As I understand it Turing degrees are defined as the equivalence classes of sets under the equivalence relation defined by $x \sim y$ iff $x$ is Turing reducible to $y$ and $y$ is Turing reducible to $x$ where a set $x$ is said to be Turing reducible to a set $y$ if there is an oracle Turing machine that decides membership in $x$ when given an oracle for membership in $y$.
In http://dl.dropbox.com/u/370127/Blog/Blog2012.pdf Yu defines a T-Vitali set to be subset $V \subset 2^\omega$ such that for every Turing degree $\textbf{x}$ the cardinality of $\textbf{x} \cap V = 1$.
He claims such sets are non-measurable. I am struggling to see how they can be anything other than a null set.
My confusion is that as far as I can see if there are two Turing degrees $\textbf{x}$ and $\textbf{y}$ such that $card(\textbf{x} \cap V)$ and $card(\textbf{y} \cap V)$ are both equal to one and $\textbf{x} \cap V$ does not equal $\textbf{y} \cap V$ then there is a third Turing degree $\textbf{z}$ such that $(\textbf{z} \cap V) = (\textbf{x} \cap V) \cup (\textbf{y} \cap V)$ which has cardinality 2.
Thus $V$ can only have one element if it exists and is thus of measure zero.
Where is the error in my reasoning?
Thanks.
I have 3 remarks on your T-Vitali sets.
(1) Every T-Vitali set T has zero inner measure. This is because $T - T$ does not contain any non zero rational (Steinhaus theorem).
(2) There are T-Vitali sets of zero Lebesgue measure so they can be measurable. This is simply because if $T \subseteq 2^{\omega}$ is any T-Vitali set then so is $\{x \oplus \overline{0} : x \in T\}$ where $x \oplus \overline{0} = 0x_00x_10x_20x_30 \dots$.
(3) There are non measurable T-Vitali sets. For example, there is a T-Vitali set which is also Bernstein.