Suppose $(X, \Sigma, \mu)$ and $(Y, \tau, \nu)$ are both complete measure spaces. Consider the following two measure spaces: $(X \times Y, \overline{\Sigma \times \tau}, \mu \times \nu)$ and $(X \times Y, \Sigma \times \tau, \mu \times \nu)$.
I know Fubini's theorem for the complete $\sigma$-algebra $\overline{\Sigma \times \tau}$:
If $f \in L^{1}(d\mu \times d\nu)$, then:
- The maps $y \mapsto f(x,y)$ are in $L^{1}(d\nu)$ for almost all $x$
- The map $x \mapsto \int \limits_{Y} f(x,y) \,d\nu$ is in $L^{1}(d\mu)$
- $\int \limits_{X \times Y} f(x,y) \,d(\mu \times \nu) = \int \limits_{X} \left [ \int \limits_{Y} f(x,y) \,d\nu \right ] \,d\mu$
My first question: in statement 1, I believe we also have that the maps are measurable for almost all $x$. Then the map in 2 can only exist almost everywhere, so shouldn't it equal an $L^{1}(d\mu)$ function almost everywhere, rather than being in $L^{1}(d\mu)$ itself?
My second question: What is different between Fubini's theorem above for $\overline{\Sigma \times \tau}$ and Fubini's theorem applied to $\Sigma \times \tau$? I know that when we are dealing with $\Sigma \times \tau$, we have that the maps $y \mapsto f(x,y)$ are measurable for all $x$, not just almost everywhere, but aside from this, I don't know what else is different.
Statement 2 means that the function defined by $\int_Y f(x,y) \, d\nu$ if the integral exists and arbitrary elsewhere is in $L^1(\mu)$.
By abuse of notation/language, this is written as "$x \mapsto \int_Y f(x,y) \,d\nu$ is in $L^1(\mu)$".
In the non-complete case, one has to define the function as $0$ (or any other fixed(!) value) if the integral does not exist, otherwise measurability may fail.
Regarding the statements that you posted, that is basically the only difference. It is true that the map $y \mapsto f(x,y)$ is now measurable for all $x$, but it does not have to be integrable for all $x$ as the example
$$ f(x,y) = \chi_{\{0\} \times \Bbb{R}} (x,y) $$
shows (with $\mu = \nu = \text{Lebesgue measure}$).