Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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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.
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Optional stopping with a dominating integrable variable

Statement

Assume AC. Let M be a martingale and τ an almost-surely finite stopping time. If Y∈L1 and ∣Mτ∧n∣≤Y almost surely for every n, define Mτ=Mτ(ω)(ω) on {τ<∞} and Mτ=0 on {τ=∞}. Then Mτ∈L1 and EMτ=EM0.

Facts & Assumptions

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

[F1]

Optional sampling for bounded stopping times gives the expectation identity at τ∧n.

[F2]

Dominated convergence passes to the finite-time limit.

[F3]

The Axiom of Choice is inherited from the martingale and optional-sampling interfaces.

Proof

1.1F2

Since τ<∞ almost surely, Mτ∧n eventually equals Mτ almost surely. On the null event {τ=∞} the stipulated cemetery value makes the stopped variable total; all limit assertions are almost-sure assertions. The assumed bound passes to ∣Mτ∣≤Y, proving integrability, and also gives ∣Mτ∧n−Mτ∣≤2Y. F2 therefore yields L1 convergence.

2.1F1F3step 1.1∎

F1 gives EMτ∧n=EM0 for every n. Pass to the L1 limit from step 1.1 to get EMτ=EM0. AC has exactly the inherited role in F3.

Depends on

Used by

Dependency tree · two levels

17 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