Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31
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.

x−a on (0,1) and on (1,∞) calibrates Lp membership

Example

Fix a>0 and let

f0(x):=x−aχ(0,1)(x),f∞(x):=x−aχ(1,∞)(x)

on R with Lebesgue measure. Then for every p>0:

  1. f0∈Lp exactly when ap<1.
  2. f∞∈Lp exactly when ap>1.

So the single power family produces both inclusion failures on R.

Facts & Assumptions

Given: Real numbers a>0 and p>0.

[L1]

Membership in Lp means finiteness of ∫∣f∣p dμ (The function space Lp(μ) for 0<p<∞).

[L2]

Positive-base power functions have the usual antiderivatives away from the logarithmic endpoint, and ∫dx/x=log⁡x (Continuity and derivatives of positive-base real powers, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[L3]

The comparison tests for improper integrals are available (Comparison tests for improper integrals).

Verification

technique · Integrate $x^{-ap}$ on $(0,1)$ and on $(1,\infty)$ using the real-power antiderivative and compare the two thresholds $ap<1$ and $ap>1$
1.1L1given

Because ∣f0∣p=x−apχ(0,1) and ∣f∞∣p=x−apχ(1,∞), [L1] reduces both claims to the improper integrals of x−ap.

2.1L2L3step 1.1

On (0,1), the antiderivative is x1−ap/(1−ap) when ap≠1, so the improper integral converges when 1−ap>0, that is, when ap<1. If ap=1, it is ∫01dx/x, which diverges by [L2]. If ap>1, then x−ap≥x−1 on (0,1), so divergence follows from [L2] and the comparison test [L3].

2.2L2L3step 1.1

On (1,∞), the same antiderivative converges when 1−ap<0, that is, when ap>1. If ap=1, it is again ∫1∞dx/x, which diverges by [L2]. If ap<1, then x−ap≥x−1 for x≥1, so divergence follows from [L2] and the comparison test [L3].

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the two threshold claims.

Depends on

Used by

Dependency tree · two levels

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Sources