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FALSE: if every horizontal and vertical section is measurable, then the set is product-measurable

Statement

If $E \subseteq X \times Y$ has measurable horizontal sections $E_{x}$ for every $x \in X$ and measurable vertical sections $E^{y}$ for every $y \in Y$ , then $E \in A \otimes B$ .

Facts & Assumptions

Given: Assume the Axiom of Countable Choice. Let $X$ be the set of countable ordinals, let $M$ be the sigma-algebra of countable and cocountable subsets of $X$ , and let $E : = {(x, y) \in X \times X : y < x} .$

[L1]

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

[L2]

A sigma-algebra is closed under complements and countable unions, and under countable choice a countable union of countable sets is countable. (Sigma-algebras, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ )

[L3]

For sigma-finite measures, the two iterated section-measure integrals of a product-measurable set agree. (For sigma-finite measures, the two section-measure integrals of a measurable set agree)

[A1]

Define $ν$ on $M$ by $ν (A) = 0$ for countable $A$ and $ν (A) = 1$ for cocountable $A$ . The same countable-union argument as in [L2] shows that $ν$ is a finite measure on $(X, M)$ .

Refutation

technique · direct

1.1

For each $x \in X$ , the section $E_{x} = {y : y < x}$ is countable by the choice of $X$ , hence measurable for $M$ . For each $y \in X$ , the section $E^{y} = {x : y < x}$ has countable complement ${x : x \leq y}$ , hence is cocountable and measurable.

givenL2

2.1

Suppose for contradiction that $E$ were product-measurable for $M \otimes M$ . Since $ν (X) = 1$ , the measure $ν$ is finite and hence sigma-finite, so [L3] would give $\int_{X} ν (E_{x}) d ν = \int_{X} ν (E^{y}) d ν .$ But step 1.1 makes $ν (E_{x}) = 0$ for every $x$ and $ν (E^{y}) = 1$ for every $y$ , so the two sides are $0$ and $1$ , a contradiction. Therefore $E$ is not product-measurable, even though all of its sections are measurable. This does not contradict [L1], which proves only the forward implication from product-measurability to section measurability.

L1L3A1step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Exercise 47 (standard reference, not scraped)