I am completing an exercise on multiple Lebesgue integrals. The problem is as follows:
Let $X=Y=\Bbb{N}$ and $\mathcal{A}=\mathcal{B}=\mathcal{P}\Bbb{N}$ with counting measures $\mu$ and $\nu$ on $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ respectively. Define $$f\colon X\times Y\to\Bbb{R}\quad\text{by}\quad f(x,y)=\begin{cases}1, &\text{if}\;x=y\\ -1, &\text{if}\;x=y+1\\ 0,&\text{otherwise.} \end{cases}$$ Show that $\int_Y\left(\int_X f\,d\mu\right)\,d\nu$ and $\int_X\left(\int_Y f\,d\nu\right)\,d\mu$ exist and are unequal.
I tried to solve this but I found them both to be zero, so I am doing something wrong. My argument is below.
First consider $$\int_X f\,d\mu=\sum_{x\in X}\int_{\{x\}}f\,d\mu=\sum_{x\in X} f(x,y)\mu(\{x\})=\sum_{x\in X} f(x,y),$$ by linearity of the integral, and then since $f$ is simple on each $\{x\}$. But summing over the natural numbers (i.e. $X$) yields $$\sum_{x\in X} f(x,y)=1+(-1)+0+0+\dots=0$$ for each fixed $y$. Then the double integral is $$\int_Y\left(\int_X f\,d\mu\right)\,d\nu=\int_Y 0\,d\nu=0.$$
But now I think this argument would be identical if done in the $Y$ case, i.e. $$\int_Y f\,d\nu=\sum_{y\in Y} f(x,y)=1+(-1)+0+0+\dots=0.$$
Any hints at where I have made a mistake are appreciated.
Consider $$ \sum_{x \geq 0}\sum_{y \geq 0} f(x,y) = \sum_{x \geq 0}\left(f(x,0) + f(x,1) + \cdots \right) = 0 + 0 + 0 + \cdots = 0. $$ However, $$ \sum_{y \geq 0}\sum_{x \geq 0} f(x,y) = \sum_{y \geq 0}\left( f(0,y) + f(1,y) + \cdots\right) = 1 + [(-1) + 1] + 0 + \dots = 1. $$ This is because $f(0,y)$ never attains the value $-1$.