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 *provably* $\Delta^1_2$, 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](http://www.cs.nyu.edu/pipermail/fom/2008-April/012792.html) that every provably $\Delta^1_2$ set of reals is Lebesgue measurable.