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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?).

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Metta World Peace
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2 Answers2

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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.

Brian M. Scott
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  • Could you please give an address for the above lemma? – ABB May 19 '18 at 19:29
  • @ABB Here's a proof sketch: For any measurable set $U$, you only "use" countably many elements in the generating set to generate $U$. Therefore, $U$ is in a sub-$\sigma$-algebra generated by countably many measurable sets, say $A_1, A_2, ...$. For each $(x, y) \in X^2$, look at which of the $A_i$'s they are in, to get two subsets of $\mathbb{N}$. This gives a map $X^2 \rightarrow P(\mathbb{N})^2$. The subsets of $X^2$ that are the preimage of subsets of $P(\mathbb{N})^2$ form a $\sigma$-algebra and contains each $A_i \times A_j$, so they are the desired boxes. – David Lui Feb 20 '23 at 08:06
  • Full details are here : https://www.drmaciver.com/2006/04/journal-of-obscure-results-1-nedomas-pathology/ – David Lui Feb 20 '23 at 08:06
  • Nice argument. Thanks. – ABB Feb 20 '23 at 10:24
  • @DavidLui: For some reason I never saw ABB’s question, so many thanks for answering it. – Brian M. Scott Feb 20 '23 at 17:01
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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.

Michael Greinecker
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  • I'm not seeing how this implies the statement. If $f= Id$ then this theorem means that ${(x,x)} \in \mathcal{X} \times \mathcal{X}$ iff $X \subseteq \sigma(A)$ for some countable $A\subset X$. But this latter statement is true which is the opposite of what we want? – lady gaga Nov 02 '21 at 19:09
  • You don't want $X$ in $\sigma(A)$, you want all singletons in $\sigma(A)$, and a countably generated $\sigma$-algebra cannot give rise to more than continuum many sets. – Michael Greinecker Nov 02 '21 at 20:34