LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27

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The simple integral is independent of the chosen representation

Statement

If a nonnegative simple measurable function $s$ admits two representations $s = \sum_{i = 1}^{m} c_{i} χ_{E_{i}} = \sum_{j = 1}^{n} d_{j} χ_{F_{j}},$ then the two coefficient sums defining $\int s d μ$ are equal. So The integral of a nonnegative simple function is well defined.

Facts & Assumptions

Given: Two simple representations of the same nonnegative simple measurable function $s$ .

[L1]

The simple integral is defined by $\int s d μ = \sum c_{j} μ (E_{j})$ with the convention $0 \cdot (+ \infty) = 0$ (The integral of a nonnegative simple function).

[L2]

A measure is countably additive on pairwise disjoint measurable families, hence finitely additive on finite measurable partitions (Measures on sigma-algebras).

Proof

technique · direct

1.1givenL1

Complete both representations to partitions of $X$ . [given, L1] Put $E_{0} = X ∖ ⋃_{i = 1}^{m} E_{i}$ and $F_{0} = X ∖ ⋃_{j = 1}^{n} F_{j}$ , with coefficients $c_{0} = d_{0} = 0$ . Both are measurable. Adding these zero terms leaves the represented function and each coefficient sum unchanged, including when a complement has infinite measure, by the convention in [L1]. The augmented families $(E_{i})_{i = 0}^{m}$ and $(F_{j})_{j = 0}^{n}$ are finite measurable partitions of $X$ .

2.1step 1.1

Refine the two partitions by their intersections. [step 1.1] For $0 \leq i \leq m$ and $0 \leq j \leq n$ set $G_{i j} = E_{i} \cap F_{j}$ . These sets are measurable and pairwise disjoint, and $E_{i} = ⨆_{j = 0}^{n} G_{i j}, F_{j} = ⨆_{i = 0}^{m} G_{i j} .$ On every nonempty $G_{i j}$ the two formulas give the same value of $s$ , so $c_{i} = d_{j}$ .

3.1L1L2step 2.1

Apply finite additivity and the nonnegative extended-real finite-sum rules. [L1, L2, step 2.1] They give $\sum_{i = 0}^{m} c_{i} μ (E_{i}) = \sum_{i = 0}^{m} \sum_{j = 0}^{n} c_{i} μ (G_{i j}) = \sum_{i = 0}^{m} \sum_{j = 0}^{n} d_{j} μ (G_{i j}) = \sum_{j = 0}^{n} d_{j} μ (F_{j}) .$ For a zero coefficient, every product with an infinite measure is $0$ by [L1]; for a positive coefficient the usual extended-real distributivity applies. Thus no subtraction of infinities occurs.

4.1step 1.1step 3.1∎

Removing the added zero terms from step 3.1 proves equality of the original coefficient sums.

Depends on

Used by

Cited to discharge well-definedness by The integral of a nonnegative simple function.

Dependency tree · two levels

6 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

John K. Hunter, Measure Theory Notes, Definition 4.1 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., §2.2 (standard reference, not scraped)