TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-29

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The region under a nonnegative measurable function is product-measurable and has measure equal to the integral

Statement

Let $(X, A, μ)$ be a sigma-finite measure space, let $λ_{1}$ be Lebesgue measure on $R$ , and let $f : X \to [0, \infty]$ be measurable. Then

$G_{f} : = {(x, t) \in X \times R : 0 \leq t < f (x)}$

belongs to $A \otimes L (R)$ and $(μ \times λ_{1}) (G_{f}) = \int_{X} f d μ .$

Facts & Assumptions

Given: A sigma-finite measure space $(X, A, μ)$ and a measurable function $f : X \to [0, \infty]$ .

[L1]

Tonelli's theorem holds for the sigma-finite product $X \times R$ . (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[L2]

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

Proof

technique · direct

1.1L2

The function $(x, t) \mapsto f (x) - t$ is measurable by [L2], so $G_{f} = {(x, t) : t \geq 0} \cap {(x, t) : f (x) - t > 0}$ is product-measurable.

2.1L1∎

For fixed $x$ , the section of $G_{f}$ is $(G_{f})_{x} = [0, f (x)),$ whose one-dimensional Lebesgue measure is exactly $f (x)$ , including the cases $f (x) = 0$ and $f (x) = \infty$ . Applying [L1] to $1_{G_{f}}$ therefore gives $(μ \times λ_{1}) (G_{f}) = \int_{X} λ_{1} ((G_{f})_{x}) d μ = \int_{X} f d μ .$ This is the claimed area formula.

Depends on

Used by

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Sources

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