I think this version of Pascal’s mugging could be rejected if you think that “infinite negative utility” is not a phrase that means anything, without appealing to probability of 0.
However, I still accept 0 and 1 as valid probabilities, because that is how probability is defined in the mathematical structures and proofs that underpin all of the probability theory we use, and as far as I know no other foundation of probability (up to isophorism)has been rigorously defined and explored.
The fact that measure#measure_space) is nonnegative, instead of positive, is a relevant fact and if you’re going to claim 0 and 1 are not probabilities, you had better be ready to re-define all of the relevant terms and re-derive all of the relevant results in probability theory in this new framework. Since no such exposition exists, you should feel free to treat any claims that 0 and 1 are not probabilities as, at best, speculation.
Now, I know those of you who have read Eliezer’s post are about to go “But wait! What about Cox’s Theorem! Doesn’t that imply that odds have to be finite?” No, it does no such thing. If you look at the Wikipedia article on Cox’s Theorem, you will see that probability must be represented by real numbers, and that this is an assumption, rather than a result. In other words, any “way of representing uncertainties” must map them to real numbers in order for Cox’s Theorem to apply, and so Cox’s Theorem only applies to odds or log odds if you assume that odds and log odds are finite to begin with. Obviously, this is circular reasoning, and no more of an argument than simply asserting that probability must be in (0,1) and stopping there.
Moreover, if you look down the page, you will see that the article explicitly states that one of Cox’s results is that probability is in… wait for it… [0,1].
So then the probability you assign to their ability to actually affect that change can be assigned a correspondingly small, nonzero number, even if you don’t want to assign 0 probability.