Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere

Statement

Let f:X[0,+] be measurable. Then fdμ=0f=0 almost everywhere.

Facts & Assumptions

Given: A nonnegative measurable function f.

[L1]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L2]

A statement holds almost everywhere when its exceptional set is contained in a measurable null set (Measure-null sets and almost-everywhere statements relative to a measure).

[L3]

The nonnegative integral is the supremum of the integrals of simple minorants (The nonnegative Lebesgue integral).

Proof

technique · direct
1.1

Assume fdμ=0. For n1 let En:={f1/n}. Then[L1, L2, given, algebra] (1/n)χEnf, so [L1] gives 1nμ(En)=(1/n)χEndμfdμ=0. Hence μ(En)=0 for every n. Since {f>0}=nEn, the exceptional set where f0 is null, so f=0 almost everywhere by [L2].

1.2

Assume f=0 almost everywhere, and let N be a measurable null set[L2, L3, given] containing {f>0}. If s=jcjχEj is a simple minorant of f, then every set Ej with cj>0 lies inside {f>0}N, so μ(Ej)=0; the remaining coefficients are 0. Therefore sdμ=0. Taking the supremum over all simple minorants in [L3] gives fdμ=0.

2.1

Step 1.1 proves the forward implication and step 1.2 proves the reverse [step 1.1, step 1.2] ∎ implication.

Depends on

Used by

Dependency tree · two levels

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Sources