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

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The indefinite integral of an integrable function is countably additive on measurable sets

Statement

If $f \in L^{1} (μ)$ and $ν_{f} (A) : = \int f χ_{A} d μ (A \in A),$ then $ν_{f}$ is countably additive on pairwise disjoint measurable families. Here $f χ_{A}$ is integrable because $∣ f χ_{A} ∣ \leq ∣ f ∣$ ; this formula defines the notation $\int_{A} f d μ$ for integrable real or complex $f$ .

Facts & Assumptions

Given: An integrable function $f$ .

[L1]

For every nonnegative measurable $h$ , the set function $A \mapsto \int_{A} h d μ$ is a measure (The indefinite integral of a nonnegative measurable function is a measure).

[L2]

Real and complex integrability are defined by positive/negative parts and by real/imaginary parts (Integrable real and complex functions, and their integrals).

[L3]

The Lebesgue integral is linear on $L^{1} (μ)$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct

1.1

For real-valued $f$ , write $f = f^{+} - f^{-}$ . Then [L1, L2, L3] $ν_{f} = ν_{f^{+}} - ν_{f^{-}},$ and both $ν_{f^{+}}$ and $ν_{f^{-}}$ are measures by [L1]. Because $f \in L^{1} (μ)$ , the total masses of those measures are finite, so subtracting their countably additive values on a disjoint family is legitimate and gives countable additivity of $ν_{f}$ .

2.1

For complex-valued $f = u + i v$ , one has [step 1.1, L2, L3] ∎ $ν_{f} = ν_{u} + i ν_{v},$ and step 1.1 applies to the real-valued functions $u$ and $v$ . Therefore $ν_{f}$ is countably additive as well.

Depends on

Used by

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Proposition 2.23 (standard reference, not scraped)