[Math] Conditions for the disintegration theorem to hold

measure-theoryprobability theory

The disintegration theorem says that under certain conditions, a probability measure $\mu$ on a measurable space $$ the existence of

Let $Y$ and $X$ be two Radon spaces (i.e. separable
metric spaces on which every probability measure is a Radon measure).
Let $μ ∈ P(Y)$, let $π : Y → X$ be a Borel-measurable function, and
let $ν ∈ P(X)$ be the pushforward measure from $Y$ to $X$ by $π$. Then
there exists a $ν$-almost everywhere uniquely determined family of
probability measures $\{μ_x\}_{x∈X} ⊆ P(Y)$ such that

  • the function $x \mapsto \mu_{x}$ is Borel measurable, in the sense that $x \mapsto \mu_{x} (B)$ is a Borel-measurable function for each
    Borel-measurable set $B ⊆ Y$;
  • $μ_x$ lives on the fiber $π^{-1}(x)$: for $ν$-almost all $x ∈ X$, $$\mu_{x} \left( Y \setminus \pi^{-1} (x) \right) = 0,$$ and so
    $\mu_x(E) = \mu_x(E \cap \pi^{-1}(x));$
  • for every Borel-measurable function $f : Y → [0, +∞]$,
    $$\int_{Y} f(y) \, \mathrm{d} \mu (y) = \int_{X} \int_{\pi^{-1} (x)} f(y) \, \mathrm{d} \mu_{x} (y) \mathrm{d} \nu (x).$$
  1. I was wondering if the probability measures can be relaxed to
    measures in the disintegration theorem?

  2. when $Y = X_1 × X_2$ and $π_i : Y → X_i$ is the natural
    projection    , we can apply the disintegration theorem, and get the
    result

    each fibre $π_1^{-1}(x1)$ can be canonically identified with $X_2$ and
    there exists a Borel family of probability measures $\{ \mu_{x_{1}} \}_{x_{1} \in X_{1}}$ in $P(X_2)$ (which is
    $(π_1)∗(μ)$-almost
    everywhere uniquely determined) such that $$ \mu = \int_{X_{1}} \mu_{x_{1}} \, \mu \left(\pi_1^{-1}(\mathrm d x_1) \right)=
    \int_{X_{1}} \mu_{x_{1}} \, \mathrm{d} (\pi_{1})_{*} (\mu) (x_{1}),
    $$

    I wonder if this result is still true if $Y$, $X_i$ are not required
    to be Radon spaces but just general measure spaces as long as $Y =
    X_1 × X_2$ and $π_i : Y → X_i$ is the natural projection?

    In other words, given two measurable spaces $X_1$ and $X_2$ and a measure on the product measurable space $X_1 \times X_2$ , what are some necessary and/or sufficient conditions for the measure on $X_1 \times X_2$ to be the composition of some measure on $X_1$ and some transition measure from $X_1$ to $X_2$?

Thanks and regards!

Best Answer

The disintegrations are really transition probabilities or proper, regular conditional probabilities. If we define a function $K:X\times\mathcal{Y}\to[0,1]$ by $K(x,B)=\mu_x(B)$, we get the corresponding transition probability.

  1. I don't have a counterexample ready for the general case, but one can extend the result to $\sigma$-finite measure spaces, essentially by solving the problem separately for each cell of a countable, measurable partition as explained here.

  2. No. Suppose you have an infinite product of the measurable spaces $(X_n,\mathcal{X}_n)$ and for each $n$ you have a measure $\mu_n$ on $\sigma(\mathcal{X}_1\times\ldots,\times\mathcal{X}_n)$ such that for all $B\in\sigma(\mathcal{X}_1\times\ldots,\times\mathcal{X}_{n-1})$ one has $\mu_{n-1}(B)=\mu_n(B\times X_n)$. If you could simply apply the disintegration theorem without further ado to product spaces, you could generate transition measures $K_n:X_{n+1}\times\sigma(\mathcal{X}_1\times\ldots,\times\mathcal{X})\to[0,1]$ that give you by the Ionescu-Tulcea-theorem a measure $\mu$ on the infinite product such that $\mu_n(B)=\mu(B\times X_{n+1}\times\ldots)$. But such an extension is in general not possible, as shown in an example by Andersen and Jessen in a 1948 paper called "On the introduction of measures in infinite product sets".

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