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==Application: disintegration of a measure==
==Application: disintegration of a measure==
Suppose <math> (E,\mathfrak F,\mu), (X,\mathfrak X,\nu)</math> are σ-finite measure spaces (μ, ν positive) and π : ''E'' → ''X'' is a measurable map. A '''disintegration of μ along π with respect to ν''' is a slew <math> X\ni x\mapsto \lambda_x</math> of positive σ-additive measures on <math> (E,\mathfrak F)</math> such that (1) λ<sub>''x''</sub> is carried by the fiber <math>\pi^{-1}(\{x\})</math>of π over ''x'':
Suppose <math> (E,\mathfrak F,\mu), (X,\mathfrak X,\nu)</math> are σ-finite measure spaces (μ, ν positive) and π : ''E'' → ''X'' is a measurable map. A '''disintegration of μ along π with respect to ν''' is a slew <math> X\ni x\mapsto \lambda_x</math> of positive σ-additive measures on <math> (E,\mathfrak F)</math> such that


#λ<sub>''x''</sub> is carried by the fiber <math>\pi^{-1}(\{x\})</math>of π over ''x'':
:<math> \{x\}\in\mathfrak X\;\;\mathrm{ and }\;\; \lambda_x\left((E\setminus \pi^{-1}(\{x\})\right)=0 \;\;\;\;\forall x\in X </math>
:::<math> \{x\}\in\mathfrak X\;\;\mathrm{ and }\;\; \lambda_x\left((E\setminus \pi^{-1}(\{x\})\right)=0 \qquad \forall x\in X </math>

and (2) for every μ-integrable function ''f'',
#for every μ-integrable function ''f'',
:::<math> \int_E f(p)\;\mu(dp)= \int_X \left(\int_{\pi^{-1}(\{x\})}f(p)\,\lambda_x(dp)\right) \;\nu(dx) \qquad (*)</math>

::in the sense that, for ν-almost all ''x'' in ''X'', ''f'' is λ<sub>''x''</sub>-integrable, the function
:<math> \int_E f(p)\;\mu(dp)= \int_X \left(\int_{\pi^{-1}(\{x\})}f(p)\,\lambda_x(dp)\right) \;\nu(dx) </math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(*)
:::<math> x\mapsto \int_{\pi^{-1}(\{x\})} f(p)\,\lambda_x(dp) </math>

::is ν-integrable, and the displayed equality (*) holds.
in the sense that, for ν-almost all ''x'' in ''X'', ''f'' is λ<sub>''x''</sub>-integrable, the function

:<math> x\mapsto \int_{\pi^{-1}(\{x\})} f(p)\,\lambda_x(dp) </math>

is ν-integrable, and the displayed equality (*) obtains.


[[Disintegration theorem|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.
[[Disintegration theorem|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.


<blockquote>'''Theorem.''' Suppose ''E'' is a polish<ref> A separable space is ''polish'' if its topology comes from a complete metric. In the present situation it would be sufficient to require that ''E'' is ''Suslin'', i.e., is the continuous Hausdorff image of a polish space.</ref> space and ''X'' a separable Hausdorff space, both equipped with their Borel σ-algebras. Let μ be a σ-finite Borel measure on ''E'' and π : ''E'' → ''X'' an <math>\mathfrak F,\mathfrak X</math>&ndash; measurable map. Then there exists a σ-finite Borel measure ν on ''X'' and a disintegration (*).
<blockquote>'''Theorem.''' Suppose ''E'' is a polish<ref> A separable space is ''polish'' if its topology comes from a complete metric. In the present situation it would be sufficient to require that ''E'' is ''Suslin'', i.e., is the continuous Hausdorff image of a polish space.</ref> space and ''X'' a separable Hausdorff space, both equipped with their Borel σ-algebras. Let μ be a σ-finite Borel measure on ''E'' and π : ''E'' → ''X'' an <math>\mathfrak F,\mathfrak X</math>&ndash;measurable map. Then there exists a σ-finite Borel measure ν on ''X'' and a disintegration (*).


If μ is finite, ν can be taken to be the pushforward<ref> The ''pushforward'' π<sub>∗</sub>μ of μ under π, also called the image of μ under π and denoted π(μ), is the measure ν on <math>\mathfrak X</math> defined by <math>\nu(A):=\mu\left(\pi^{-1}(A)\right), A\in\mathfrak X</math>.</ref> π<sub>∗</sub>μ, ''and then the λ<sub>''x''</sub> are probabilities.</blockquote>
If μ is finite, ν can be taken to be the pushforward<ref> The ''pushforward'' π<sub>∗</sub>μ of μ under π, also called the image of μ under π and denoted π(μ), is the measure ν on <math>\mathfrak X</math> defined by <math>\nu(A):=\mu\left(\pi^{-1}(A)\right), A\in\mathfrak X</math>.</ref> π<sub>∗</sub>μ, and then the λ<sub>''x''</sub> are probabilities.</blockquote>


'''Proof.''' Because of the polish nature of ''E'' there is a sequence of compact subsets of ''E'' that are mutually disjoint, whose union has negligible complement, and on which π is continuous. This observation reduces the problem to the case that both ''E'' and ''X'' are compact and π is continuous, and ν = π<sub>∗</sub>μ. Complete <math>\mathfrak X</math> under ν and fix a strong lifting ''T'' for <math>(X,\mathfrak X,\nu)</math>. Given a bounded μ-measurable function ''f'', let <small><math>\lfloor f\rfloor</math></small> denote its conditional expectation under π, i.e., the [[Radon–Nikodym theorem|Radon-Nikodym derivative]] of<ref> ''f''μ is the measure that has density ''f'' with respect to μ</ref> π<sub>∗</sub>(''f''μ) with respect to π<sub>∗</sub>μ. Then set, for every ''x'' in ''X'', <math>\lambda_x(f):=T(\lfloor f\rfloor)(x).</math> 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
'''Proof.''' Because of the polish nature of ''E'' there is a sequence of compact subsets of ''E'' that are mutually disjoint, whose union has negligible complement, and on which π is continuous. This observation reduces the problem to the case that both ''E'' and ''X'' are compact and π is continuous, and ν = π<sub>∗</sub>μ. Complete <math>\mathfrak X</math> under ν and fix a strong lifting ''T'' for <math>(X,\mathfrak X,\nu)</math>. Given a bounded μ-measurable function ''f'', let <small><math>\lfloor f\rfloor</math></small> denote its conditional expectation under π, i.e., the [[Radon–Nikodym theorem|Radon-Nikodym derivative]] of<ref> ''f''μ is the measure that has density ''f'' with respect to μ</ref> π<sub>∗</sub>(''f''μ) with respect to π<sub>∗</sub>μ. Then set, for every ''x'' in ''X'', <math>\lambda_x(f):=T(\lfloor f\rfloor)(x).</math> 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


:<math> \lambda_x(f\cdot\varphi\circ\pi)=\varphi(x) \lambda_x(f) \;\;\;\; \forall x\in X\;,\;\varphi\in C_b(X)\;,\;f\in L^\infty(E,\mathfrak F,\mu)</math>
:<math> \lambda_x(f\cdot\varphi\circ\pi)=\varphi(x) \lambda_x(f) \qquad \forall x\in X, \varphi\in C_b(X), f\in L^\infty(E,\mathfrak F,\mu)</math>


and take the infimum over all positive φ in ''C<sub>b</sub>''(''X'') with φ(''x'') = 1; it becomes apparent that the support of λ<sub>''x''</sub> lies in the fiber over ''x''.
and take the infimum over all positive φ in ''C<sub>b</sub>''(''X'') with φ(''x'') = 1; it becomes apparent that the support of λ<sub>''x''</sub> lies in the fiber over ''x''.

Revision as of 20:40, 9 June 2013

Lifting theory was first introduced in a series of pioneering papers by John von Neumann.[1] A lifting on a measure space is a linear and multiplicative inverse

of the quotient map

In other words, a lifting picks from every equivalence class [f] of bounded measurable functions modulo negligible functions a representative— which is henceforth written T([f]) or T[f] or simply Tf — in such a way that

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 is complete.[2] Then admits a lifting if and only if there exists a collection of mutually disjoint integrable sets in whose union is E. In particular, if is the completion of a σ-finite[3] measure or of an inner regular Borel measure on a locally compact space, then 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 that is complete and that E is equipped with a completely regular Hausdorff topology such that the union of any collection of negligible open sets is again negligible – this is the case if is σ-finite or comes from a Radon measure. Then the support of μ, Supp(μ), can be defined as the complement of the largest negligible open subset, and the collection 'Cb(E, τ) of bounded continuous functions belongs to .

A strong lifting for is a lifting such that Tφ = φ on Supp(μ) for all φ in Cb(E, τ). This is the same as requiring that[4] for all open sets U in τ.

Theorem. If is σ-finite and complete and τ has a countable basis then admits a strong lifting.

Proof. Let T0 be a lifting for and a countable basis for τ. For any point p in the negligible set

let Tp be any character[5] on that extends the character φ ↦ φ(p) of Cb(E, τ). Then for p in E and [f] in L define:

T is the desired strong lifting.

Application: disintegration of a measure

Suppose are σ-finite measure spaces (μ, ν positive) and π : EX is a measurable map. A disintegration of μ along π with respect to ν is a slew of positive σ-additive measures on such that

  1. λx is carried by the fiber of π over x:
  1. for every μ-integrable function f,
in the sense that, for ν-almost all x in X, f is λx-integrable, the function
is ν-integrable, and the displayed equality (*) 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 E is a polish[6] space and X a separable Hausdorff space, both equipped with their Borel σ-algebras. Let μ be a σ-finite Borel measure on E and π : EX an –measurable map. Then there exists a σ-finite Borel measure ν on X and a disintegration (*). If μ is finite, ν can be taken to be the pushforward[7] πμ, and then the λx are probabilities.

Proof. Because of the polish nature of E there is a sequence of compact subsets of E that are mutually disjoint, whose union has negligible complement, and on which π is continuous. This observation reduces the problem to the case that both E and X are compact and π is continuous, and ν = πμ. Complete under ν and fix a strong lifting T for . Given a bounded μ-measurable function f, let denote its conditional expectation under π, i.e., the Radon-Nikodym derivative of[8] π(fμ) with respect to πμ. Then set, for every x in X, 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 φ in Cb(X) with φ(x) = 1; it becomes apparent that the support of λx lies in the fiber over x.

References

  • Bellow, Alexandra (1969), Topics in the theory of lifting, Springer, Berlin, Heidelberg, New York
  1. ^ Bellow, Alexandra (1969), Topics in the theory of lifting, Springer, Berlin, Heidelberg, New York
  2. ^ A subset NE is locally negligible if it intersects every integrable set in in a subset of a negligible set of . is complete if every locally negligible set is negligible and belongs to .
  3. ^ i.e., there exists a countable collection of integrable sets – sets of finite measure in – that covers the underlying set E.
  4. ^ U, Supp(μ) are identified with their indicator functions.
  5. ^ A character on a unital algebra is a multiplicative linear functional with values in the coefficient field that maps the unit to 1.
  6. ^ A separable space is polish if its topology comes from a complete metric. In the present situation it would be sufficient to require that E is Suslin, i.e., is the continuous Hausdorff image of a polish space.
  7. ^ The pushforward πμ of μ under π, also called the image of μ under π and denoted π(μ), is the measure ν on defined by .
  8. ^ fμ is the measure that has density f with respect to μ

See also