 Solution

Both terms of each one-dimensional section are integrable by a positive scaling substitution, since $f\in L^1(0,\infty )$. For fixed $y>0$,For fixed $x>0$,The strict inequalities hold because $a>1$ and the measurable integrand is strictly positive on each nonempty interval. If $g$ were integrable on the stated product domain, Fubini's theorem would make both iterated integrals finite and equal. But the first order of integration would give a strictly negative value, and the other a strictly positive value. This contradiction proves $g$ is not integrable on that product.