# 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 $S=[0,1]\times [0,1]$ in the Euclidean plane $\mathbb {R} ^{2}$. Consider the probability measure $\mu$ defined on $S$ by the restriction of two-dimensional Lebesgue measure $\lambda ^{2}$ to $S$. That is, the probability of an event $E\subseteq S$ is simply the area of $E$. We assume $E$ is a measurable subset of $S$.
Consider a one-dimensional subset of $S$ such as the line segment $L_{x}=\{x\}\times [0,1]$. $L_{x}$ has $\mu$-measure zero; every subset of $L_{x}$ 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 the one-dimensional Lebesgue measure $\lambda ^{1}$, 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 topological space $(X,T)$.) The assumptions of the theorem are as follows:
* Let $Y$ and $X$ be two Radon spaces (i.e. a topological space such that every Borel probability measure on it is inner regular, e.g. separably metrizable spaces; in particular, every probability measure on it is outright a Radon measure).
* Let $\mu \in {\mathcal {P}}(Y)$.
* Let $\pi :Y\to X$ be a Borel-measurable function. Here one should think of $\pi$ as a function to "disintegrate" $Y$, in the sense of partitioning $Y$ into $\{\pi ^{-1}(x)\ |\ x\in X\}$. For example, for the motivating example above, one can define $\pi ((a,b))=a$, $(a,b)\in [0,1]\times [0,1]$, which gives that $\pi ^{-1}(a)=a\times [0,1]$, a slice we want to capture.
* Let $\nu \in {\mathcal {P}}(X)$ be the pushforward measure $\nu =\pi _{*}(\mu )=\mu \circ \pi ^{-1}$. This measure provides the distribution of $x$ (which corresponds to the events $\pi ^{-1}(x)$).
The conclusion of the theorem: There exists a $\nu$-almost everywhere uniquely determined family of probability measures $\{\mu _{x}\}_{x\in X}\subseteq {\mathcal {P}}(Y)$, which provides a "disintegration" of $\mu$ into $\{\mu _{x}\}_{x\in X}$, 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 $B\subseteq Y$;
* $\mu _{x}$ "lives on" the fiber $\pi ^{-1}(x)$: for $\nu$-almost all $x\in X$, $\mu _{x}\left(Y\setminus \pi ^{-1}(x)\right)=0,$ and so $\mu _{x}(E)=\mu _{x}(E\cap \pi ^{-1}(x))$;
* for every Borel-measurable function $f:Y\to [0,\infty ]$, $\int _{Y}f(y)\,\mathrm {d} \mu (y)=\int _{X}\int _{\pi ^{-1}(x)}f(y)\,\mathrm {d} \mu _{x}(y)\,\mathrm {d} \nu (x).$ In particular, for any event $E\subseteq Y$, taking $f$ to be the indicator function of $E$, $\mu (E)=\int _{X}\mu _{x}(E)\,\mathrm {d} \nu (x).$

## 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 there exists a Borel family of probability measures $\{\mu _{x_{1}}\}_{x_{1}\in X_{1}}$ in ${\mathcal {P}}(X_{2})$ (which is $(\pi _{1})_{*}(\mu )$-almost everywhere uniquely determined) such that $\mu =\int _{X_{1}}\mu _{x_{1}}\,\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right)=\int _{X_{1}}\mu _{x_{1}}\,\mathrm {d} (\pi _{1})_{*}(\mu )(x_{1}),$ which is in particular $\int _{X_{1}\times X_{2}}f(x_{1},x_{2})\,\mu (\mathrm {d} x_{1},\mathrm {d} x_{2})=\int _{X_{1}}\left(\int _{X_{2}}f(x_{1},x_{2})\mu (\mathrm {d} x_{2}\mid x_{1})\right)\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right)$ and $\mu (A\times B)=\int _{A}\mu \left(B\mid x_{1}\right)\,\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right).$
The relation to conditional expectation is given by the identities $\operatorname {E} (f\mid \pi _{1})(x_{1})=\int _{X_{2}}f(x_{1},x_{2})\mu (\mathrm {d} x_{2}\mid x_{1}),$ $\mu (A\times B\mid \pi _{1})(x_{1})=1_{A}(x_{1})\cdot \mu (B\mid x_{1}).$

### 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 $\Sigma \subset \mathbb {R} ^{3}$, 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 ∂Σ is the same as the disintegration of $\lambda ^{3}$ on $\partial \Sigma$.

### Conditional distributions

The disintegration theorem can be applied to give a rigorous treatment of conditional probability distributions in statistics, while avoiding purely abstract formulations of conditional probability. The theorem is related to the Borel–Kolmogorov paradox, for example.