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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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The indefinite integral of a nonnegative measurable function is a measure

Statement

Let f:X→[0,+∞] be measurable and define νf(A):=∫Af dμ(A∈A). Then νf is a measure on (X,A).

Facts & Assumptions

Given: A nonnegative measurable function f.

[L1]

The set function A↦∫Af dμ is defined by A↦∫fχA dμ (Integral over a measurable subset).

[L2]

Monotone convergence holds for the nonnegative integral (Monotone convergence for the integral).

[L3]

The nonnegative integral is additive on measurable sets (Additivity of the nonnegative Lebesgue integral).

[L4]

A measure must vanish at the empty set and be countably additive on disjoint measurable sequences (Measures on sigma-algebras).

Proof

technique · direct
1.1L1L2given

One has νf(∅)=0. If (An) is a pairwise disjoint sequence, put Bn:=⋃k<nAk. Then χBn↑χ⋃kAk, so fχBn↑fχ⋃kAk, including where f=+∞ under the convention 0⋅∞=0. By [L2], νf(⋃kAk)=lim⁡nνf(Bn).

2.1step 1.1L3algebra

Because the sets Ak are disjoint, fχBn=∑k<nfχAk. Repeated use of [L3] gives νf(Bn)=∑k<nνf(Ak). Taking the increasing limit in step 1.1 proves countable additivity.

3.1step 1.1step 2.1L4∎

Steps 1.1 and 2.1 verify the two conditions in [L4], so νf is a measure.

Depends on

Used by

Dependency tree · two levels

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Sources