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

1.1

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.

F1F2F3

2.1

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 $\int_{A} Y = \int_{A} X$ .

step 1.1

3.1

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.

F4F5step 1.1step 2.1∎

Depends on

Used by

Kolmogorov zero-one law from martingale convergence Corollary
A dyadic martingale converges to the original L1 variable Example
An Lp-bounded martingale with an Lp terminal value Example

Dependency tree · two levels

22 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

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