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.


2 Answers 2


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


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