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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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One existing iterated integral does not imply multiple Riemann integrability

Statement refuted

False claim. If one ordinary iterated Riemann integral of a bounded function on a rectangle exists, then the function is Riemann integrable on the rectangle.

Facts & Assumptions

Given: On [1,1]×[0,1][-1,1]\times[0,1], let f(x,y)=xf(x,y)=x for rational yy and f(x,y)=0f(x,y)=0 for irrational yy.

[L2]

For an integrable function, Riemann--Fubini makes the lower and upper section-integral envelopes integrable with the same multiple integral (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

Counterexample

technique · direct
1.1

For each fixed yy, the xx-section is either xxx\mapsto x or zero; both have integral 00 on [1,1][-1,1]. Hence the xx-first iterated integral exists and is 00.

given
1.2

For fixed x0x\ne0, density in [L1] makes every lower and upper Darboux sum of the yy-section equal to the interval length times min(x,0)\min(x,0) and max(x,0)\max(x,0) respectively. Thus the section is nonintegrable, and integrating its lower and upper integral functions in xx gives 1/2-1/2 and 1/21/2.

L1given
2.1

If ff were multiply Riemann integrable, [L2] would force those two envelope integrals to agree. They do not, so ff is not Riemann integrable although the iteration in step 1.1 exists. This refutes the claim.

L2step 1.1step 1.2

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Direct dependencies and their dependencies through the next three levels: 87 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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