CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Point evaluation at $0$ is not well defined on $L^{p} [0, 1]$

Statement refuted

Point evaluation at $0$ defines a map on $L^{p} ([0, 1])$ .

Facts & Assumptions

Given: Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $[0, 1]$ carry Lebesgue measure, and let $1 \leq p \leq \infty$ .

[L1]

Elements of $L^{p}$ are equivalence classes modulo almost-everywhere equality, so a pointwise operation on representatives descends only if it is independent of the chosen representative (The space $L^{p} (μ)$ as the quotient by null functions).

[L2]

The singleton ${0}$ has Lebesgue measure $0$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

Counterexample

technique · Compare the zero function with the indicator of the singleton $\{0\}$: they define the same $L^p$ class but have different values at $0$

1.1

Let $u : = 0$ and $v : = 1_{{0}}$ on $[0, 1]$ . [given, construct] Then $u (0) = 0, v (0) = 1,$ so point evaluation at $0$ distinguishes these two representatives.

givenconstruct

2.1

By [L2], the set ${0}$ is null, so $u = v$ almost everywhere. [L1, L2, step 1.1] Therefore [L1] says that $u$ and $v$ define the same element of $L^{p} ([0, 1])$ .

L1L2step 1.1

3.1

A well-defined map on $L^{p} ([0, 1])$ cannot assign two different values to [L1, step 1.1, step 2.1] the same class. Steps 1.1 and 2.1 show that evaluation at $0$ would have to do exactly that, so it does not descend to $L^{p} ([0, 1])$ .

L1step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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., Section 2.5 (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 7.4 (standard reference, not scraped)

Point evaluation at 00 is not well defined on Lp[0,1]Lp[0,1]