Disintegration theorem

In mathematics, the disintegration theorem is a result in measure theory and probability theory. It rigorously defines the idea of a non-trivial "restriction" of a measure to a measure zero subset of the measure space in question. It is related to the existence of conditional probability measures. In a sense, "disintegration" is the opposite process to the construction of a product measure.

Motivation
Consider the unit square in the Euclidean plane R², S = [0, 1] &times; [0, 1]. Consider the probability measure μ defined on S by the restriction of two-dimensional Lebesgue measure λ² to S. That is, the probability of an event E ⊆ S is simply the area of E.

Consider a one-dimensional subset of S such as the line segment Lx = {x} &times; [0, 1]. Lx has μ-measure zero; every subset of Lx is a μ-null set; since the Lebesgue measure space is a complete measure space,


 * $$E \subseteq L_{x} \implies \mu (E) = 0.$$

While true, this is somewhat unsatisfying. It would be nice to say that μ "restricted to" Lx is the one-dimensional Lebesgue measure λ1, rather than the zero measure. The probability of a "two-dimensional" event E could then be obtained as an integral of the one-dimensional probabilities of the vertical "slices" E ∩ Lx: more formally, if μx denotes one-dimensional Lebesgue measure on Lx, then


 * $$\mu (E) = \int_{[0, 1]} \mu_{x} (E \cap L_{x}) \, \mathrm{d} x$$

for any "nice" E ⊆ S. The disintegration theorem makes this argument rigorous in the context of measures on metric spaces.

Statement of the theorem
(Hereafter, P(X) will denote the collection of Borel probability measures on a metric space (X, d).)

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(X), let π : Y → X be a Borel-measurable function, and let ν ∈ P(X) be the pushforward measure π∗(μ). Then there exists a ν-almost everywhere uniquely determined family of probability measures {μx}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 ⊆ X;
 * μx "lives on" the fibre π&minus;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;&minus;1(x));


 * for every Borel-measurable function f : Y → [0, +∞],


 * $$\int_{Y} f(y) \, \mathrm{d} \mu (y) = \int_{X} \int_{\pi^{-1} (x)} f(y) \, \mathrm{d} \mu_{x} (y) \mathrm{d} \nu (x).$$


 * In particular, for any event E &sube; Y, taking f to be the indicator function of E,


 * $$\mu (E) = \int_{X} \mu_{x} \left( E \cap \pi^{-1}(x) \right) \, \mathrm{d} \nu (x).$$

Product spaces
The original example was a special case of the problem of product spaces, to which the disintegration theorem applies.

When Y is written as a Cartesian product Y = X1 &times; X2 and πi : Y → Xi is the natural projection, then each fibre π1&minus;1(x1) can be canonically identified with X2 and there exists a Borel family of probability measures $$\{ \mu_{x_{1}} \}_{x_{1} \in X_{1}}$$ in P(X2) (which is (π1)∗(μ)-almost everywhere uniquely determined) such that


 * $$\mu = \int_{X_{1}} \mu_{x_{1}} \, \mathrm{d} (\pi_{1})_{*} (\mu) (x_{1}).$$

Vector calculus
The disintegration theorem can also be seen as justifying the use of a "restricted" measure in vector calculus. For instance, in Stokes' theorem as applied to a vector field flowing through a compact surface Σ ⊂ R³, it is implicit that the "correct" measure on Σ is the disintegration of three-dimensional Lebesgue measure λ³ on Σ, and that the disintegration of this measure on ∂Σ is the same as the disintegration of λ³ on ∂Σ.