TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Taking out what is known

Statement

Assume AC for existence. If $X \in L^{1} (P)$ and $Z$ is bounded real $G$ -measurable, then $E [Z X ∣ G] = Z E [X ∣ G]$ almost surely. The identity also holds for finite real $G$ -measurable $Z$ whenever $Z X$ and $Z E [X ∣ G]$ are integrable. In fact $X, Z X \in L^{1}$ imply the latter integrability.

Facts & Assumptions

Given: AC, real integrable X, and a real G-measurable factor Z; first assume Z bounded, then assume Z finite and ZX integrable (or explicitly both products integrable).

[F1]

Conditional versions are integrable and satisfy all event integrals. (Conditional expectation as an ae class)

[F2]

Linearity, expectation preservation and the conditional modulus bound hold. (Basic algebra and order properties of conditional expectation)

[F3]

The event identities characterize the version up to almost-sure equality. (Conditional expectation is unique almost surely)

[F4]

Nonnegative measurable functions have increasing nonnegative simple approximations. (Every nonnegative measurable function is the increasing limit of simple measurable functions)

[F5]

Pointwise almost-everywhere convergence with an integrable majorant permits convergence of event integrals. (Dominated convergence)

[F6]

Integrals of increasing nonnegative functions converge to the integral of the limit. (Monotone convergence for the integral)

Proof

technique · direct

1.1

Write $U = E [X ∣ G]$ . If $Z = 1_{B}$ with $B \in G$ , then for every $A \in G$ , $\int_{A} Z U = \int_{A \cap B} U = \int_{A \cap B} X = \int_{A} Z X$ . The products are integrable and $Z U$ is $G$ -measurable, so uniqueness gives the identity. Finite sums of such indicators give the identity for bounded simple $Z$ by linearity.

F1F2F3

2.1

If $∣ Z ∣ \leq M$ , apply [F4] to $Z^{+}$ and $Z^{-}$ and subtract their approximations to obtain simple $Z_{n} \to Z$ with $∣ Z_{n} ∣ \leq M$ . Then $∣ Z_{n} X ∣ \leq M ∣ X ∣$ and $∣ Z_{n} U ∣ \leq M ∣ U ∣$ , both integrable majorants. By [F5] in each event identity of step 1.1, $\int_{A} Z U = \int_{A} Z X$ . The limit $Z U$ is measurable and integrable, so [F3] proves the bounded case.

step 1.1F4F5F3

3.1

For finite measurable $Z$ with $Z X \in L^{1}$ , set $V = E [∣ X ∣ ∣ G] \geq 0$ and $R_{n} = ∣ Z ∣ \land n$ . The bounded case and expectation preservation imply $E [R_{n} V] = E [R_{n} ∣ X ∣]$ . MCT gives $E [∣ Z ∣ V] = E ∣ Z X ∣ < \infty$ . Since $∣ U ∣ \leq V$ almost surely, $E ∣ Z U ∣ \leq E ∣ Z X ∣ < \infty$ .

step 2.1F2F6

4.1

Let $Z_{n} = \max (- n, \min (Z, n))$ . For each event $A$ the bounded case gives $\int_{A} Z_{n} U = \int_{A} Z_{n} X$ . Now $Z_{n} U \to Z U$ and $Z_{n} X \to Z X$ , dominated by the integrable $∣ Z U ∣$ and $∣ Z X ∣$ respectively. DCT and uniqueness yield $E [Z X ∣ G] = Z U$ . This proves both the stated two-product extension and its stronger integrability observation.

step 2.1step 3.1F5F3

5.1

The same indicator identity gives locality: if $X = Y$ almost surely on $B \in G$ , then $1_{B} E [X ∣ G] = E [1_{B} X ∣ G] = E [1_{B} Y ∣ G] = 1_{B} E [Y ∣ G]$ as classes. No assertion is made about arbitrary values on exceptional points.

step 1.1F1∎

Source notes

Durrett Theorem 4.1.14, printed pp.212–213; locality Theorem 4.1.2, printed p.206. The absolute-product integrability estimate is explicitly proved by ordinary MCT before the signed DCT limit; conditional MCT is not used.

Depends on

Used by

Nonnegative predictable transforms preserve submartingale gains Corollary
Taking out an unbounded factor needs integrability Counterexample
Conditional variance Definition
Product martingale from independent mean one factors Example
Conditional variance is well-defined and has the second-moment formula Lemma
Martingale differences are orthogonal in l2 Lemma
Bounded predictable transforms preserve martingales Theorem
Conditional expectation is the l2 orthogonal projection Theorem
Square minus predictable quadratic variation is a martingale Theorem
Uniform integrability of conditional expectations of one variable Theorem

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Durrett, Probability: Theory and Examples, 5th ed. (standard reference, not scraped)