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

Statement

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 $ν$ and a measurable set $A$ with $0 < ν (A) < + \infty$ .

[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

1.1

Define $R_{1} : = A$ . If $R_{n}$ is not positive, choose a measurable subset $B_{n} \subseteq R_{n}$ with $ν (B_{n}) < 0$ , set $δ_{n} : = \inf {ν (E) : E \in A, E \subseteq R_{n}},$ and choose $A_{n} \subseteq R_{n}$ so that either $δ_{n} \leq ν (A_{n}) \leq δ_{n} / 2 < 0 when δ_{n} > - \infty,$ or $ν (A_{n}) \leq - n when δ_{n} = - \infty .$ If $R_{n}$ is positive, put $A_{n} = \emptyset$ and $δ_{n} = 0$ . In every case define $R_{n + 1} : = R_{n} ∖ A_{n}$ . Then the $A_{n}$ are pairwise disjoint subsets of $A$ and each $ν (A_{n}) \leq 0$ .

L1L2choose

2.1

Put $B : = ⋃_{n \geq 1} A_{n}$ and $P : = A ∖ B$ . Because $B \subseteq A$ , [L2] makes $ν (B)$ finite, and [L3] makes the real series $\sum_{n \geq 1} ν (A_{n})$ absolutely convergent. Since every nonzero term is nonpositive, only finitely many satisfy $ν (A_{n}) \leq - 1$ ; therefore the $δ_{n} = - \infty$ branch of step 1.1 occurs only finitely often. For all large $n$ one then has $δ_{n} > - \infty$ and $0 \geq δ_{n} \geq 2 ν (A_{n}) .$ Hence $\sum_{n \geq 1} ∣ δ_{n} ∣$ converges by comparison with $2 \sum_{n \geq 1} ∣ ν (A_{n}) ∣$ , so $δ_{n} \to 0$ .

L2L3step 1.1

3.1

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.

L1step 1.1step 2.1

3.2

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

step 2.1algebra

4.1

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

step 3.1step 3.2∎

Depends on

Used by

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

Dependency tree · two levels

6 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)