Question about statement of Fubini's theorem This is a question on the statement of Fubini's theorem for measurable sets.
The theorem looks like this:

Let $(X \times Y, \overline{\Sigma \times \tau}, \lambda = \mu \times \nu)$ be a complete measure space.  If $E \in \overline{\Sigma \times \tau}$ with $\lambda(E) < \infty$, then
  
  
*
  
*$E_{x} \in \tau$ a.e. $d\mu(x)$ (that is, $\chi_{E_{x}}(y)$ is a measurable function for almost all $x$)
  
*The map $x \rightarrow \nu(E_{x})$ is measurable a.e. $d\mu(x)$ (i.e., the map $x \rightarrow \int \limits_{X} \chi_{E_{x}}(y) \,d\nu$ is measurable a.e. $d\mu(x)$)
  
*$\lambda(E) = \int \limits_{X} \nu(E_{x}) \,d\mu$ (i.e., $\int \limits_{X \times Y} \chi_{E} \,d\lambda = \int \limits_{X} \left [ \int \limits_{Y} \chi_{E_{x}}(y) \,d\nu \right ] \,d\mu$ -- the integral is iterated)
  

My questions are: 


*

*Since $E_{x}$ is only a measurable set for almost all $x$, then the map $\nu(E_{x})$ only exists for almost all $x$.  Then, in 2, are we showing that this map $\nu(E_{x})$ which exists almost everywhere equals a measurable function almost everywhere?

*In statement 3, when we integrate $\nu(E_{x})$, do we really mean we are integrating the measurable function equal to $\nu(E_{x})$ almost everywhere?


I know by the time one is at an advanced level, these questions don't seem necessary, but they are absolutely necessary for me at my level.  I really dislike talking about the integral of a non-measurable function, even though the function equals a measurable one almost everywhere.  If you have advice on this, too, in addition to answers to my questions, I'd love to hear it.



Edit: Is it wrong to make the following changes to the theorem statement?: For 2, we would say "the map $x \rightarrow \nu(E_{x})$ equals a measurable function $\hat{f}$ almost everywhere $d\mu(x)$.  Then, for 3, we would say $\lambda(E) = \int \limits_{X} \hat{f} \,d\mu$, rather than writing it as the integral of a function that isn't necessarily defined everywhere?



 A: Your interpretation in 1 and 2 is correct. Regarding the point 

I really dislike talking about the integral of a non-measurable function, even though the function equals a measurable one almost everywhere.

I share your dislike. This is why I think that spaces $X$ and $Y$ should themselves be assumed complete. On a complete measure space, a function that is a.e. equal to a measurable function is itself measurable. 
It's true that $\nu(E_x)$ is only defined almost everywhere: for some values of $x$, the set $E_x$ may be nonmeasurable. So, at some point (preferably, before Fubini's theorem) it should be said that on a complete measure space $(X,\mu)$, we can interpret $\int_X f\,d\mu$ even if $f$ is only defined on some set $A\subset X$ such that $\mu(X\setminus A)=0$. Namely, we interpret $\int_X f\,d\mu$ as $\int_X f\chi_A\,d\mu$, setting the function to be $0$ on the complement of $A$. Or any other number, does not matter. 
In a complete measure space, negligible sets can be safely neglected.
