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I am reading N. H. Bingham's (2010) paper: ''Finite additivity vs. countable additivity''.

Page 8, he states:

"One area where the distinction between finite and countable additivity shows up most clearly is in the question of a uniform distribution over the integers. In the countably additive case, no such distribution can exist (the total mass would be infinity or zero depending on whether singletons had positive or zero measure). In the finitely additive case, such distributions do exist (all finite sets having zero measure)''.

I understand his point for the countable additive case, but do not understand why, in the finitely additive case, a uniform distribution over the integers exists. Could somebody explain? Thank you.

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    $\begingroup$ Does “uniform distribution” here mean only that all singletons have the same measure (in which case you can just take any nonprincipal ultrafilter), or are there more requirements? $\endgroup$
    – Emil Jeřábek
    Commented Aug 7, 2014 at 15:19
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    $\begingroup$ In the article your quote ends with a reference to stat.cmu.edu/tr/tr814/tr814.pdf. Does that answer your question? $\endgroup$
    – Ben Barber
    Commented Aug 7, 2014 at 15:24

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This refers I think to invariant means on $\mathbb{Z}$, namely translation-invariant ways to assign a "size" between 0 and 1 to every subset of $\mathbb{Z}$. These are equivalent to translation-invariant finitely additive measures, so that, for example, the even integers will necessarily have size 1/2. Such invariant means exist in any amenable group. For a basic and elementary account of this circle of ideas, starting from scratch, see Chapter 4 of Cellular Automata and Groups by Ceccherini-Silberstein and Coornaert.

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