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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Uniform integrability of conditional expectations of one variable

Statement

Assume AC. For fixed real $X \in L^{1} (P)$ , the classes $E [X ∣ G]$ , as $G$ ranges over all sub-sigma-algebras of $F$ , form a uniformly integrable family.

Facts & Assumptions

Given: AC and one fixed real $X \in L^{1} (P)$ ; G ranges over sub-sigma-algebras of F.

[F1]

Versions have the same event integrals as their input. (Conditional expectation as an ae class)

[F2]

The modulus is bounded by the conditional mean of the absolute input; ordinary expectations are preserved. (Basic algebra and order properties of conditional expectation)

[F3]

For fixed integrable X, sufficiently small measure events have uniformly small integrals of |X|. (Absolute continuity of the integral)

[F4]

Uniform integrability means the supremum of absolute tail integrals tends to zero. (A uniformly integrable family)

Proof

technique · direct

1.1

Fix one G, a version $Y = E [X ∣ G]$ , and $K > 0$ . Put $A = {∣ Y ∣ > K} \in G$ . Since $K 1_{A} \leq ∣ Y ∣ 1_{A}$ and $E ∣ Y ∣ \leq E ∣ X ∣$ , integration gives $P (A) \leq E ∣ X ∣ / K$ . Also $∣ Y ∣ \leq E [∣ X ∣ ∣ G]$ almost surely, so the defining event integral gives $\int_{A} ∣ Y ∣ \leq \int_{A} E [∣ X ∣ ∣ G] = \int_{A} ∣ X ∣$ .

F1F2

2.1

Given $ε > 0$ , choose $δ > 0$ using [F3] for the fixed X. Choose $K > E ∣ X ∣ / δ$ (any positive K works when the numerator is zero). Then step 1.1 gives $P (A) < δ$ and hence $\int_{{∣ Y ∣ > K}} ∣ Y ∣ < ε$ . The same K works for every G and every version, so [F4] proves uniform integrability. The proof fixed G arbitrarily and did not select versions simultaneously over all sigma-algebras.

step 1.1F3F4∎

Source notes

Van der Vaart, Martingales, Diffusions and Financial Mathematics, Lemma 1.21 and its full proof, printed p.6 (PDF index 11). The local argument uses the same tail event with absolute continuity of the fixed input integral.

Depends on

Used by

Nothing in the library uses this result yet.

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 (standard reference, not scraped)