CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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The diagonal under Lebesgue times counting measure shows that Tonelli needs sigma-finiteness

Statement refuted

Tonelli's theorem holds without any sigma-finiteness hypothesis.

Counterexample

technique · direct

Let $X = Y = [0, 1]$ , let $μ$ be Lebesgue measure on $X$ , let $ν$ be counting measure on $Y$ , and let $D : = {(x, y) \in [0, 1]^{2} : x = y} .$

Facts & Assumptions

Given: Lebesgue measure $μ$ on $[0, 1]$ , counting measure $ν$ on $[0, 1]$ , and the diagonal set $D$ .

[A1]

For every $x, y \in [0, 1]$ , the horizontal and vertical sections of the diagonal are $D_{x} = {x}$ and $D^{y} = {y}$ .

Verification

1.1

For fixed $x \in [0, 1]$ , the section $D_{x} = {x}$ has counting measure $1$ , so $\int_{X} ν (D_{x}) d μ (x) = \int_{0}^{1} 1 d x = 1.$

givenA1

2.1

For fixed $y \in [0, 1]$ , the section $D^{y} = {y}$ has Lebesgue measure $0$ , so $\int_{Y} μ (D^{y}) d ν (y) = \int_{[0, 1]} 0 d ν = 0.$ The iterated integrals are unequal, so Tonelli fails once the counting-measure factor is not sigma-finite.

A1∎

Depends on

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Gerald B. Folland, Real Analysis, 2nd ed., Exercise 46 (standard reference, not scraped)