AD implies that each set of Turing degrees either contains a cone or is disjoint from one. It follows that the set of Turing degrees has a countably additive measure: $\mu(A) = 1$ if $A$ contains a cone and $\mu(A)=0$ otherwise.
In Classical Recursion Theory by Odifreddi it is said, that the measure on $\omega_1$ can be defined by giving a set $A$ of countable ordinals the same measure as the set of degrees containing well-orderings with ordinal in $A$.
If I understood correctly that means $\mu'(A) = \mu(\{{\bf a}\in\mathcal D : x\in {\bf a}\text{ for some well-ordering isomorphic to some ordinal from } A\})$ for each $A\subset \omega_1$.
For me it is not quite clear, why this is a measure, in particular why $\mu'(\omega_1)=1$. It should follow from some fact like ''each Turing degree contains a well-ordering of natural numbers''. But I don't know, if it is some theorem or a simple fact.
Edit:
The problem was with the definition of the measure, which is incorrect, since both $Succ \cap \omega_1$ and $Lim\cap\omega_1$ have a measure one in this case.
The right definition is $\mu'(A) = \mu\{{\bf a} : \omega_1^X = \alpha \text{ for } X\in {\bf a} \text{ and } \alpha\in A\}$. Then $\mu'(\omega_1)=1$, since $\{{\bf a} : \omega_1^X = \alpha \text{ for } X\in {\bf a}\text{ and } \alpha\in \omega_1\}=\mathcal D$, because set of well-orderings in each Turing degree is bounded in $\omega_1$. And the other properties of measure can be pulled back from $\mu$.
Is this right?