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 integrable time and bounded increments

Statement

Assume AC. Let M be a martingale with MnMn1C almost surely for every n1, for deterministic C<. If τ is a stopping time with Eτ<, define Mτ=0 on the null event {τ=}. Then MτL1 and EMτ=EM0.

Facts & Assumptions

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

[F1]

Optional sampling for bounded stopping times gives EMτn=EM0.

[F2]

Expectation of a nonnegative or integrable random variable defines Eτ as an extended nonnegative integral; Markov's inequality for random variables bounds P(τm) by Eτ/m for each m>0.

[F3]

Dominated convergence passes the stopped values in L1.

[F4]

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

Proof

1.1

For every positive integer m, F2 gives P(τ=)P(τm)Eτ/m. Since Eτ<, letting m shows τ< almost surely. On that event, MτnMτC(τn)+Cτ, because the difference telescopes over at most (τn)+ increments. It tends pointwise to zero and has the integrable dominator Cτ.

F2
2.1

F3 gives MτnMτ in L1; in particular Mτ is integrable. F1 gives EMτn=EM0 for every n, so taking the L1 limit proves the equality. Bounded increments and integrability of τ are used exactly in step 1.1. AC has only the role in F4.

F1F3F4step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources