To try and settle this debate on uniform measures, the best definition
of measure I could find was at
http://www.probability.net/. Unfortunately, this site is rather
difficult to get into. However, a measure is a function m defined over
the subsets of the set O in question (eg O=Z in the case of integers). It
has the following two properties:

m(\empty) = 0

for all A\subsetO, A_n\subsetO such that A=\union_n=1^\infty A_n and A_i
\intersect A_j =\empty \forall i,j => m(A)=\sum_n=1^\infty A_n.

(called the countably additive property). In less formal terms, it
means you get the same number for your measure, no matter how you
slice the set into disjoint subsets.


Furthermore, if m(A)\in[0,\infty], the measure is called a positive
measure.

It can be readily seen that the cardinality function over the powerset
of the integers satisfies these properties, and hence is a
measure. Furthermore, it is correctly a uniform measure, as each set
element contributes equal weight to the set's measure.


I still think the problem has to do with confusing measure with
probability distribution, which must additionally be normalisable (ie
m(A)\in[0,1]). There is clearly no uniform probability distribution
over the integers, or any set that is not compact for that matter.

                                        Cheers

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