= Solution
Both terms of each one-dimensional section are integrable by a positive scaling substitution, since $f\in L^1(0,\infty)$. For fixed $y>0$,
$$
\int_0^1g(x,y)\,dx
=\frac1y\left(\int_0^yf(t)\,dt-\int_0^{ay}f(t)\,dt\right)
=-\frac1y\int_y^{ay}f(t)\,dt<0.
$$
For fixed $x>0$,
$$
\int_1^\infty g(x,y)\,dy
=\frac1x\left(\int_x^\infty f(t)\,dt-\int_{ax}^\infty f(t)\,dt\right)
=\frac1x\int_x^{ax}f(t)\,dt>0.
$$
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 \b[$g$ is not integrable on that product].
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