LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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On a finite-measure space, a bounded functional on $L^{p}$ defines a finite signed measure

Statement

Let $(X, A, μ)$ be a finite measure space, let $1 \leq p < \infty$ , and let $Λ : L^{p} (μ) \to R$ be a bounded linear functional. Define $ν (E) : = Λ ([1_{E}]) (E \in A) .$ Then $ν$ is a finite signed measure on $(X, A)$ .

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ , an exponent $1 \leq p < \infty$ , and a bounded linear functional $Λ : L^{p} (μ) \to R$ .

[L1]

A bounded linear functional on $L^{p} (μ)$ is linear and continuous with respect to the $L^{p}$ norm (A bounded linear functional on $L^{p} (μ)$ and its operator norm).

[L2]

A real-valued countably additive set function with finite values is a signed measure (A signed measure is countably additive and takes at most one infinite value).

[L3]

Dominated convergence applies to integrable majorants (Dominated convergence).

Proof

technique · Define $\nu(E)=\Lambda([\mathbf 1_E])$. Finite measure puts every indicator in $L^p$, and countable additivity comes from $L^p$ convergence of partial indicator sums plus continuity of $\Lambda$

1.1L1given

For every measurable $E \subseteq X$ , $\int ∣ 1_{E} ∣^{p} d μ = μ (E) \leq μ (X) < \infty,$ so $[1_{E}] \in L^{p} (μ)$ and $ν (E)$ is a finite scalar. Also $ν (\emptyset) = Λ (0) = 0$ .

1.2L3givenconstruct

Let $(E_{n})$ be pairwise disjoint measurable sets, and put $E : = ⋃_{n = 1}^{\infty} E_{n}, s_{N} : = \sum_{n = 1}^{N} 1_{E_{n}} = 1_{⋃_{n = 1}^{N} E_{n}} .$ Then $s_{N} \to 1_{E}$ pointwise and $∣ s_{N} - 1_{E} ∣^{p} = 1_{⋃_{n > N} E_{n}} \leq 1_{E} .$ Because $μ (X) < \infty$ , the majorant $1_{E}$ is integrable, so [L3] gives $∥ [s_{N}] - [1_{E}] ∥_{p}^{p} = \int ∣ s_{N} - 1_{E} ∣^{p} d μ ⟶ 0.$

2.1L1step 1.1algebra

If $A, B \in A$ are disjoint, then $1_{A \cup B} = 1_{A} + 1_{B}$ . By linearity of $Λ$ , $ν (A \cup B) = Λ ([1_{A} + 1_{B}]) = Λ ([1_{A}]) + Λ ([1_{B}]) = ν (A) + ν (B) .$ So $ν$ is finitely additive on disjoint measurable sets.

3.1L1L2step 1.1step 1.2step 2.1∎

By continuity of $Λ$ from [L1], finite additivity from step 2.1, and the convergence from step 1.2, $ν (E) = Λ ([1_{E}]) = \lim_{N \to \infty} Λ ([s_{N}]) = \lim_{N \to \infty} \sum_{n = 1}^{N} ν (E_{n}) .$ So $ν$ is countably additive. Together with step 1.1, [L2] shows that $ν$ is a finite signed measure.

Depends on

Used by

Dependency tree · two levels

16 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

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 6.15 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Theorem 15.11 (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 7.14 (standard reference, not scraped)

On a finite-measure space, a bounded functional on LpLp defines a finite signed measure