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Jordan decomposition of a signed measure into unique mutually singular positive parts

Statement

Let $ν$ be a signed measure on $(X, A)$ . Then there exist positive measures $ν^{+}, ν^{-} : (X, A) \to [0, + \infty]$ such that $ν = ν^{+} - ν^{-},$ and $ν^{+} ⊥ ν^{-}$ .

These measures are unique: if $ν = μ - η$ with positive measures $μ ⊥ η$ , then $μ = ν^{+}$ and $η = ν^{-}$ .

Facts & Assumptions

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

[L1]

Hahn decomposition gives measurable sets $P, N$ with $P ⊔ N = X$ , $P$ positive, and $N$ negative, unique up to null sets. (Hahn decomposition for signed measures, unique up to total-variation-null sets)

[L2]

Mutual singularity means that the two set functions vanish on measurable subsets of complementary measurable pieces. (Mutual singularity for signed or complex measures)

[L3]

A measure is a nonnegative countably additive set function on a sigma-algebra. (Measures on sigma-algebras)

Proof

technique · direct

1.1

Choose a Hahn decomposition $X = P ⊔ N$ from [L1]. Define [L1, L3] $ν^{+} (A) : = ν (A \cap P), ν^{-} (A) : = - ν (A \cap N) (A \in A) .$ Because $P$ is positive and $N$ is negative, these values lie in $[0, + \infty]$ . Their countable additivity is inherited from that of $ν$ , so [L3] makes $ν^{+}$ and $ν^{-}$ positive measures. Also $ν (A) = ν (A \cap P) + ν (A \cap N) = ν^{+} (A) - ν^{-} (A)$ for every measurable $A$ .

2.1

The defining pieces in step 1.1 also show mutual singularity: every [L1, L2, step 1.1] measurable subset of $N$ has $ν^{+}$ -value $0$ , and every measurable subset of $P$ has $ν^{-}$ -value $0$ . Thus [L2] gives $ν^{+} ⊥ ν^{-}$ .

2.2

Suppose $ν = μ - η$ with positive measures $μ ⊥ η$ . By [L2], [L1, L2, step 1.1] choose $P^{'}, N^{'}$ with $P^{'} ⊔ N^{'} = X$ , $μ$ vanishing on subsets of $N^{'}$ , and $η$ vanishing on subsets of $P^{'}$ . Then every measurable subset of $P^{'}$ has $ν$ -value $μ (E) \geq 0$ , so $P^{'}$ is positive, and every measurable subset of $N^{'}$ has $ν$ -value $- η (E) \leq 0$ , so $N^{'}$ is negative. Hence $(P^{'}, N^{'})$ is a Hahn decomposition, so [L1] makes $P △ P^{'}$ null.

3.1

Because $μ$ vanishes on subsets of $N^{'}$ and null subsets of $P^{'}$ have [L1, L2, step 1.1, step 2.2] $μ$ -value $0$ as well, step 2.2 gives $μ (A) = μ (A \cap P^{'}) = ν (A \cap P^{'}) = ν (A \cap P) = ν^{+} (A) .$ The same argument on $N^{'}$ gives $η (A) = - ν (A \cap N) = ν^{-} (A)$ . Thus the Jordan decomposition is unique.

4.1

Steps 1.1, 2.1, and 3.1 prove existence, mutual singularity, and [step 1.1, step 2.1, step 3.1] ∎ uniqueness.

Depends on

Used by

Dependency tree · two levels

11 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 6.21 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Theorem 12.8 (standard reference, not scraped)