TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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Finite and countable subadditivity of measures

Statement

Let $μ$ be a measure and let $(E_{k})_{k \in N}$ be measurable. Then

$μ (⋃_{k \in N} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k}) .$

For every $m \in N$ one also has

$μ (⋃_{k < m} E_{k}) \leq \sum_{k < m} μ (E_{k}),$

including $m = 0$ , where both sides are $0$ .

Facts & Assumptions

Given: A measure $μ$ and a sequence $(E_{k})$ of measurable sets.

[L1]

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ (Measures are monotone).

[L3]

A nonnegative extended series is the supremum of its finite partial sums, beginning with the empty sum $0$ (Series in the nonnegative extended real line).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1

Define $F_{k} : = E_{k} ∖ ⋃_{j < k} E_{j}$ . Then every $F_{k}$ is measurable, the $F_{k}$ are pairwise disjoint, and $F_{k} \subseteq E_{k}$ .

given

1.2

The unions of the two sequences agree: if $x \in ⋃_{k} E_{k}$ , then the nonempty set ${k : x \in E_{k}}$ has a least member $r$ , and the definition gives $x \in F_{r}$ ; the reverse inclusion follows from $F_{k} \subseteq E_{k}$ .

givenL4

2.1

Countable additivity, monotonicity, and the definition of a nonnegative series give $μ (⋃_{k} E_{k}) = \sum_{k} μ (F_{k}) \leq \sum_{k} μ (E_{k})$ .

step 1.1step 1.2L1L2L3

3.1

For $m \in N$ , apply step 2.1 to the sequence $E_{0}, \dots, E_{m - 1}, \emptyset, \emptyset, \dots$ ; its union and sum are the displayed finite union and finite sum, and when $m = 0$ they are both empty and equal to $0$ .

step 2.1L1L3∎

Depends on

Used by

Dependency tree · two levels

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Sources

S. Axler, Measure, Integration & Real Analysis, Theorem 2.58 (standard reference, not scraped)