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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

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 $ν ≪ μ$ .

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]

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

1.1L2given

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

2.1L3step 1.1∎

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

Depends on