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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 , counting measure on is sigma-finite if and only if is at most countable. In particular, counting measure on is not sigma-finite.
Facts & Assumptions
Given: A set , its counting measure , and the Axiom of Countable Choice.
Counting measure assigns finite measure exactly to finite sets (Counting measure on an arbitrary set) and is a measure (Counting measure is a measure).
Sigma-finiteness means that the whole space is a countable union of measurable finite-measure sets (Finite, sigma-finite, and semifinite measures).
Under countable choice, a countable union of at most countable sets is at most countable (The Axiom of Countable Choice (), Countable unions of at most countable sets, assuming ).
The real line is uncountable ( is uncountable (Cantor's nested intervals, 1874)).
Verification
For the forward implication, suppose counting measure on is sigma-finite. By [L2], with , so every is finite by [L1] and is at most countable by [L3].
For the reverse implication when is finite, the constant cover has finite counting measure; when is countably infinite, a bijection gives finite initial sets with union . The empty set is covered by the constant empty sequence.
Steps 1.1 and 1.2 prove the equivalence. Since is uncountable by [L4], the forward implication shows that its counting measure is not sigma-finite.
Depends on
Used by
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Dependency tree · two levels
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Sources
- S. Axler, Measure, Integration & Real Analysis, §2C (standard reference, not scraped)
- G. Folland, Real Analysis, 2nd ed., §1.3 (standard reference, not scraped)