Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to think of a probability measure on the set of rationals between 0 and 1 ($X:=\mathbb{Q}\cap[0,1]$). I want to achieve something like a uniform measure, i.e. every number should have the same probability.

Of course from the fact, that there are infinitely many atoms, it follows, that the measure has to be zero for every rational, which in turn yields a zero probability for any event, even the whole space, which is bad.

So is something wrong about my idea of uniformity?

What kind of measures can be defined on $X$?

How could one design a "completely random drawing of rationals"?

(Do we appreciate what we have with the reals, when we talk about uniformly distributed random number?)

share|cite|improve this question
related I don't know if anything else can be said here in addition. Note that $(0,1)$ can be given a uniform measure, whereas an isomorphic $\Bbb R$ can't. The point is that the "uniformity" requires some additional structure besides of measurability. E.g. Haar measures can be thought of as uniform measure over groups. – Ilya Feb 22 '13 at 17:35
Suppose you assigned to every interval a measure equal to its length. Then singletons would have measure $0$. And the measure would be finitely additive and you could close the class of measureable sets under finite unions, finite intersection, and complements, and still have finite additivity. Countable additivity is out because the measure of an interval would differ from the sum of the measures of the singletons in it. It might be of interest to ask whether you have countable additivity for some sort of "well behaved" sequences of measurable sets. The hard part would be – Michael Hardy Feb 22 '13 at 17:43
. . . ..finding the precise notion of good behavior that is suitable for the occasion. – Michael Hardy Feb 22 '13 at 17:44
what about use Peano–Jordan measure? – tom Feb 22 '13 at 18:09
@tom: which is a restriction of Lebesgue measure? – Ilya Feb 22 '13 at 19:59
up vote 1 down vote accepted

[Comment promoted to answer at request of OP]

No countably infinite set supports a uniform measure (other than the measure where every set has measure zero, and the one where each element has the same non-zero measure and the whole set has infinite measure). Drawing a rational uniformly at random is as impossible as drawing an integer uniformly at random.

share|cite|improve this answer
Drawing a rational uniformly at random is as impossible as drawing an integer uniformly at random. Why, since there a measurable isomorphism between these sets? – Ilya Feb 25 '13 at 14:36
@Ilya, since there is a set isomorphism between these sets. No countably infinite set supports a (finite, nonzero) uniform measure. – Gerry Myerson Feb 25 '13 at 22:34

Not really answering the question, but there is a section of this paper on non-Archimedean probability

on fair lotteries over the rationals. Some insight on the nature of your question can probably be gained by this, even though your question is about ordinary probability distributions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.