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

Statement

Let $ν$ be a signed measure on $(X, A)$ . Then there exist measurable sets $P, N$ such that $P \cap N = \emptyset, P \cup N = X,$ $P$ is positive for $ν$ , and $N$ is negative for $ν$ .

If $(P^{'}, N^{'})$ is another such pair, then $P △ P^{'}$ is null for $ν$ and hence has total variation $0$ .

Facts & Assumptions

Given: A signed measure $ν$ on $(X, A)$ .

[L1]

A measurable set is positive, negative, or null according to the signs of the signed measures of all its measurable subsets. (Positive, negative, and null sets for a signed measure)

[L2]

A measurable set of positive finite signed measure contains a positive subset whose signed measure is at least as large. (A set of positive finite signed measure contains a positive subset of at least the same mass)

[L3]

A set is null for a signed measure exactly when its total variation there is $0$ . (A set is null for a signed measure exactly when its total variation is zero there)

Proof

technique · direct

1.1

Replacing $ν$ by $- ν$ swaps positive and negative sets, so it is enough [L1, choose] to treat the case in which $ν (E) < + \infty$ for every measurable $E$ . Let $m : = \sup {ν (A) : A \in A, A is positive} .$ Because $\emptyset$ is positive, $m \geq 0$ . Choose positive sets $A_{n}$ with $ν (A_{n}) \to m$ , and put $P : = ⋃_{n} A_{n}$ .

2.1

The union $P$ is positive: if $E \subseteq P$ is measurable, define [L1, step 1.1] $B_{0} : = E \cap A_{0}$ and $B_{n} : = E \cap (A_{n} ∖ ⋃_{k < n} A_{k})$ for $n \geq 1$ . Then the $B_{n}$ are pairwise disjoint measurable subsets of the positive sets $A_{n}$ , so each $ν (B_{n}) \geq 0$ by [L1], and $E = ⋃_{n} B_{n}$ . Countable additivity gives $ν (E) = \sum_{n} ν (B_{n}) \geq 0$ , so [L1] makes $P$ positive. Because each $A_{n} \subseteq P$ , one has $ν (A_{n}) \leq ν (P) \leq m$ ; letting $n \to \infty$ yields $ν (P) = m < + \infty$ .

3.1

Let $N : = X ∖ P$ . If $N$ were not negative, [L1] would give a [L1, L2, step 2.1] measurable $E \subseteq N$ with $ν (E) > 0$ . By [L2], $E$ would contain a positive subset $Q$ with $ν (Q) \geq ν (E) > 0$ . Then $P \cup Q$ would be a positive set, $Q$ would be disjoint from $P$ , and $ν (P \cup Q) = ν (P) + ν (Q) > m,$ contradicting the definition of $m$ . Hence $N$ is negative.

4.1

If $(P^{'}, N^{'})$ is another Hahn decomposition, then [L1, L3, step 3.1] $P ∖ P^{'} \subseteq P \cap N^{'}$ and $P^{'} ∖ P \subseteq P^{'} \cap N$ . Thus each of $P ∖ P^{'}$ and $P^{'} ∖ P$ is both positive and negative, hence null by [L1]. Their union is $P △ P^{'}$ , so [L3] gives $∣ ν ∣ (P △ P^{'}) = 0$ .

5.1

Steps 2.1 through 4.1 give a positive set $P$ , a negative set $N = X ∖ P$ , [step 2.1, step 3.1, step 4.1] ∎ and uniqueness up to total-variation-null sets.

Depends on

Used by

Dependency tree · two levels

7 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, Theorem 6.18 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Theorem 12.5 (standard reference, not scraped)