TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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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 $Y \in L^{1}$ and $∣ M_{τ \land n} ∣ \leq Y$ almost surely for every $n$ , define $M_{τ} = M_{τ (ω)} (ω)$ on ${τ < \infty}$ and $M_{τ} = 0$ on ${τ = \infty}$ . Then $M_{τ} \in L^{1}$ and $E M_{τ} = E M_{0}$ .

Facts & Assumptions

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

[F1]

Optional sampling for bounded stopping times gives the expectation identity at $τ \land n$ .

[F2]

Dominated convergence passes to the finite-time limit.

[F3]

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

Proof

1.1

Since $τ < \infty$ almost surely, $M_{τ \land n}$ eventually equals $M_{τ}$ almost surely. On the null event ${τ = \infty}$ the stipulated cemetery value makes the stopped variable total; all limit assertions are almost-sure assertions. The assumed bound passes to $∣ M_{τ} ∣ \leq Y$ , proving integrability, and also gives $∣ M_{τ \land n} - M_{τ} ∣ \leq 2 Y .$ F2 therefore yields $L^{1}$ convergence.

2.1

F1 gives $E M_{τ \land n} = E M_{0}$ for every $n$ . Pass to the $L^{1}$ limit from step 1.1 to get $E M_{τ} = E M_{0}$ . AC has exactly the inherited role in F3.

F1F3step 1.1∎

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

Durrett, Probability: Theory and Examples, 5th ed., optional stopping criteria in §4.8 (standard reference, not scraped)