Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A subset of a set of finite signed measure also has finite signed measure

Statement

Let ν be a signed measure on (X,A), let E∈A, and suppose ν(E)∈R. Then every measurable subset F⊆E also satisfies ν(F)∈R.

Facts & Assumptions

Given: A signed measure ν, a measurable set E with finite value ν(E), and a measurable subset F⊆E.

[L1]

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

Proof

technique · direct
1.1

The sets F and E∖F are disjoint and have union E, so [L1] gives ν(E)=ν(F)+ν(E∖F).

2.1

If ν(F)=+∞, then the at-most-one-infinite-sign clause in [L1] [L1, step 1.1] forces ν(E∖F)≠−∞, so the right side of step 1.1 is +∞, contradicting the finiteness of ν(E). The same argument with the signs reversed rules out ν(F)=−∞. Therefore ν(F)∈R.

3.1

The subset F was arbitrary, so every measurable subset of E has finite [step 2.1] ∎ signed measure.

Depends on

Used by

Dependency tree · two levels

4 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