Assume two $\sigma-$finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$. Consider a $\mathcal{M}\times\mathcal{N}$-measurable function $f$, and we are interested in computing $$ \int f d\left(\mu \times \nu\right) $$ where $\mu \times \nu$ is the product measure. I know the results of Fubini's theorem. Lets assume that one first observes that $f(x,y)\in L^{1}(\mu)$ for fixed $y$ and $f(x,y)\in L^{1}(\nu)$ for fixed $x$. Then, he naively computes $$ \int \left[ \int f(x,y) d\mu(x) \right]d\nu(y) $$ and $$ \int \left[ \int f(x,y) d\nu(y) \right]d\mu(x) $$ . and, he notices that the value of these two integral exist and are equal.
I have two questions:
1) Whether we can find a case as I described above?
2) Provided that we can find such an example, can we conclude that $f \in L^{1}(\mu \times \nu)$, and value of $ \int f d\left(\mu \times \nu\right) $ equal to the $\int \left[ \int f(x,y) d\nu(y) \right]d\mu(x)=\int \left[ \int f(x,y) d\mu(x) \right]d\nu(y)$.
EDIT: Answer: Assume the following function
$$ f=\begin{cases} 0 & (x,y)=(0,0)\\\frac{xy}{(x^2+y^2)^2} & \mbox{else}\end{cases} $$ It is exactly an example which satisfies the above conditions.