...How do we know which UTM is making up this infinite array?

I have to admit this question is much easier to answer from the opposite point of view. That is, we can take the point of view that only the product of the utility function by the Solomonoff measure is meaningful rather than each factor separately (like advocated by Coscott). This way changing the UTM can be re-interpreted as changing the utility function. This approach is similar to what you suggested in the sense that if we allow for arbitrary bounded utility functions, the Solomonoff factor only sets the asymptotic behavior of the product. However, if we constrain our utilities to be computable, we still cannot do away with the utility boost of "easily definable places".

But the probability approaches 0 as times goes to infinity and we test the artifact on more and more problems, right?

OK, I attacked this question from the wrong angle. Allow me to backtrack.

Consider an infinite binary sequence x. Let's call x a "testable hypothesis" if there is a program P that given any infinite binary sequence y produces a sequence of probabilities Pn(y) s.t. Pn(x) converges to 1 whereas P_n(y) converges to 0 for y =/= x. Then, it can be proved that x is computable (I can spell out the proof if you want).

An FAI likely has to make decisions that depend on whether some program halts or not. (For example, it needs to decide whether or not to invest resources into finding polynomial time solutions to NP-complete problems, but doesn't know whether such a program would halt.)

If the utility function is upper semicontinuous as I suggested in the post, the FAI won't have to answer such questions. In particular, it is sufficient to decide whether a polynomial time solution to NP-complete problems can be found in some finite time T because finding it later won't be worth the resource investment anyway.