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!


The disintegrations are really transition propbabilities 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 eparately 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.

  • $\begingroup$ No. It is fixed now. Thank you! $\endgroup$ – Michael Greinecker Feb 15 '12 at 14:22
  • $\begingroup$ +1 Thanks! For part 2, I don't quite understand yet. My question is: given a measure on the product space, whether there exists one measure on each component space, such that the relation in terms of integrals in the disintegration theorem holds. Is your reply the other way around, i.e. given marginal measures, whether the product measure exists, s.t. the relation in terms of integrals in the disintegration theorem holds? (I have edited my part 2 in hope of being more clear.) $\endgroup$ – Tim Mar 5 '12 at 0:38
  • $\begingroup$ Even if you know the marginals, that doesn't give you the disintegrations connecting them. You asked for the latter. If one knows the first marginal and all conditional probabilities that allow you to construct the other "finite dimensional distributions", one can construct a process on the infinite product space. Since this is impossible by the mentioned example, there exists a product such that we cannot determine the probability of the second coordinate given the first. $\endgroup$ – Michael Greinecker Mar 5 '12 at 7:23

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