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A set of positive finite signed measure contains a positive subset of at least the same mass

Statement

Assume the Axiom of Choice. Let $ν$ be a signed measure on $(X, A)$ and let $A \in A$ satisfy $0 < ν (A) < + \infty$ . Then there exists a positive set $P \subseteq A$ such that $ν (P) \geq ν (A) .$

Facts & Assumptions

Given: A signed measure $ν$ , a measurable set $A$ with $0 < ν (A) < + \infty$ , and the Axiom of Choice (The Axiom of Choice).

[L1]

A measurable set is positive when every measurable subset has nonnegative signed measure. (Positive, negative, and null sets for a signed measure)

[L2]

Every measurable subset of $A$ has finite signed measure. (A subset of a set of finite signed measure also has finite signed measure)

[L3]

If a disjoint union has finite signed measure, then the resulting real series converges absolutely. (If a disjoint union has finite signed measure, then the signed-measure series converges absolutely)

Proof

technique · direct

1.1givenL1L2choose

For each $n \geq 1$ and measurable $R \subseteq A$ , put $δ (R) = \inf {ν (E) : E \in A, E \subseteq R}$ . If $R$ is positive, prescribe the subset $\emptyset$ . Otherwise $δ (R) < 0$ : if it is finite, the subsets $E \subseteq R$ with $ν (E) \leq δ (R) / 2$ form a nonempty family; if it is $- \infty$ , use the nonempty family with $ν (E) \leq - n$ . Apply AC to this set-indexed family of nonempty admissible-subset families to fix a selector $E (n, R)$ . Define recursively $R_{1} = A$ , $A_{n} = E (n, R_{n})$ , $δ_{n} = δ (R_{n})$ and $R_{n + 1} = R_{n} ∖ A_{n}$ . The $A_{n}$ are pairwise disjoint, have finite nonpositive values by [L2], and satisfy $δ_{n} \leq ν (A_{n}) \leq δ_{n} / 2$ when $δ_{n}$ is finite, and $ν (A_{n}) \leq - n$ otherwise. This selector is the precise use of AC; no minimal choice principle is asserted.

2.1L2L3step 1.1

Put $B : = ⋃_{n \geq 1} A_{n}$ and $P : = A ∖ B$ . Because $B \subseteq A$ , [L2] makes $ν (B)$ finite, and [L3] makes $\sum_{n \geq 1} ν (A_{n})$ absolutely convergent. Thus $ν (A_{n}) \to 0$ , and only finitely many terms satisfy $ν (A_{n}) \leq - 1$ . The $δ_{n} = - \infty$ branch therefore occurs only finitely often. For some $n_{0}$ and every $n \geq n_{0}$ , $δ_{n}$ is finite and $0 \geq δ_{n} \geq 2 ν (A_{n})$ . Comparison proves convergence of the tail $\sum_{n \geq n_{0}} ∣ δ_{n} ∣$ , and in particular $δ_{n} \to 0$ . No assertion about summing earlier infinite values is needed.

3.1L1step 1.1step 2.1

If $E \subseteq P$ is measurable, then $E \subseteq R_{n}$ for every $n$ , so $ν (E) \geq δ_{n}$ by definition of $δ_{n}$ . Letting $n \to \infty$ in step 2.1 gives $ν (E) \geq 0$ , so [L1] shows that $P$ is positive.

3.2step 2.1algebra

Because every term $ν (A_{n})$ is nonpositive, step 2.1 gives $ν (B) = \sum_{n \geq 1} ν (A_{n}) \leq 0.$ Hence $ν (P) = ν (A) - ν (B) \geq ν (A)$ .

4.1step 3.1step 3.2∎

Steps 3.1 and 3.2 give a positive subset $P \subseteq A$ with $ν (P) \geq ν (A)$ .

Depends on

Used by

Hahn decomposition for signed measures, unique up to total-variation-null sets Theorem

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

John K. Hunter, Measure Theory, Lemma 6.17 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Proposition 12.4 (standard reference, not scraped)