# What is this disintegration-like theorem?

This is cross-posted at MSE.

I'm looking for a reference for the following result. It seems like it must be known, or follow quickly from something known, but I have not been able to find it in any of the textbooks I have.

Let $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space. Let $$\mathcal G$$ be a sub-$$\sigma$$-field. Let $$\mu$$ be a regular conditional probability for $$\mathbb P$$ given $$\mathcal G$$. By this I mean that $$\mu: \mathcal F \times \Omega \to [0,1]$$ is a probability measure (denoted $$\mu_\omega$$) in its first argument, $$\mathcal G$$ measurable in its second argument and satsifies $$\int_G \mu_\omega(A)\mathbb P(d\omega) = \mathbb P(A \cap G)$$ for all $$A \in \mathcal F$$ and $$G \in \mathcal G$$. Let $$X: \Omega \times \Omega \to \mathbb R$$ be bounded and $$\mathcal G \otimes \mathcal F$$ measurable. Let $$EX(\omega) = \int X(\omega, v)\mu_\omega(dv).$$

Theorem. $$EX$$ is $$\mathcal G$$ measurable and $$\int EX(\omega)\mathbb P(d\omega) = \int X(\omega,\omega)\mathbb P(d\omega).$$

• This seems false. Consider the case of finite measure spaces. Sep 7 '19 at 9:50
• @YuvalPeres Sorry, there was a typo. The claim is about probability spaces. Why do you think it's false? Sep 7 '19 at 13:06
• Usually one speaks of a regular conditional probability for a random variable, not for a measure. Could you write down precisely the properties that $\mu$ is assumed to have? Sep 7 '19 at 13:25
• @NateEldredge Please see the edit. Sep 7 '19 at 13:29
• You should probably try to prove this first for $X = 1_{A \times B}$ with $A \in \mathcal{G}$ and $B \in \mathcal{F}$. Then extend to $X = 1_M$ for $M \in \mathcal{G}\otimes \mathcal{F}$ using Dynkins $\pi-\lambda$ theorem. Finally, extend to general measurable (nonnegative) $X$ by representing such a function as an increasing limit of simple functions. Sep 7 '19 at 19:31

The measure-theoretic formulation of conditional probability/expectation is cumbersome, so it is often easy to lose connection between intuitive facts and the formal demonstrations of them. I personally often have difficulty with arguing for seemingly obvious facts.

In this case, I think more than your theorem is true: your function $$EX(\omega)$$ is a regular version of the conditional expectation of $$f(\omega):=X(\omega,\omega)$$ given $$\mathcal{G}$$, i.e., $$\mathbb{E}[f\,|\,\mathcal{G}]$$. By definition, this means that $$EX(\omega)$$ is $$\mathcal{G}$$-measurable and \begin{align} \underbrace{\int_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega)}_{\int_G \mathbb{E}[f\,|\,\mathcal{G}]\,\mathrm{d}\mathbb{P}(\omega)} &= \underbrace{\int_G X(\omega,\omega)\, \mathrm{d}\mathbb{P}(\omega)}_{\int_G f(\omega) \mathrm{d}\mathbb{P}(\omega)} \;, \end{align} for every $$G\in\mathcal{G}$$.

This is almost immediate from your definition. Note that

$$\quad$$($$\star$$) for each $$\omega$$, we have $$X(\omega,v)=X(v,v)$$ for $$\mu_\omega$$-almost every $$v$$ (see @1 below).

Thus, $$EX(\omega)=\int f(v)\,\mathrm{d}\mu_\omega(v)$$. That $$\int f(v)\,\mathrm{d}\mu_\omega(v)$$ is $$\mathcal{G}$$-measurable for every $$f$$ is standard (see @2 below). Furthermore, \begin{align} \int_G EX(\omega)\,\mathrm{d}\mathbb{P}(\omega) &= \int_G\int f(v)\,\mathrm{d}\mu_\omega(v)\mathrm{d}\mathbb{P}(\omega) \\ &= \int_G f(v)\,\mathrm{d}\mathbb{P}(v) \;. \end{align}

Further details:

@1: I believe this can be shown using the usual measure theory trick. Namely, first show that ($$\star$$) holds if $$X$$ is the indicator of a rectangle $$G\times F$$ with $$G\in\mathcal{G}$$ and $$F\in\mathcal{F}$$. Then, show that the family of sets $$A\in\mathcal{G}\otimes\mathcal{F}$$ for which $$X(\cdot,\cdot):=1_A$$ satisfies ($$\star$$) is a $$\sigma$$-algebra. Then, show that $$(\star)$$ holds if $$X$$ is a simple function. Finally, show that if $$X$$ is the uniform limit of an increasing sequence of functions satisfying ($$\star$$), then $$X$$ itself satisfies ($$\star$$).

@2: The argument for this again uses the usual trick. We already know (by definition) that $$\int f(v)\,\mathrm{d}\mu_\omega(v)$$ is $$\mathcal{G}$$-measurable if $$f$$ is an indicator function. First, show that the same holds for simple function. Then, show that if $$f$$ is the uniform limit of an increasing sequence $$f_1\leq f_2\leq\cdots$$ such that $$\int f_1(v)\,\mathrm{d}\mu_\omega(v)$$ is $$\mathcal{G}$$-measurable, then $$\int f(v)\,\mathrm{d}\mu_\omega(v)$$ is also $$\mathcal{G}$$-measurable.

• Thanks. A couple comments. (1) The argument about the $\mathcal G$ measurability seems circular; at some point this must be shown (though I don't doubt it's true). (2) I'm not sure I see why $(\star)$ is true. (3) The final expression has a typo I think. Sep 7 '19 at 23:50
• Yes, sorry for the typo. See the update for (1) and (2). Sep 8 '19 at 13:21
• Thanks. I did manage to convince myself of $(\star)$. Somehow it's still not "intuitive" to me in the way it seems to be to you and others, but I guess that means I just have to think about it more. Sep 8 '19 at 20:57
• It's probably also worth commenting that $(\star)$ holds only for $\mathbb P$ almost every $\omega$ (at least as far as I can see). The argument still goes through, of course. Sep 9 '19 at 2:31

Let me be short in details, only giving the essential picture (though I'd be happy to clarify if asked to). Let me also make clear: I am not suggesting that the picture below is the most economical when it comes down to writing things up.

Assume given a measurable map $$f:X\to Y$$ of standard Borel spaces $$X, Y$$, consider the following commutative diagram

$$\require{AMScd}$$ $$\begin{CD} X @>id\times id>> X\times X @>p_2>> X \\ @| @V id\times fVV @V f VV \\ X @> id \times f>> X\times Y @>p_2>> Y @>>> \{\text{point}\} \end{CD}$$

Endowing $$X$$ with a probability Radon measure, we push it down the diagram and endow all spaces with the corresponding measures. Now the diagram lives in the category of of standard Borel spaces equipped with a Radon probability measures, morphisms being classes of measure preserving measurable maps, defined up to a.e identification. In fact, all spaces in this diagram but $$Y$$ and $$\{\text{point}\}$$ are isomorphic, and all morphisms but the ones around $$Y$$ are isomprhisms (consider the composition $$X\to X\times Y \overset{p_1}{\to} X$$ in order to observe that $$id\times f$$ is an isomorphism).

Next, we apply the functor $$L^1$$ from the above described category to Banach spaces and contracting maps, taking a measured space to its $$L^1$$-space and a morphism to the associated conditional expectation map. The above commutative diagram gives a commutative diagram of $$L^1$$ spaces.

Your conditional expectation formula becomes obvious now by letting $$X$$ be the von-Neumann spectrum of $$L^\infty(\Omega, \mathcal F, \mathbb P)$$ and $$Y$$ be the spectrum of the subalgera of $$\mathcal G$$-measurable functions.