Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Finite and countable subadditivity of measures

Statement

Let μ be a measure and let (Ek)k∈N be measurable. Then

μ(⋃k∈NEk)≤∑k=0∞μ(Ek).

For every m∈N one also has

μ(⋃k<mEk)≤∑k<mμ(Ek),

including m=0, where both sides are 0.

Facts & Assumptions

Given: A measure μ and a sequence (Ek) of measurable sets.

[L1]

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

[L2]

If A⊆B are measurable, then μ(A)≤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
1.1given

Define Fk:=Ek∖⋃j<kEj. Then every Fk is measurable, the Fk are pairwise disjoint, and Fk⊆Ek.

1.2givenL4

The unions of the two sequences agree: if x∈⋃kEk, then the nonempty set {k:x∈Ek} has a least member r, and the definition gives x∈Fr; the reverse inclusion follows from Fk⊆Ek.

2.1step 1.1step 1.2L1L2L3

Countable additivity, monotonicity, and the definition of a nonnegative series give μ(⋃kEk)=∑kμ(Fk)≤∑kμ(Ek).

3.1step 2.1L1L3∎

For m∈N, apply step 2.1 to the sequence E0,…,Em−1,∅,∅,…; its union and sum are the displayed finite union and finite sum, and when m=0 they are both empty and equal to 0.

Depends on

Used by

…and 59 more results.

Dependency tree · two levels

16 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