Here is the question: Let $X = Y = \mathbb{N}, S = \Sigma= P(\mathbb{N})$, the power set of the natural numbers. Let $\mu = \nu = $ the counting measure. Let $h: X \times Y \rightarrow \mathbb{R}$ be defined as $h(x,y) = 1$ if x = y, $h(x,y) = -1$ if x = y + 1 and $h(x,y) = 0 $ otherwise. Is h a $\mu \times \nu$ integrable function?
Now I drew out a picture of what the region looks like. It is basically a lattice on the space $\mathbb{N} \times \mathbb{N}$. I am thinking like double integrals in cartesian coordinates. Since we are dealing with the counting measure, we are summing up the points on the graph. The sum of the collumns are the same but the first collumn is 1. However, the sum of the rows is the same, which is zero. So do I use Fubini Theorem and conclude by saying that f is not $\mu \times \nu $ is not integrable? Thank you very much for your help!
By definition, "$h$ integrable" means that both its positive and negative parts have finite integral (even if you only care about the existence of the integral, you need at least one to be finite). Here you have, if $A=\{x=y\}$ and $B=\{x=y+1\}$, $$\int_{X\times Y} h_+\,d\mu\times d\nu=(\mu\times\nu)(A)=\infty,$$ $$\int_{X\times Y} h_-\,d\mu\times d\nu=(\mu\times\nu)(B)=\infty.$$ So $h$ is not integrable.
As you mention, you can also say that the iterated integrals differ, and conclude by Fubinin that $h$ is not integrable.