TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Generating a sigma-algebra commutes with taking traces

Statement

Let $E \subseteq P (X)$ and $Y \subseteq X$ , and put $E ∣_{Y} : = {E \cap Y : E \in E}$ . Then

$σ_{X} (E) ∣_{Y} = σ_{Y} (E ∣_{Y}) .$

Facts & Assumptions

Given: A family $E \subseteq P (X)$ and a subset $Y \subseteq X$ .

[L1]

A trace of a sigma-algebra is a sigma-algebra on the traced subset (The trace of a sigma-algebra is a sigma-algebra on the traced subset).

[L2]

A generated sigma-algebra is the smallest sigma-algebra containing its generators (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal).

[L3]

The trace is $A ∣_{Y} = {A \cap Y : A \in A}$ (The trace of a sigma-algebra on a subset).

Proof

technique · direct

1.1L1L2L3

By [L1], $σ_{X} (E) ∣_{Y}$ is a sigma-algebra on $Y$ , and it contains $E ∣_{Y}$ . Hence [L2] gives $σ_{Y} (E ∣_{Y}) \subseteq σ_{X} (E) ∣_{Y}$ .

1.2L3algebra

Let $G : = {A \subseteq X : A \cap Y \in σ_{Y} (E ∣_{Y})}$ . The identities $(X ∖ A) \cap Y = Y ∖ (A \cap Y)$ and $(⋃_{n} A_{n}) \cap Y = ⋃_{n} (A_{n} \cap Y)$ show that $G$ is a sigma-algebra on $X$ ; it contains $E$ .

2.1step 1.1step 1.2L2L3∎

By [L2], $σ_{X} (E) \subseteq G$ , so tracing gives $σ_{X} (E) ∣_{Y} \subseteq σ_{Y} (E ∣_{Y})$ . Together with step 1.1 this proves equality.

Depends on

Used by

The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra 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

T. Tao, An Introduction to Measure Theory, Exercise 1.4.12 (standard reference, not scraped)