Are you sure you get to choose a specific normalized probability function? Because then you can just number your possible outcomes by integers n and set p(n) to 1/U(n) * 1/2^n, which seems too easy to have been missed.

It might be slightly more complicated in practice, since S doesn't have to be countable if the uncountable parts can be integrated over. This doesn't violate the linked theorem because each item in the "sum" is not greater than zero, it's infinitesimal. But you could do pretty much the same thing to the probability densities inside any integrals.

So I suspect that the result is only proved for when you don't get to choose your own probability function, and that that comes up in the paper. But I agree that due to some effects you can end up with probability and utility functions that naturally are biased against high-utility situations, and this would be a counterexample to the result's applicability in those special cases.