TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Levy upward convergence of conditional expectations

Statement

Assume AC. If $(F_{n})$ is increasing, $F_{\infty} = σ (⋃_{n} F_{n})$ , and $X \in L^{1}$ , then $E [X ∣ F_{n}] ⟶ E [X ∣ F_{\infty}]$ almost surely and in $L^{1}$ .

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Conditional expectation process is a martingale makes $M_{n} = E [X ∣ F_{n}]$ a martingale.

[F2]

Uniform integrability of conditional expectations of one variable makes $(M_{n})$ uniformly integrable.

[F3]

Closed martingale characterization supplies an almost-sure and $L^{1}$ limit $Y$ .

[F4]

The monotone class generated by an algebra equals the sigma-algebra it generates identifies a measure equality first checked on the algebra $⋃_{n} F_{n}$ .

[F5]

The Axiom of Choice is used only for the conditional expectations and countable representatives.

Proof

1.1F1F2F3

By F1, F2, F3 there is $Y \in L^{1}$ such that $M_{n} \to Y$ almost surely and in $L^{1}$ . As an almost-sure limit of $F_{\infty}$ -measurable variables, $Y$ has an $F_{\infty}$ -measurable version.

2.1step 1.1

The union $A = ⋃_{n} F_{n}$ is an algebra because the filtration is increasing. If $A \in A$ , then $A \in F_{N}$ for some $N$ , and for every $n \geq N$ , $\int_{A} M_{n} d P = \int_{A} X d P .$ Passing to the $L^{1}$ limit yields $\int_{A} Y = \int_{A} X$ .

3.1F4F5step 1.1step 2.1∎

Let $D = {A \in F_{\infty} : \int_{A} Y = \int_{A} X}$ . Integrability makes $D$ a monotone class, and step 2.1 gives $A \subseteq D$ . F4 yields $σ (A) = F_{\infty} \subseteq D$ . Thus $Y$ has exactly the defining event integrals of $E [X ∣ F_{\infty}]$ , proving the result. AC has the role stated in F5.

Depends on

Used by

Dependency tree · two levels

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, Corollary 2.24 and proof, p. 16 (standard reference, not scraped)