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Signed and Complex Measures Hahn and Jordan

1 · Prerequisites

2 · Summary

This page fixes the signed-measure convention used later in the measure-theory track: a signed measure may take one infinite sign but not both, null sets are defined by vanishing on every measurable subset, and total variation is the countable-partition object whose measure property must be proved.

Hahn and Jordan decomposition then organize the whole page. Their consequences include continuity for signed measures, the total-variation formulas, bounded integration against signed or complex measures, the finite-total-variation normed-space package, completeness, and the density examples that the next measure pages reuse.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

A signed measure is countably additive and takes at most one infinite value

Definition

Let $(X, A)$ be a measurable space. A signed measure on $(X, A)$ is a function $ν : A ⟶ \overline{R}$ such that:

$ν (\emptyset) = 0$ ;
$ν$ takes at most one infinite sign: either $ν (E) \neq - \infty$ for every $E \in A$ , or $ν (E) \neq + \infty$ for every $E \in A$ ;
for every pairwise disjoint sequence $(E_{n})_{n \in N}$ in $A$ , the series on the right is defined in the extended reals and $ν (⋃_{n \in N} E_{n}) = \sum_{n = 0}^{\infty} ν (E_{n})$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

The second and third clauses are load-bearing, not stylistic: together they rule out the undefined form $(+ \infty) + (- \infty)$ and require the disjoint series to exist before countable additivity is asserted.

The next proposition proves the extra fact that if $ν (⋃_{n} E_{n})$ is finite, then the real series $\sum_{n} ν (E_{n})$ converges absolutely in the sense of Absolutely convergent and conditionally convergent series, and the general starting index.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Positive, negative, and null sets for a signed measure

Definition

Let $ν$ be a signed measure on $(X, A)$ .

A measurable set $P$ is positive for $ν$ when $ν (E) \geq 0$ for every measurable $E \subseteq P$ .
A measurable set $N$ is negative for $ν$ when $ν (E) \leq 0$ for every measurable $E \subseteq N$ .
A measurable set $Z$ is null for $ν$ when $ν (E) = 0$ for every measurable $E \subseteq Z$ .

The null-set clause is the strong one used throughout the page: it asks for vanishing on every measurable subset, not only on the ambient set itself.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

A complex measure is a finite-valued countably additive set function

Definition

Let $(X, A)$ be a measurable space. A complex measure on $(X, A)$ is a function $ν : A ⟶ C$ such that:

$ν (\emptyset) = 0$ ;
for every pairwise disjoint sequence $(E_{n})_{n \in N}$ in $A$ , $ν (⋃_{n \in N} E_{n}) = \sum_{n = 0}^{\infty} ν (E_{n})$ in $C$ .

The codomain is the field $C$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ), so a complex measure is finite-valued by definition: there is no complex number called $+ \infty$ or $- \infty$ to allow.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The real and imaginary parts of a complex measure are finite signed measures, and nu = Re nu + i Im nu

Statement

Let $ν$ be a complex measure on $(X, A)$ . Then the set functions $Re ν (E) : = Re (ν (E)), Im ν (E) : = Im (ν (E))$ are finite signed measures on $(X, A)$ , and $ν (E) = Re ν (E) + i Im ν (E) (E \in A) .$

Facts & Assumptions

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

[L1]

A complex measure is a finite-valued countably additive set function on a sigma-algebra. (A complex measure is a finite-valued countably additive set function)

[L2]

Every complex number $z$ has real and imaginary parts and satisfies $z = Re z + i Im z$ . (Real and imaginary parts, complex conjugation, and modulus)

[L3]

A signed measure is countably additive and takes at most one infinite sign. (A signed measure is countably additive and takes at most one infinite value)

Proof

technique · direct

1.1L1L2

Because $ν (E) \in C$ for every $E \in A$ , [L2] makes $Re ν (E)$ and $Im ν (E)$ honest real numbers for every measurable $E$ . In particular neither set function takes an infinite value.

1.2L1L2

If $(E_{n})$ is pairwise disjoint, then [L1] gives $ν (⋃_{n} E_{n}) = \sum_{n = 0}^{\infty} ν (E_{n}) .$ Taking real parts and imaginary parts termwise yields $Re ν (⋃_{n} E_{n}) = \sum_{n = 0}^{\infty} Re ν (E_{n}), Im ν (⋃_{n} E_{n}) = \sum_{n = 0}^{\infty} Im ν (E_{n}) .$ Also $Re ν (\emptyset) = Im ν (\emptyset) = 0$ .

2.1L2L3step 1.1step 1.2∎

Step 1.1 supplies the finiteness clause and step 1.2 supplies countable additivity, so [L3] shows that $Re ν$ and $Im ν$ are finite signed measures. The decomposition $ν (E) = Re ν (E) + i Im ν (E)$ is exactly the identity from [L2] applied to the complex number $ν (E)$ .

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

The total variation |nu|(E) from countable measurable partitions

Definition

Let $ν$ be a signed measure or a complex measure on $(X, A)$ . For $E \in A$ , define its total variation on $E$ by $∣ ν ∣ (E) : = \sup {\sum_{n = 0}^{\infty} ∣ ν (E_{n}) ∣ : (E_{n})_{n \in N} is a countable measurable partition of E},$ where, for a signed measure, the term $∣ ν (E_{n}) ∣$ means the ordinary absolute value when $ν (E_{n}) \in R$ and means $+ \infty$ when $ν (E_{n}) \in {+ \infty, - \infty}$ ; for a complex measure it is the usual complex modulus. A countable measurable partition of $E$ means:

each $E_{n}$ lies in $A$ ;
the sets $E_{n}$ are pairwise disjoint;
$E = ⋃_{n \in N} E_{n}$ .

The sum on the right is a nonnegative extended series, so it is always defined in $[0, + \infty] \subseteq \overline{R}$ .

For signed measures, a later proposition shows that finite partitions already suffice. For complex measures, that finite-partition shortcut is not built into the definition here.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

A set is null for a signed measure exactly when its total variation is zero there

Statement

Let $ν$ be a signed measure on $(X, A)$ and let $E \in A$ . Then $E$ is null for $ν$ if and only if $∣ ν ∣ (E) = 0$ .

Facts & Assumptions

Given: A signed measure $ν$ and a measurable set $E$ .

[L1]

A null set for a signed measure means: every measurable subset of it has signed measure $0$ . (Positive, negative, and null sets for a signed measure)

[L2]

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

Proof

technique · direct

1.1L1L2

Assume $E$ is null. If $(E_{n})$ is a countable measurable partition of $E$ , [L1, L2] then every $E_{n} \subseteq E$ has $ν (E_{n}) = 0$ by [L1], so its partition sum in [L2] is $0$ . Hence every admissible sum is $0$ , and therefore $∣ ν ∣ (E) = 0$ .

1.2L1L2

Assume instead that $∣ ν ∣ (E) = 0$ . Let $F \subseteq E$ be measurable. Then [L1, L2] $F$ and $E ∖ F$ form a measurable partition of $E$ , so [L2] gives $0 = ∣ ν ∣ (E) \geq ∣ ν (F) ∣ + ∣ ν (E ∖ F) ∣ \geq ∣ ν (F) ∣ .$ Thus $ν (F) = 0$ . Since $F \subseteq E$ was arbitrary, [L1] shows that $E$ is null.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove both implications. [step 1.1, step 1.2].

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Mutual singularity for signed or complex measures

Definition

Let $μ$ and $ν$ be signed measures or complex measures on the same measurable space $(X, A)$ . They are mutually singular, written $μ ⊥ ν,$ when there are measurable sets $P, N \in A$ with $P \cap N = \emptyset, P \cup N = X,$ such that:

every measurable subset of $N$ has $μ$ -value $0$ ;
every measurable subset of $P$ has $ν$ -value $0$ .

For positive measures this is equivalent to the usual condition $μ (N) = 0$ and $ν (P) = 0$ , because positivity turns vanishing on the ambient set into vanishing on all measurable subsets.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A subset of a set of finite signed measure also has finite signed measure

Statement

Let $ν$ be a signed measure on $(X, A)$ , let $E \in A$ , and suppose $ν (E) \in R$ . Then every measurable subset $F \subseteq E$ also satisfies $ν (F) \in R$ .

Facts & Assumptions

Given: A signed measure $ν$ , a measurable set $E$ with finite value $ν (E)$ , and a measurable subset $F \subseteq E$ .

[L1]

A signed measure takes at most one infinite sign and is additive on disjoint measurable unions. (A signed measure is countably additive and takes at most one infinite value)

Proof

technique · direct

1.1L1

The sets $F$ and $E ∖ F$ are disjoint and have union $E$ , so [L1] gives $ν (E) = ν (F) + ν (E ∖ F) .$

2.1L1step 1.1

If $ν (F) = + \infty$ , then the at-most-one-infinite-sign clause in [L1] [L1, step 1.1] forces $ν (E ∖ F) \neq - \infty$ , so the right side of step 1.1 is $+ \infty$ , contradicting the finiteness of $ν (E)$ . The same argument with the signs reversed rules out $ν (F) = - \infty$ . Therefore $ν (F) \in R$ .

3.1step 2.1

The subset $F$ was arbitrary, so every measurable subset of $E$ has finite [step 2.1] ∎ signed measure.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

If a disjoint union has finite signed measure, then the signed-measure series converges absolutely

Statement

Let $ν$ be a signed measure on $(X, A)$ and let $(E_{n})_{n \in N}$ be pairwise disjoint measurable sets. If $ν (⋃_{n \in N} E_{n}) \in R,$ then the real series $\sum_{n = 0}^{\infty} ν (E_{n})$ converges absolutely.

Facts & Assumptions

Given: A signed measure $ν$ , a pairwise disjoint measurable sequence $(E_{n})$ , and the finite value $ν (⋃_{n} E_{n}) \in R$ .

[L1]

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

[L2]

A signed measure is countably additive on every disjoint measurable sequence. (A signed measure is countably additive and takes at most one infinite value)

[L3]

Unconditional convergence of a real series means that every rearrangement converges to the same sum. (Rearrangement of a series along a bijection of $N$ , and unconditional convergence)

[L4]

For a series of real numbers, unconditional convergence is equivalent to absolute convergence. (For a series of real numbers, unconditional convergence and absolute convergence are the same property)

Proof

technique · direct

1.1L1L3L4

Put $E = ⋃_{n} E_{n}$ . Each $E_{n} \subseteq E$ , so [L1] makes every [L1, L3, L4] $ν (E_{n})$ a real number. Thus $\sum_{n} ν (E_{n})$ is a real series to which [L3] and [L4] apply.

1.2L2L3

Let $σ : N \to N$ be a bijection. The sequence [L2, L3] $(E_{σ (n)})$ is again pairwise disjoint and has the same union $E$ , so [L2] gives $\sum_{n = 0}^{\infty} ν (E_{σ (n)}) = ν (E) = \sum_{n = 0}^{\infty} ν (E_{n}) .$ Hence the series is unconditionally convergent in the sense of [L3].

2.1L4step 1.1step 1.2∎

Step 1.2 and [L4] imply that $\sum_{n} ν (E_{n})$ converges absolutely.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)Open item page →

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 $ν (P) \geq ν (A)$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)Open item page →

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

Statement

Assume the Axiom of Choice. 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 $ν$ .

Separately, without any choice axiom, if two such decompositions $(P, N)$ and $(P^{'}, N^{'})$ are given, then $P △ P^{'}$ is null for $ν$ and hence has total variation $0$ . This conditional comparison does not assume the general existence assertion.

Facts & Assumptions

Given: A signed measure $ν$ on $(X, A)$ in the sense of A signed measure is countably additive and takes at most one infinite value, and the Axiom of Choice (The Axiom of Choice).

[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]

Under AC, 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

1.1givenL1choose

A signed measure omits at least one infinite sign. Replacing $ν$ by $- ν$ swaps positive and negative sets, so first work in the case $ν (E) < + \infty$ for every measurable $E$ . Put $m = \sup {ν (A) : A \in A, A positive} \in [0, + \infty]$ ; the empty set is admissible. If $m < + \infty$ , for every $n \geq 1$ the family of positive sets with $ν (A) > m - 1 / n$ is nonempty. If $m = + \infty$ , use instead the nonempty family with $ν (A) > n$ . AC selects $A_{n}$ from these families simultaneously. Set $P = ⋃_{n \geq 1} A_{n}$ . This is the optimizing-sequence use of AC; the positive-subset supplier [L2] has its separate stated AC use.

2.1L1step 1.1algebra

For measurable $E \subseteq P$ , set $B_{n} = E \cap (A_{n} ∖ ⋃_{1 \leq j < n} A_{j})$ . These are disjoint measurable subsets of positive sets, with union $E$ . Thus countable additivity gives $ν (E) = \sum_{n} ν (B_{n}) \geq 0$ . Hence $P$ is positive and $ν (P)$ is finite in the chosen sign convention. For each $n$ , positivity of $P ∖ A_{n}$ gives $ν (P) \geq ν (A_{n})$ . The branch $m = + \infty$ would imply $ν (P) > n$ for every $n$ , impossible. Therefore $m$ is finite and $m - 1 / n < ν (A_{n}) \leq ν (P) \leq m$ , whence $ν (P) = m$ .

3.1L1L2step 2.1algebra

Put $N = X ∖ P$ . If a measurable $E \subseteq N$ had $ν (E) > 0$ , its value would be finite by the chosen sign convention. By [L2] it would contain a positive subset $Q$ with $ν (Q) \geq ν (E) > 0$ . Disjoint positive sets have positive union, by splitting each measurable subset between them; hence $P \cup Q$ would be positive and $ν (P \cup Q) = m + ν (Q) > m$ , a contradiction. Thus every measurable subset of $N$ has nonpositive value, so $N$ is negative.

4.1L1L3step 3.1algebra∎

If the original sign was reversed, interchange the two sets just constructed; they then form a positive/negative partition for the original measure. For any two decompositions of that measure, $P ∖ P^{'}$ lies in both a positive set $P$ and a negative set $N^{'}$ , and similarly $P^{'} ∖ P$ lies in $P^{'} \cap N$ . Every measurable subset of either difference therefore has value zero. Splitting a measurable subset of their disjoint union into its two parts proves that $P △ P^{'}$ is null. By [L3] its total variation is zero. This uniqueness argument itself makes no sequence choices.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)Open item page →

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

Statement

Assume the Axiom of Choice. Let $ν$ be a signed measure on $(X, A)$ . Then there exist positive measures $ν^{+}, ν^{-} : (X, A) \to [0, + \infty]$ , at least one finite on $X$ , 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 $η = ν^{-}$ .

The following conditional assertion is choice-free: if a Hahn partition $X = P ⊔ N$ is supplied, the formulas $ν^{+} (E) = ν (E \cap P)$ and $ν^{-} (E) = - ν (E \cap N)$ construct these unique measures. Comparing two supplied mutually singular positive decompositions also requires no AC.

Facts & Assumptions

Given: A signed measure $ν$ on $(X, A)$ (A signed measure is countably additive and takes at most one infinite value). Assume AC for the general existence assertion (The Axiom of Choice); for the conditional assertion a Hahn partition is supplied.

[L1]

Under AC, 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

1.1givenL1L3

For the general existence assertion, obtain a Hahn partition from [L1]; this invokes AC through the Hahn theorem's optimizing choices and its positive-subset lemma. For the conditional assertion start with the supplied partition instead. Define $ν^{+} (E) = ν (E \cap P)$ and $ν^{-} (E) = - ν (E \cap N)$ . The signs follow from Positive, negative, and null sets for a signed measure, and countable additivity and the empty-set value follow from those of $ν$ , so these are positive measures by [L3].

2.1step 1.1L2algebra

At least one of $ν^{+} (X) = ν (P)$ and $ν^{-} (X) = - ν (N)$ is finite, since otherwise $ν$ would take both infinite signs. By positivity the same side is finite on every measurable $E$ . Thus the difference is always defined, and finite additivity gives $ν (E) = ν^{+} (E) - ν^{-} (E)$ . The defining formulas give $ν^{+} (N) = ν^{-} (P) = 0$ , hence mutual singularity by [L2].

3.1step 2.1L2algebra

Suppose $ν = μ - η$ with positive mutually singular measures, with the displayed difference defined on every measurable set. Take one witnessing partition $P^{'}, N^{'}$ from [L2]. On subsets of $P^{'}$ one has $ν = μ \geq 0$ , and on subsets of $N^{'}$ one has $ν = - η \leq 0$ . Thus this is another Hahn partition. A measurable subset of $P ∖ P^{'}$ lies in positive $P$ and negative $N^{'}$ , so has value zero; the same holds on $P^{'} ∖ P$ . Splitting between these differences shows they are null sets. This comparison uses only the supplied partitions, not Hahn existence or further AC.

4.1step 1.1step 3.1algebra∎

For every measurable $E$ , $μ (E) = μ (E \cap P^{'}) = ν (E \cap P^{'}) = ν (E \cap P) = ν^{+} (E)$ : the first equality uses vanishing on $N^{'}$ , and the middle change of sets discards only the null differences from step 3.1. Similarly $η (E) = - ν (E \cap N) = ν^{-} (E)$ . There is no subtraction of two infinite quantities. Hence the parts are unique; steps 1.1–4.1 also prove the choice-free conditional assertion because AC was used only to obtain the initial arbitrary Hahn partition.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Continuity from below, and from above when one set has finite signed measure

Statement

Let $ν$ be a signed measure on $(X, A)$ .

If $E_{0} \subseteq E_{1} \subseteq \dots$ are measurable and $E = ⋃_{n} E_{n}$ , then $ν (E) = \lim_{n \to \infty} ν (E_{n}) .$
If $E_{0} \supseteq E_{1} \supseteq \dots$ are measurable, $E = ⋂_{n} E_{n}$ , and $ν (E_{n_{0}}) \in R$ for some $n_{0}$ , then $ν (E) = \lim_{n \to \infty} ν (E_{n}) .$

Facts & Assumptions

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

[L1]

Jordan decomposition gives positive measures $ν^{+}, ν^{-}$ with $ν = ν^{+} - ν^{-}$ . (Jordan decomposition of a signed measure into unique mutually singular positive parts)

[L2]

Measures are continuous from below on increasing measurable sequences. (Continuity from below for measures)

[L3]

Measures are continuous from above on decreasing measurable sequences once one term has finite measure. (Continuity from above when one set has finite measure)

[L4]

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

Proof

technique · direct

1.1

Let $E_{n} ↑ E$ . By [L1], write $ν = ν^{+} - ν^{-}$ . Then [L2] gives $ν^{+} (E) = \lim_{n} ν^{+} (E_{n}), ν^{-} (E) = \lim_{n} ν^{-} (E_{n}) .$ Subtracting these two equalities yields $ν (E) = ν^{+} (E) - ν^{-} (E) = \lim_{n} (ν^{+} (E_{n}) - ν^{-} (E_{n})) = \lim_{n} ν (E_{n}) .$

1.2L1L3L4

Let $E_{n} ↓ E$ and assume $ν (E_{n_{0}}) \in R$ for some $n_{0}$ . Choose a Hahn decomposition $X = P ⊔ N$ from [L1]. Because $E_{n_{0}} \cap P$ and $E_{n_{0}} \cap N$ are measurable subsets of the finite signed-measure set $E_{n_{0}}$ , [L4] shows that both $ν (E_{n_{0}} \cap P) = ν^{+} (E_{n_{0}})$ and $ν (E_{n_{0}} \cap N) = - ν^{-} (E_{n_{0}})$ are finite. Hence [L3] gives $ν^{+} (E) = \lim_{n} ν^{+} (E_{n}), ν^{-} (E) = \lim_{n} ν^{-} (E_{n}) .$ Subtracting again yields $ν (E) = \lim_{n} ν (E_{n})$ .

2.1L1step 1.1

The displayed subtractions are defined because for a signed measure at most one of $ν^{+} (E)$ and $ν^{-} (E)$ can be infinite, and the same holds for each $E_{n}$ .

3.1step 1.1step 1.2step 2.1∎

Steps 1.1 through 2.1 prove continuity from below and from above under the stated finiteness hypothesis.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

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

1.1L2L1

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 $∣ ν ∣ (E) \leq ν^{+} (E) + ν^{-} (E)$ .

2.1L2L1step 1.1

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.

3.1L2step 2.1

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 $\inf_{F \subseteq E} ν (F) = - ν^{-} (E) .$

4.1step 2.1step 3.1∎

Steps 2.1 and 3.1 prove all four displayed formulas.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Every complex measure has finite total variation

Statement

If $ν$ is a complex measure on $(X, A)$ , then $∣ ν ∣ (X) < + \infty$ . More generally, $∣ ν ∣ (E) < + \infty$ for every measurable $E$ .

Facts & Assumptions

Given: A complex measure $ν$ on $(X, A)$ 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]

The set functions $α : = Re ν$ and $β : = Im ν$ are finite signed measures and $ν = α + i β$ . (The real and imaginary parts of a complex measure are finite signed measures, and nu = Re nu + i Im nu)

[L3]

For a signed measure $ρ$ with Jordan parts $ρ^{+}, ρ^{-}$ , $∣ ρ ∣ = ρ^{+} + ρ^{-}$ . (For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal)

Proof

technique · direct

1.1L2algebra

Put $α : = Re ν$ and $β : = Im ν$ . By [L2], these are finite signed measures and $∣ ν (A) ∣ = ∣ α (A) + i β (A) ∣ \leq ∣ α (A) ∣ + ∣ β (A) ∣$ for every measurable $A$ .

2.1L1L3step 1.1

Let $(E_{n})$ be any countable measurable partition of $E$ . Step 1.1 and the one-piece lower bound in the definition of variation give $\sum_{n} ∣ ν (E_{n}) ∣ \leq \sum_{n} ∣ α (E_{n}) ∣ + \sum_{n} ∣ β (E_{n}) ∣ \leq \sum_{n} ∣ α ∣ (E_{n}) + \sum_{n} ∣ β ∣ (E_{n}) .$ By [L3], $∣ α ∣ = α^{+} + α^{-}$ and $∣ β ∣ = β^{+} + β^{-}$ , so countable additivity of the four positive Jordan parts turns the right side into $∣ α ∣ (E) + ∣ β ∣ (E)$ .

3.1L1L2L3step 2.1∎

The quantities $∣ α ∣ (E)$ and $∣ β ∣ (E)$ are finite by [L2] and [L3]: both Jordan parts of a finite signed measure are finite on $E$ . Thus step 2.1 gives the partition-independent bound $\sum_{n} ∣ ν (E_{n}) ∣ \leq ∣ α ∣ (E) + ∣ β ∣ (E) < + \infty .$ Taking the supremum over all countable measurable partitions in [L1] proves $∣ ν ∣ (E) < + \infty$ . Applying this with $E = X$ gives $∣ ν ∣ (X) < + \infty$ .

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)Open item page →

The total variation of a signed or complex measure is a positive measure

Statement

Let $ν$ be a signed measure or a complex measure on $(X, A)$ . Then $∣ ν ∣$ is a measure on $(X, A)$ .

Facts & Assumptions

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

[L1]

The total variation $∣ ν ∣ (E)$ is defined by a supremum of nonnegative partition sums over countable measurable partitions of $E$ . (The total variation |nu|(E) from countable measurable partitions)

[L2]

A measure is a nonnegative set function with value $0$ at $\emptyset$ and countable additivity on pairwise disjoint measurable families. (Measures on sigma-algebras)

[L3]

A finite family of nonempty sets has a choice function (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1L1L2

The set function $∣ ν ∣$ is nonnegative by [L1]. Also $∣ ν ∣ (\emptyset) = 0$ , because every member of any countable measurable partition of $\emptyset$ is empty and its partition sum is $0$ .

1.2L1L3construct

Let $(E_{m})$ be pairwise disjoint measurable sets and put $E = ⋃_{m} E_{m}$ . Fix $N$ and suppose $∣ ν ∣ (E_{m}) < + \infty$ for $m \leq N$ . For any $ε > 0$ , the definition of each supremum gives a countable partition of $E_{m}$ whose sum exceeds $∣ ν ∣ (E_{m}) - ε / (N + 1)$ . Choose these finitely many partitions using [L3], enumerate all their cells in one countable partition, and append $E ∖ ⋃_{m \leq N} E_{m}$ as one more cell. This is a countable measurable partition of $E$ , so [L1] gives $∣ ν ∣ (E) \geq \sum_{m \leq N} ∣ ν ∣ (E_{m}) - ε$ . Let $ε ↓ 0$ . If some $∣ ν ∣ (E_{m}) = + \infty$ , for each finite threshold choose one partition of that $E_{m}$ with sum above the threshold and append $E ∖ E_{m}$ ; then $∣ ν ∣ (E) = + \infty$ . Hence in all cases $∣ ν ∣ (E) \geq \sum_{m \leq N} ∣ ν ∣ (E_{m})$ for every $N$ , and taking the supremum of finite partial sums gives $∣ ν ∣ (E) \geq \sum_{m} ∣ ν ∣ (E_{m})$ . No partitions were chosen across infinitely many $m$ .

1.3L1algebra

Conversely, let $(B_{j})$ be a countable measurable partition of $E$ . For each $j$ , countable additivity of $ν$ on $B_{j} = ⨆_{m} (B_{j} \cap E_{m})$ and the triangle inequality for finite partial sums and their limits give $∣ ν (B_{j}) ∣ \leq \sum_{m} ∣ ν (B_{j} \cap E_{m}) ∣$ . For each $m$ , the cells $(B_{j} \cap E_{m})_{j}$ partition $E_{m}$ , so $\sum_{j} ∣ ν (B_{j} \cap E_{m}) ∣ \leq ∣ ν ∣ (E_{m})$ by [L1]. Summing the first inequality over $j$ and interchanging two nonnegative series yields $\sum_{j} ∣ ν (B_{j}) ∣ \leq \sum_{m} ∣ ν ∣ (E_{m})$ . Taking the supremum over all partitions $(B_{j})$ gives $∣ ν ∣ (E) \leq \sum_{m} ∣ ν ∣ (E_{m})$ .

2.1L2step 1.1step 1.2step 1.3∎

Steps 1.2 and 1.3 prove countable additivity. Together with step 1.1 and [L2], this shows that $∣ ν ∣$ is a measure.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-30Open item page →

Complex simple functions as finite sums of measurable indicators

Definition

Let $(X, A)$ be a measurable space. A function $s : X \to C$ is a complex simple function when there are pairwise disjoint measurable sets $E_{1}, \dots, E_{m} \in A$ and coefficients $c_{1}, \dots, c_{m} \in C$ such that $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}} .$

Equivalently, $s$ is measurable and has finite range. The disjoint representation above can always be taken to be the canonical one given by the nonempty level sets of $s$ .

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

The simple integral against a signed or complex measure

Definition

Let $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ be the canonical disjoint representation of a complex simple function on $(X, A)$ using only its nonzero level sets, so every $c_{j} \neq 0$ . Let $ν$ be a signed measure or a complex measure on $(X, A)$ .

Assume that $∣ ν ∣ (E_{j}) < + \infty$ for every $j$ .

Define the simple integral of $s$ against $ν$ by $\int s d ν : = \sum_{j = 1}^{m} c_{j} ν (E_{j}) .$ If $F \in A$ is measurable and $∣ ν ∣ (E_{j} \cap F) < + \infty$ for every $j$ , define likewise $\int_{F} s d ν : = \sum_{j = 1}^{m} c_{j} ν (E_{j} \cap F) .$

The finiteness hypotheses make every $ν (E_{j})$ and $ν (E_{j} \cap F)$ a finite real or complex number: the one-piece partition of the relevant set contributes at least its single term to the defining supremum for total variation. Because the canonical representation is unique up to deleting empty level sets, the value above is well defined.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Simple integrals are bounded by total variation

Statement

Let $ν$ be a signed measure or a complex measure on $(X, A)$ , let $F \in A$ , and let $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ be the canonical disjoint representation of a complex simple function using only its nonzero level sets. Assume $∣ ν ∣ (E_{j} \cap F) < + \infty$ for every $j$ . Then $∣ \int_{F} s d ν ∣ \leq \int_{F} ∣ s ∣ d ∣ ν ∣ .$ In particular, if $∣ s ∣ \leq M$ on $F$ and $∣ ν ∣ (F) < + \infty$ , then $∣ \int_{F} s d ν ∣ \leq M ∣ ν ∣ (F) .$

Facts & Assumptions

Given: A signed measure or complex measure $ν$ , a measurable set $F$ , and the canonical nonzero-level-set representation $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ of a complex simple function, with $∣ ν ∣ (E_{j} \cap F) < + \infty$ for every $j$ .

[L1]

The simple integral against $ν$ is computed from a disjoint measurable level-set representation. (The simple integral against a signed or complex measure)

[L2]

The integral of a nonnegative simple function against a positive measure is the weighted sum over a disjoint representation. (The integral of a nonnegative simple function)

[L3]

The total variation $∣ ν ∣$ is a measure. (The total variation of a signed or complex measure is a positive measure)

Proof

technique · direct

1.1L1

Write the canonical disjoint representation of $s$ as [L1] $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ . For each $j$ , the one-piece partition of $E_{j} \cap F$ gives $∣ ν (E_{j} \cap F) ∣ \leq ∣ ν ∣ (E_{j} \cap F) < + \infty$ , so [L1] makes $\int_{F} s d ν$ well defined and gives $\int_{F} s d ν = \sum_{j = 1}^{m} c_{j} ν (E_{j} \cap F) .$ By the triangle inequality, $∣ \int_{F} s d ν ∣ \leq \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν (E_{j} \cap F) ∣ \leq \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν ∣ (E_{j} \cap F) .$

2.1L2L3step 1.1

Because $∣ ν ∣$ is a measure by [L3], the sets $E_{j} \cap F$ are disjoint [L2, L3, step 1.1] and measurable, and [L2] gives $\int_{F} ∣ s ∣ d ∣ ν ∣ = \sum_{j = 1}^{m} ∣ c_{j} ∣ ∣ ν ∣ (E_{j} \cap F) .$ Substituting this into step 1.1 proves the first inequality. If $∣ s ∣ \leq M$ on $F$ and $∣ ν ∣ (F) < + \infty$ , then [L3] gives $∣ ν ∣ (E_{j} \cap F) \leq ∣ ν ∣ (F) < + \infty$ for every $j$ , so the displayed finiteness hypothesis is automatic. Moreover $∣ s ∣ 1_{F} \leq M 1_{F}$ , so monotonicity of the simple integral with respect to the positive measure $∣ ν ∣$ gives $\int_{F} ∣ s ∣ d ∣ ν ∣ \leq M ∣ ν ∣ (F)$ .

3.1step 1.1step 2.1∎

The displayed inequalities follow from steps 1.1 and 2.1.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Total variation is the supremum of simple integrals over unit-bounded test functions

Statement

Let $ν$ be a signed measure or complex measure on $(X, A)$ and let $E \in A$ satisfy $∣ ν ∣ (E) < + \infty$ . Then every complex simple function $s$ on $(X, A)$ with $∣ s ∣ \leq 1$ has a defined simple integral over $E$ , and $∣ ν ∣ (E) = \sup {∣ \int_{E} s d ν ∣ : s is a complex simple function and ∣ s ∣ \leq 1} .$

Facts & Assumptions

Given: A signed measure or complex measure $ν$ and a measurable set $E$ with $∣ ν ∣ (E) < + \infty$ .

[L1]

Simple integrals are bounded by total variation: $∣ \int_{E} s d ν ∣ \leq \int_{E} ∣ s ∣ d ∣ ν ∣$ . (Simple integrals are bounded by total variation)

[L2]

The simple integral over $E$ is computed from the measurable level-set representation of $s$ . (The simple integral against a signed or complex measure)

[L3]

Every complex number $z \neq 0$ has unit-modulus phase $\overline{z} / ∣ z ∣$ . (Real and imaginary parts, complex conjugation, and modulus)

[L4]

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

Proof

technique · direct

1.1L1L4

Let $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ be the canonical disjoint representation of a complex simple function using only its nonzero level sets. Every countable measurable partition of $E_{j} \cap E$ extends to one of $E$ by adding $E ∖ E_{j}$ , so [L4] gives $∣ ν ∣ (E_{j} \cap E) \leq ∣ ν ∣ (E) < + \infty$ for each $j$ . Thus [L1] applies to $F = E$ and gives [L1] $∣ \int_{E} s d ν ∣ \leq \int_{E} ∣ s ∣ d ∣ ν ∣ \leq ∣ ν ∣ (E) .$ Therefore the displayed supremum is at most $∣ ν ∣ (E)$ .

1.2L3L4choose

Fix $ε > 0$ . By [L4], choose a countable measurable partition [L3, L4, choose] $E = ⨆_{n \geq 0} E_{n}$ such that $\sum_{n = 0}^{\infty} ∣ ν (E_{n}) ∣ > ∣ ν ∣ (E) - ε .$ Because $∣ ν ∣ (E) < + \infty$ , every term $∣ ν (E_{n}) ∣$ is finite. Choose $N$ so that the first $N + 1$ terms already satisfy $\sum_{n = 0}^{N} ∣ ν (E_{n}) ∣ > ∣ ν ∣ (E) - 2 ε .$ For each $0 \leq n \leq N$ with $ν (E_{n}) \neq 0$ , define $c_{n} : = \overline{ν (E_{n})} / ∣ ν (E_{n}) ∣$ , and put $c_{n} : = 0$ when $ν (E_{n}) = 0$ . After deleting the zero-coefficient terms, the simple function $s : = \sum_{n = 0}^{N} c_{n} 1_{E_{n}}$ satisfies $∣ s ∣ \leq 1$ .

2.1L2step 1.2

Using [L2], [L2, step 1.2] $\int_{E} s d ν = \sum_{n = 0}^{N} c_{n} ν (E_{n}) = \sum_{n = 0}^{N} ∣ ν (E_{n}) ∣,$ so step 1.2 gives $∣ \int_{E} s d ν ∣ > ∣ ν ∣ (E) - 2 ε .$ Because $ε > 0$ was arbitrary, the supremum is at least $∣ ν ∣ (E)$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the equality.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Every L^1 function admits dominated complex simple approximations

Statement

Let $(X, A, μ)$ be a measure space and let $f \in L^{1} (μ)$ . Then there exists a sequence $(s_{n})$ of complex simple functions such that:

$∣ s_{n} ∣ \leq 2 ∣ f ∣$ for every $n$ ;
$\int ∣ f - s_{n} ∣ d μ \to 0$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and an integrable function $f : X \to C$ .

[L1]

An integrable complex function has measurable real and imaginary parts and integrable modulus. (Integrable real and complex functions, and their integrals)

[L2]

The positive and negative parts satisfy $u = u^{+} - u^{-}$ and $∣ u ∣ = u^{+} + u^{-}$ . (The positive and negative parts of a function)

[L3]

Arithmetic and lattice operations preserve measurability. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[L4]

Every nonnegative measurable function admits increasing simple approximations. (Every nonnegative measurable function admits an explicit increasing sequence of simple approximations)

[L5]

Monotone convergence passes increasing limits through the integral. (Monotone convergence for the integral)

Proof

technique · direct

1.1L1L2L3

Write $f = u + i v$ with $u = Re f$ and $v = Im f$ . [L1, L2, L3] By [L1], the real functions $u$ and $v$ are measurable and satisfy $∣ u ∣ \leq ∣ f ∣$ and $∣ v ∣ \leq ∣ f ∣$ , hence are integrable. By [L2] and [L3], the four functions $u^{\pm}$ and $v^{\pm}$ are nonnegative measurable.

2.1L4L2L3step 1.1

Apply [L4] to choose increasing nonnegative simple functions [L2, L3, L4, step 1.1] $u_{n}^{\pm} ↑ u^{\pm}$ and $v_{n}^{\pm} ↑ v^{\pm}$ . Put $u_{n} : = u_{n}^{+} - u_{n}^{-}, v_{n} : = v_{n}^{+} - v_{n}^{-}, s_{n} : = u_{n} + i v_{n} .$ Then each $s_{n}$ is a complex simple function, and $∣ s_{n} ∣ \leq u_{n}^{+} + u_{n}^{-} + v_{n}^{+} + v_{n}^{-} \leq ∣ u ∣ + ∣ v ∣ \leq 2 ∣ f ∣ .$

3.1L5L2step 2.1

By [L5], the four increasing simple approximations in step 2.1 satisfy [L2, L5, step 2.1] $\int u_{n}^{\pm} d μ \to \int u^{\pm} d μ$ and $\int v_{n}^{\pm} d μ \to \int v^{\pm} d μ$ . Therefore $\int ∣ u - u_{n} ∣ d μ = \int (u^{+} - u_{n}^{+}) d μ + \int (u^{-} - u_{n}^{-}) d μ \to 0,$ and similarly $\int ∣ v - v_{n} ∣ d μ \to 0$ . Hence $\int ∣ f - s_{n} ∣ d μ \leq \int ∣ u - u_{n} ∣ d μ + \int ∣ v - v_{n} ∣ d μ \to 0.$

4.1step 2.1step 3.1

Steps 2.1 and 3.1 give the required dominated complex simple [step 2.1, step 3.1] ∎ approximations.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-30Open item page →

Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|)

Definition

Let $ν$ be a signed measure or complex measure on $(X, A)$ . Because $∣ ν ∣$ is a measure (The total variation of a signed or complex measure is a positive measure), define $L^{1} (ν) : = L^{1} (∣ ν ∣)$ using the published meaning of $L^{1}$ from The class $L^{1} (μ)$ of integrable functions.

If $f \in L^{1} (ν)$ , choose complex simple functions $s_{n}$ as in Every L^1 function admits dominated complex simple approximations with $\int ∣ f - s_{n} ∣ d ∣ ν ∣ \to 0$ . The simple-integral bound Simple integrals are bounded by total variation makes $(\int s_{n} d ν)$ a Cauchy sequence in $C$ , and its limit is independent of the chosen approximating sequence. Define $\int f d ν : = \lim_{n \to \infty} \int s_{n} d ν .$

For a measurable set $E \in A$ , define $\int_{E} f d ν : = \int f 1_{E} d ν .$

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Integrals against signed or complex measures are bounded by total variation

Statement

Let $ν$ be a signed measure or complex measure on $(X, A)$ and let $f \in L^{1} (ν)$ . Then $∣ \int f d ν ∣ \leq \int ∣ f ∣ d ∣ ν ∣ .$ More generally, for every measurable $E$ , $∣ \int_{E} f d ν ∣ \leq \int_{E} ∣ f ∣ d ∣ ν ∣ .$

Facts & Assumptions

Given: A signed measure or complex measure $ν$ , a function $f \in L^{1} (ν)$ , and a measurable set $E$ .

[L1]

Integration against $ν$ is defined as the limit of simple integrals along an $L^{1} (∣ ν ∣)$ -approximating sequence. (Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|))

[L2]

Simple integrals satisfy $∣ \int_{E} s d ν ∣ \leq \int_{E} ∣ s ∣ d ∣ ν ∣$ . (Simple integrals are bounded by total variation)

Proof

technique · direct

1.1L1L2

By [L1], choose complex simple functions $s_{n}$ with [L1, L2] $\int ∣ f - s_{n} ∣ d ∣ ν ∣ \to 0$ and $\int f d ν = \lim_{n} \int s_{n} d ν .$ Applying [L2] to $s_{n} - s_{m}$ shows that $(\int s_{n} d ν)$ is Cauchy.

2.1L2L1step 1.1

By [L2], [L1, L2, step 1.1] $∣ \int s_{n} d ν ∣ \leq \int ∣ s_{n} ∣ d ∣ ν ∣ \leq \int ∣ f ∣ d ∣ ν ∣ + \int ∣ f - s_{n} ∣ d ∣ ν ∣ .$ Letting $n \to \infty$ in step 1.1 yields $∣ \int f d ν ∣ \leq \int ∣ f ∣ d ∣ ν ∣ .$ Applying the same argument to $f 1_{E}$ gives the measurable-subset version.

3.1step 2.1∎

Step 2.1 proves both inequalities. [step 2.1].

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

A complex L^1 density defines a complex measure whose total variation is |h| dmu

Statement

Let $(X, A, μ)$ be a measure space and let $h \in L^{1} (μ)$ . Define $ν (E) : = \int_{E} h d μ (E \in A) .$ Then $ν$ is a complex measure on $(X, A)$ , and for every measurable $E$ , $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ .$

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and a function $h \in L^{1} (μ)$ .

[L1]

For an integrable function, the measurable-set integral $\int_{E} h d μ$ is defined. (Integrable real and complex functions, and their integrals, Integral over a measurable subset)

[L2]

If $g \in L^{1} (μ)$ , then $∣ \int g d μ ∣ \leq \int ∣ g ∣ d μ$ . (The modulus of an integral is bounded by the integral of the modulus)

[L3]

Total variation is the supremum of the simple integrals against unit-bounded simple test functions. (Total variation is the supremum of simple integrals over unit-bounded test functions)

[L4]

Every $L^{1}$ function admits dominated complex simple approximations. (Every L^1 function admits dominated complex simple approximations)

[L5]

Arithmetic operations preserve measurability. (Closure properties of measurable functions used by the integral)

[L6]

For a nonnegative measurable function $f$ , the set function $A \mapsto \int_{A} f d μ$ is a measure. (The indefinite integral of a nonnegative measurable function is a measure)

[L7]

The Lebesgue integral is complex-linear on $L^{1} (μ)$ . (The Lebesgue integral is linear on $L^{1} (μ)$ )

Proof

technique · direct

1.1L1L2

The set function $ν$ is finite-valued because $∣ ν (E) ∣ = ∣ \int_{E} h d μ ∣ \leq \int_{E} ∣ h ∣ d μ \leq \int ∣ h ∣ d μ < + \infty$ by [L1] and [L2]. If $(E_{n})$ is a pairwise disjoint measurable sequence, then $\sum_{n = 0}^{N} 1_{E_{n}} ↑ 1_{\cup_{n} E_{n}}$ , so monotone convergence applied to the positive and negative parts of the real and imaginary parts of $h$ gives $ν (⋃_{n} E_{n}) = \sum_{n} ν (E_{n}) .$ Thus $ν$ is a complex measure.

1.2L4L5L6

Define $u (x) : = {\begin{cases} \overline{h (x)} / ∣ h (x) ∣, & h (x) \neq 0, \\ 0, & h (x) = 0. \end{cases}$ Then $∣ u ∣ \leq 1$ and $u h = ∣ h ∣$ . By [L5], the function $u$ is measurable. Apply [L6] to $ρ (F) : = \int_{F} ∣ h ∣ d μ$ and obtain a finite measure $ρ$ on $(X, A)$ . Applying [L4] to $u$ on $(E, A ⁣ ↾_{E}, ρ)$ gives complex simple functions $s_{n}$ with $∣ s_{n} ∣ \leq 2$ and $\int_{E} ∣ u - s_{n} ∣ d ρ \to 0.$ For each $n$ , define the clipped simple function $t_{n} : = \frac{s_{n}}{\max {1, ∣ s_{n} ∣}} .$ Then $∣ t_{n} ∣ \leq 1$ , and because $∣ u ∣ \leq 1$ one has $∣ t_{n} - u ∣ \leq ∣ t_{n} - s_{n} ∣ + ∣ s_{n} - u ∣ \leq 2 ∣ s_{n} - u ∣ .$ Hence $\int_{E} ∣ u - t_{n} ∣ d ρ \to 0$ .

2.1L1step 1.1

For any measurable $E$ and any countable measurable partition $E = ⨆_{n} E_{n}$ , the inequality in step 1.1 applied on each piece gives $\sum_{n} ∣ ν (E_{n}) ∣ \leq \sum_{n} \int_{E_{n}} ∣ h ∣ d μ = \int_{E} ∣ h ∣ d μ .$ Taking the supremum over partitions shows $∣ ν ∣ (E) \leq \int_{E} ∣ h ∣ d μ$ .

2.2L2L7step 1.2

Write the canonical representation of $t_{n}$ on $E$ as $t_{n} = \sum_{j = 1}^{m_{n}} c_{n, j} 1_{A_{n, j}} .$ Because $∣ t_{n} h ∣ \leq ∣ h ∣$ , each $t_{n} h$ lies in $L^{1} (μ)$ . By the definition of $ν$ and the linearity of the Lebesgue integral, $\int_{E} t_{n} d ν = \sum_{j = 1}^{m_{n}} c_{n, j} ν (A_{n, j}) = \sum_{j = 1}^{m_{n}} c_{n, j} \int_{A_{n, j}} h d μ = \int_{E} t_{n} h d μ .$ Therefore $∣ \int_{E} t_{n} d ν - \int_{E} ∣ h ∣ d μ ∣ = ∣ \int_{E} (t_{n} - u) h d μ ∣ \leq \int_{E} ∣ t_{n} - u ∣ ∣ h ∣ d μ = \int_{E} ∣ t_{n} - u ∣ d ρ \to 0.$ So $\int_{E} t_{n} d ν \to \int_{E} ∣ h ∣ d μ$ .

3.1L3step 2.1step 2.2

Since each $t_{n}$ is a unit-bounded complex simple function on $E$ , [L3] gives $∣ ν ∣ (E) \geq ∣ \int_{E} t_{n} d ν ∣$ for every $n$ . Letting $n \to \infty$ and using step 2.2 shows $∣ ν ∣ (E) \geq \int_{E} ∣ h ∣ d μ$ . Together with step 2.1, this proves $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ$ .

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1, and 3.1 prove that $ν$ is a complex measure and that its total variation is $\int_{E} ∣ h ∣ d μ$ on every measurable set $E$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

The space of finite total variation signed measures

Definition

Fix a measurable space $(X, A)$ . Write $M (X, A) : = {ν : ν is a signed measure on (X, A) and ∣ ν ∣ (X) < + \infty} .$

For $ν \in M (X, A)$ , define its variation norm candidate by $∥ ν ∥ : = ∣ ν ∣ (X) .$

The next theorem proves that $M (X, A)$ is a real vector space under pointwise addition and scalar multiplication and that $∥ \cdot ∥$ is a norm in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

The finite-total-variation signed measures form a real normed space

Statement

Fix a measurable space $(X, A)$ . With pointwise addition and scalar multiplication, $M (X, A)$ is a real vector space. Moreover, $∥ ν ∥ : = ∣ ν ∣ (X)$ defines a norm on it.

Facts & Assumptions

Given: A measurable space $(X, A)$ .

[L1]

$M (X, A)$ consists of the signed measures with $∣ ν ∣ (X) < + \infty$ . (The space of finite total variation signed measures)

[L2]

Total variation is the supremum of unit-bounded simple integrals. (Total variation is the supremum of simple integrals over unit-bounded test functions)

[L3]

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

[L4]

A normed space is a real vector space together with a norm satisfying separation, absolute homogeneity, and the triangle inequality. (Vector space over a field, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms)

Proof

technique · direct

1.1L1L2L4

If $ν, μ \in M (X, A)$ and $a \in R$ , then the [L1, L2, L4] pointwise set functions $ν + μ$ and $a ν$ are again signed measures because their values are finite on every measurable set and countable additivity is preserved termwise. Also [L2] gives $∣ ν + μ ∣ (X) \leq ∣ ν ∣ (X) + ∣ μ ∣ (X), ∣ a ν ∣ (X) = ∣ a ∣ ∣ ν ∣ (X),$ so $ν + μ$ and $a ν$ remain in $M (X, A)$ . Thus $M (X, A)$ is closed under the pointwise operations, and the vector-space axioms are inherited from the real-valued function space on $A$ .

1.2L1L3L4

The formula $∥ ν ∥ = ∣ ν ∣ (X)$ is nonnegative by definition. If $∥ ν ∥ = 0$ , [L1, L3, L4] then [L3] makes $X$ null for $ν$ , so every measurable set has $ν$ -value $0$ and therefore $ν$ is the zero measure. Conversely the zero measure has variation $0$ . Thus the separation axiom of [L4] holds.

2.1L2L4step 1.1

Step 1.1 already proved absolute homogeneity and the triangle inequality: [L2, L4, step 1.1] $∥ a ν ∥ = ∣ a ∣ ∥ ν ∥, ∥ ν + μ ∥ \leq ∥ ν ∥ + ∥ μ ∥ .$ Hence [L4] shows that $∥ \cdot ∥$ is a norm.

3.1step 1.1step 1.2step 2.1

Steps 1.1 through 2.1 prove that $M (X, A)$ is a real [step 1.1, step 1.2, step 2.1] ∎ normed space.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30Open item page →

Finite-total-variation signed measures are complete

Statement

Fix a measurable space $(X, A)$ . The normed space $M (X, A)$ of finite-total-variation signed measures is complete.

Facts & Assumptions

Given: A Cauchy sequence $(ν_{n})$ in $M (X, A)$ .

[L1]

The norm on $M (X, A)$ is $∥ ν ∥ = ∣ ν ∣ (X)$ . (The space of finite total variation signed measures, The finite-total-variation signed measures form a real normed space)

[L2]

If $∣ ρ ∣ (X)$ is finite, then $∣ ρ (E) ∣ \leq ∣ ρ ∣ (E) \leq ∣ ρ ∣ (X)$ for every measurable $E$ . (Total variation is the supremum of simple integrals over unit-bounded test functions)

[L3]

Positive measures are continuous from above on decreasing measurable sets once one term has finite measure. (Continuity from above when one set has finite measure)

Proof

technique · direct

1.1L1L2

By [L1] and [L2], for every measurable $E$ the scalar sequence $(ν_{n} (E))$ is Cauchy in $R$ , because $∣ ν_{n} (E) - ν_{m} (E) ∣ \leq ∣ ν_{n} - ν_{m} ∣ (X) = ∥ ν_{n} - ν_{m} ∥ .$ Define $ν (E) : = \lim_{n} ν_{n} (E)$ .

2.1L1L2L3step 1.1

Let $(E_{k})$ be pairwise disjoint and put $E = ⋃_{k} E_{k}$ . Fix $ε > 0$ . Choose $m$ so that $∥ ν_{n} - ν_{m} ∥ < ε$ for all $n \geq m$ . Because $∣ ν_{m} ∣$ is a finite positive measure, [L3] gives $∣ ν_{m} ∣ (⋃_{k \geq N} E_{k}) \to 0$ , so choose $N$ with that tail below $ε$ . Then for $n \geq m$ , $∣ ν_{n} (E) - \sum_{k = 0}^{N - 1} ν_{n} (E_{k}) ∣ = ∣ ν_{n} (⋃_{k \geq N} E_{k}) ∣ \leq ∥ ν_{n} - ν_{m} ∥ + ∣ ν_{m} ∣ (⋃_{k \geq N} E_{k}) < 2 ε .$ Passing $n \to \infty$ gives $∣ ν (E) - \sum_{k = 0}^{N - 1} ν (E_{k}) ∣ \leq 2 ε,$ so $ν (E) = \sum_{k} ν (E_{k})$ . Thus $ν$ is a signed measure.

3.1L1step 1.1step 2.1

For any countable measurable partition $(A_{j})$ of a measurable set $F$ , Fatou's lemma for nonnegative series gives $\sum_{j} ∣ ν (A_{j}) ∣ \leq {lim inf}_{n \to \infty} \sum_{j} ∣ ν_{n} (A_{j}) ∣ \leq \sup_{n} ∣ ν_{n} ∣ (F),$ so $∣ ν ∣ (F) < + \infty$ and therefore $ν \in M (X, A)$ . Likewise, for fixed $n$ and any partition $(A_{j})$ of $X$ , $\sum_{j} ∣ ν_{n} (A_{j}) - ν (A_{j}) ∣ \leq {lim inf}_{m \to \infty} \sum_{j} ∣ ν_{n} (A_{j}) - ν_{m} (A_{j}) ∣ \leq {lim inf}_{m \to \infty} ∥ ν_{n} - ν_{m} ∥ .$ Taking the supremum over partitions gives $∥ ν_{n} - ν ∥ \leq {lim inf}_{m \to \infty} ∥ ν_{n} - ν_{m} ∥,$ and the right side tends to $0$ because $(ν_{n})$ is Cauchy.

4.1step 3.1∎

Step 3.1 shows that $ν_{n} \to ν$ in norm, so every Cauchy sequence in $M (X, A)$ converges there.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)Open item page →

A real L^1 density defines a finite signed measure with its canonical Hahn and Jordan data

Statement

Let $(X, A, μ)$ be a measure space and let $f \in L^{1} (μ)$ be real-valued. Define $ν (E) : = \int_{E} f d μ (E \in A) .$ Then $ν$ is a finite signed measure. Its canonical Hahn sets are $P : = {f > 0}, N : = {f \leq 0},$ its Jordan parts are $ν^{+} (E) = \int_{E} f^{+} d μ, ν^{-} (E) = \int_{E} f^{-} d μ,$ and its total variation is $∣ ν ∣ (E) = \int_{E} ∣ f ∣ d μ .$

These explicit constructions require no choice axiom. Hahn partitions are unique only up to signed-null sets; the Jordan measures are unique.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and a real-valued function $f \in L^{1} (μ)$ .

[L1]

A nonnegative measurable density defines a positive measure. (The indefinite integral of a nonnegative measurable function is a measure)

[L2]

For a real integrable function, the positive and negative parts satisfy $f = f^{+} - f^{-}$ and $∣ f ∣ = f^{+} + f^{-}$ . (Integrable real and complex functions, and their integrals)

[L3]

Arithmetic and threshold operations preserve measurability. (Closure properties of measurable functions used by the integral)

[L4]

Given Hahn decompositions, their positive pieces differ by a signed-null set; given a Hahn partition, its Jordan measures are constructed by restriction and are unique. These are the separately stated choice-free conditional clauses, not the AC-qualified general existence assertions. (Hahn decomposition for signed measures, unique up to total-variation-null sets, Jordan decomposition of a signed measure into unique mutually singular positive parts)

[L5]

The nonnegative integral is additive. (Additivity of the nonnegative Lebesgue integral)

[L6]

Signed measures, positive/negative sets and partition-defined total variation have their usual local meanings. (A signed measure is countably additive and takes at most one infinite value, Positive, negative, and null sets for a signed measure, The total variation |nu|(E) from countable measurable partitions)

Proof

technique · direct

1.1givenL1L2L3L6algebra

Put $α (E) = \int_{E} f^{+} d μ$ and $β (E) = \int_{E} f^{-} d μ$ . By [L1]–[L3] these are positive measures, and their total masses are finite because $f$ is integrable. By [L2], $ν (E) = α (E) - β (E)$ is finite. For disjoint measurable $E_{n}$ , the series $\sum_{n} α (E_{n})$ and $\sum_{n} β (E_{n})$ have finite sums, so they can be subtracted termwise to obtain countable additivity of $ν$ ; moreover $\sum_{n} ∣ ν (E_{n}) ∣ \leq α (X) + β (X) < \infty$ . Also $ν (\emptyset) = 0$ . Thus [L6] makes $ν$ a finite signed measure.

2.1L3L4L6step 1.1

By [L3], $P = {f > 0}$ and $N = {f \leq 0}$ are measurable and partition $X$ . The density $f^{-}$ vanishes on $P$ and $f^{+}$ vanishes on $N$ , so on measurable subsets of $P$ one has $ν = α \geq 0$ , and on subsets of $N$ one has $ν = - β \leq 0$ . These signs prove directly that $(P, N)$ is a Hahn partition. Only the conditional comparison in [L4] is needed to say that every other supplied Hahn partition differs by a signed-null set.

3.1L4step 1.1step 2.1algebra

Since $α$ vanishes on $N$ and $β$ vanishes on $P$ , the conditional construction in [L4] identifies them with the unique Jordan measures: $ν (E \cap P) = α (E)$ and $- ν (E \cap N) = β (E)$ . No arbitrary Hahn partition is selected.

4.1L2L5L6step 2.1step 3.1algebra∎

For any countable measurable partition $E = ⨆_{j} E_{j}$ , one has $\sum_{j} ∣ ν (E_{j}) ∣ \leq \sum_{j} (α (E_{j}) + β (E_{j})) = α (E) + β (E)$ . Conversely, the explicit partition $E \cap P, E \cap N$ (with empty later parts) has variation sum exactly $α (E) + β (E)$ by step 2.1. Taking the supremum in [L6] proves $∣ ν ∣ (E) = α (E) + β (E) = \int_{E} ∣ f ∣ d μ$ by [L2] and [L5]. Together with steps 1.1–3.1 this proves every assertion without using the general Hahn/Jordan existence clauses.

5 · Examples, counterexamples and false statements

None yet.