If a probability is strictly positive, is it discrete? 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.
 A: 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.
A: 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.
