Here begins a new series, akin to the “Silly Proofs” series. Obscure results which are cool, but which you probably haven’t heard of.

Suppose we’ve got a pair of measurable spaces (sets with a \sigma algebra on them) \(X, Y\). We make \(X \times Y\) by taking the \sigma algebra generated by sets of the form \(A \times B\). in the case where we have topologies on \(X\) and \(Y\) and are giving them their Borel algebras, we might suppose this agrees with the Borel algebra of the product. Alas, ’tis not so! It does in the second countable case, but \in general not:

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 which is not contained in the product algebra.

The proof proceeds by way of two lemmas:

Lemma: Let \(X\) be a set, \(\mathcal{A} \subseteq P(X)\) and \(U \in \sigma(\mathcal{A})\). There exist \(A_1, \ldots, A_n, \ldots\) with \(U \in \sigma(A_1, \ldots, A_n, \ldots)\)

Proof: The set of \(U\) satisfying the conclusion of the theorem is a \(\sigma\) algebra containing \(\mathcal{A}\).

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

Proof:

By the preceding lemma we can find \(A_n\) with \(U \in \sigma (\{ A_m \times A_n \})\)

For \(x \in \{0, 1\}^{\mathbb{N}}\), define \(B_x = \bigcap C_n\) where \(C_n = A_n\) if \(x_n = 1\) and \(A_n^c\) otherwise.

Sets of the form \(B_x \times B_y\) then form a partition of of \(X^2\). Thus the sets which can be written as a union of sets of the form \(B_x \times B_y\) form a \(\sigma\) algebra. This contains each of the \(A_m \times A_n\), and so contains \(U\). Thus \(U = \bigcup { B_x \times B_y : B_x \times B_y \subseteq U }\). There are at most \(2^{\aleph_0}\) sets in this union. Hence the desired result.

Finally we have the proof of the theorem:

Let \(D\) be the diagonal. Suppose \(D\) is measurable. Then \(D\) is the union of at most \(2^{\aleph_0}\) sets of the form \(A \times B\). Because \(|D| > 2^{\aleph_0}\) at least one of these sets must contain two points. Say \((u, u)\) and \((v, v)\). But then it also contains \((u, v) \not\in D\). This is a contradiction.

QED

To be honest, this theorem doesn’t seem that useful to me. But knowing about it lets you avoid a potential pitfall – when you’re dealing with measures on large spaces (e.g. on \(\{0, 1\}^{\kappa}\), which it’s really important to be able to do when you’re playing with certain forcing constructions) things are significantly less well behaved than you might hope.