TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Every section of a product-measurable function is measurable

Statement

Let $(X, A)$ and $(Y, B)$ be measurable spaces, and let $f : X \times Y \to \overline{R}$ be $(A \otimes B, B (\overline{R}))$ -measurable. Then $f_{x} : Y \to \overline{R}$ is $B$ -measurable for every $x \in X$ , and $f^{y} : X \to \overline{R}$ is $A$ -measurable for every $y \in Y$ .

Facts & Assumptions

Given: Measurable spaces $(X, A)$ and $(Y, B)$ , and a product-measurable function $f : X \times Y \to \overline{R}$ .

[L1]

A function into $\overline{R}$ is measurable once the preimages of a generating family for $B (\overline{R})$ are measurable. (A generating family on the codomain suffices to test measurability)

[L2]

Sections of product-measurable sets are measurable. (Every section of a product-measurable set is measurable)

[A1]

For every Borel set $U \subseteq \overline{R}$ , $f_{x}^{- 1} (U) = (f^{- 1} (U))_{x}, (f^{y})^{- 1} (U) = (f^{- 1} (U))^{y} .$

Proof

technique · direct

1.1L1A1L2

Fix $x \in X$ and let $G$ be a generating family for $B (\overline{R})$ . For each $U \in G$ , the set $f^{- 1} (U)$ is product-measurable, so [L2] and [A1] give $f_{x}^{- 1} (U) = (f^{- 1} (U))_{x} \in B$ . By [L1], $f_{x}$ is therefore $B$ -measurable.

2.1step 1.1A1L1L2∎

Fix $y \in Y$ . The same argument gives $(f^{y})^{- 1} (U) = (f^{- 1} (U))^{y} \in A$ for every generator $U \in G$ , so [L1] implies that $f^{y}$ is $A$ -measurable. Thus every horizontal and vertical section of $f$ is measurable.

Depends on

Used by

Dependency tree · two levels

10 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

John K. Hunter, Measure Theory, Theorem 5.15 (standard reference, not scraped)