Lifting theory

In mathematics, lifting theory was first introduced by John von Neumann in a pioneering paper from 1931, in which he answered a question raised by Alfréd Haar.^[1] The theory was further developed by Dorothy Maharam (1958)^[2] and by Alexandra Ionescu Tulcea and Cassius Ionescu Tulcea (1961).^[3] Lifting theory was motivated to a large extent by its striking applications. Its development up to 1969 was described in a monograph of the Ionescu Tulceas.^[4] Lifting theory continued to develop since then, yielding new results and applications.

Definitions

A lifting on a measure space ${\displaystyle (X,\Sigma ,\mu )}$ is a linear and multiplicative operator

which is a right inverse of the quotient map

where ${\displaystyle {\mathcal {L}}^{\infty }(X,\Sigma ,\mu )}$ is the seminormed L^p space of measurable functions and ${\displaystyle L^{\infty }(X,\Sigma ,\mu )}$ is its usual normed quotient. In other words, a lifting picks from every equivalence class ${\displaystyle [f]}$ of bounded measurable functions modulo negligible functions a representative— which is henceforth written ${\displaystyle T([f])}$ or ${\displaystyle T[f]}$ or simply ${\displaystyle Tf}$ — in such a way that ${\displaystyle T[1]=1}$ and for all ${\displaystyle p\in X}$ and all ${\displaystyle r,s\in \mathbb {R} ,}$

Liftings are used to produce disintegrations of measures, for instance conditional probability distributions given continuous random variables, and fibrations of Lebesgue measure on the level sets of a function.

Existence of liftings

Theorem. Suppose ${\displaystyle (X,\Sigma ,\mu )}$ is complete.^[5] Then ${\displaystyle (X,\Sigma ,\mu )}$ admits a lifting if and only if there exists a collection of mutually disjoint integrable sets in ${\displaystyle \Sigma }$ whose union is ${\displaystyle X.}$ In particular, if ${\displaystyle (X,\Sigma ,\mu )}$ is the completion of a σ-finite^[6] measure or of an inner regular Borel measure on a locally compact space, then ${\displaystyle (X,\Sigma ,\mu )}$ admits a lifting.

The proof consists in extending a lifting to ever larger sub-σ-algebras, applying Doob's martingale convergence theorem if one encounters a countable chain in the process.

Strong liftings

Suppose ${\displaystyle (X,\Sigma ,\mu )}$ is complete and ${\displaystyle X}$ is equipped with a completely regular Hausdorff topology ${\displaystyle \tau \subseteq \Sigma }$ such that the union of any collection of negligible open sets is again negligible – this is the case if ${\displaystyle (X,\Sigma ,\mu )}$ is σ-finite or comes from a Radon measure. Then the support of ${\displaystyle \mu ,}$ ${\displaystyle \operatorname {Supp} (\mu ),}$ can be defined as the complement of the largest negligible open subset, and the collection ${\displaystyle C_{b}(X,\tau )}$ of bounded continuous functions belongs to ${\displaystyle {\mathcal {L}}^{\infty }(X,\Sigma ,\mu ).}$

A strong lifting for ${\displaystyle (X,\Sigma ,\mu )}$ is a lifting

such that ${\displaystyle T\varphi =\varphi }$ on ${\displaystyle \operatorname {Supp} (\mu )}$ for all ${\displaystyle \varphi }$ in ${\displaystyle C_{b}(X,\tau ).}$ This is the same as requiring that^[7] ${\displaystyle TU\geq (U\cap \operatorname {Supp} (\mu ))}$ for all open sets ${\displaystyle U}$ in ${\displaystyle \tau .}$

Theorem. If ${\displaystyle (\Sigma ,\mu )}$ is σ-finite and complete and ${\displaystyle \tau }$ has a countable basis then ${\displaystyle (X,\Sigma ,\mu )}$ admits a strong lifting.

Proof. Let ${\displaystyle T_{0}}$ be a lifting for ${\displaystyle (X,\Sigma ,\mu )}$ and ${\displaystyle U_{1},U_{2},\ldots }$ a countable basis for ${\displaystyle \tau .}$ For any point ${\displaystyle p}$ in the negligible set

let ${\displaystyle T_{p}}$ be any character^[8] on ${\displaystyle L^{\infty }(X,\Sigma ,\mu )}$ that extends the character ${\displaystyle \phi \mapsto \phi (p)}$ of ${\displaystyle C_{b}(X,\tau ).}$ Then for ${\displaystyle p}$ in ${\displaystyle X}$ and ${\displaystyle [f]}$ in ${\displaystyle L^{\infty }(X,\Sigma ,\mu )}$ define: ${\displaystyle T}$ is the desired strong lifting.

Application: disintegration of a measure

Suppose ${\displaystyle (X,\Sigma ,\mu )}$ and ${\displaystyle (Y,\Phi ,\nu )}$ are σ-finite measure spaces (${\displaystyle \mu ,\mu }$ positive) and ${\displaystyle \pi :X\to Y}$ is a measurable map. A disintegration of ${\displaystyle \mu }$ along ${\displaystyle \pi }$ with respect to ${\displaystyle \nu }$ is a slew ${\displaystyle Y\ni y\mapsto \lambda _{y}}$ of positive σ-additive measures on ${\displaystyle (\Sigma ,\mu )}$ such that

1. ${\displaystyle \lambda _{y}}$ is carried by the fiber ${\displaystyle \pi ^{-1}(\{y\})}$ of ${\displaystyle \pi }$ over ${\displaystyle y}$, i.e. ${\displaystyle \{y\}\in \Phi }$ and ${\displaystyle \lambda _{y}\left((X\setminus \pi ^{-1}(\{y\})\right)=0}$ for almost all ${\displaystyle y\in Y}$
2. for every ${\displaystyle \mu }$-integrable function ${\displaystyle f,}$
   $${\displaystyle \int _{X}f(p)\;\mu (dp)=\int _{Y}\left(\int _{\pi ^{-1}(\{y\})}f(p)\,\lambda _{y}(dp)\right)\nu (dy)\qquad (*)}$$

   
   in the sense that, for ${\displaystyle \nu }$-almost all ${\displaystyle y}$ in ${\displaystyle Y,}$ ${\displaystyle f}$ is ${\displaystyle \lambda _{y}}$-integrable, the function
   $${\displaystyle y\mapsto \int _{\pi ^{-1}(\{y\})}f(p)\,\lambda _{y}(dp)}$$

   
   is ${\displaystyle \nu }$-integrable, and the displayed equality ${\displaystyle (*)}$ holds.

Disintegrations exist in various circumstances, the proofs varying but almost all using strong liftings. Here is a rather general result. Its short proof gives the general flavor.

Theorem. Suppose ${\displaystyle X}$ is a Polish space^[9] and ${\displaystyle Y}$ a separable Hausdorff space, both equipped with their Borel σ-algebras. Let ${\displaystyle \mu }$ be a σ-finite Borel measure on ${\displaystyle X}$ and ${\displaystyle \pi :X\to Y}$ a ${\displaystyle \Sigma ,\Phi -}$measurable map. Then there exists a σ-finite Borel measure ${\displaystyle \nu }$ on ${\displaystyle Y}$ and a disintegration (*). If ${\displaystyle \mu }$ is finite, ${\displaystyle \nu }$ can be taken to be the pushforward^[10] ${\displaystyle \pi _{*}\mu ,}$ and then the ${\displaystyle \lambda _{y}}$ are probabilities.

Proof. Because of the polish nature of ${\displaystyle X}$ there is a sequence of compact subsets of ${\displaystyle X}$ that are mutually disjoint, whose union has negligible complement, and on which ${\displaystyle \pi }$ is continuous. This observation reduces the problem to the case that both ${\displaystyle X}$ and ${\displaystyle Y}$ are compact and ${\displaystyle \pi }$ is continuous, and ${\displaystyle \nu =\pi _{*}\mu .}$ Complete ${\displaystyle \Phi }$ under ${\displaystyle \nu }$ and fix a strong lifting ${\displaystyle T}$ for ${\displaystyle (Y,\Phi ,\nu ).}$ Given a bounded ${\displaystyle \mu }$-measurable function ${\displaystyle f,}$ let ${\displaystyle \lfloor f\rfloor }$ denote its conditional expectation under ${\displaystyle \pi ,}$ that is, the Radon-Nikodym derivative of^[11] ${\displaystyle \pi _{*}(f\mu )}$ with respect to ${\displaystyle \pi _{*}\mu .}$ Then set, for every ${\displaystyle y}$ in ${\displaystyle Y,}$ ${\displaystyle \lambda _{y}(f):=T(\lfloor f\rfloor )(y).}$ To show that this defines a disintegration is a matter of bookkeeping and a suitable Fubini theorem. To see how the strongness of the lifting enters, note that

and take the infimum over all positive ${\displaystyle \varphi }$ in ${\displaystyle C_{b}(Y)}$ with ${\displaystyle \varphi (y)=1;}$ it becomes apparent that the support of ${\displaystyle \lambda _{y}}$ lies in the fiber over ${\displaystyle y.}$

References