You won't be able to find a countable union of sets of just one point which covers [0,1], so the third axiom you state is not violated. In fact, like the law s

You won't be able to find a countable union of sets of just one point which covers [0,1], so the third axiom you state is not violated. In fact, like the law says, if you take such a countable union, the measure of the set will be 0 (e.g. the rationals). If you're going to hold that the sum of measure for all points is 1, you're not going to be able to make any continuous distributions at all. The thing is, the assignment of measure does not work like this, as in you don't just add up the measure of all the points to get the measure of the whole set. Instead, measure is also assigned to subsets of [0,1] (in particular too all the sets in the sigma algebra of your choosing). There are some laws which prevent you from whatever probabilities to everything (such as the one you mentioned about countable unions), but there is nothing that says it has to be the sum of the probability of the points.