TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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The indefinite integral of a nonnegative measurable function is a measure

Statement

Let $f : X \to [0, + \infty]$ be measurable and define $ν_{f} (A) : = \int_{A} f d μ (A \in A) .$ Then $ν_{f}$ is a measure on $(X, A)$ .

Facts & Assumptions

Given: A nonnegative measurable function $f$ .

[L1]

The set function $A \mapsto \int_{A} f d μ$ is defined by $A \mapsto \int f χ_{A} d μ$ (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

1.1

One has $ν_{f} (\emptyset) = 0$ . If $(A_{n})$ is a pairwise disjoint sequence,[L1, L2, given] put $B_{n} : = ⋃_{k < n} A_{k}$ . Then $χ_{B_{n}} ↑ χ_{⋃_{k} A_{k}}$ , so $f χ_{B_{n}} ↑ f χ_{⋃_{k} A_{k}}$ and [L2] gives $ν_{f} ⁣ (⋃_{k} A_{k}) = \lim_{n} ν_{f} (B_{n}) .$

2.1

Because the sets $A_{k}$ are disjoint, repeated use of [L3] gives [step 1.1, L3, algebra] $ν_{f} (B_{n}) = \sum_{k < n} ν_{f} (A_{k}) .$ Substituting this into step 1.1 proves countable additivity.

3.1

Steps 1.1 and 2.1 verify the two conditions in [L4], so $ν_{f}$ is a [step 1.1, step 2.1, L4] ∎ measure.

Depends on

Used by

Dependency tree · two levels

15 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, ch. 7 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., §2.2 (standard reference, not scraped)