Past exam of the integration course of the University of Oxford / 2017 / 3 / b

On sigma-finite measure spaces, including the real line with Lebesgue measure, Tonelli's theorem says that a nonnegative measurable function $H$ has equal product and iterated integrals:with infinity allowed. Finiteness is not a prerequisite.

Fubini's theorem says that if a measurable signed or complex $H$ is absolutely integrable on the product, then its sections are integrable for almost every value of the other variable, the iterated integral functions are integrable, and the same equalities hold with finite values. For completed Lebesgue measure, exceptional sections need not themselves be measurable; the integrals use the almost-everywhere defined sections. Thus Tonelli can first establish absolute integrability using $|H|$, after which Fubini permits changing the order for $H$.

Because $f$ is continuous, $g(x,y)=e^{-x^2}f(e^{-x^2/2}y)$ is continuous on ${\mathbb{R}}^2$, so $|g|$ is a nonnegative measurable function. Apply Tonelli's theorem before assuming integrability. For each fixed $x$, set $z=e^{-x^2/2}y$, so $dy=e^{x^2/2}dz$. Lebesgue substitution on an interval givesTherefore the Gaussian integral yieldsThis proves that $g$ is Lebesgue integrable on ${\mathbb{R}}^2$. Fubini's theorem now permits changing the order for $g$ itself. The same substitution without absolute values gives the stronger exact identityTaking absolute values and using the triangle inequality for an integral provesThe constant is sharp: any nonzero nonnegative continuous integrable $f$ gives equality. This is a weighted dilation kernel with $a(x)=e^{-x^2}$ and $b(x)=e^{-x^2/2}$. The inner $x$ integral in the requested order is guaranteed to be integrable for almost every $y$, which is all the iterated integral requires.