I'm working on the following problem:
Let $\mathcal{M}$ be a $\sigma$-algebra on subsets of $X$ and $\mathcal{N}$ be a $\sigma$-algebra on $Y$. Discuss the validity of the following statement: $A \times B \in \mathcal{M} \otimes \mathcal{N}$ iff $A \in \mathcal{M}$ and $B \in \mathcal{N}$.
I have the forward direction. My book is saying the reverse direction is false. I have a "proof" that it's true. I trust the book, so what's wrong with my "proof"?
To show $A \times B \in \mathcal{M} \otimes \mathcal{N}$ check that the $x$-sections and $y$-sections are $\mathcal{N}$ and $\mathcal{M}$ respectively. $(A\times B)_x = \{y \in Y : (x,y) \in A\times B\} = B$ and $B$ is in $\mathcal{N}$ by assumption. Same for the y-section. Thus we are done.
For the forward direction:
Assume $A \times B \in \mathcal{M} \otimes \mathcal{N}$. The x-section $(A\times B)_x = \{y \in Y : (x,y) \in A\times B\} = B$. $B \in \mathcal{N}$ beacuse of Proposition 2.34 in Folland. (This reads: if $E \in \mathcal{M} \otimes \mathcal{N}$ then $E_x \in \mathcal{N}$ for all $x \in X$ and $E_y \in \mathcal{M}$ for all $y \in Y$) Same argument for the y-section.
You have proved that $A \in \mathcal M$ and $B \in \mathcal N$ implies $A\times B \in \mathcal M \otimes \mathcal N$. I don't see any proof of the converse. It is the converse that is false. For example if we take both $X$ and $Y$ to be the real line with the Borel sigma algebra then $\Delta=\{(x,x):x \in \mathbb R\}$ belongs to $\mathcal M \otimes \mathcal N$ but it cannot be written as $A\times B$.