Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?).

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.