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For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal

Statement

Let $ν$ be a signed measure on $(X, A)$ ), with Jordan decomposition $ν = ν^{+} - ν^{-}$ . Then for every measurable $E$ :

$∣ ν ∣ (E) = ν^{+} (E) + ν^{-} (E);$
the same value is the supremum over finite measurable partitions of $E$ ;
$ν^{+} (E) = \sup {ν (F) : F \subseteq E, F \in A};$
$ν^{-} (E) = - \inf {ν (F) : F \subseteq E, F \in A} .$

Facts & Assumptions

Given: A signed measure $ν$ , its Jordan decomposition $ν = ν^{+} - ν^{-}$ , and a measurable set $E$ .

[L1]

The total variation $∣ ν ∣ (E)$ is the supremum of the countable partition sums $\sum_{n} ∣ ν (E_{n}) ∣$ . (The total variation |nu|(E) from countable measurable partitions)

[L2]

Jordan decomposition gives positive measures $ν^{+}, ν^{-}$ and a Hahn decomposition $X = P ⊔ N$ with $ν^{+} (A) = ν (A \cap P)$ and $ν^{-} (A) = - ν (A \cap N)$ . (Jordan decomposition of a signed measure into unique mutually singular positive parts)

Proof

technique · direct

1.1

Let $(E_{n})$ be a countable measurable partition of $E$ . Using [L2] and the [L1, L2] triangle inequality, $∣ ν (E_{n}) ∣ = ∣ ν^{+} (E_{n}) - ν^{-} (E_{n}) ∣ \leq ν^{+} (E_{n}) + ν^{-} (E_{n})$ for each $n$ . Summing and using countable additivity of the positive measures $ν^{+}$ and $ν^{-}$ gives $\sum_{n} ∣ ν (E_{n}) ∣ \leq ν^{+} (E) + ν^{-} (E) .$ Hence [L1] yields $∣ ν ∣ (E) \leq ν^{+} (E) + ν^{-} (E)$ .

2.1

The two-piece partition $E = (E \cap P) ⊔ (E \cap N)$ from [L2] gives [L1, L2, step 1.1] $∣ ν (E \cap P) ∣ + ∣ ν (E \cap N) ∣ = ν^{+} (E) + ν^{-} (E) .$ Therefore $∣ ν ∣ (E) \geq ν^{+} (E) + ν^{-} (E)$ , and together with step 1.1 this proves equality. Because this equality is already realized by a finite partition, finite partitions suffice for signed measures.

3.1

If $F \subseteq E$ is measurable, then [L2] gives [L2, step 2.1] $ν (F) = ν^{+} (F) - ν^{-} (F) \leq ν^{+} (F) \leq ν^{+} (E),$ so $\sup_{F \subseteq E} ν (F) \leq ν^{+} (E)$ . Taking $F = E \cap P$ gives equality: $ν (F) = ν^{+} (E)$ . The same argument with $E \cap N$ gives $\inf_{F \subseteq E} ν (F) = - ν^{-} (E) .$

4.1

Steps 2.1 and 3.1 prove all four displayed formulas.

step 2.1step 3.1∎

Depends on

Used by

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

Richard F. Bass, Real Analysis for Graduate Students, Exercises 12.6 and 12.7 (standard reference, not scraped)
John K. Hunter, Measure Theory, sentence after Theorem 6.21 (standard reference, not scraped)