\Omega), 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 f(\Omega) 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 H(x) = \lim H_n(x) holds for all x for which the limit exists. There are at least two ways of taking such an H:

  1. Let H be \lim H_n where the limit exits, and 0 where the limit does not exist; or
  2. Let H be \limsup H_n everywhere.

Now it only remains to show g = H \circ f. For each \omega \in \Omega, we have:
g(\omega) = \lim g_n (\omega) = \lim H_n \circ f (\omega) = H \circ f (\omega).

(Another work around: Define a measurable H such that H = \lim H_n holds on f(\Omega) by letting H = \sup_n H_n everywhere. Then proceed with this H.)

6. proof for standard Borel space case

If (Y, \mathcal B) 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.

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4 Responses to Proofs of Doob-Dynkin Lemma

  1. Pingback: 在进行线性回归时,为什么最小二乘法是最优方法? | 学习热

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  3. edesig says:

    “You may know that the unit interval is isomorphic to the Cantor space (as measurable spaces)” the proof of 4 works when the sigma-algebra over the Cantor space is its power set. If this measure space would be isomorphic to the unit interval having the Borel-sets over it, then the Borel unit interval were isomorphic to the unit interval having its power set over it. Am I wrong?

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