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 ---------------------------------------------------------------------------- Dr. Russell Standish Director High Performance Computing Support Unit, Phone 9385 6967, 8308 3119 (mobile) UNSW SYDNEY 2052 Fax 9385 6965, 0425 253119 (") Australia [EMAIL PROTECTED] Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks International prefix +612, Interstate prefix 02 ----------------------------------------------------------------------------