I am reading this article, "A non-measurable tail set" by Blackwell and Diaconis. In this article we don't have axiom of choice, we just suppose that there exists a free ultrafilter on $\omega$. Here's something I don't understand:
$X_i=\{0,1\}$, $X=\prod_{i\in \omega} X_i$
...
Let a transformation $T : X \rightarrow X$ be called a tail transformation if there is an $n$ such that for all $x$, $T(x)$ and $x$ only differ in their first $n$ coordinates. The collection of tail transformations is countable and for any set $B\subset X$, $\bigcup_T T(B)$ is a tail set. Suppose $E$ has a measurable subset $B$ of positive measure. [...] $B^*$ is a measurable tail set [...]
I don't understand why the collection of tail transformation is countable, for me, even the collection of transformation $T$ such that $T(x)$ and $x$ only differ in the first coordinate should be at least as large as $X$, because for every $x\in X$, I can define $T_x$ to be the transformation that alters the first coordinate of $x$ while leave other $y\in X$ unchanged.
I believe you are correct, but I think that what they were trying to say is that there is a countable set of tail-maps that is inducing the "closure" of $B$: For any $n\in\omega$, let $P_n$ be the set of functions $n\to 2$ and define for each $\pi\in P_n$, $T^{\pi}$ be the transformation that sends $x$ to the sequence $(y_0,...,y_{n-1},x_n,...)$ where $y_i=\pi(i)$, and let $T^{n}$ be $\{T^{\pi}\}_{\pi\in P_n}$.
So $T^\omega=\bigcup T^n$ is countable. Let $B\subseteq X$ be any set, then $\bigcup_{T\in T^\omega} T(B)=\bigcup_{T\text{ is tail map}} T(B)$
The LHS is clearly a subset of the RHS, and indeed if $x\in \bigcup_{T\text{ is tail map}} T(B)$, there is $y\in B$, and a tail map $T'$ such that $T'y=x$, let $m$ be a natural that witness the fact that they are tail-equivalent, then there is $T^{\circ}\in T^m$ such that $T^\circ y=x$ (for $z\ne y$ we may have $T^\circ z\ne T' z$, but we don't care about that).
So the argument for $B^*$ being measurable still works