Probability of choosing a given rational number between 0 and 1 out of $\mathbb{Q} \cap [0,1]$? I have never taken a course involving measure theory but I have taken some very elementary analysis and calculus. From my understanding, one of the consequences of the reals being uncountable is that there’s probability 0 of choosing an arbitrary real number between 0 and 1 in the set of reals $[0,1]$. Since the reals are uncountable, it doesn’t contradict the theorem that probabilistic events are closed under countable union. But if we switch it to being the rationals between 0 and 1 and pose the same question, how can the probability still be 0 without it contradicting the fact that these events are only closed under countable union? I must be misunderstanding something here. 
 A: Let $X=\mathbb{Q}\cap [0,1]$. 

If the sample space is $X$, and if a specified distribution on $X$ is such that all singleton subsets of $X$ are events, then it's not possible for all singleton subsets of $X$ to have probability zero, otherwise $X$ could be expressed as a disjoint union of a countably infinite number of sets with probability zero, contrary to the requirement that the probability of $X$ itself is $1$.

For the same reason, there is no "uniform" distribution on $X$, since if all singleton subsets of $X$ had probability $p > 0$, then $X$ could be expressed as a disjoint union of a countably infinite number of sets with probability $p$, which would imply that the probability of $X$ is infinite, contradiction.

However, there are valid distributions on $X$. For example, for a given enumeration $x_1,x_2,x_3,...$ of $X$, and any infinite sequence $p_1,p_2,p_3,...$ of nonnegative real numbers whose sum is $1$, there is a unique distribution on $X$ such that $P(\{x_n\})=p_n$.
A: If you want a probability measure on $\mathbb{Q}\cap[0,1]$ then the probability of every rational number in the set indeed can not be $0$. If it was then by countable additivity the "probability" of the entire set would be zero as well. This is what happens if we simply restrict the Lebesgue measure to this set.
The catch is that there is no probability measure on $\mathbb{Q} \cap [0,1]$ that is invariant, i.e. assigns equal probability to every number, at all. We already know why this probability can not be $0$, but it can not be non-zero either. If every number in $\mathbb{Q}\cap[0,1]$ had the same non-zero probability then again by countable additivity the "probability" of the entire set would be infinite.
In some applications it is natural to ask for "random" numbers from a countable set, implying some kind of "uniform distribution". The closest one can find in such cases is to define a finitely additive probability measure by using the so-called asymptotic densities.
