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

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Finite-total-variation signed measures are complete

Statement

Fix a measurable space $(X, A)$ . The normed space $M (X, A)$ of finite-total-variation signed measures is complete.

Facts & Assumptions

Given: A Cauchy sequence $(ν_{n})$ in $M (X, A)$ .

[L1]

The norm on $M (X, A)$ is $∥ ν ∥ = ∣ ν ∣ (X)$ . (The space of finite total variation signed measures, The finite-total-variation signed measures form a real normed space)

[L2]

If $∣ ρ ∣ (X)$ is finite, then $∣ ρ (E) ∣ \leq ∣ ρ ∣ (E) \leq ∣ ρ ∣ (X)$ for every measurable $E$ . (Total variation is the supremum of simple integrals over unit-bounded test functions)

[L3]

Positive measures are continuous from above on decreasing measurable sets once one term has finite measure. (Continuity from above when one set has finite measure)

Proof

technique · direct

1.1

By [L1] and [L2], for every measurable $E$ the scalar sequence $(ν_{n} (E))$ is Cauchy in $R$ , because $∣ ν_{n} (E) - ν_{m} (E) ∣ \leq ∣ ν_{n} - ν_{m} ∣ (X) = ∥ ν_{n} - ν_{m} ∥ .$ Define $ν (E) : = \lim_{n} ν_{n} (E)$ .

L1L2

2.1

Let $(E_{k})$ be pairwise disjoint and put $E = ⋃_{k} E_{k}$ . Fix $ε > 0$ . Choose $m$ so that $∥ ν_{n} - ν_{m} ∥ < ε$ for all $n \geq m$ . Because $∣ ν_{m} ∣$ is a finite positive measure, [L3] gives $∣ ν_{m} ∣ (⋃_{k \geq N} E_{k}) \to 0$ , so choose $N$ with that tail below $ε$ . Then for $n \geq m$ , $∣ ν_{n} (E) - \sum_{k = 0}^{N - 1} ν_{n} (E_{k}) ∣ = ∣ ν_{n} (⋃_{k \geq N} E_{k}) ∣ \leq ∥ ν_{n} - ν_{m} ∥ + ∣ ν_{m} ∣ (⋃_{k \geq N} E_{k}) < 2 ε .$ Passing $n \to \infty$ gives $∣ ν (E) - \sum_{k = 0}^{N - 1} ν (E_{k}) ∣ \leq 2 ε,$ so $ν (E) = \sum_{k} ν (E_{k})$ . Thus $ν$ is a signed measure.

L1L2L3step 1.1

3.1

For any countable measurable partition $(A_{j})$ of a measurable set $F$ , Fatou's lemma for nonnegative series gives $\sum_{j} ∣ ν (A_{j}) ∣ \leq {lim inf}_{n \to \infty} \sum_{j} ∣ ν_{n} (A_{j}) ∣ \leq \sup_{n} ∣ ν_{n} ∣ (F),$ so $∣ ν ∣ (F) < + \infty$ and therefore $ν \in M (X, A)$ . Likewise, for fixed $n$ and any partition $(A_{j})$ of $X$ , $\sum_{j} ∣ ν_{n} (A_{j}) - ν (A_{j}) ∣ \leq {lim inf}_{m \to \infty} \sum_{j} ∣ ν_{n} (A_{j}) - ν_{m} (A_{j}) ∣ \leq {lim inf}_{m \to \infty} ∥ ν_{n} - ν_{m} ∥ .$ Taking the supremum over partitions gives $∥ ν_{n} - ν ∥ \leq {lim inf}_{m \to \infty} ∥ ν_{n} - ν_{m} ∥,$ and the right side tends to $0$ because $(ν_{n})$ is Cauchy.

L1step 1.1step 2.1

4.1

Step 3.1 shows that $ν_{n} \to ν$ in norm, so every Cauchy sequence in $M (X, A)$ converges there.

step 3.1∎

Depends on

Used by

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Sources

Sheldon Axler, Measure, Integration & Real Analysis, Theorem 9.14 (standard reference, not scraped)