# Conditions for the disintegration theorem to hold

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.