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Lebesgue components of a positive finite measure are positive

Statement

Let $ν$ be a finite positive measure and $μ$ a positive measure on the same measurable space. If $ν = ν_{a c} + ν_{s}$ with $ν_{a c} ≪ μ$ and $ν_{s} ⊥ μ$ , then $ν_{a c}$ and $ν_{s}$ are positive measures.

Facts & Assumptions

Given: A finite positive measure $ν$ , a positive measure $μ$ , and a decomposition $ν = ν_{a c} + ν_{s}$ with $ν_{a c} ≪ μ$ and $ν_{s} ⊥ μ$ .

[F1]

Mutual singularity for signed or complex measures: singularity supplies a measurable carrier $N$ whose measurable subsets are $μ$ -null, while $ν_{s}$ vanishes on every measurable subset of $X ∖ N$ .

[F2]

Absolute continuity of a signed or complex measure with respect to a positive measure: $ν_{a c} ≪ μ$ makes $ν_{a c}$ vanish on every measurable $μ$ -null set.

[F3]

Measures on sigma-algebras and Finite, sigma-finite, and semifinite measures: a measure has nonnegative values on measurable sets, and a finite measure has finite values on all of them.

Proof

technique · direct

1.1F1F2F3givenchoose

Choose the singular carrier $N$ from [F1]. For any measurable $E$ , the singular part vanishes on $E ∖ N$ , so the decomposition gives $ν_{a c} (E ∖ N) = ν (E ∖ N) \geq 0$ . Also $ν_{a c} (E \cap N) = 0$ by [F1] and [F2]. Additivity therefore gives $ν_{a c} (E) = ν_{a c} (E ∖ N) \geq 0$ . All displayed component values are finite because the sum is the finite value $ν$ and the other component is zero on the set being evaluated.

2.1F1F2F3givenalgebra∎

On $E \cap N$ , [F2] gives $ν_{a c} (E \cap N) = 0$ , so the decomposition yields $ν_{s} (E \cap N) = ν (E \cap N) \geq 0$ . The singular part vanishes on $E ∖ N$ by [F1]; hence $ν_{s} (E) = ν_{s} (E \cap N) \geq 0$ . Since $E$ was arbitrary, both components are positive measures.

Depends on

Used by

Likelihood ratio as a Radon–Nikodym derivative Definition
Likelihood-ratio change of measure Theorem

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

Richard F. Bass, Real Analysis for Graduate Students, Theorem 13.5 and proof, §13.3 (book pagination) (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 6.27 and proof, §6.8 (notes pp. 75–77) (standard reference, not scraped)