ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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The $L_{l o c}^{1}$ class of $1_{Q}$ has every point as a Lebesgue point

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $f = 1_{Q}$ on $R$ . Then the $L_{l o c}^{1}$ class of $f$ is the zero class, and its Lebesgue set is all of $R$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and the Dirichlet function $f = 1_{Q}$ on $R$ .

[L1]

The Lebesgue set of a class consists of the points where some representative has vanishing averaged oscillation. (Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class)

[L2]

Every countable subset of $R$ is Lebesgue null. (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ )

Verification

technique · direct

1.1

The set $Q$ is countable, so [L2] gives [L2, given] $1_{Q} = 0$ almost everywhere. Thus the $L_{l o c}^{1}$ class of $f$ is the same as the class of the zero function.

L2given

2.1

For the zero representative and every $x \in R$ , [L1, step 1.1, algebra] $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ 0 - 0 ∣ d λ = 0 (r > 0) .$ Therefore every point is a Lebesgue point of the zero representative, so by [L1] the Lebesgue set of the class is all of $R$ .

L1step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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: Modern Techniques and Their Applications, 2nd ed., Theorem 3.20 (standard reference, not scraped)

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