How to prove a lemma required for the Banach Tarski Paradox? I tried to teach myself the proof of the Banach Tarski Paradox by reading Terence Tao's paper on the subject; the link to the paper is here: https://www.math.ucla.edu/~tao/preprints/Expository/banach-tarski.pdf
My question concerns Lemma 1.3.
Let $C$ be a countable subset of of $S^2$. Now if we choose a rotation $R$ from $SO(3)$ at random, then, with probability 1, we can say that $C, RC, R^2 C, R^3 C, \ldots$ are all disjoint sets. Why can we say this? How do we prove that such an $R$ are exists, and furthermore, that almost all elements of $SO(3)$ satisfy this condition? I can't seem to come up with any good reason for this.
 A: First of all, remember that "almost all" is measure-theoretic shorthand for "all but measure zero" - note that there is a natural measure on $SO(3)$, the "Haar measure" (https://en.wikipedia.org/wiki/Haar_measure), so this makes sense. 
With this in mind, the key to the proof of the lemma is the following standard result: the union of countably many measure-zero sets is measure zero. The argument goes as follows:


*

*Write $C$ as $\{c_i: i\in\mathbb{N}\}$; we can do this since $C$ is countable.

*Now, for $i, j, k, l\in\mathbb{N}$ with $i\not=k$, let $B_{i, j, k, l}$ be the set of $(i, j)$-bad rotations: $$B_{i, j, k, l}=\{R\in SO(3): R^i(c_j)=R^k(c_l)\}.$$ It's enough to show that each $B_{i, j, k, l}$ has measure zero.

*For simplicity, let's consider the case where $i=0, k=1$ - so we're trying to show that $B=\{R: c_j=R(c_l)\}$ has measure zero. But this is easy: either $l=j$, in which case $B$ is the set of rotations fixing $c_j$, which has measure zero; or $l\not=j$, in which case every element $b$ of $B$ can be written as a composition $b=f\circ g$, where $f$ is a fixed rotation sending $c_l$ to $c_j$ and $g$ is a rotation fixing $c_j$ (it doesn't matter what $f$ we choose, here), and since there are only measure-zero many such $g$, we know $B$ must have measure zero.

*The more general case of arbitrary $B_{i, j, k, l}$ is not much more complicated, so I leave it as an exercise.

*So each $B_{i, j, k, l}$ has measure zero, so the set $B_\infty=\bigcup B_{i, j, k, l}$ has measure zero; so its complement, $G$, has full measure. But by definition $G$ consists of all rotations $R$ such that for all $i, j, k, l$, with $i\not=k$, we have $R^i(c_j)\not=R^k(c_l)$. This is just a complicated way of saying that $G$ consists of all rotations $R$ such that for all $i\not=k$ we have $R^i(C)\cap R^k(C)=\emptyset$, that is, that $G$ is the set of all rotations with the desired property.

Technically there's a piece missing here - how do I know that the whole space $SO(3)$ doesn't have measure $0$? This is something we have to show, but for now it's probably best to just take it for granted.
