Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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Null sets are closed under countable unions and, in a complete space, under arbitrary subsets

Statement

In every measure space, a countable union of measurable null sets is measurable and null, and every measurable subset of a null set is null. If the measure space is complete, every subset of a measurable null set is measurable and null. Thus the null sets of a complete measure space form a sigma-ideal.

Facts & Assumptions

Given: A measure space (X,A,μ).

[L1]

Countable subadditivity bounds the measure of a countable union by the sum of the member measures (Finite and countable subadditivity of measures).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

A complete measure space contains every subset of every measurable null set in its sigma-algebra (Complete measure spaces).

Proof

technique · direct
1.1

If NkA and μ(Nk)=0 for every k, then kNk is measurable and [L1] gives 0μ(kNk)k0=0.

givenL1
1.2

If SA and SN for a measurable null set N, then 0μ(S)μ(N)=0.

givenL2
2.1

If the space is complete and SN for a measurable null set N, then [L3] first makes S measurable and step 1.2 makes it null.

step 1.2L3
3.1

Step 1.1 gives closure under countable unions, and step 2.1 gives closure under arbitrary subsets in a complete space, including the empty union and the empty subset.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

7 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