I just had an (intellectual) panic attack.
- A continuous random variable that follows a uniform in a closed interval $U(a,b)$: a comfortably familiar statistical concept.
- A continuous uniform r.v. having support over the extended reals (half or whole): not an r.v. proper, but a basic Bayesian concept for an improper prior, useful and applicable.
- A discrete uniform taking a finite number of values: let's throw a geodesic dome, no big deal.
But what about a function that has as its domain all the rationals that are included in a closed interval with integer bounds (start with the $[0,1]$ if you wish)? And we want to use it in a probabilistic framework, requiring that each possible value has equal probability with all the others?
The number of possible values are countably infinite (which characterizes a lot of discrete distributions), but how to express the probability of a single value given that we want probabilities equal?
Can we say-show-prove that such an entity is (is not) a random variable?
If not, is this another incarnation (perhaps already well-known) of an "improper prior"?
Is it possible that this entity is in some well-defined sense, however special, "equivalent" to a continuous uniform r.v.? Or I just committed a cardinal(ity) sin?
It appears that the fact that the domain is a closed interval does not let me let go. Bounded things are usually manageable.
The questions are many in order to be indicative of the internal maelstrom- I am not asking to get answers to each one of them.
At any time that I may come up with any insights, I will update.
UPDATE: the present question just acquired a constructivist sequel here.
Best Answer
This "random variable" is similar to the idea of having a flat prior on the entire real line (your second example).
To show that there can be no random variable $X$ such that $P(X=q)=c$ for all $q\in \mathbb{Q}\cap[0,1]$ and constant $c$, we use the $\sigma$-additive property of random variables: the countable union of disjoint events has probability equal to the (possibly infinite) sum of probabilites of the events. So, if $c=0$, the probability $P(X\in\mathbb{Q}\cap[0,1])=0$, as it is the sum of countably many zeros. If $c>0$, then $P(X\in\mathbb{Q}\cap[0,1])=\infty$. However a proper random variable taking values in $\mathbb{Q}\cap[0,1]$ must be such that $P(X\in\mathbb{Q}\cap[0,1])=1$, so there is no such random variable.
The key here, as you may already be aware, is that if the space is composed of finitely many points, then we can use $c>0$ and have no problem with the sum, and if the space has uncountably many points you can have $c=0$ and the $\sigma$-additivity isn't violated when integrating over the space because it is a statement about countable things. However you're going to problems when you want a uniform distribution over a countably infinite set.
In the context of a Bayesian prior, though, you can of course just say that $P(X=q)\propto 1$ for all $q\in \mathbb{Q}\cap[0,1]$ if you're willing to use the improper prior.