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

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Levy downward convergence of conditional expectations

Statement

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

Facts & Assumptions

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

[F1]

Tower property of conditional expectation verifies the reverse-martingale identities.

[F2]

Reverse martingale convergence identifies the reverse limit.

[F3]

The Axiom of Choice states AC, assumed here because F1 and F2 use conditional expectations and chosen representatives.

Proof

1.1F1

Put $X_{n} = E [X ∣ G_{n}]$ . If $m \leq n$ , then $G_{n} \subseteq G_{m}$ , and F1 gives $E [X_{m} ∣ G_{n}] = E [E [X ∣ G_{m}] ∣ G_{n}] = E [X ∣ G_{n}] = X_{n} .$ Thus $(X_{n}, G_{n})$ is a reverse martingale.

2.1F1F2F3∎

F2 gives convergence almost surely and in $L^{1}$ to $E [X_{0} ∣ G_{\infty}]$ . Since $X_{0} = E [X ∣ G_{0}]$ and $G_{\infty} \subseteq G_{0}$ , another tower identity identifies this with $E [X ∣ G_{\infty}]$ . AC is used exactly through F3.

Depends on

Used by

A reverse martingale and the tail sigma-algebra Example
Brownian-filtration martingale representation Theorem

Dependency tree · two levels

12 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, Theorem 2.30 and Exercise 2.31, pp. 17–18 (standard reference, not scraped)