3rd Probability Axiom: Where did I go wrong? Today in class we're starting probability, and we went over three axioms of probability, which coincide with those here: http://en.wikipedia.org/wiki/Probability_axioms
The third axiom states: For a countable set of events $A_1, A_2, A_3, \ldots$, which are all disjoint:
$$P(A_1 \cup A_2 \cup A_3 \cup \ldots) = P(A_1) + P(A_2) + P(A_3) + \ldots$$
I thought I had found a counter-example, but since this seems to be pretty darn well accepted, I know I'm wrong somewhere.
Say I choose a positive integer. Let $A_n$ be the event that I picked $n$. Clearly the left-hand side is 1, but $\forall n ~ P(A_n) = 0$ (almost never), so the right-hand side should be 0, I think.
What's the catch?
 A: There isn't a catch, per se; the conclusion is simply that there does not exist any probability measure $P$ on the positive integers that is "uniform", i.e. such that $P(\text{choosing }n)$ is equal for all $n$, for precisely the reason you observed (I believe you are implicitly assuming this to be the case).
Suppose that $P$ were a probability measure on $\mathbb{N}$ (the positive integers) such that for any $n\in\mathbb{N}$, we have $$P(\{n\})=\alpha$$
for some constant $\alpha$. If $\alpha>0$, then because $P$ is countably additive, we have
$$P(\mathbb{N})=P(\{1\})+P(\{2\})+\cdots=\alpha+\alpha+\cdots=\infty,$$
but this is not equal to $1$; but if $\alpha=0$, then similarly we conclude that $P(\mathbb{N})=0$, which is also a problem. Thus such a $P$ cannot exist.

However, there very well can exist probability measures on $\mathbb{N}$ that are not uniform. A standard example is the measure $P$ defined by
$$P(\{n\})=\frac{1}{2^n}.$$
Then we have
$$P(\mathbb{N})=P(\{1\})+P(\{2\})+\cdots=\frac{1}{2}+\frac{1}{4}+\cdots=1$$
and all is well. With this $P$, it is not true that $P(A_n)=0$ for all $n$, so the reasoning you followed leading to a contradiction doesn't apply.
A: I believe you are mistaking two things. 


*

*The $P(A_n) = 0$ is not quite correct, as $P(A_n) \gt 0$ for at least some of the n.

*$\sum P(A_N) \ne 0$ as it is an infinite sum, and by other axioms of probability all $P(A_n) \ge 0$ such that $\sum P(A_N) = 1$ else the defined probabilities are not actually a probability distribution.


Edit to clarify: It may help to think of it this way.
Suppose $P(A_n) = 0$ for all n. Then by the rule of compliments, the $P(A_n^c) = 1 - P(A_n) = 1 - 0 = 1$, which implies that $P(A_j) \ne 0$ for some $j\ne n$, a contradiction to your original assumption
