# 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.

$$y \in (F^{c})_x$$ 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 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.