Let $\alpha$ be an admissible ordinal and suppose $X \subset L_{\alpha}$. Then $X$ is regular (or amenable) if for every $x \in L_{\alpha}$ we have $X \cap x \in L_{\alpha}$.
Working in $\alpha$-recursion theory, Sacks has shown that every non-regular $\alpha$-r.e. set has the same $\alpha$-degree as some regular $\alpha$-r.e. set; hence, if $\alpha$ is admissible, we need not worry about irregular r.e. sets.
Regular sets play an important role in higher recursion theory: we want to be able to consider every subset of $L_{\alpha}$ (or equivalently $\alpha$, since it’s admissible) to be allowed as an oracle. This fails if $X \subset L_{\alpha}$ is not regular.
I’ve seen multiple times that if $V=L$ then every subset of $\alpha$ is regular (given some conditions on $\alpha$; I don’t know what those conditions are, however); this is important as then $L_{\alpha}$ is a suitable domain for computability theory.
My questions are:
(1) what exactly goes wrong if $X \subset L_{\alpha}$ is not regular? I presume the theory of r.e. degrees does not behave as it should, but I have never seen an argument how exactly this manifests.
(2) where can I find a proof of the latter fact that $V=L$ implies all subsets of $\alpha$ are regular (provided some conditions on $\alpha$; what are those conditions btw?)? I looked in Sacks’ and Barwise’ books, but to no avail—I believe I’m not sure what exact statement I’m looking for.
Any advice is very much appreciated!
PS: one mention of the relationship between $V=L$ and regularity of subsets is given in Chong’s “Techniques of admissible recursion theory” on page 9: “Assuming $V=L$, every subset of an infinite cardinal $\kappa$ is $\kappa$-regular.”