Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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Assuming countable choice, counting measure is sigma-finite exactly on countable sets

Example

Assume the Axiom of Countable Choice. For a set X, counting measure on (X,P(X)) is sigma-finite if and only if X is at most countable. In particular, counting measure on R is not sigma-finite.

Facts & Assumptions

Given: A set X, its counting measure #, and the Axiom of Countable Choice.

[L1]

Counting measure assigns finite measure exactly to finite sets (Counting measure on an arbitrary set) and is a measure (Counting measure is a measure).

[L2]

Sigma-finiteness means that the whole space is a countable union of measurable finite-measure sets (Finite, sigma-finite, and semifinite measures).

[L3]

Under countable choice, a countable union of at most countable sets is at most countable (The Axiom of Countable Choice (ACω), Countable unions of at most countable sets, assuming ACω).

[L4]

Verification

technique · direct
1.1

For the forward implication, suppose counting measure on X is sigma-finite. By [L2], X=nEn with #(En)<+, so every En is finite by [L1] and X is at most countable by [L3].

givenL1L2L3
1.2

For the reverse implication when X is finite, the constant cover En=X has finite counting measure; when X is countably infinite, a bijection e:NX gives finite initial sets En={e(k):k<n} with union X. The empty set is covered by the constant empty sequence.

givenL1L2
2.1

Steps 1.1 and 1.2 prove the equivalence. Since R is uncountable by [L4], the forward implication shows that its counting measure is not sigma-finite.

step 1.1step 1.2L4

Depends on

Used by

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