Improper Double Integral Counterexample Let $f: \mathbf{R}^2\to \mathbf{R}$. I want to integrate $f$ over the entire first quadrant, call $D$. Then by definition we have
$$\int \int_D f(x,y) dA =\lim_{R\to[0, \infty]\times[0, \infty]}\int \int_R f(x,y) dA$$
where $R$ is a rectangle. 
I remember vaguely that the above is true if $f$ is positive. In other words, if $f$ is positive, then the shape of the rectangle does not matter.
So this brings me to my question: give a function $f$ such that the shape of the rectangles DO MATTER when evaluating the improper double integral.

Let $g$ denote an integrable odd function. Define $f$ by $f(x,y)=g(y-x)$. Then, for every nonnegative $z$, the integral of $f$ on the rectangle $(0,x)\times(0,x+z)$ converges to a finite limit $\ell(z)$ when $x\to+\infty$. In general, $\ell(z)$ does depend on $z$, since
$$
\ell(z)=\int_0^{+\infty}\min\{t,z\}g(t)\,\mathrm dt.
$$
