Nice question!
The answer is no, because if a set of reals $A$ is infinite time semi-decidable, then it is $\Delta^1_2$, and not only $\Delta^1_2$, but absolutely $\Delta^1_2$, using the fact that every infinite time computation either halts in some countable ordinal number of steps or repeats (in the suitably strong sense of repeating that we mention in the main paper on ITTMs). That is, for the $\Sigma^1_2$ side, a real $x$ is in $A$ just in case there is a well-founded computation according to program $p$ accepting $x$; and for the $\Pi^1_2$ side, $x$ is not in $A$ just in case there is a well-founded computation according to program $p$ that either rejects $x$ or else is strongly repeating. But it is an old result of Bob Solovay (see Kanamori exercise 14.4) that every absolutely $\Delta^1_2$ set of reals is Lebesgue measurable.
This argument appears in my joint article:
- Sam Coskey, Joel David Hamkins, Infinite time decidable equivalence relation theory, Notre Dame Journal of Formal Logic, vol. 52, iss. 2, pp. 203-228, 2011.
In that article, Sam and I undertake to develop the analogue of Borel equivalence relation theory, but using infinite time computable reductions in place of Borel reductions. So it is a nice blend of descriptive set theory and infinitary computability. Along the way we prove:
Theorem 2.6 Every infinite time eventually computable function is absolutely $\Delta^1_2$.
Corollary 2.7 Every infinite time eventually decidable set is measurable. Every infinite time eventually computable function is a measurable function.
Note that eventually decidable is a weaker property than semi-decidable, but there are some subtle issues with partial functions, as mentioned in the article.