Does an infinite sample space $\Omega$ imply that the probability of an elementary event is equal to zero? Suppose $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, where $\mathbb{P}$ is a probabilitiy measure such that each elementary event has the same probability. It would seem natural to me, that the two implications


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*$\Omega$ countably infinite $ \, \Longrightarrow \, $ $\forall \, \omega \in \Omega \colon \mathbb{P}(\{\omega\})=0$

*$\Omega$ uncountably infinite $ \, \Longrightarrow \, $ $\forall \, \omega \in \Omega \colon \mathbb{P}(\{\omega\})=0$


are both correct. Anyway, I am not sure if they are true or how to prove them.

Edit: If we additionally assume that each elementary event has the same probability then both assertions are correct. To see this, let $$\Bbb P(\{ω\})=ε$$ for any $ω\in Ω$ and some $ε>0$ fixed. Choose infinite but countably many of the $ω$'s (with index say $I$, so that $I$ is a countably infinite discrete set) and sum up their probabilities $$\sum_{ω\in Ι}\Bbb P({ω})=|I|ε=+\infty$$ contradicting the fact that $\Bbb P(I)$ must be finite (less or equal than $1$).

If we let different elementary events have different probabilities, then neither is correct. Two well known counterexamples of distributions with countable infinite support are the geometric distribution $$\Bbb P(X=k)=p(1-p)^k$$ for any $k\in \Bbb N$ and $0<p<1$ and the Poisson distribution $$\Bbb P(X=k)=e^{-λ}\frac{λ^k}{k!}$$ for any $k\in \Bbb N$ and $λ>0$. The second is also not correct, since you can assign positive probability only to some $ω\in Ω$. For example, let $Ω=[0,1]$ and $\Bbb P(0)=P(1)=1/2$ and $\Bbb P(ω)=0$ else. 
Actually what is correct in this case is the following: There are at most countably many $ω \in Ω$ such that $\Bbb P(\{ω\})>0$. This cannot occur when all elementary events have equal probability as requested.
