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

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The measure defined by a bounded $L^{p}$ functional is absolutely continuous with respect to $μ$

Statement

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

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ , a bounded linear functional $Λ$ on $L^{p} (μ)$ , and the induced measure $ν (E) = Λ ([1_{E}])$ .

[L2]

In $L^{p} (μ)$ , functions equal almost everywhere define the same class (The space $L^{p} (μ)$ as the quotient by null functions).

[L3]

A bounded linear functional sends the zero vector to $0$ (A bounded linear functional on $L^{p} (μ)$ and its operator norm).

Proof

technique · If $\mu(E)=0$, then $\mathbf 1_E$ is the zero class in $L^p$, so the induced set function must vanish on $E$

1.1

Let $E \in A$ with $μ (E) = 0$ . Then $1_{E} = 0$ almost [L2, given] everywhere, so [L2] gives $[1_{E}] = [0] in L^{p} (μ) .$

L2given

2.1

Applying $Λ$ and then [L3] yields [L3, step 1.1] $ν (E) = Λ ([1_{E}]) = Λ ([0]) = 0.$ Therefore $ν ≪ μ$ .

L3step 1.1∎

Depends on

Used by

Dependency tree · two levels

10 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)
John K. Hunter, Measure Theory, Theorem 7.14 (standard reference, not scraped)

The measure defined by a bounded LpLp functional is absolutely continuous with respect to μμ