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].