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Assuming countable choice, zero on countable sets and infinity on cocountable sets is a non-semifinite measure

Example

Assume the Axiom of Countable Choice and let $X$ be uncountable. On the sigma-algebra $A$ of countable and cocountable subsets of $X$ , define

$μ (A) : = {\begin{cases} 0, & A is countable, \\ + \infty, & X ∖ A is countable . \end{cases}$

Then $μ$ is a measure, but it is not semifinite.

Facts & Assumptions

Given: An uncountable set $X$ and the Axiom of Countable Choice.

[L1]

A sigma-algebra is closed under complements and countable unions (Sigma-algebras), and a measure is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

[L2]

Countable means finite or countably infinite (Finite, countably infinite, countable, uncountable), and under countable choice a countable union of at most countable sets is at most countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L3]

Semifiniteness requires every positive-measure measurable set to contain a positive finite-measure measurable subset (Finite, sigma-finite, and semifinite measures).

Verification

technique · direct

1.1

The countable-cocountable family $A$ is a sigma-algebra: complements exchange its two classes, a countable union of countable members is countable by [L2], and a union containing a cocountable member is cocountable.

givenL1L2

1.2

The whole space has measure $+ \infty$ , while every measurable set of finite measure has measure $0$ ; hence $X$ contains no measurable subset of positive finite measure.

given

2.1

In a pairwise disjoint sequence from $A$ , at most one member is cocountable. If all members are countable, their union is countable by [L2] and both sides of countable additivity are $0$ ; if one is cocountable, the union is cocountable and both sides are $+ \infty$ .

step 1.1L1L2

3.1

Steps 1.1 and 2.1 prove that $μ$ is a measure, and step 1.2 violates the semifiniteness condition [L3].

step 1.1step 2.1step 1.2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

G. Folland, Real Analysis, 2nd ed., §1.3 (standard reference, not scraped)