PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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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 $ν (E)$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources