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.
 A: 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.
A: If $\Omega$ is countable and the sets $\{\omega\}$ are measurable then: $$\mathbb P(\Omega)=\sum_{\omega\in\Omega}\mathbb P(\{\omega\})\tag1$$
If secondly $\mathbb P(\{\omega\})=c$ for each $\omega\in\Omega$ then (1) shows that $c=0$ leads to $\mathbb P(\Omega)=0$ hence contradicts $\mathbb P(\Omega)=1$.
If $c>0$ and $\Omega$ is not finite then (1) also leads to a contradiction: $P(\Omega)=+\infty\neq1$.
So the mentioned conditions can only be satisfied if $\Omega$ is finite.

If $\Omega$ is uncountably infinite then again $c>0$ leads to a contradiction. 
For any countably infinite $A\subset\Omega$ we find $\mathbb P(A)=\sum_{\omega\in A}\mathbb P(\{\omega\})=+\infty>1$.
In that case we must have $c=0$ wich will not lead necessarily to a contradiction.
A: Both implications are false. Note that for any $x \in \Omega$, the Dirac measure defined as 
$$\delta_x(A)=\begin{cases} 0, & x \notin A \\ 1, & x \in A \end{cases}$$
for any measurable set $A \in \mathcal{F}$ is probability measure, regardless of the set $\Omega$.
Answer to the edited question: If you restrict your attention to just the probability measures that give each singleton the same probability, then there is no such measure in the first case, since
$$\bigcup_{x \in \Omega} \{x\}=\Omega$$
Probability $P(\Omega)=1$, but if $P(\{x\})=\varepsilon \neq 0$ for all $x \in \Omega$, the sum $\sum_{x \in \Omega} P(\{x\})$ diverges.
To see that second implication is true, observe any infinite countable subset $I \subset \Omega$. Then, again,
$$\bigcup_{x \in I} \{x\} \subset \Omega$$
We know that $P(\Omega)=1$, but sum $\sum_{x \in I} P(\{x\})$ diverges if $P(\{x\}) \neq 0$, for all $x \in I$. Hence, $P(\{x\})=0$, for all $x \in \Omega$.
