PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

The restriction of a measure to a measurable set is a measure

Statement

If $μ$ is a measure on $(X, A)$ and $E \in A$ , then $μ_{E} (A) = μ (A \cap E)$ is a measure on $(X, A)$ .

Facts & Assumptions

Given: A measure $μ$ on $(X, A)$ and a measurable set $E$ .

[L1]

The restricted set function is $μ_{E} (A) = μ (A \cap E)$ on the original sigma-algebra (Restriction of a measure to a measurable set).

[L2]

A measure vanishes at the empty set and is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

Proof

technique · direct

1.1givenL1L2

One has $μ_{E} (\emptyset) = μ (\emptyset \cap E) = 0$ .

1.2given

If $(A_{k})$ is pairwise disjoint in $A$ , then $(A_{k} \cap E)$ is pairwise disjoint and $⋃_{k} (A_{k} \cap E) = (⋃_{k} A_{k}) \cap E$ .

2.1step 1.1step 1.2L1L2∎

Countable additivity of $μ$ applied to step 1.2 gives $μ_{E} (⋃_{k} A_{k}) = \sum_{k} μ (A_{k} \cap E) = \sum_{k} μ_{E} (A_{k})$ ; together with step 1.1 this proves that $μ_{E}$ is a measure, including $E = \emptyset$ and $E = X$ .

Depends on

Dependency tree · 2 levels

Sources

G. Folland, Real Analysis, 2nd ed., §1.3, Exercise 10