# Image of probability measures under measurable mappings

Given two probability measures on two probability spaces, ($$\mu, X$$) and ($$\gamma, Y$$), what's the sufficient and necessary condition such that there is a measurable mapping $$f:X\rightarrow Y$$, such that $$f^*\mu = \gamma$$?

## 3 Answers

There is a complete classification of probability spaces up to a measure-preserving isomorphism.

Specifically, consider a category whose objects are triples (X,Σ,μ), where X is a set, Σ is a σ-algebra of measurable subsets on Σ, and μ is a probability measure on (X,Σ).

If we want to have a nice description of morphisms in terms of equivalence classes of point-set maps, we must also require that (X,Σ,μ) is a compact probability space in the sense of Marczewski's 1953 paper “On compact measures”, but without countability assumptions. We also assume completeness, since this does not change isomorphism classes, but makes it easier to define morphisms.

Morphisms (X,Σ,μ)→(X',Σ',μ') are equivalence classes of maps of sets f:X→X' such that f*Σ'⊂Σ and μf*=μ'. The equivalence relation identifies f and g if for all σ∈Σ' the symmetric difference f*σ⊕g*σ has measure 0.

The better-known equivalence relation of equivalence almost everywhere reduces to the above one if the involved spaces are countably separated (like the real line), but in general one must use the above definition.

The isomorphism classes of objects in the resulting category admit a complete classification.

First, any object canonically decomposes as a coproduct (disjoint union) of ergodic spaces, i.e., spaces for which the automorphism group has no nontrivial invariant subsets.

Secondly, the ergodic spaces admit a complete classification as follows. First, the discrete spaces are ergodic. Secondly, the nondiscrete ergodic spaces are isomorphic to the infinite product of I copies of {0,1} (a discrete space), where I is an infinite set, whose cardinality is the only invariant of the resulting space.

The classification of isomorphism classes (first stated for Boolean algebras Σ/N) is known as the von Neumann–Maharam theorem. A 1965 theorem by Cassius Ionescu Tulcea (with subsequent improvements by Vesterstrøm–Wils, Edgar, Graf, Fremlin, Rinkewitz) shows that isomorphisms classes of such Boolean algebras coincide with appropriately defined isomorphism classes of probability spaces.

(This claim also holds if we remove the requirement that (X,Σ,μ) is compact, in which case we must first localize the category at all morphisms f such that the induced map of Boolean algebras Σ'/N'→Σ/N is an isomorphism. Completeness can also be removed if the notion of measurability is adjusted accordingly: elements in f*Σ' must be symmetric differences of elements in Σ and a subset of a measure 0 set.)

Thus, the original question can be answered as follows: there is a measurable mapping f:X→Y such that f*μ=γ if and only if the decompositions of X and Y into their ergodic components have the same measures for each type of ergodic summand, and the discrete parts are isomorphic.

• It is not that $\Sigma$ is a compact class, but rather that there exists a compact class $\mathcal{K} \subseteq \Sigma$ such that $\mu$ is inner regular with respect to $\mathcal{K}$ (i.e. that $(X,\Sigma,\mu)$ is compact in the terminology used in Fremlin's Measure Theory volume 3). This cannot be replaced in the general case by any property of $\Sigma$, because there are $\sigma$-algebras that admit both compact and non-compact measures (e.g. the countable-cocountable $\sigma$-algebra), though of course all probability measures on a standard Borel space are compact. – Robert Furber May 9 '20 at 17:06
• You have also omitted the requirement that $(X,\Sigma,\mu)$ have a lifting. Shelah proved that it is relatively consistent to ZFC that $([0,1],\mathrm{Bo}([0,1]), \lambda)$, $\lambda$ being the Lebesgue measure restricted to Borel sets, have no lifting. Since complete probability spaces always have a lifting, it follows that the Borel restriction of Lebesgue measure is not isomorphic to its completion in this model. (Von Neumann proved, on the other hand, that this space does have a lifting under the continuum hypothesis). – Robert Furber May 9 '20 at 17:11
• And you may call it the "von Neumann-Maharam theorem", but from my perspective that only refers to the existence of an isomorphism of measure algebras as complete Boolean algebras, not of measure spaces. The existence of an isomorphism of measure spaces requires substantial extra work due to Fremlin. – Robert Furber May 9 '20 at 17:15
• @RobertFurber: I certainly meant a compact measure, not just a set-theoretic notion. I also added the completeness condition that I left out. – Dmitri Pavlov May 9 '20 at 17:58
• @RobertFurber: I must also say that the existence of point-set maps of measurable spaces is not due to Fremlin, but rather is a theorem by Cassius Ionescu Tulcea (see his 1965 paper in Comptes Rendus). There were several subsequent improvements, see my paper arxiv.org/abs/2005.05284, where I try to list (hopefully) all contributors. – Dmitri Pavlov May 14 '20 at 1:54

Most likely just standard probability spaces are sufficient for your purposes. All non-atomic such spaces are pairwise isomorphic.

One sufficient condition is that the source space is nonatomic and the target space has the Borel sets of a Polish space as the underlying $$\sigma$$-algebra. See here for pointers on how one can prove this.