PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Simple integrals are bounded by total variation

Statement

Let $ν$ be a signed measure or a complex measure on $(X, A)$ , let $F \in A$ , and let $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ be the canonical disjoint representation of a complex simple function using only its nonzero level sets. Assume $∣ ν ∣ (E_{j} \cap F) < + \infty$ for every $j$ . Then $∣ \int_{F} s d ν ∣ \leq \int_{F} ∣ s ∣ d ∣ ν ∣ .$ In particular, if $∣ s ∣ \leq M$ on $F$ and $∣ ν ∣ (F) < + \infty$ , then $∣ \int_{F} s d ν ∣ \leq M ∣ ν ∣ (F) .$

Facts & Assumptions

Given: A signed measure or complex measure $ν$ , a measurable set $F$ , and the canonical nonzero-level-set representation $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ of a complex simple function, with $∣ ν ∣ (E_{j} \cap F) < + \infty$ for every $j$ .

[L1]

The simple integral against $ν$ is computed from a disjoint measurable level-set representation. (The simple integral against a signed or complex measure)

[L2]

The integral of a nonnegative simple function against a positive measure is the weighted sum over a disjoint representation. (The integral of a nonnegative simple function)

[L3]

The total variation $∣ ν ∣$ is a measure. (The total variation of a signed or complex measure is a positive measure)

Proof

technique · direct

1.1L1

Write the canonical disjoint representation of $s$ as [L1] $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ . For each $j$ , the one-piece partition of $E_{j} \cap F$ gives $∣ ν (E_{j} \cap F) ∣ \leq ∣ ν ∣ (E_{j} \cap F) < + \infty$ , so [L1] makes $\int_{F} s d ν$ well defined and gives $\int_{F} s d ν = \sum_{j = 1}^{m} c_{j} ν (E_{j} \cap F) .$ By the triangle inequality, $∣ \int_{F} s d ν ∣ \leq \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν (E_{j} \cap F) ∣ \leq \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν ∣ (E_{j} \cap F) .$

2.1L2L3step 1.1

Because $∣ ν ∣$ is a measure by [L3], the sets $E_{j} \cap F$ are disjoint [L2, L3, step 1.1] and measurable, and [L2] gives $\int_{F} ∣ s ∣ d ∣ ν ∣ = \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν ∣ (E_{j} \cap F) .$ Substituting this into step 1.1 proves the first inequality. If $∣ s ∣ \leq M$ on $F$ and $∣ ν ∣ (F) < + \infty$ , then [L3] gives $∣ ν ∣ (E_{j} \cap F) \leq ∣ ν ∣ (F) < + \infty$ for every $j$ , so the displayed finiteness hypothesis is automatic. Moreover $∣ s ∣ 1_{F} \leq M 1_{F}$ , so monotonicity of the simple integral with respect to the positive measure $∣ ν ∣$ gives $\int_{F} ∣ s ∣ d ∣ ν ∣ \leq M ∣ ν ∣ (F)$ .

3.1step 1.1step 2.1∎

The displayed inequalities follow from steps 1.1 and 2.1.

Depends on

Used by

Dependency tree · two levels

13 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, Exercise 12.2 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, Chapter 9A (standard reference, not scraped)