Does it? As far as I know, all it says is that the utility function exists. Maybe it's bounded or maybe not -- VNM does not say.
VNM main theorem proves that if you have a set of preferences consistent with some requirements, then an utility function exists such that maximizing its expectation satisfies your preferences.
If you are designing an agent ex novo, you can choose a bounded utility function. This restricts the set of allowed preferences, in a way that essentially prevents Pascal's Mugging.
I don't think it would because the bounds are arbitrary and if you make them wide enough, Pascal's Mugging will still work perfectly well.
Yes, but if the expected utility for common scenarios is not very far from the bounds, then Pascal's Mugging will not apply.
you can choose a bounded utility function. This restricts the set of allowed preferences
How does that work? VNM preferences are basically ordering or ranking. What kind of VNM preferences would be disallowed under a bounded utility function?
if the expected utility for common scenarios is not very far from the bounds, then Pascal's Mugging will not apply
Are you saying that you can/should set the bounds narrowly? You lose your ability to correctly react to rare events, then -- and black swans are VERY influential.
Summary: the problem with Pascal's Mugging arguments is that, intuitively, some probabilities are just too small to care about. There might be a principled reason for ignoring some probabilities, namely that they violate an implicit assumption behind expected utility theory. This suggests a possible approach for formally defining a "probability small enough to ignore", though there's still a bit of arbitrariness in it.