A question about $\sigma$-Algebra determine whether the diagonal it is measurable. Let $X = Y$ be uncountable sets. $S = T$ the $\sigma$- algebra of all
sets $A$ such that $A$ or $A^c$
Is it countable. Let W be the $\sigma$-algebra defined on
$X × X$ where $E \in W$ if $E$ or $E^c$ is countable.
Determine what relationship is
meets between $S \times T$ and $W$. Who is larger? Determine whether the diagonal
it is measurable.
 A: I understand that by $S \times T$ you mean the product $\sigma$-algebra of $S$ with $T$.
Let us prove that $W \subsetneq S\times S$.
Let $E$ be any countable subset of $X \times X$. Then $E=\{(x_i,y_i) \in X \times X: i\in \mathbb{N}\}$. Note that, for all $i \in  \mathbb{N}$, $\{(x_i,y_i)\} = \{x_i\}\times \{y_i\}$. So $\{(x_i,y_i)\} \in S\times S$. So
$$ E=\bigcup_{i \in \mathbb{N}} \{(x_i,y_i)\} \in S\times S $$
Let $E$ be any set such that $E^c$ is countable. Then, by what we have just proved, $E^c \in S\times S $. So, $E \in S\times S $.
So if $E$ is any subset of $X \times X$ such that $E$ or $E^c$ is countable, then  $E \in S\times S $.
In other words, $W \subseteq S\times S$.
Now let us prove that $W \neq S\times S$. Let $a$ be any element of $X$. It is immediate that $$\{a\} \times (X\setminus \{a\}) \in  S\times S $$ Since $X$ is uncountable, we have that $\{a\} \times (X\setminus \{a\})$ is uncountable and $(\{a\} \times (X\setminus \{a\}))^c$ is also uncountable. So $$\{a\} \times (X\setminus \{a\}) \notin W $$
The diagonal:
Let $\Delta=\{(x,x) : x \in X\}$. Let us prove that $\Delta \notin S\times S$.
The simplest way to do it is to note that we can define a measure $\mu$ in $(X, S)$ by $\mu(E)=0$ if $E$ is countable and $\mu(E)=1$ if $E$ is co-countable. It is easy to check that $\mu$ is in fact a measure.
Now consider the product measure $\mu \times \mu$ and its outer measure $(\mu \times \mu)^*$. It is easy to see that $(\mu \times \mu)^*(\Delta)=1$ and $(\mu \times \mu)^*(\Delta^c)=1$. So we have
$$(\mu \times \mu)^*(X \times X)=1 < 2 = (\mu \times \mu)^*(\Delta) +  (\mu \times \mu)^*(\Delta^c)$$
So $\Delta$ is not measurable in the Carathéodory's extension sense. So $\Delta \notin S\times S$.
Remark: To see that $(\mu \times \mu)^*(\Delta)=1$, note that, if $\Delta$ is covered by a countable union of rectangles $A \times B$ (where $A, B \in S$), at least for one of such rectangles $A$ and $B$ must be co-countable.
