TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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Total variation is the supremum of simple integrals over unit-bounded test functions

Statement

Let $ν$ be a signed measure or complex measure on $(X, A)$ and let $E \in A$ satisfy $∣ ν ∣ (E) < + \infty$ . Then every complex simple function $s$ on $(X, A)$ with $∣ s ∣ \leq 1$ has a defined simple integral over $E$ , and $∣ ν ∣ (E) = \sup {∣ \int_{E} s d ν ∣ : s is a complex simple function and ∣ s ∣ \leq 1} .$

Facts & Assumptions

Given: A signed measure or complex measure $ν$ and a measurable set $E$ with $∣ ν ∣ (E) < + \infty$ .

[L1]

Simple integrals are bounded by total variation: $∣ \int_{E} s d ν ∣ \leq \int_{E} ∣ s ∣ d ∣ ν ∣$ . (Simple integrals are bounded by total variation)

[L2]

The simple integral over $E$ is computed from the measurable level-set representation of $s$ . (The simple integral against a signed or complex measure)

[L3]

Every complex number $z \neq 0$ has unit-modulus phase $\overline{z} / ∣ z ∣$ . (Real and imaginary parts, complex conjugation, and modulus)

[L4]

The total variation $∣ ν ∣ (E)$ is the supremum of countable partition sums $\sum_{n} ∣ ν (E_{n}) ∣$ . (The total variation |nu|(E) from countable measurable partitions)

Proof

technique · direct

1.1L1L4

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. Every countable measurable partition of $E_{j} \cap E$ extends to one of $E$ by adding $E ∖ E_{j}$ , so [L4] gives $∣ ν ∣ (E_{j} \cap E) \leq ∣ ν ∣ (E) < + \infty$ for each $j$ . Thus [L1] applies to $F = E$ and gives [L1] $∣ \int_{E} s d ν ∣ \leq \int_{E} ∣ s ∣ d ∣ ν ∣ \leq ∣ ν ∣ (E) .$ Therefore the displayed supremum is at most $∣ ν ∣ (E)$ .

1.2L3L4choose

Fix $ε > 0$ . By [L4], choose a countable measurable partition [L3, L4, choose] $E = ⨆_{n \geq 0} E_{n}$ such that $\sum_{n = 0}^{\infty} ∣ ν (E_{n}) ∣ > ∣ ν ∣ (E) - ε .$ Because $∣ ν ∣ (E) < + \infty$ , every term $∣ ν (E_{n}) ∣$ is finite. Choose $N$ so that the first $N + 1$ terms already satisfy $\sum_{n = 0}^{N} ∣ ν (E_{n}) ∣ > ∣ ν ∣ (E) - 2 ε .$ For each $0 \leq n \leq N$ with $ν (E_{n}) \neq 0$ , define $c_{n} : = \overline{ν (E_{n})} / ∣ ν (E_{n}) ∣$ , and put $c_{n} : = 0$ when $ν (E_{n}) = 0$ . After deleting the zero-coefficient terms, the simple function $s : = \sum_{n = 0}^{N} c_{n} 1_{E_{n}}$ satisfies $∣ s ∣ \leq 1$ .

2.1L2step 1.2

Using [L2], [L2, step 1.2] $\int_{E} s d ν = \sum_{n = 0}^{N} c_{n} ν (E_{n}) = \sum_{n = 0}^{N} ∣ ν (E_{n}) ∣,$ so step 1.2 gives $∣ \int_{E} s d ν ∣ > ∣ ν ∣ (E) - 2 ε .$ Because $ε > 0$ was arbitrary, the supremum is at least $∣ ν ∣ (E)$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the equality.

Depends on

Used by

Dependency tree · two levels

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Exercise 12.3 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, Chapter 9A (standard reference, not scraped)