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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?)

Johannes Gerer
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    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. – SBF Feb 22 '13 at 17:35
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    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? – SBF Feb 22 '13 at 19:59
  • No it is not. For example $\lambda(\mathbb{Q}\cap[0,1]) = 0$ but $m(\mathbb{Q}\cap[0,1]) = 1$. Where $\lambda$ is Lebesgue measure and $m$ is Peano-Jordan measure. – tom Feb 22 '13 at 20:20
  • @tom But $\mathbb Q\cap[0,1]$ is not Jordan measurable, is it? – Did Feb 22 '13 at 22:45
  • 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. – Gerry Myerson Feb 23 '13 at 06:29
  • Oh, that makes sense. Would you mind putting that in an answer? – Johannes Gerer Feb 23 '13 at 14:10
  • @Ilya: I think the notion of uniformity you and OP have in mind is properly generalized if you consider measure $G$-spaces for a given group. $\bf R$ has no probability measure invariant under its action on itself by addition, and ${\bf Q}\cap [0,1]$ has no measure invariant under the obvious action of the group of finitely supported permutations, which is what OP asked about. – tomasz Jul 27 '13 at 13:34

3 Answers3

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[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.

Gerry Myerson
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  • 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? – SBF 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
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Perhaps you should read the recent work of Professor Aerts and Redei on the classical interpretation of probability where to avoid issues either with the classical interpretation (Bertrand's paradox) or countable additivity, t ascribe something a uniform probability measure over an infinite number of atomic/elementary events .

Miklos Redei, Z Gyenis makes use of something like the Haar measure; see https://www.google.com/search?q=redei+bertrands+paradox

see also see https://link.springer.com/chapter/10.1007/978-3-319-23015-3_20

They use this to discuss the classical interpretation of probability and the issues that arise with a uniform measure and the principal of indifference with regard to Bertrand's paradox, or otherwise countably additivity and normalizatiom.

Its a bit on issue to satisfy both to preserving a

(A) uniform discrete measure (principal of indifference) (measure) invariance,

(B)Whilst also avoiding what is a violation of called 'labelling irrelevance'

Labelling irrelevance is the ontic counterpart to (A) and its violation is the counter-part to frequentism's reference class problem for the classical interpretation, i.e., Bertrand's paradox; where one wishes to preserve the same probability of the individual events (B)under relabeling whilst maintaining a uniform metric (A) (PI) ; that is without this changing the uniform metric, and vice versa. I.e., maintaining (B) without having the invariance of the uniform metric being violated to ensure ie that when one looks at the reference of tables produced with side length n, in an uncountable number of them, that if looks at n^2, their area, that one has to ascribe a non uniform metric (violating (A)) so that the probabilities densities of the side lengths remain the same, or one keeps (A), and the density remains uniform (but the individual probability densities of that side length change, as its now a uniform measure over N^2).

having to change of the metric) whilst at the same time trying to avoiding issues normalization or countable additivity. Its an effort to get around Bertrand's paradox, on the one hand and issues pertaining to countable additivity and normalization on the other, as far I can gather

And when there are countably infinite many events all with the same probability (much worse when there uncountably many) one can cannot,have a discrete uniform or equal probability for each; whilst if there are un-countably many events, not only do do all have probability zero, and other issues arise due to bertrands paradox (when one now has to consider finite partitions, or reference classes, and the uniform measure over that set of events is not preserved.

'The Uniform measure/or principal of indifference, or translsation in-variance' is preserved, (preservation of the uniform metric) often at the expense of labelling irrelevance,(the probabilities of the actual events or reference classes remaining the same under relabelling). And there often a fundamental conflict between the two .

Different event might get ascribed different probabilities, and thus the probability of the events are not ontic/ labelling irrelevant or invariant, at the expense of maintainng the uniform measure/Principal of indifference, under translation-that all events have the same probability or a uniform metric, translation in-variance.

One can maintain it, but (the uniform metric) but how it gets distributed amongst the events (or the probability density of the actual events changes; aka betrands paradox).

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

Vieri Benci, Leon Horsten, Sylvia Wenmackers: Non-Archimedean Probability, https://arxiv.org/abs/1106.1524

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.

Martin Sleziak
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John Starrett
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