PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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A set is null for a signed measure exactly when its total variation is zero there

Statement

Let $ν$ be a signed measure on $(X, A)$ and let $E \in A$ . Then $E$ is null for $ν$ if and only if $∣ ν ∣ (E) = 0$ .

Facts & Assumptions

Given: A signed measure $ν$ and a measurable set $E$ .

[L1]

A null set for a signed measure means: every measurable subset of it has signed measure $0$ . (Positive, negative, and null sets for a signed measure)

[L2]

The total variation $∣ ν ∣ (E)$ is the supremum of the partition sums $\sum_{n} ∣ ν (E_{n}) ∣$ over countable measurable partitions of $E$ . (The total variation |nu|(E) from countable measurable partitions)

Proof

technique · direct

1.1

Assume $E$ is null. If $(E_{n})$ is a countable measurable partition of $E$ , [L1, L2] then every $E_{n} \subseteq E$ has $ν (E_{n}) = 0$ by [L1], so its partition sum in [L2] is $0$ . Hence every admissible sum is $0$ , and therefore $∣ ν ∣ (E) = 0$ .

1.2

Assume instead that $∣ ν ∣ (E) = 0$ . Let $F \subseteq E$ be measurable. Then [L1, L2] $F$ and $E ∖ F$ form a measurable partition of $E$ , so [L2] gives $0 = ∣ ν ∣ (E) \geq ∣ ν (F) ∣ + ∣ ν (E ∖ F) ∣ \geq ∣ ν (F) ∣ .$ Thus $ν (F) = 0$ . Since $F \subseteq E$ was arbitrary, [L1] shows that $E$ is null.

2.1

Steps 1.1 and 1.2 prove both implications.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

5 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

Richard F. Bass, Real Analysis for Graduate Students, Exercise 12.1 (standard reference, not scraped)