## Journal of Obscure Results 1: Nedoma’s Pathology

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) . We make by taking the sigma algebra generated by sets of the form . In the case where we have topologies on and 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 be a measurable space with . Then the product algebra on does not contain the diagonal.

In particular, if 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 be a set, and . There exist with

Proof: The set of satisfying the conclusion of the theorem is a algebra containing .

Lemma: Let be measurable. Then is the union of at most sets of the form .

Proof:

By the preceding lemma we can find with

For , define where if and otherwise.

Sets of the form then form a partition of of . Thus the sets which can be written as a union of sets of the form form a algebra. This contains each of the , and so contains . Thus . There are at most sets in this union. Hence the desired result.

Finally we have the proof of the theorem:

Let be the diagonal. Suppose is measurable. Then is the union of at most sets of the form . Because at least one of these sets must contain two points. Say and . But then it also contains . 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 , 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.