[waps]

Why isn't the answer P = { Inf -> 1, otherwise 0 } ?

Axiom 1: P(E) elem N => P(E) >= 0, for all E

Trivially satisfied

Axiom 2: P(Omega) = 1

Satisfied: Omega = N { Inf } elem N

Axiom 3: Sigma additivity. Trivially satisfied since it either includes { Inf } or it doesn't, making the outcome 0 or 1.

Where is the problem ?

I think it's pretty clear that this is the only possible solution, because since N is not closed, there is no way to keep a uniform density other than 0.

This does not seem like it's a very useful solution, but it does seem to satisfy the axioms.

[jerf]

In addition to anonymoushn's correct point, I would also observe that even if we do treat infinity as a specific real number, the integration of a function zero at everywhere except one specific point from -inf to +inf is still 0. And further if we are treating it as a real number, you've now violated the uniformity constraint; P(0) != P(inf), therefore it is no longer a uniform distribution, the thing you are putatively building.

For an analysis of this and some other interesting things, see the Dirac delta (psuedo-)function: http://en.wikipedia.org/wiki/Dirac_delta_function

[anonymoushn]

Infinity isn't a positive real :(

[waps]

In most definitions I've seen of it, it is. In fact, it's a natural number.