Asssume we have a radioactive particle with a half-life of 1. Then there is a countably additive probability function, whose domain includes all open intervals (x, y) and is closed under union and complementation, such that Pr(S) is the probability that the particle’s decay time is in S.

In cases where Pr(T) is non-zero, we can define Pr(S|T) in the usual way as Pr(S&T)/Pr(T). But even in cases where Pr(T) is zero, we might like to be able to have Pr(S|T) as defined.

Let T then be the set of rational numbers. (Note that if the domain of Pr is closed under countable union and complementation, then T will be in the domain.) Now we might wonder what Pr( |T) looks like. That is, we might wonder what Pr looks like when we conditionalise on T.

I think, and if I’m wrong here I’d welcome having this pointed out, that these conditional probabilities are not defined. And not because Pr(T)=0. In lots of cases probability conditional on a zero-probability event can be sensibly defined. But in this case, if there were such a thing as Pr( |T), then for any rational number *x*, Pr({*x*)|T) would be 0. And that would lead to a failure of countable additivity.

I imagine all of this is well known, but I hadn’t realised the consequences of this. Let D be the smallest set of sets of positive reals that includes all open intervals (x, y) and is closed under countable union and complementation with respect to the reals. Then there is no *conditional* probability function from D x D\{} -> [0, 1] such that for any open interval (x, y), Pr((x, y)|R) is the chance that the particle will decay in (x, y). (By R here I mean the set of all reals.) If there is any function that has this last property, it must be defined over a narrower domain than D x D\{} -> [0, 1].

Why is it that “if there were such a thing as Pr( |T), then for any rational number x, Pr({x)|T) would be 0.”? I could see that it would be extremely plausible that this would be the case, but you seem to have given a convincing argument against it. It might seem somewhat strange for different singletons to get different probabilities conditional on T, but it doesn’t contradict anything you’ve said in your description of the situation, at least not in any obvious way.

Additionally, we might assume that our probability function is countably additive, but this example points out that countable additivity doesn’t entail conditional countable additivity on every condition. If these were the credences of some rational agent, and we thought that rational agents always updated by conditionalization, and any (non-empty) proposition could be the one conditionalized on, then conditional countable additivity would follow from a requirement that rational agents always satisfy countable additivity. But this is a chance function, and just because the chances at any time must satisfy countable additivity doesn’t entail that all conditional chances satisfy countable additivity as well. That only follows for cases where we conditionalize on a possible set of all events that occur between the time of the chance function and some future time (assuming that chances evolve by conditionalization).