I see that when Banach-Taraski paradox emerged we solved this problem by stating that not every subset is measurable so we restrict ourselves to nice sets which are measurable. But How? I'm confused a little about what nice sets should mean? why the axioms of a sigma algebra ensure that Banach-Taraski paradox will not occur?
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The measure of countable (in particular finite) unions is rather important here – Henry Feb 01 '19 at 18:11
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It's not the axioms of $\sigma$-algebra which guarantee it, but rather properties of measure - measurable sets have well-defined notion of volume which is preserved by isometries, so the intuitive argument "you can increase the volume" works. – Wojowu Feb 01 '19 at 18:11
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https://math.stackexchange.com/questions/2365031/the-banach-tarski-paradox-and-the-notion-of-measure – Feb 01 '19 at 18:19
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The axioms of a $\sigma$-algebra itself do not guarantee that the sets can't produce the Banach-Tarski paradox. After all, the full power set of $\mathbb R^3$ is a $\sigma$-algebra and obviously allows Banach-Tarski.
What prevents it is the defining feature of the Lebesgue $\sigma$-algebra (the measurable sets). For any set $E\subseteq \mathbb R^n$ to be in the Lebesgue $\sigma$-algebra, its Lebesgue outer measure $\lambda^*(E)$ must satisfy $\lambda^*(A) = \lambda^*(A\cap E) + \lambda^*(A \setminus E)$ for every $A\subseteq \mathbb R^n$. This is the sense in which they are "nice" with respect to measure--they never divide sets in a way that their volume doesn't add up to the original. From this it can be proved that the Lebesgue outer measure is countably additive when restricted to the measurable sets. This prevents making a Banach-Tarski-like result out of measurable sets--they always add up to the volume of their union.
eyeballfrog
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