TITLE: "Uniform probability" on a set of naturals
QUESTION [16 upvotes]: It's an obvious and well-known fact that there is no uniform probability measure on a set of natural numbers (i.e. the one that gives the same probability to each singleton).
On a recent probability seminar one professor mentioned that this nonexistance is one of the main objections to measure-theoretic formulation of probability and that even Kolmogorov admitted the problem in it.
What are possible solutions (i.e. alternative definitions of probability) to this problem?
I'm looking either for an informative answer or a good reference.

REPLY [6 votes]: I think the usual way is to define a sequence $f_n$ of probability measures whose limit has whichever property you consider to be the essence of uniformity (invariance under translation, the evens have probability 1/2, finite additivitiy, et cetera).
For example, let $f_n(A) = |A \cap \{1,2,\dots,n\} | / n$. Each $f_n$ is a probability measure, and the limit $\lim f_n(A)$ (if it exists) is called the natural density of $A$. Another example, let $f_n(A) = \zeta(1+1/n)^{-1} \sum_{k\in A} k^{-1-1/n}$. The limit $\lim f_n(A)$ (if it exists) is the logarithmic density of $A$.
Another approach is to take an unlimited hyperinteger (as in, nonstandard analysis) $N$, and to take the measure $f(A)$ to be the standard part of $|{}^\ast A\cap \{1,2,\dots,N\}|/N$. This has a tendency to be meaningless, however, unless you can prove a result that works for all unlimited hyperintegers. Essentially, this is the same difficulty that arises in Denis Serre's answer.