counter-example of Fubini’s theorem for the Lebesgue integral

The following observation demonstrates the necessity of the integrability assumptionPlanetmathPlanetmath in Fubini’s theorem. Let


denote the upper, right quadrant. Let RQ be the region in the quadrant bounded by the lines y=x,y=x-1, and let let SQ be a similar region, but this time bounded by the lines y=x-1,y=x-2. Let


where χ denotes a characteristic functionMathworldPlanetmathPlanetmathPlanetmath.

Observe that the Lebesgue measureMathworldPlanetmath of R and of S is infiniteMathworldPlanetmathPlanetmath. Hence, f is not a Lebesgue-integrable function. However for every x0 the function


is integrable. Indeed,

g(x)={-x for 0x1,x-2 for 1x2,0 for x2.

Similarly, for y0, the function


is integrable. Indeed,


Hence, the values of the iterated integrals


are finite, but do not agree. This does not contradict Fubini’s theorem because the value of the planar Lebesgue integralMathworldPlanetmath


where μ(x,y) is the planar Lebesgue measure, is not defined.

Title counter-example of Fubini’s theorem for the Lebesgue integral
Canonical name CounterexampleOfFubinisTheoremForTheLebesgueIntegral
Date of creation 2013-03-22 18:18:15
Last modified on 2013-03-22 18:18:15
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 6
Author rmilson (146)
Entry type Example
Classification msc 28A35