LemmaStatement: 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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The stopping-time sigma-algebra is a sigma-algebra

Statement

For every stopping time $τ$ , $F_{τ}$ is a $σ$ -algebra contained in $F$ . For deterministic $τ = n$ it equals $F_{n}$ . If $σ \leq τ$ pointwise, then $F_{σ} \subseteq F_{τ}$ .

Facts & Assumptions

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

[F1]

Sigma-algebra at a stopping time gives the defining intersection tests.

[F2]

Sigma-algebras supplies closure operations.

Proof

1.1F1F2

Since ${τ \leq n} \in F_{n}$ , $Ω$ passes every test. If $A$ passes, then $A^{c} \cap {τ \leq n} = {τ \leq n} ∖ (A \cap {τ \leq n}) \in F_{n} .$ If all $A_{j}$ pass, distributivity gives $(⋃_{j} A_{j}) \cap {τ \leq n} = ⋃_{j} (A_{j} \cap {τ \leq n}) \in F_{n} .$ Thus $F_{τ}$ is a $σ$ -algebra, and containment in $F$ is part of F1.

1.2F1

If $τ \equiv n$ , all tests below $n$ are vacuous and the test at $n$ is $A \in F_{n}$ ; upward nesting then supplies every later test. Hence $F_{τ} = F_{n}$ .

2.1F1F2∎

Suppose $σ \leq τ$ and $A \in F_{σ}$ . For each $n$ , $A \cap {τ \leq n} = ⋃_{k = 0}^{n} (A \cap {σ \leq k}) \cap {τ = k} .$ On ${τ = k}$ the condition $σ \leq k$ is automatic, so the equality is exact. Each term belongs to $F_{k} \subseteq F_{n}$ , proving $A \in F_{τ}$ . The order hypothesis is not weakened to an untracked almost-sure order.

Depends on

Used by

Discrete strong Markov property Theorem
Optional stopping under uniform integrability Theorem

Dependency tree · two levels

5 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

van der Vaart, Martingales, Diffusions and Financial Mathematics, Exercises 2.37–2.39 and Lemma 2.41, pp. 20–21 (standard reference, not scraped)