Can you find a real-valued random variable $X$ that is almost surely rational but also such that, for any $a, b ∈ \mathbb{Q} \cap [0, 1]$ with $a < b$, $p(X ∈ [a, b]) = b - a$? If not, can you prove that no such $X$ exists?
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Why doesn't the obvious construction work? That is, define a uniform distribution on $\mathbb Q \cap [0,1]$ with respect to its Borel measure and let $X$ be the identity map. – whuber Nov 05 '17 at 21:45
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1@whuber Maybe I'm just being dense (pun not intended), but what do you mean by its Borel measure? I know you don't mean the usual measure on $[0, 1]$ restricted to $ℚ ∩ [0, 1]$, because in that case, $ℚ ∩ [0, 1]$ itself would be a null set. – Kodiologist Nov 05 '17 at 23:31
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No it is not possible.
If such a random variable exists, we would have $\Pr(X=q)=0$ for every $q \in \mathbb{Q} \cap [0,1]$, because we can write the singleton $\{q\}$ as a decreasing intersection of intervals $[a_n, b_n]$ whose length $b_n-a_n$ goes to $0$, and then $\Pr(X = q) = \lim \Pr(X \in [a_n,b_n])=0$.
Now, the set $\mathbb{Q} \cap [0,1]$ is countable, that is, it is equal to a sequence ${\{q_i\}}_{i \in \mathbb{N}}$. One should have $\sum_{i \geq 0} \Pr(X=q_i)=\Pr(X \in \mathbb{Q} \cap [0,1])=1$ and finally we would get $0=1$.

Stéphane Laurent
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1Very nice. I would summarize it as: you can't build a uniform distribution on a countably infinite set. – Kodiologist Nov 06 '17 at 18:11