The probability of picking a rational number in the segment $ [0,1] $ I'm interested in something that came to my mind regarding the probability function.
We know that the probability function is additive over countable sets.
Now, let's take the set $ A $ to be the set of all rational numbers in the segment $ [0,1] $. Assume I want to calculate the probability to pick $ 0.5 $.
My initial thought would be that the probability would be $ 0 $, because there are infinity rational numbers in the segment $ [0,1] $, and intuitively I cannot see a reason that the probability to pick $ 0.5 $ would be different from the probability to pick $ 0.23 $. So I'll assume that the probability of any rational number in the segment should be equal, thus, it has to be $ 0 $.
In the other hand, if it is indeed $0$ probability for any rational number, since $ A $ is a countable set we can sum the probabilities of all the rational numbers in $ [0,1] $ and get that the sum equal to $ 0$. That's of course a contradiction because $ P(A)=1 $.
So I guess my first assumption that the probability of any rational number is equal, is incorrect. It means that some rational numbers has a higher probability to be picked than others.
Can someone explain, intuitively, how can it be? And is there a way to calculate the probability of each rational number to be picked?
Thanks in advance.
 A: What you have written down is half of a correct proof by contradiction that there exists no probability measure on $A$ such that every rational number has the same probability (the other half is to show that if the probability of some rational number is positive then the measure can't have total measure $1$). All you use about $A$ is that it's countably infinite so in fact the same is true of any other countably infinite set, say the natural numbers.
What this means that there is no such thing as "the" probability of anything on a countably infinite set. We have to pick a probability measure and there are many and the probability measure must inevitably privilege some members over others. The probabilities can be any countably infinite set $p_i, i \in I$ of nonnegative real numbers such that
$$\sum_{i \in I} p_i = 1$$
which means in particular that the $p_i$ must converge to $0$ (what this means if we don't pick an ordering on $I$ is that for any $\epsilon > 0$ there exists a finite subset $S \subset I$ such that if $i \not\in S$ then $p_i < \epsilon$).
