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.

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.
