TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Every section of a product-measurable set is measurable

Statement

Let $(X, A)$ and $(Y, B)$ be measurable spaces. If $E \in A \otimes B$ , then $E_{x} \in B$ for every $x \in X$ and $E^{y} \in A$ for every $y \in Y$ .

Facts & Assumptions

Given: Measurable spaces $(X, A)$ and $(Y, B)$ , and a set $E \in A \otimes B$ .

[L1]

The product sigma-algebra $A \otimes B$ is generated by the measurable rectangles. (The product sigma-algebra and its finite iterates)

[L2]

The generated sigma-algebra is the smallest sigma-algebra containing the generating family. (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal)

[A1]

For fixed $x \in X$ , $(A \times B)_{x} = {\begin{cases} B, & x \in A, \\ \emptyset, & x \notin A, \end{cases}$ and for fixed $y \in Y$ , $(A \times B)^{y} = {\begin{cases} A, & y \in B, \\ \emptyset, & y \notin B . \end{cases}$

[A2]

For fixed $x$ , $((X \times Y) ∖ E)_{x} = Y ∖ E_{x}, {(⋃_{n \geq 1} E_{n})}_{x} = ⋃_{n \geq 1} (E_{n})_{x},$ and the analogous formulas hold for vertical sections.

Proof

technique · direct

1.1

Fix $x \in X$ and let $M_{x}$ be the family of sets $F \subseteq X \times Y$ such that $F_{x} \in B$ . By [A1], every measurable rectangle belongs to $M_{x}$ . By [A2], $M_{x}$ is a sigma-algebra on $X \times Y$ . Since [L1] says $A \otimes B$ is generated by the measurable rectangles, [L2] gives $A \otimes B \subseteq M_{x}$ . Therefore $E_{x} \in B$ .

A1A2L1L2

2.1

Fix $y \in Y$ and let $N_{y}$ be the family of sets $F \subseteq X \times Y$ such that $F^{y} \in A$ . The same argument shows that $N_{y}$ is a sigma-algebra containing all measurable rectangles, hence all of $A \otimes B$ . Therefore $E^{y} \in A$ . Since $x$ and $y$ were arbitrary, every horizontal and vertical section of $E$ is measurable.

step 1.1L1L2A1A2∎

Depends on

Used by

A nonmeasurable subset of a null line shows that the product of complete measures need not be complete Counterexample
FALSE: if every horizontal and vertical section is measurable, then the set is product-measurable False statement
FALSE: the product Lebesgue sigma-algebra is the full Euclidean Lebesgue sigma-algebra False statement
FALSE: the product of two complete measure spaces is complete False statement
For sigma-finite measures, the section-measure functions are measurable Proposition
Every section of a product-measurable function is measurable Theorem

Dependency tree · two levels

8 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, Proposition 5.2 (standard reference, not scraped)