In Rosenthal's A First Look At Rigorous Probability Theory one of the phrases about a random variable $X$ having a Uniform Distribution from $0$ to $1$ is the following:
...But now suppose we ask, what is the probability that $X$ is rational? ... More fundamentally, are all probabilites such as these necessarily even defined ?
The next part of the book then proves that there doesn't exist a definition of probability defined on all subsets in $[0,1]$ that satisfy "reasonable" properties.
However, I would argue that the above question: what is the probability that $X$ is rational, is defined, and equal to $0$.
My reasoning is that because the rational numbers are countable, then the subset of all rational numbers between $0$ and $1$ must also be countable (this follows from a proof I've done in real analysis that states a subset of a countable set is countable). Therefore, the probability that $X$ is rational is therefore $$P(X \in \Bbb{Q} \cap [0,1] ) = \sum_{r \in \Bbb{Q} \cap [0,1] } P(r) = 0$$
With the last equality holding because the probability of a single point in a continuous distribution is $0$ and the probability of a union of countably infinite disjoint events is the sum of the probability of each event.
Does my argument make sense? And if so, then what sort of subset in $[0,1]$ would it not be possible to define the probability that $X$ lies in the given subset?
Thanks for any clarification given.
Well, you're not arguing against the question, you merely have a (correct) answer for it: In the usual uniform probability measure on $[0,1]$, the measure of $\mathbb Q\cap[0,1]$ is zero.
But merely that you know an answer doesn't invalidate asking the question in the first place. It looks like the book you're quoting from poses that question as a way to motivate developing measure theory rigorously instead of just getting by with hand-wavy intuition.
As Hans Lundmark notes in a comment, the classical example of a non-measurable set would be a Vitali set. All such examples are strange beasts that cannot be completely specified in finite space, because proving they exist requires the axiom of choice. (There are, under certain plausible conditions, models of standard set theory except the axiom of choice where all the subsets of $[0,1]$ that the model knows are in fact measurable -- but they can then be pretty weird in other ways).