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.

Motivation

Consider the probability measure $\mu$ 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 $\mu$-measure zero; every subset of $L$ is a $\mu$-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 $\mu$ "restricted to" $L_{x}$ 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 $\mu _{x}$ denotes one-dimensional Lebesgue measure on $L_{x}$, 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.

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 $\nu$-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 $B\subseteq X$;
• $\mu _{x}$ "lives on" $\pi ^{-1}(x)$: for $\nu$-almost all $x\in X$,

$\mu _{x}\left(Y\setminus \pi ^{-1}(x)\right)=0;$

• for every Borel-measurable function $f:Y\to [0,+\infty ]$,

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

Applications

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 $X_{2}$ 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

$\mu =\int _{X_{1}}\mu _{x_{1}}\,\mathrm {d} (\pi _{1})_{*}(\mu )(x_{1}).$

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 $\Sigma$ is the disintegration of three-dimensional Lebesgue measure $\lambda ^{3}$ on $\Sigma$, and that the disintegration of this measure on $\partial \Sigma$ is the same as the disintegration of $\lambda ^{3}$ on $\partial \Sigma$.

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.{{cite book}}: CS1 maint: multiple names: authors list (link)