CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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The graph of a measurable function R^n to R is Lebesgue null

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $f : R^{n} \to R$ be Lebesgue measurable. Then its graph $Γ_{f} : = {(x, t) \in R^{n} \times R : t = f (x)}$ is Lebesgue measurable and has $(n + 1)$ -dimensional Lebesgue measure zero.

Facts & Assumptions

Given: The Axiom of Countable Choice, a positive integer $n$ , and a Lebesgue measurable function $f : R^{n} \to R$ .

[L1]

Arithmetic operations preserve measurability. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[L2]

Tonelli's theorem holds on sigma-finite products. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[L3]

Assuming countable choice, countable subsets of Euclidean space are Lebesgue null. (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ )

Proof

technique · direct

1.1

The function $H (x, t) : = t - f (x)$ is measurable by [L1], so $Γ_{f} = H^{- 1} ({0})$ is measurable.

2.1

For each $x \in R^{n}$ , the section $(Γ_{f})_{x}$ is the singleton ${f (x)}$ , which is countable and therefore Lebesgue null by [L3]. Applying [L2] to $1_{Γ_{f}}$ yields $λ_{n + 1} (Γ_{f}) = \int_{R^{n}} λ_{1} ((Γ_{f})_{x}) d λ_{n} = \int_{R^{n}} 0 d λ_{n} = 0.$ Hence the graph is Lebesgue null.

L2L3∎

Depends on

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Exercise 50 (standard reference, not scraped)