Let
* (X, A, lambda) and (Y, B, mu) be sigma-finite measure spaces, and
* T : X --> Y be a measurable function.
A family of measures {lambda_t : A –> RR | t in Y} is called a (T, mu)-disintegration of lambda, if
1. forall t in Y: lambda_t is sigma-finite,
2. forall_mu t in Y: lambda_t(T^-1(Y \ {t})) = 0,
3. forall measurable f : X --> RR, such that f >= 0:
* t |--> int f d(lambda_t) is measurable, and
* int f lambda = int int f d(lambda_t) d(mu(t))
The {lambda_t} are called the disintegrating measures and the mu is called the mixing measure. If the property 2 holds for t in Y, we say that lambda_t is concentrated on the set T^-1({t}). The idea is to split X into the level-sets of T, and then give an individual measure lambda_t for each level-set t such that we can split integration over X into a double-integration over lambda_t and mu. For the outer integral to make sense, the inner integral must be a measurable function of t.
Open: Why sigma-finite everywhere? Open: Why f >= 0? Open: Why not: forall_mu t in Y: lambda_t is sigma-finite?
A disintegration does not always exist. Here’s a sufficient condition. Let
* (X, A, lambda) and (Y, B, mu) be sigma-finite measure spaces,
* T : X --> Y be a measurable function,
* (X, d) be a metric space,
* lambda be a Radon measure,
* T(lambda) << mu,
* B be countably generated, and
* B contain all singleton sets.
Then lambda has a (T, mu)-disintegration.
Interestingly, this sufficient condition uses topological properties to guarantee the existence of a disintegration. Chang et al. mention that there exists more general existence theorems, but deem this version adequate.