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

1.1L1L3L4

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.

1.2L2L3

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].

2.1L4step 1.1step 1.2∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources