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

Tower property of conditional expectation

Statement

Assume AC for existence. If H⊆G⊆F and X∈L1(P), then E[E[X∣G]∣H]=E[X∣H] and E[E[X∣H]∣G]=E[X∣H] almost surely. Also, if a version of E[X∣G] is H-measurable, it is a version of E[X∣H].

Facts & Assumptions

Given: AC, real integrable X, and H⊆G⊆F; for the last clause a G-conditional version is H-measurable.

[F1]

Conditional versions are measurable and integrable and have the defining event integrals. (Conditional expectation as an ae class)

[F2]

Versions for the same input and sigma-algebra agree almost surely. (Conditional expectation is unique almost surely)

Proof

technique · direct
1.1F1F2

Set U=E[X∣G] and take a version V=E[U∣H]. For A∈H⊆G, ∫AV=∫AU=∫AX. Since V is H-measurable and integrable, [F2] identifies it with E[X∣H].

2.1F1F2step 1.1∎

A version W=E[X∣H] is already G-measurable and integrable. It has its own event integrals ∫AW=∫AW on every A∈G, so it is a version of E[W∣G]; [F2] gives the second identity. Finally if U itself is H-measurable, its G event identities restrict to H, so [F2] gives the additional assertion.

Source notes

Durrett Theorems 4.1.12–4.1.13, printed p.212; van der Vaart Lemma 1.9(v), printed p.4.

Depends on

Used by

…and 5 more results.

Dependency tree · two levels

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Sources