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Let $(\Omega, \mathcal{F}, P)$ be a probability space.

Call $P$ strictly positive if $P(F)>0$ for all $F \in \mathcal{F} \setminus \{\emptyset\}$.

Call $F \in \mathcal{F}$ an atom if $P(F)>0$ and $P(A)=0$ for all strict, measurable subsets $A$ of $F$.

Note that if $P$ is strictly positive and $F$ is an atom, then the only strict, measurable subset of $F$ is $\emptyset$.

Call $P$ discrete if there exists a countable set $\mathcal{D}$ of pairwise disjoint atoms such that $\sum_{F \in \mathcal{D}}P(F)=1$.

If $P$ is strictly positive, does it follow that $P$ is discrete?

If $\Omega$ is countable, then $\mathcal{F}$ is generated by a countable partition, the cells of which are atoms, and the result follows.

If $\Omega$ is uncountable, then either $\mathcal{F}$ is finite or uncountable; $\mathcal{F}$ cannot be countably infinite. In the case where $\mathcal{F}$ is finite, it seems obvious to me that the result holds and I will omit a proof. The case that I am stuck on is the final one where $\mathcal{F}$ is uncountable.

aduh
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2 Answers2

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First, note that any two atoms are either equal or disjoint. For if $F_1, F_2$ are atoms with $F_1 \cap F_2 \ne \emptyset$, then $F_1 \cap F_2^c$ is a strict measurable subset of $F_1$, hence empty, meaning $F_1 \subseteq F_2$, and the reverse inclusion by symmetry.

Next, note that the number of atoms is at most countable; see Is a family of disjoints atoms in $\sigma$-finite neasurable space at most countable?. So the union $A$ of all atoms is measurable. If $P(\Omega \setminus A) = 0$ then $\Omega = A$ and the space is discrete, so we are done. Hence assume $P(\Omega \setminus A) > 0$. By rescaling this reduces us to the case of an atomless probability space.

However, on an atomless probability space, there exists a random variable $U : \Omega \to \mathbb{R}$ whose distribution under $P$ is $U(0,1)$; see How to split an integral exactly in two parts. Now the events $\{U = x\}$, as $x$ ranges over $[0,1]$, are all measurable and have probability zero, so they must all be empty. But their union has probability 1, a contradiction.

Nate Eldredge
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  • Thanks very much, Nate. I'd really like to find an elementary argument that doesn't appeal to the existence of $U$, but I think I can figure it out now with the ideas in your answer. – aduh Nov 15 '17 at 14:23
  • @aduh: I think it is always going to come down to the fact that an atomless probability space contains events of every probability, and from there the existence of $U$ is easy. – Nate Eldredge Nov 15 '17 at 14:28
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    Here is an idea. I want to show that if $(\Omega, \mathcal{F}, P)$ is atomless, then it contains a nonempty $P$-null set. If $\mathcal{F}$ contains a singleton ${ \omega}$, then it's clear that it must have measure $0$ and we're done. So suppose $\mathcal{F}$ contains no singletons. Still, $P$ induces an outer measure $P^$ on $\mathcal{P}(X)$ and it's clear that $P^({\omega})=0$. Now we find a decreasing sequence $(A_n)$ above $\omega$ in $\mathcal{F}$ such that $P(\cap_nA_n) = P^*({ \omega})=0$ and we're done. The rest of the answer to my question now follows from what you said above. – aduh Nov 15 '17 at 15:12
  • @aduh: You're right, that is simpler. Would you like to post it as an answer? – Nate Eldredge Nov 16 '17 at 13:51
  • Sure, I guess I'll post it for posterity. I meant to accept your answer though, since the argument is perfectly good and the existence of $U$ seems like a very useful thing to remember. Thanks again! – aduh Nov 16 '17 at 15:06
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As Nate Eldredge shows in his answer, the problem reduces to showing that strictly positive probabilities are not atomless. Here is a way of showing it without introducing a random variable.

We will show that if $(\Omega, \mathcal{F}, P)$ is atomless, then there is a nonempty $P$-null set, so $P$ isn't strictly positive. If $\mathcal{F}$ contains a singleton $\{\omega\}$, $\omega \in \Omega$, then $P(\{\omega\})=0$ (if not, then $\{\omega\}$ is an atom) and we're done. So assume $\mathcal{F}$ contains no singletons. Now, $P$ induces an outer measure $P^*$ on all subsets of $\Omega$ in the usual way and for any singleton $P^*(\{ \omega\})=0$ (if not, then there's some atom $A \supset \{\omega\}$). We can now approximate $\{\omega\}$ from above. That is, there exists a decreasing sequence $A_1 \supset A_2 \supset...$ of supersets of $\{\omega\}$ such that $P(\cap_nA_n) = P^*(\{ \omega\})=0$ (for example, let $A_n$ be such that $P(A_n) \leq P^*(\{\omega\}) + 1/n$). Thus, $\cap_n A_n$ is a nonempty $P$-null set.

aduh
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