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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

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

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    $\begingroup$ Implication 1) leads to $\mathbb P(\Omega)=\sum_{\omega\in\Omega}\mathbb P(\{\omega\})=0$ contradicting that $P(\Omega)=1$. $\endgroup$ – drhab Mar 30 '16 at 10:04
  • $\begingroup$ I forgot an important assumption: each elementary event should have the same probability. $\endgroup$ – user148364 Mar 30 '16 at 10:05
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    $\begingroup$ In that case it cannot happen that $\Omega$ is countably infinite. $\endgroup$ – drhab Mar 30 '16 at 10:07
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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.

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  • $\begingroup$ I think we have a problem in the second case as $\mathbb{P}$ is only $\sigma$-additive. It does not work for uncountable unions. $\endgroup$ – user148364 Mar 30 '16 at 10:21
  • $\begingroup$ @user148364 Yes, you are right. This is false, since it may well be that $P(ω)=0$ for any $ω\in Ω$ if $P$ is atomless. Let me fix this (shame on me now). $\endgroup$ – Jimmy R. Mar 30 '16 at 10:23
  • $\begingroup$ @user148364 I think that I have a proof now that works for any case. Just go back to a countable sum. If any $ω$ has positive probability then this suffices for a contradiction. What do you think? $\endgroup$ – Jimmy R. Mar 30 '16 at 10:45
  • $\begingroup$ You can add to your last sentence that at least one $P(\{\omega\})>0$, hence they are not all equal as requested by the OP, and your first sentence is wrong. $\endgroup$ – Jean-Claude Arbaut Mar 30 '16 at 10:48
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    $\begingroup$ @Jimmy R. Yes it sounds clear to me. Thanks. $\endgroup$ – user148364 Mar 30 '16 at 11:22
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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.

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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 $x \in \Omega$.

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  • $\begingroup$ I do not think that the sum diverges in every case. $\{\omega_0\} \subseteq \Omega$ is a countable subset of $\Omega$ and then the sum is equal to $\mathbb{P}(\{\omega_0\})$. $\endgroup$ – user148364 Mar 30 '16 at 10:38
  • $\begingroup$ @user148364 Yes of course, by countable I meant infinite and countable. $\endgroup$ – Zoran Loncarevic Mar 30 '16 at 10:41
  • $\begingroup$ How would the first implication be true? You would have by $\sigma$-additivity that $P(\Omega)=0$. $\endgroup$ – Jean-Claude Arbaut Mar 30 '16 at 10:50
  • $\begingroup$ @Jean-ClaudeArbaut Well, in the first case, the set of uniform probability measures on countable set $\Omega$ is empty, so if you add the condition that $P$ is such measure to the left side of implication, it is trivially true. $\endgroup$ – Zoran Loncarevic Mar 30 '16 at 10:56
  • $\begingroup$ True, but misleading. $\endgroup$ – Jean-Claude Arbaut Mar 30 '16 at 10:58

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