This is for a graduate level mathematical probability course.

So, to make my life easier, I'm going to Romanize all the greek letters here. Starting with a countably infinite set X, take the collection F of all subsets A of X such that either A or A-complement is finite. We define a set function P(E) on sets E in F as P(E)=0 if E is finite and P(E)=1 if E-complement is finite.

I'm supposed to show that P is finitely additive but not sigma additive. I've easily shown that P is not sigma additive, and I've shown that for any finite collection of sets E, P(union of the sets)=0 if and only if p(each set)=0. I thought I was done, but then I thought of the following which kind of stumped me:

Take two infinite sets, say A and B in F. . then A union B is infinite, so we have P(A union B)=1, but P(A)=P(B)=1, so P(A union B) does NOT seem to equal P(A) + P(B)....Am I doing something wrong here, or is this function not actually finitely additive?

## (homework) Finite additivity problem

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### Re: (homework) Finite additivity problem

Forgive my potential ignorance; I'm reading up on this on the wiki page and making guesses. But does finite additivity requires that the sets be pairwise disjoint? If so, then you've forgotten that two co-finite sets cannot be disjoint.

(In other words, not all infinite sets are co-finite. For instance, the set of all odd natural numbers is infinite but not co-finite, but the set of all natural numbers whose decimal expansion contains a digit more than once is co-finite because there are fewer than ten billion exceptions.)

(In other words, not all infinite sets are co-finite. For instance, the set of all odd natural numbers is infinite but not co-finite, but the set of all natural numbers whose decimal expansion contains a digit more than once is co-finite because there are fewer than ten billion exceptions.)

### Re: (homework) Finite additivity problem *Solved, thanks*

Doh, that's it exactly! Thanks.

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