Banach-Tarski Paradox and Infinite Sample Spaces

Question

Recently, I was reading a paper on Stochastic Processes and I saw a footnote on page 9 of the PDF that stated:

When $Ω$ is infinite, not all of its subsets can be considered events, due to very strange technical reasons. We will disregard that fact for the rest of the course. If you feel curious as to why that is the case, google Banach-Tarski paradox, and try to find a connection.

(In this case $\Omega$ is a sample space)

I was trying to think of ways this relates to the paradox. Does it have to do with the fact that I can create an infinite number of subsets? And how does this relation imply that some subsets aren't events?

Thanks

Answer 1

Ultimately both of these are related by the surprising fact that we cannot define a universal length function on the real line. This naturally gets extended to $\mathbb{R}^n$. The idea of a universal length function can be encapsulated by the following theorem:

Let $\mathcal{P}(\mathbb{R})$ be the power set of the reals and let $f:\mathcal{P}(\mathbb{R}) \to [0,\infty]$ be a function with the following properties:

1. For intervals of the form $(a,b)$ we have $f(a,b) = b-a$.
2. For any countable sequence of disjoint subsets $\{A_n\}$ we have that $$ f\left (\bigcup_{i=1}^\infty A_n\right ) \;\; =\;\; \sum_{i=1}^\infty f(A_n). $$
3. The value of $f$ should not change under translation: for any $A \in \mathcal{P}(\mathbb{R})$ and $x \in \mathbb{R}$ define the set $x + A = \{x + a\; | \; a \in A\}$. Then $f$ satisfies $f(x+A) = f(A)$.

Then, no function $f$ with these listed properties exists.

The crux of the proof for this theorem comes down to the fact that certain uncountable sets cannot possess this notion of length. Therefore in functional analysis (and hence, probability) we focus on subsets of $\mathcal{P}(\mathbb{R})$ which will allow us to define a notion of length, and these are called $\sigma$-algebras. This is the precise reason why the Banach-Tarski paradox works. The classical BTP is that one can take a sphere of a fixed radius, and split it into two subsets which are both spheres identical to the first sphere. The reason why this works is because we can pick subsets that bypass the notion of length/volume, because they are subsets that are not part of a $\sigma$-algebra in which a notion of length/volume exists.