TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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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

1.1

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 $μ (P_{n}) = ν (P_{n})$ .

givenL4

1.2

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

givenL3

2.1

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 $n$ .

step 1.1step 1.2L1

3.1

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.

step 2.1L2∎

Depends on

Used by

Two four-point probability measures agree on a generating family that is not a pi-system Counterexample
FALSE: agreement on an arbitrary generating family determines a measure False statement

Dependency tree · two levels

14 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

D. Pollard, A User's Guide to Measure Theoretic Probability, §10 (standard reference, not scraped)