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I know that https://en.wikipedia.org/wiki/Natural_density

But, I want to know if there is uniform probability in every subset of natural numbers.

It means,

  1. every subset $A \subset \mathbb{N}$, $P(A)$ is well-defined, $0 \leq P(A) \leq 1$

  2. $P( \emptyset )=0$, $P(\mathbb{N})=1$

  3. $A=\sqcup A_i$ (disjoint union), $P(A) = P(A_1 )+...+P(A_n )$

  4. $P(A+n)=P(A)$ (Here, $A+n = \{ a+n \in \mathbb{N} ~|~ a \in A\}$ )

Or is there a counter example that shows that such a uniform probability distribution does not exist?

추민서
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  • There can't be a uniform probability on an infinite countable set. Probability is countably additive. – lulu Oct 01 '22 at 11:49
  • @lulu I means 'similar' probability. – 추민서 Oct 01 '22 at 11:51
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    I don't know what that means. Either $P(n)$ is $0$ or it isn't (for some fixed natural number $n$). If it is $0$, then countable additivity means that $P(\mathbb N)=0$. If it is $>0$ then countable additivity means $P(\mathbb N)$ is infinite. Neither are possible. – lulu Oct 01 '22 at 11:52
  • @lulu I think it's worth considering the non-countably additive similarity probabilities. – 추민서 Oct 01 '22 at 11:55
  • If you have some non-standard axiom system in mind, you need to tell us what it is. We can hardly guess. – lulu Oct 01 '22 at 11:55
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    Note, though, that even with just finite additivity (as in your third assumption), if $P(1)=\epsilon>0$ then for sufficiently large $N$ we have $\sum_{i=1}^N P(i)>1$, another contradiction. Note that your fourth assumption tells us that $P(i)=P(1)$ for all $i$. – lulu Oct 01 '22 at 11:56
  • In there, I think finite set has zero probability. – 추민서 Oct 01 '22 at 11:57
  • Let us continue this discussion in chat. – 추민서 Oct 01 '22 at 11:59
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    I think your question needs clarification. Of course one could say that all finite sets have measure $0$, and all infinite sets have measure $1$. So, sure. It's possible. See this duplicate. The Axiom of Choice no doubt lets you find others. – lulu Oct 01 '22 at 12:04
  • here is a possibly interesting paper on the subject. – lulu Oct 01 '22 at 12:13
  • the group of relative integers is amenable: https://en.wikipedia.org/wiki/Amenable_group so there is a finitely additive measure on it – Albert Oct 01 '22 at 12:17
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    see also https://mathoverflow.net/questions/95954/how-to-construct-a-continuous-finite-additive-measure-on-the-natural-numbers and https://math.stackexchange.com/questions/485443/finitely-additive-not-countably-additive-on-bbb-n. this is definitely not an easy question – Albert Oct 01 '22 at 12:26
  • @lulu I upvoted your comment before finding an error with it (and I can't undo it now); in order to counteract my misplaced vote, I will point out what's wrong: the even and odd numbers are both infinite, and assigning 1 to each would lead to $P(\mathbb N)=2$. – D.R. Oct 01 '22 at 19:50
  • @D.R. you are correct. There isn't such a simple way to define a finitely additive measure that assigns a measure to all sets. – lulu Oct 01 '22 at 19:53

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