pr.probability - Polish spaces in probability

One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size $leq 2^{aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let $X$ be a measure space with $|X| > 2^{aleph_0}$. The diagonal in $X^2$ is not measurable.

We'll prove this by way of a theorem:

Let $U subseteq X^2$ be measurable. $U$ can be written as a union of at most $2^{aleph_0}$ spaces of the form $A times B$.

Proof: First note that we can find some countable collection $A_i$ such that $U subseteq sigma(A_i times A_j)$ (proof: The set of $V$ such that we can find such $A_i$ is a sigma algebra containing the basis sets).

For $x in {0, 1}^mathbb{N}$ define $B_x = bigcap { A_i : x_i = 1 } cap bigcap { A_i^c : x_i = 0 }$.

Consider all sets which can be written as a (possibly uncountable) union of $B_x times B_y$ for some $y$. This is a sigma algebra and obviously contains all the $A_i times A_j$, so contains $A$.

But now we're done. There are at most $2^{aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{aleph_0}$ sets of the form $A times B$.

QED

Corollary: The diagonal is not measurable.

Evidently the diagonal cannot be written as a union of at most $2^{aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{aleph_0}$.

• Next nt.number theory - Twin Prime Conjecture Reference
• Previous tag removed - Strong Bezout's Identity?