Past exam of the options II course of the University of Oxford / 2008 printed variant / D / D4

On a product of sigma-finite measure spaces, Tonelli's theorem equates product and iterated integrals for a nonnegative measurable function, allowing infinite values. Fubini's theorem says that an absolutely integrable measurable function has integrable sections almost everywhere, their section integrals are integrable, and both iterated integrals equal the product integral. Absolute integrability may be checked first by applying Tonelli to the absolute value.

For fixed $n>0$, Tonelli and the allowed exponential integral giveThe point $x=0$ is irrelevant to the integral. Fubini therefore permits reversing the signed integrals. Evaluating the elementary inner integral yieldsThe first term has integral $\pi /2$. The absolute value of the remainder is at mostConsequently the improper integral tends to $\pi /2$. The interchange was justified on each finite strip, without asserting absolute integrability on the infinite quadrant.

It is not Lebesgue integrable, because its absolute integral diverges. On $I_k=[k\pi +\pi /6,k\pi +5\pi /6]$, one has $|\sin x|\ge 1/2$ and $x\le (k+1)\pi$. ThusThese disjoint-interval lower bounds have a divergent harmonic sum. The finite improper value from part (i) is therefore conditional convergence, not Lebesgue integrability.