Disintegration theorem

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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.

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[edit] Motivation

Consider the probability measure μ defined on the square S := [0, 1] \times [0, 1] \subsetneq \mathbb{R}^{2} by the restriction of two-dimensional Lebesgue measure to S. That is, the probability of an event E \subseteq S is simply the area of E.

Consider a one-dimensional subset of S such as the line segment L_{x} := \{ x \} \times [0, 1]. L has μ-measure zero; every subset of L 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 one-dimensional Lebesgue measure, 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 \cap L_{x}: 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 \subseteq S.

The disintegration theorem makes this argument rigorous in the context of measures on metric spaces.

[edit] Statement of the theorem

(Hereafter, \mathcal{P} (X) will denote the collection of Borel probability measures on a metric space X.)

Let Y and X be two Radon spaces (i.e. separable metric spaces on which every probability measure is a Radon measure). Let \mu \in \mathcal{P} (Y), let \pi : Y \to X be a Borel-measurable function, and let \nu := \pi_{*} (\mu) \in \mathcal{P} (X). Then there exists a ν-almost everywhere uniquely determined family of probability measures \{ \mu_{x} \}_{x \in X} \subseteq \mathcal{P} (Y) such that

  • 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 \subseteq X;
  • μx "lives on" π − 1(x): for ν-almost all x \in X,
\mu_{x} \left( Y \setminus \pi^{-1} (x) \right) = 0;
  • for every Borel-measurable function f : Y \to [0, + \infty],
\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).

[edit] Applications

[edit] 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 = X_{1} \times X_{2} and \pi_{i} : Y \to X_{i} is the natural projection, then each fibre \pi_{1}^{-1} (x_{1}) can be canonically identified with X2 and find a Borel family of probability measures \{ \mu_{x_{1}} \}_{x_{1} \in X_{1}} \subseteq \mathcal{P} (X_{2}) (which is (\pi_{1})_{*} \left( \mu \right)-almost everywhere uniquely determined) such that

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

[edit] 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 \Sigma \subsetneq \mathbb{R}^{3}, 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 \partial \Sigma is the same as the disintegration of λ3 on \partial \Sigma.

[edit] Reference

  • Ambrosio, L., Gigli, N. & Savaré, G. (2005). Gradient Flows in Metric Spaces and in the Space of Probability Measures. ETH Zürich, Birkhäuser Verlag, Basel. ISBN 3-764-32428-7.
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