Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Optional stopping with integrable time and bounded increments

Statement

Assume AC. Let M be a martingale with ∣Mn−Mn−1∣≤C almost surely for every n≥1, for deterministic C<∞. If τ is a stopping time with Eτ<∞, define Mτ=0 on the null event {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 EMτ∧n=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(τ≥m) 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

1.1F2

For every positive integer m, F2 gives P(τ=∞)≤P(τ≥m)≤Eτ/m. Since Eτ<∞, letting m→∞ shows τ<∞ almost surely. On that event, ∣Mτ∧n−Mτ∣≤C(τ−n)+≤Cτ, because the difference telescopes over at most (τ−n)+ increments. It tends pointwise to zero and has the integrable dominator Cτ.

2.1F1F3F4step 1.1∎

F3 gives Mτ∧n→Mτ in L1; in particular Mτ is integrable. F1 gives EMτ∧n=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.

Depends on

Used by

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