Measure Question: Is m Finitely Additive?

[homology]

Yep its me again, with another dumb question.

Say you have a set I with an ultrafilter F on it. Now I came across the following in a text on nonstandard analysis:

let m be the measure induced by F defined as m(A) =1 if A is an element of F and zero otherwise.

I know this is going to be stupid, but it doesn't seem as though m is even finitely additive. Suppose A and B are elements of F then AUB is in F since if they weren't then their complement A'^B' (A' the complement of A and B' the complement of B and ^ for intersection) would be. But if A'^B' is in F then we could intersect A with A'^B' and get the empty set, which is not an element of F. So anyways, since AUB is in F the "induced measure" m(AUB)=1, but the m(A)+m(B)=2, so what gives?

Thanks

 

[matt grime]

A and B are not disjoint.

 

[homology]

Doh! Of course, they can't be if they belong to F. So then the fact that m is finitely additive is vacuously satisfied. There are no pairwise disjoint elements of F. So likewise for the countable additivity clause. So then m is a measure because F has not pairwise...i see.

Thanks Matt Grime!