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

1.1

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 $∣ ν ∣ (E)$ .

L1L4

1.2

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 $∣ s ∣ \leq 1$ .

L3L4choose

2.1

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 $∣ ν ∣ (E)$ .

L2step 1.2

3.1

Steps 1.1 and 2.1 prove the equality.

step 1.1step 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 12.3 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, Chapter 9A (standard reference, not scraped)