LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Bounded variation of a vector measure is a finite measure

Statement

If $ν : A \to X$ is a norm-countably additive vector measure of bounded variation, then $∣ ν ∣$ is a finite positive countably additive measure. Moreover, for every $x^{*} \in X^{*}$ ,

$∣ x^{*} \circ ν ∣ (E) \leq ∥ x^{*} ∥ ∣ ν ∣ (E)$

for every measurable $E$ .

Facts & Assumptions

[L1]

Vector-measure variation is the supremum of norm sums over finite measurable partitions, and bounded variation means finite total variation (Banach-valued vector measure and variation).

[L2]

A bounded functional satisfies $∣ x^{*} (x) ∣ \leq ∥ x^{*} ∥ ∥ x ∥$ (The dual space X^* of a normed space and its dual norm).

Proof

technique · direct

Given: A bounded-variation vector measure $ν$ and a bounded functional $x^{*}$ .

1.1givenL1

Prove finite additivity of variation. For disjoint $A, B$ , joining finite partitions of $A$ and $B$ shows $∣ ν ∣ (A) + ∣ ν ∣ (B) \leq ∣ ν ∣ (A \cup B)$ (use partitions within $ε$ of each supremum). Conversely, intersect any finite partition of $A \cup B$ with $A$ and $B$ ; finite additivity of $ν$ and the triangle inequality show that its norm sum is at most $∣ ν ∣ (A) + ∣ ν ∣ (B)$ . Taking the supremum gives equality.

1.2L1L2

Prove the functional domination estimate. For every finite partition $(A_{j})$ of $E$ , [L2] gives $\sum_{j} ∣ x^{*} (ν (A_{j})) ∣ \leq ∥ x^{*} ∥ \sum_{j} ∥ ν (A_{j}) ∥$ . Taking suprema as in [L1] proves the displayed inequality, including $x^{*} = 0$ and $E = \emptyset$ .

2.1L1step 1.1

Prove countable additivity. Let $E = ⨆_{n \geq 1} E_{n}$ . Finite additivity gives $\sum_{n = 1}^{N} ∣ ν ∣ (E_{n}) \leq ∣ ν ∣ (E)$ , hence $\sum_{n} ∣ ν ∣ (E_{n}) \leq ∣ ν ∣ (E)$ . For the reverse inequality, take any finite partition $(A_{j})$ of $E$ . Norm countable additivity gives $ν (A_{j}) = \sum_{n} ν (A_{j} \cap E_{n})$ , so $\sum_{j} ∥ ν (A_{j}) ∥ \leq \sum_{n} \sum_{j} ∥ ν (A_{j} \cap E_{n}) ∥ \leq \sum_{n} ∣ ν ∣ (E_{n})$ . Taking the supremum over $(A_{j})$ proves the reverse inequality.

3.1L1step 1.2step 2.1∎

Conclude finiteness and all boundary cases. [L1, step 1.2, step 2.1] The empty partition gives $∣ ν ∣ (\emptyset) = 0$ , step 2.1 gives countable additivity, and bounded variation in [L1] gives $∣ ν ∣ (E) \leq ∣ ν ∣ (Ω) < \infty$ . Thus $∣ ν ∣$ is a finite positive measure, and step 1.2 supplies the asserted scalar-variation bound.

Depends on

Used by

Dependency tree · two levels

8 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

Gilles Pisier, Martingales in Banach Spaces (standard reference, not scraped)