CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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A nonnegative integral over a null set vanishes

Statement

Let $f : X \to [0, + \infty]$ be measurable and let $E$ be measurable with $μ (E) = 0$ . Then $\int_{E} f d μ = 0.$

Facts & Assumptions

Given: A nonnegative measurable function $f$ and a measurable null set $E$ .

[L1]

The set function $A \mapsto \int_{A} s d μ$ is a measure whenever $s$ is nonnegative simple (The indefinite integral of a nonnegative simple function is a measure).

[L2]

The integral over a measurable set is defined by $\int_{E} f d μ = \int f χ_{E} d μ$ (Integral over a measurable subset).

[L3]

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

Proof

technique · direct

1.1L1L2given

Let $s = \sum_{j} c_{j} χ_{A_{j}}$ be a nonnegative simple minorant of $f χ_{E}$ , with $c_{j} \geq 0$ . If $c_{j} > 0$ , then $A_{j} \subseteq E$ because $s = 0$ outside $E$ ; hence $μ (A_{j}) = 0$ . The simple-integral formula, equivalently the finite-sum calculation in [L1], gives $\int s d μ = \sum_{j} c_{j} μ (A_{j}) = 0$ , including when the original representation overlaps.

2.1step 1.1L2L3∎

Taking the supremum over all such simple minorants in [L3] gives $\int_{E} f d μ = \int f χ_{E} d μ = 0$ .

Depends on

Dependency tree · 2 levels

Sources

Richard F. Bass, Real Analysis for Graduate Students, Proposition 6.3(4)