Let $(X,\Sigma_1),~(Y,\Sigma_2)$ be two measurable space (possibly without measure defined on it).
Let $A\subseteq X,~B\subseteq Y$ and $E=A\times B$.
Then I find in some resources that said if $f:E\to\overline{\Bbb R}$ is measurable, then $x\mapsto f(x,y)$ is measurable for every $x\in A$. And at the same times other authors state the conclusion differently: $x\mapsto f(x,y)$ is measurable for almost everywhere $x\in A$. Is the former correct? If so, why some authors chose to state a somehow weaker version of the theorem? It's so weird.
Question two: Is the whole story different related to the space $X, Y$? When $X=\Bbb R^n,~Y=\Bbb R^m$, does it matter with respect to our question?
When $f$ is measurable w.r.t. the product sigma algebra $\Sigma_1 \times \Sigma_2$ we can say that $x \to f(x,y)$ is mesurable for every $y$. In the special case you mentioned it is often assumed that $f$ is Lebesgue measurable and not Borel measurable (which is same as measurability w.r.t. the product Borel sigma algebra). Here you are taking the completion of the product sigma algebra. In this case you can only say that $x \to f(x,y)$ is measurable for almost all $y$.