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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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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 graph of a continuous function RR is Lebesgue null in R2

Example

Assume the Axiom of Countable Choice and let f:RR be continuous. Then its graph

Γf:={(x,f(x)):xR}

is Lebesgue null in R2.

Facts & Assumptions

Given: The Axiom of Countable Choice and a continuous function f:RR.

[L1]

A subset of Rm has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers (A subset of Rm has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers).

[L2]

Assuming countable choice, a box in Rn with parameters aibi is Lebesgue measurable of measure i<n(biai) (A box in Rn with parameters aibi is Lebesgue measurable of measure i<n(biai), whichever of its faces are included).

[F1]

Let μ be a measure and let (Ek)kN be measurable. Then μ(kEk)kμ(Ek) (Finite and countable subadditivity of measures).

[F4]

For every real δ>0 there is a natural number N1 with 1/N<δ (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[F5]

If r<1 then the series rk converges; in particular n=02n=2 (For r<1, k0rk=1/(1r), and for r1 the series diverges).

Verification

technique · direct
1.1

Fix a real η>0. For each integer m, the interval [m,m+1] is compact by [F3], so [L3] gives a real δm>0 such that x,y[m,m+1] and xy<δm imply f(x)f(y)<η2m3. By [F4] choose a natural number Nm1 with 1/Nm<δm, and put tj:=m+j/Nm for 0jNm. Then for every j with 1jNm and every x[tj1,tj] one has xtj11/Nm<δm, hence f(x)f(tj1)<η2m3.

F2F3L3F4
2.1

For each such subinterval, the corresponding graph piece lies in the closed box [tj1,tj]×[f(tj1)η2m3,f(tj1)+η2m3], whose width is tjtj1 and whose height is η2m2. Therefore [L2] gives a box cover of Γf([m,m+1]×R) whose total area is j=1Nm(tjtj1)η2m2=η2m2.

step 1.1L2algebra
3.1

Summing these covers over all integers m and using [F1], the whole graph is covered by countably many closed boxes with total area at most ηmZ2m2=η(22+2n=12n2)=3η/4<η, the geometric-series identity coming from [F5]. Since η>0 was arbitrary, [L1] gives that Γf is Lebesgue null.

step 2.1L1F1F5algebra

Depends on

Used by

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Dependency tree · two levels

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Sources