 Solution

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.