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 $(1, \infty)$ calibrates $L^{p}$ membership

Example

Fix $a > 0$ and let

$f_{0} (x) : = x^{- a} χ_{(0, 1)} (x), f_{\infty} (x) : = x^{- a} χ_{(1, \infty)} (x)$

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

$f_{0} \in L^{p}$ exactly when $a p < 1$ .
$f_{\infty} \in L^{p}$ exactly when $a p > 1$ .

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

Facts & Assumptions

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

[L1]

Membership in $L^{p}$ means finiteness of $\int ∣ f ∣^{p} d μ$ (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

[L2]

Positive-base power functions have the usual antiderivatives away from the logarithmic endpoint, and $\int d x / x = \log x$ (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

Proof technique: Integrate $x^{- a p}$ on $(0, 1)$ and on $(1, \infty)$ using the real-power antiderivative and compare the two thresholds $a p < 1$ and $a p > 1$ .

1.1

Because $∣ f_{0} ∣^{p} = x^{- a p} χ_{(0, 1)}$ and $∣ f_{\infty} ∣^{p} = x^{- a p} χ_{(1, \infty)}$ , [L1] reduces both claims to the improper integrals of $x^{- a p}$ .

L1given

2.1

On $(0, 1)$ , the antiderivative is $x^{1 - a p} / (1 - a p)$ when $a p \neq 1$ , so the improper integral converges when $1 - a p > 0$ , that is, when $a p < 1$ . If $a p = 1$ , it is $\int_{0}^{1} d x / x$ , which diverges by [L2]. If $a p > 1$ , then $x^{- a p} \geq x^{- 1}$ on $(0, 1)$ , so divergence follows from [L2] and the comparison test [L3].

L2L3step 1.1

2.2

On $(1, \infty)$ , the same antiderivative converges when $1 - a p < 0$ , that is, when $a p > 1$ . If $a p = 1$ , it is again $\int_{1}^{\infty} d x / x$ , which diverges by [L2]. If $a p < 1$ , then $x^{- a p} \geq x^{- 1}$ for $x \geq 1$ , so divergence follows from [L2] and the comparison test [L3].

L2L3step 1.1

3.1

Steps 2.1 and 2.2 prove the two threshold claims.

step 2.1step 2.2∎

Depends on

Used by

L¹ is not a subset of L² on the line Counterexample
L² is not a subset of L¹ on the line Counterexample

Dependency tree · two levels

22 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 L. Wheeden and Antoni Zygmund, Measure and Integral, Chapter 8 (standard reference, not scraped)
John K. Hunter, Measure Theory, Chapter 17 (standard reference, not scraped)

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