# How to show the diagonal of product of Hausdorff spaces is not in the product of its Borel-$\sigma$ algebras?

Let $X$ be a Hausdorff space, with $|X| > \mathfrak{c}$. $\mathcal{B}(X)$, $\mathcal{B}(X \times X)$ are Borel-$\sigma$ Algebras on $X$ and $X\times X$ respectively. $\mathcal{B}(X)⊗\mathcal{B}(X)$ is the product of Borel Algebra of $X$.Let the diagonal of $X \times X$ be$$\Delta = \{(x,x):x \in X\}$$

Then how to show that $\Delta \notin \mathcal{B}(X)⊗\mathcal{B}(X)$

I ran into this claim in this post, and particularly in Gerald Edgar's answer in which $X$ is discrete(Is it necessary?).

-

Discreteness is not necessary: there is a link in one of the comments to this page, which has a proof of the following result:

Theorem (Nedoma’s Pathology): Let $X$ be a measurable space with $|X|>2^{\aleph_0}$. Then the product algebra on $X^2$ does not contain the diagonal. In particular, if $X$ is Hausdorff, then the diagonal is a closed set in the product topology that is not contained in the product algebra.

The crucial lemma:

Lemma: Let $U\subseteq X^2$ be measurable. Then $U$ is the union of at most $2^{\aleph_0}$ boxes, sets of the form $A\times B$.

Once you have the lemma, the argument is easy: any box contained in the diagonal is a singleton.

-

You can also get this as a simple corollary of the following result of general interest:

Theorem: Let $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ be measurable spaces and $f:X\to Y$ be a measurable function. Then the graph of $f$ is in $\mathcal{X}\otimes\mathcal{Y}$ if and only if there is a countably generated $\sigma$-algebra $\mathcal{C}\subseteq\mathcal{Y}$ such that $\{y\}\in\mathcal{C}$ for all $y\in f(X)$.

For a proof, see proposition 2.1. here. The result follows now using the fact that every countably generated $\sigma$-algebra is generated by a real-valued random variable. Just take $f$ to be the identity.

-