TheoremStatement: AI-adaptedProof: 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.

Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system

Statement

Let $P$ be a pi-system on $X$ generating $A$ , and let $μ, ν$ be measures on $(X, A)$ that agree on $P$ . Suppose there is an increasing sequence $(P_{n})$ in $P$ with

$X = ⋃_{n} P_{n}, μ (P_{n}) = ν (P_{n}) < + \infty (n \in N) .$

Then $μ = ν$ on $A$ .

Facts & Assumptions

Given: Measures $μ, ν$ , a generating pi-system $P$ , and an increasing finite-measure exhaustion $(P_{n})$ as in the Statement.

[L1]

Finite measures agreeing on a generating pi-system and on the whole space are equal (Finite measures agreeing on a generating pi-system and on the whole space are equal).

[L2]

Measures are continuous from below (Continuity from below for measures).

[L3]

A pi-system is closed under binary intersections (Pi-systems).

[L4]

For a measurable $P_{n}$ , the set function $A \mapsto μ (A \cap P_{n})$ is a measure on the original sigma-algebra (The restriction of a measure to a measurable set is a measure).

Proof

technique · direct

1.1givenL4

Fix $n$ and define $μ_{n} (A) : = μ (A \cap P_{n})$ and $ν_{n} (A) : = ν (A \cap P_{n})$ on $A$ . By [L4] these are measures, and their total masses are the common finite value $μ (P_{n}) = ν (P_{n})$ .

1.2givenL3

If $Q \in P$ , then $Q \cap P_{n} \in P$ by [L3], so $μ_{n} (Q) = μ (Q \cap P_{n}) = ν (Q \cap P_{n}) = ν_{n} (Q)$ .

2.1step 1.1step 1.2L1

The finite uniqueness lemma applied to steps 1.1 and 1.2 gives $μ (A \cap P_{n}) = ν (A \cap P_{n})$ for every $A \in A$ and every $n$ .

3.1step 2.1L2∎

For fixed $A$ , the sets $A \cap P_{n}$ increase to $A$ ; continuity from below and step 2.1 give $μ (A) = \sup_{n} μ (A \cap P_{n}) = \sup_{n} ν (A \cap P_{n}) = ν (A)$ . Thus the measures agree everywhere.

Depends on

Dependency tree · 2 levels

Sources

D. Pollard, A User's Guide to Measure Theoretic Probability, §10