 Solution

Use $u=x-y$, $v=x+y$. It sends the diamond-shaped open square onto $(-1,1{)}^2$, and the inverse Jacobian has absolute determinant $1/2$. On this rectangle $|u|<1$ and $v+2>1$, so the original function is measurable, bounded by one and supported on a finite-area set, proving integrability. The transformed integral separates:The two one-dimensional integrals follow from the fundamental theorem of calculus; Fubini's theorem applies by the established integrability.