TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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Countable additivity and continuity of finitely additive set functions

Statement

Let $ϕ : A \to [0, + \infty]$ be a finitely additive nonnegative set function on a sigma-algebra. The following are equivalent:

$ϕ$ is countably additive, and hence is a measure;
whenever $E_{n} ↑ E$ , one has $ϕ (E) = \sup_{n} ϕ (E_{n})$ .

If in addition $ϕ (X) < + \infty$ , these conditions are also equivalent to:

whenever $E_{n} ↓ \emptyset$ , one has $\inf_{n} ϕ (E_{n}) = 0$ .

Facts & Assumptions

Given: A finitely additive nonnegative set function $ϕ$ on a sigma-algebra $A$ over $X$ .

[L1]

Finite additivity means $ϕ (\emptyset) = 0$ and $ϕ (A \cup B) = ϕ (A) + ϕ (B)$ for disjoint measurable $A, B$ (Finitely additive nonnegative set functions).

[L2]

Countable additivity together with the empty-set condition is exactly the definition of a measure (Measures on sigma-algebras).

[L3]

A nonnegative extended series is the supremum of its finite partial sums (Series in the nonnegative extended real line).

Proof

technique · direct

1.1given

For the implication from countable additivity to continuity from below, let $E_{n} ↑ E$ and define $D_{0} = E_{0}$ , $D_{n + 1} = E_{n + 1} ∖ E_{n}$ ; the $D_{n}$ are disjoint, their union is $E$ , and their first $n + 1$ terms have union $E_{n}$ .

1.2givenL1

For the implication from continuity from below to countable additivity, let $(A_{n})$ be disjoint and put $B_{n} = ⋃_{k < n} A_{k}$ ; then $B_{n} ↑ ⋃_{k} A_{k}$ and finite additivity gives $ϕ (B_{n}) = \sum_{k < n} ϕ (A_{k})$ .

1.3givenL1

For the finite-total-mass implications, assume $ϕ (X) < + \infty$ ; then every value of $ϕ$ is finite by finite additivity and nonnegativity.

2.1step 1.1L2L3

Under countable additivity, [L2] and the decomposition in step 1.1 give $ϕ (E) = \sum_{n} ϕ (D_{n}) = \sup_{n} ϕ (E_{n})$ , proving condition 1 implies condition 2.

2.2step 1.2L3

Under condition 2, step 1.2 and [L3] give $ϕ (⋃_{n} A_{n}) = \sup_{n} ϕ (B_{n}) = \sum_{n} ϕ (A_{n})$ , proving condition 2 implies condition 1.

2.3step 1.3L1algebra

For condition 2 implies condition 3 under finite total mass, if $E_{n} ↓ \emptyset$ then $X ∖ E_{n} ↑ X$ ; finite additivity and condition 2 give $ϕ (X) - \inf_{n} ϕ (E_{n}) = \sup_{n} ϕ (X ∖ E_{n}) = ϕ (X)$ , hence the infimum is $0$ .

2.4step 1.3L1algebra

For condition 3 implies condition 2 under finite total mass, if $E_{n} ↑ E$ then $E ∖ E_{n} ↓ \emptyset$ ; finite additivity and condition 3 give $ϕ (E) - \sup_{n} ϕ (E_{n}) = \inf_{n} ϕ (E ∖ E_{n}) = 0$ .

3.1step 2.1step 2.2step 2.3step 2.4∎

Steps 2.1 and 2.2 prove the first equivalence, while steps 2.3 and 2.4 separately prove both directions involving condition 3 under the stated finite-total-mass hypothesis.

Depends on

Used by

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Dependency tree · two levels

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Sources

G. Folland, Real Analysis, 2nd ed., §1.3, Exercise 11 (standard reference, not scraped)