Real Analysis, Folland Proposition $2.34$ 
Proposition $2.34$ :
  a. if $E\in \mathcal{M}\times \mathcal{N}$, then $E_x\in\mathcal{N}$ for all $x\in X$ and $E^y\in\mathcal{M}$ for all $y\in Y$.

In the proof of the above, we define
$$\mathcal{R}:=\{F\subseteq X\times Y|\enspace F_x\in\mathcal{N}\text{ for all }x\in X\text{ and }F^y\in\mathcal{M}\text{ for all }y\in Y\}$$
We do the following two things:
(i) $\mathcal{R}$ is a $\sigma$-algebra
(ii) $\mathcal{M}\times\mathcal{N}\subseteq\mathcal{R}$
Now, to show that $\mathcal{R}$ is a $\sigma$-algebra


*

*$\phi$, $X\times Y\in\mathcal{R}$This is true because $\phi\in\mathcal{N}$ and $\phi\in\mathcal{M}$. Similarly, $Y\in\mathcal{N}$ and $X\in\mathcal{M}$.

*For $F\in \mathcal{R}$, then need to show: $F^c\in\mathcal{R}$
$(F^c)_x\stackrel{?}{=}(F_x)^c$ and $(F^c)_y\stackrel{?}{=}(F_y)^c$

*If $\{F_j\}_{j\ge1}\in\mathcal{R}$, then need to show that $\bigcup_{j=1}^\infty F_j\in\mathcal{R}$
$(\bigcup_{j=1}^\infty F_j)_x\stackrel{?}{=}\bigcup_{j=1}^\infty(F_x)_j$ and $\bigcup_{j=1}^\infty(F_j)_y\stackrel{?}{=}\bigcup_{j=1}^\infty(F_y)_j$
Can anyone help me to understand the question mark(?) in the above? Thanks.
 A: $y \in (F^{c})_x$ iff $(x,y) \in F^{c}$ iff $(x,y) \notin F$ iff  it is not true that $(x,y) \in F$ iff $y \notin F_x$. Similarly for $(F^{c})^y$.
$y \in (\bigcup_{j=1}^{\infty} F_j)_x$ iff $(x,y) \in \bigcup_{j=1}^{\infty} F_j)$ iff $(x,y) \in F$ for some $j$ iff $y \in \bigcup_{j=1}^{\infty} (F_j)_x$
A: A point $y$ is in $(F^\complement)_x$ if, by definition, $(x,y) \in F^\complement$, which means that $(x,y) \notin F$, which implies $y \notin F_x$ so $y \in (F_x)^\complement$, again by definition. So $(F^\complement)_x \subseteq (F_x)^\complement$, and the reverse inclusion follows the same ideas.
In $X$, we get the same $(F^\complement)_y = (F_y)^\complement$ for section at any $y$.
So it's just a matter of element chasing. Same for the union:
$$y \in \left(\bigcup_{j=0}^\infty F_j\right)_x$$ iff (definition of the section)
$$(x,y) \in \bigcup_{j=0}^\infty F_j$$ iff (definition of union)
$$\exists j\in \Bbb N: (x,y) \in F_j$$ iff (definition of the section)
$$\exists j\in \Bbb N: y \in (F_j)_x$$ iff (definition of union)
$$y \in  \bigcup_{j=0}^\infty (F_j)_x$$
so we have equality of these subsets of $Y$.
If now $F \in \mathcal{R}$, we see $F^\complement \in \mathcal{R}$: let $x \in X$ be arbitrary, then $F_x \in \mathcal{N}$ by definition of $\mathcal{R}$, so $(F_x)^\complement \in \mathcal{N}$ as $\mathcal{N}$ is a $\sigma$-algebra.
And so $(F^\complement)_x = (F_x)^\complement \in \mathcal{N}$ too, which we have to show. A similar argument shows that $(F^\complement)_y \in \mathcal{M}$ for all $y \in Y$ as well, and so by definition, $F^\complement \in \mathcal{R}$.
The union argument is analogous.
