False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24

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FALSE: the covering construction agrees with every finitely additive source function

Statement

If a nonnegative set function on an algebra is finitely additive and vanishes at the empty set, then the countable-covering infimum agrees with it on the source algebra.

Facts & Assumptions

Given: The finite-cofinite algebra $A_{0}$ of $N$ and $μ_{0} (A) = 0$ for finite $A$ , $μ_{0} (A) = + \infty$ for cofinite $A$ .

[F1]

The set function induced by $μ_{0}$ assigns $E \subseteq X$ the infimum of $\sum_{k} μ_{0} (A_{k})$ over all countable algebra covers $E \subseteq ⋃_{k} A_{k}$ . (The outer set function induced by a premeasure)

[F2]

An algebra of subsets of $X$ contains $\emptyset$ and is closed under complements and finite unions. (Algebras of subsets)

Refutation

technique · direct

1.1F2algebra

The finite-cofinite family is an algebra by [F2]. For disjoint $A, B$ , two cofinite sets cannot both occur; if both are finite, all three values are $0$ , and if one is cofinite, the union and that set have value $+ \infty$ , so $μ_{0}$ is finitely additive.

2.1step 1.1F1algebra∎

Apply the covering formula of [F1] to this source function. The singleton sequence $({k})_{k \in N}$ covers $N$ with total cost $0$ , so the covering infimum at $N$ is $0$ , while $μ_{0} (N) = + \infty$ ; hence finite additivity does not force agreement.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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

T. Tao, An Introduction to Measure Theory, Exercises 1.7.4(iii) and 1.7.6 (standard reference, not scraped)