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Conditional integration through a regular conditional law

Statement

Let K be a specified regular conditional distribution of X:(Ω,F,P)(E,S) given G. For measurable f:E[0,], the measurable function If(ω)=f(x)K(ω,dx) satisfies

HIfdP=Hf(X)dP(HG).

This assertion is choice-free. Under AC, it says If=E[f(X)G] as nonnegative almost-sure classes. If f:ER is measurable and Ef(X)<, put D={If<}. Then DG, P(D)=1, and the signed integral on D, extended by zero off D, is a real integrable conditional-expectation version of f(X). Its identification with the AC-based conditional-expectation class uses AC only for that class convention.

Facts & Assumptions

Given: The hypotheses and conventions in the statement.

[F1]

Kernel event evaluations satisfy every conditioning-event identity. Regular conditional distribution.

[F2]

A probability kernel integrates nonnegative measurable functions measurably and gives measurable zero-filled signed integrals on the absolute-integrability set. Measurability of integration against a kernel.

[F3]

Increasing nonnegative limits pass through every section integral and every event integral. Monotone convergence for the integral.

[F4]

Nonnegative measurable functions have an explicit increasing simple approximation. Every nonnegative measurable function is the increasing limit of simple measurable functions.

[F5]

Under AC nonnegative conditional classes are characterized by all conditional event identities. Conditional monotone convergence.

[F6]

The real integrable version is unique up to almost-sure equality. Conditional expectation is unique almost surely.

[F7]

AC is used only for the inherited conditional-class existence convention (RN selections and nonnegative truncations). The Axiom of Choice.

[F8]

For real |f|, the absolute expectation also equals its integral against the law of X. Change of variables for expectation.

Proof

technique · direct
1.1

For f=1A, the assertion is [F1]. If f=j=1maj1Aj is nonnegative simple with disjoint measurable Aj and finite coefficients, then If=jajK(,Aj), so finite additivity of integrals proves the identity by summing the indicator identities. It includes the empty sum, giving I0=0, and f=1, giving I1=1.

F1
2.1

For general nonnegative f, choose the prescribed simple approximants snf of [F4]. For every ω, [F3] for the probability K(ω,) gives Isn(ω)If(ω). The function (ω,x)f(x) is product-measurable since inverse images are Ω×f1(B); thus [F2] ensures G-measurability of If. Applying [F3] on each H on both sides of step 1.1 proves the displayed identity, allowing infinity. No representative selection or AC occurred. If using conditional-class notation, [F5] identifies this characterized nonnegative function with the class, under [F7].

step 1.1F2F3F4F5F7
3.1

For real f with C=Ef(X)<, step 2.1 at f and H=Ω gives IfdP=C. Also C=fdPX by [F8], checking the equivalent law integrability condition. The set D={If<} is measurable. For every positive integer n, nP(Dc)If=C, hence P(Dc)=0. On D, both If+,If are finite and their difference J is the signed integral. Fill J by zero off D; it is real measurable by [F2], and JIf on D, so J is integrable. The two nonnegative identities from step 2.1 have finite integrals, at most C, so subtraction gives HJdP=Hf(X)dP. Removing Dc changes neither integral, since it is measurable null. Thus J is a conditional version; [F6] gives its almost-sure uniqueness, and [F7] supplies only the inherited class notation. No infinity is subtracted from infinity.

step 2.1F2F6F7F8

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