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I was just curious, since the B-T paradox is a measure theoretic result, if there are any consequences of this paradox in probability theory? Also, is there is a way of stating the B-T paradox in the language of probability theory?

I am ultimately interested in finding an application of the B-T paradox in physics which leads to an experimental prediction.

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The B-T paradox ultimately boils down to the non-amenability of the free group of rank 2. I don't see how it would be applicable to probability. – Mustafa Gokhan Benli Aug 18 '10 at 17:19
The B-T paradox does not arise from any contradiction, Harry. It is only a paradox in that it states the truth of something which we find unintuitive, but that is only a reflection of the fact that our axioms do not perfectly model our intuitions. – Mariano Suárez-Alvarez Aug 18 '10 at 17:20
The premise of this question is not only false, but the exact opposite of the true state of affairs: the Banach-Tarski paradox is $\textit{not}$ a "measure-theoretic result". On the contrary, as has already been pointed out, it shows that certain geometric facts $\textit{cannot}$ be fit into measure-theoretic paradigm. I am voting to close this as "not a real question". – Victor Protsak Aug 18 '10 at 17:28
With regard to your second remark, I do not believe there are applications of B-T in physics. Physics uses mathematics as a framework to build models that correspond to the physical world, but those models will certainly exclude non-measurable objects like the sets involved in B-T, for the simple reason that such objects do not well describe the physical world. So I think no accepted physical model is going to make any predictions that involve B-T in any way, much less ones that you could test experimentally. – Nate Eldredge Aug 18 '10 at 17:52
Victor, please see my comment to Artem's answer below. I think the question, "Are there any consequences of this paradox in probability theory?" is quite valid, even though I have no idea how to answer it. I suspect an interesting answer will deal with chopping up balls or other convex sets in the Banach space of continuous functions, the natural space where stochastic processes lie. Matt, search through the literature on Gaussian measures on Banach spaces. Here's a preliminary Google search:… – Tom LaGatta Aug 18 '10 at 18:40
up vote 4 down vote accepted

I thought the whole point of having a $\sigma$-algebra for your probability space was to avoid non-measurable sets like the ones used in the proof of BT. Hence, it would seem that the BT paradox would be impossible to state in probability theory on account of the sets you need not being present in your algebra... but I might be mistaken, can someone else comment more?

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Correction: all subsets of $\mathbb{R}^3$ form a $\sigma$-algebra, but it's impossible to define a $\sigma$-finite measure on it (due to a variant of the Vitali construction). This was basically the content of my comment above. – Victor Protsak Aug 18 '10 at 17:51
Artem, we frequently deal with non-measurable sets in probability theory. For example, when we have a stochastic process $\xi_t$, we are concerned with the filtration (increasing sequence of $\sigma$-algebras) $\mathcal F_t$ generated by the process at times $s \le t$. Consequently, an event such as $\{ \xi_2 \ge 0 \}$ is not measurable with respect to the $\sigma$-algebra $\mathcal F_1$. My point with this comment is not to answer Matt's question; rather, it's to say that you can't easily dismiss it by saying that we only work with measurable sets in probability. – Tom LaGatta Aug 18 '10 at 18:33
...and let us not forget conditional expectations.. – efq Aug 18 '10 at 20:15

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