Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D / D4

For product measure spaces, Tonelli's theorem states that for a nonnegative measurable function the product-space integral equals either iterated integral, allowing the value $+\infty$. Fubini's theorem states that if a measurable function has finite integral of its absolute value, then its sections are integrable almost everywhere, both iterated integrals exist and equal the product-space integral. The absolute-integrability hypothesis is what permits this conclusion for a signed function. These statements apply to the Lebesgue product spaces used here, which are sigma-finite.

The function $g$ is measurable, and for fixed positive $y$ both terms are integrable over $0<x<1$ by linear substitution and integrability of $f$. Those substitutions giveThe strict sign follows from positivity of $f$ on a nonempty interval, since $a>1$. Similarly each section in $y$ is absolutely integrable for fixed $x>0$, andIf $g$ were absolutely integrable on the stated rectangle, Fubini would make the two finite iterated integrals equal. The first would be strictly negative and the second strictly positive, which is impossible. Therefore $\boxed{g\notin L^1((0,1)\times (1,\infty ))}$. The one-dimensional section integrability does not rescue the missing product-space hypothesis.

The function $h$ is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral isThe radial factor is $1/2$ by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves $h\in L^1$.

Fubini may now factor the Cartesian integral. Substituting $u=x^2$, $v=y^2$ in the respective factors givesBoth are finite: their powers at zero exceed $-1$, and the exponential controls infinity. Comparing with the polar expression proveswhich is exactly the product of integrals requested.