PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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If a disjoint union has finite signed measure, then the signed-measure series converges absolutely

Statement

Let $ν$ be a signed measure on $(X, A)$ and let $(E_{n})_{n \in N}$ be pairwise disjoint measurable sets. If $ν (⋃_{n \in N} E_{n}) \in R,$ then the real series $\sum_{n = 0}^{\infty} ν (E_{n})$ converges absolutely.

Facts & Assumptions

Given: A signed measure $ν$ , a pairwise disjoint measurable sequence $(E_{n})$ , and the finite value $ν (⋃_{n} E_{n}) \in R$ .

[L1]

A subset of a set of finite signed measure also has finite signed measure. (A subset of a set of finite signed measure also has finite signed measure)

[L2]

A signed measure is countably additive on every disjoint measurable sequence. (A signed measure is countably additive and takes at most one infinite value)

[L3]

Unconditional convergence of a real series means that every rearrangement converges to the same sum. (Rearrangement of a series along a bijection of $N$ , and unconditional convergence)

[L4]

For a series of real numbers, unconditional convergence is equivalent to absolute convergence. (For a series of real numbers, unconditional convergence and absolute convergence are the same property)

Proof

technique · direct

1.1

Put $E = ⋃_{n} E_{n}$ . Each $E_{n} \subseteq E$ , so [L1] makes every [L1, L3, L4] $ν (E_{n})$ a real number. Thus $\sum_{n} ν (E_{n})$ is a real series to which [L3] and [L4] apply.

1.2

Let $σ : N \to N$ be a bijection. The sequence [L2, L3] $(E_{σ (n)})$ is again pairwise disjoint and has the same union $E$ , so [L2] gives $\sum_{n = 0}^{\infty} ν (E_{σ (n)}) = ν (E) = \sum_{n = 0}^{\infty} ν (E_{n}) .$ Hence the series is unconditionally convergent in the sense of [L3].

2.1

Step 1.2 and [L4] imply that $\sum_{n} ν (E_{n})$ converges absolutely.

L4step 1.1step 1.2∎

Depends on

Used by

A set of positive finite signed measure contains a positive subset of at least the same mass Lemma

Dependency tree · two levels

14 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

Richard F. Bass, Real Analysis for Graduate Students, Definition 12.1 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, 9.3 (standard reference, not scraped)