Is the diagonal set a measurable rectangle? Let $\Sigma$ denotes the  Borel $\sigma$-algebra of $\mathbb{R}$ and $\Delta=\{(x,y)\in\mathbb{R}^2: x=y\}$. I am trying to clarity the definitions of $\Sigma\times\Sigma$ (the sets which contains all measurable rectangles) and $\Sigma\otimes\Sigma$ (the $\sigma$-algebra generated by the collection of all measurable rectangles). My question is (1) does $\Delta$ belong to $\Sigma\times\Sigma$? (2) does $\Delta$ belong to $\Sigma\otimes\Sigma$?
I am thinking that (1) would be no (since a measurable rectangle can be arbitrary measurable sets which are not required to be intervals?) and (2) would be yes (can we write $\Delta$ like countable unions of some open intervals? I cannot find a one at time).
 A: Yes.  Draw this line in the plane.  Now you can easily see how it is a diagnonal of squares whose side length is $1/n$.  The union of these squares is in $\Sigma$; denote it by $U_n$.  Notice that the diagonal line is
$$\cap_n U_n.$$
A: Hints:
$(1)$: Assume towards contradiction that $\Delta\in\Sigma\times\Sigma$, i.e. $\Delta=A\times B$ for some Borel sets $A,B$. If $A\neq B$ then there exists $x\in A\setminus B$ or $x\in B\setminus A$. Hence $(x,x)\in\Delta$ but... ? If $A=B$ then obviously $A\neq \mathbb{R}$. So there exists $x\notin A$, so $(x,x)$ is in ... but ...?
$(2)$: Show that $\Delta$ is a closed subset of $\mathbb{R}^{2}$ by considering the preimage of $\{0\}$ via the continuous function $(x,y)\mapsto x-y$. Show that the complement of $\Delta$ (as an open set) is a countable union of rectangles, where each rectangle is a product of two intervals with rational endpoints. Use the fact that $\Sigma\otimes\Sigma$ is a $\sigma$-algebra.
A: Let $x\neq y$. If there is a measurable rectangle containing $(x,x)$ and $(y,y)$, there must be a set $A$ containing both $x$ and $y$ and a set $B$ doing the same such that $(x,x)$ and $(y,y)$ are in $A\times B$. But then $(x,y)\in A\times B$.
