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I was wondering if there is any known examples of measures on the set of rational numbers besides Lebesgue measure. In particular, an example of a probability measure on $\mathbb{Q}$ would be nice to see.

Here is a somewhat naive attempt: Let $X = \mathbb{Q}, \Sigma = 2^X $, the sigma algebra which is the set of all subsets of $\mathbb{Q}$. Now for $E \subset \Sigma$ define $m(E) = \lim_{n \to \infty} \frac{|E \cap \{ q_1, q_2, ..., q_n \}|}{n}$ where $(q_j)_{j=1}^\infty$ is an enumeration of the rationals and $|A|$ denotes the cardinality of a finite set $A$. Is this a probability measure?

Mustafa Said
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  • How do you define the limit if it doesn't exist? – N. S. Oct 31 '14 at 20:09
  • can you give me an example of a set for which it does not exist? – Mustafa Said Oct 31 '14 at 20:10
  • For any fixed set $E \in \mathcal{P(\mathbb{Q})}$ the sequence $a_n = \frac{|E \cap {q_1, ..., q_n }|}{n}$ is a bounded and monotonic sequence, so the limit exists. – Mustafa Said Oct 31 '14 at 20:15
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    It is not monotonic. For example, consider the following set of indexes $$ \cup_{ n \geq 1} {2^{2n}, 2^{2n}+1,..., 2^{2n+1}-1 }$$ For this set the limit doesn't exist. Look at your fraction when the denominator is an odd vs. an even power of 2. – N. S. Oct 31 '14 at 20:17
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    To make things clearer, your sequence for which you wrote the limit is increasing from $n$ to $n+1$ if $q_{n+1} \in E$ and decreasing if $q_{n+1} \notin E$. So by having long strings of numbers in $E$ followed by long strings of numbers not in $E$, you can make the sequence oscilating, and this is exactly what I did in the above example. – N. S. Oct 31 '14 at 20:27

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A simple example of such a measure is the following, but this is a boring example.

Let $\{ q_n \}_{n=1}^ \infty$ be an enumeration of the rationals.

Define $$m=\sum_{k=1}^\infty \frac{1}{2^k} \delta_{q_k} \,.$$ This measure can be defined alternatelly $$m(E) = \sum_{ q_n \in E} \frac{1}{2^n}$$

N. S.
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  • Could you please show why this measure you propose is a probability measure on $\mathbb{Q}$? In other words, why does it satisfy countable additivity and why does it yield results between [0,1], 0 for the empty space and 1 for the entire space? Thank you very much. – EoDmnFOr3q Nov 02 '20 at 21:12
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No, your suggestion is not a probability measure as it is not additive over countable unions of disjoint sets.

If $t$ is a natural number then $m(\{q_t\})= \lim_{n \to \infty} \frac1n=0$ and so $\sum_t m(\{q_t\})=0$. But $m( \mathbb{Q})= \lim_{n \to \infty}\frac{n}{n}=1$.

You need a discrete measure.

For example $\mathbb{Q} \cap (0,1)$ you could use $m \left( \left\{ \frac{a}{b} \right\}\right) =\frac{\zeta(k)}{\zeta(k-1) - \zeta(k) } \left(\frac{1}{b}\right)^k$ for some $k\gt2$ and $a$ and $b$ coprime; this can be extended to all rationals.

Henry
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  • Is $\zeta$ the riemann zeta function? How do you prove that your $m$ is a measure. A reference would be nice. – Mustafa Said Oct 31 '14 at 20:28
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    It gives a positive value to each rational in $(0,1)$ and $\sum \left(\frac{1}{b}\right)^k = \frac{\zeta(k-1) - \zeta(k) }{\zeta(k)}$ where $\zeta$ is the Riemann zeta function, so the total probability is $1$. See http://math.stackexchange.com/questions/189813/is-this-graph-based-on-rationals-familiar for a picture. – Henry Oct 31 '14 at 21:27
  • See Wikipedia saying $\displaystyle \sum_{n=1}^\infty \frac{\varphi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}$, pointing at Hardy & Wright (1979), An Introduction to the Theory of Numbers, theorem 288. – Henry Oct 31 '14 at 21:36
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Note that $m({q}) = 0$ for every $q \in \mathbb{Q}$, whereas $m(\mathbb{Q}) = 1$, so $m$ cannot be $\sigma$-additive. This argument shows that there are no probability measures on a denumerable set that assign the same measure to every singleton.

Moreover, the limit in your definition does not always exist.

There are, however, interesting examples of finitely additive measures (or probability charges). The example you mention if in fact to being a probability charge, but one must be careful enough to consider the cases where the limit does not exist see this article.

Henry Brown
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    A related article on finitely additive measures, from much earlier: http://matwbn.icm.edu.pl/ksiazki/fm/fm65/fm6513.pdf – John Hughes Oct 31 '14 at 20:22