For every $\omega$ in an atomless set with positive measure, there's a sequence with positive measure decreasing to $\omega$. Let $(\Omega, \mathcal F, P)$ be a probability space. Assume that singletons are measurable.
A set $A \in \mathcal F$ is an atom if $P(A)>0$ and every measurable subset of $A$ has probability $0$ or $P(A)$. A set $A \in \mathcal F$ is atomless if no measurable subset of $A$ is an atom.
I'm trying to prove the following.
Claim. If $A \in \mathcal F$ is atomless and $P(A)>0$, then for every $\omega \in A$, there is a decreasing sequence $A_1 \supset A_2 \supset ...$ of measurable sets such that $\bigcap_n A_n = \{\omega\}$ and $P(A_n)>0$ for all $n$.
Attempt. I know (though this is non-trivial) that for every atomless set $A$, and every $x \in [0,P(A)]$, there is a subset $B$ of $A$ with $P(B)=x$. Using that I can easily find a decreasing sequence $(B_n)$ such that $P(B_n)>0$ and $P(\cap_n B_n) = 0$. Let $C=\bigcap_n B_n$, and let
$$A_n = (B_n \setminus C) \cup \{\omega\}.$$
Then $(A_n)$ has the desired properties.
Is this correct? Is there a simpler proof (one that doesn't use the non-trivial fact mentioned in the first sentence of my attempt)?
 A: Your proof is correct. But the result can be proved in a simpler way, without using the non-trivial fact that for every atomless set $A$, and every $x \in [0,P(A)]$, there is a subset $B$ of $A$ with $P(B)=x$. In fact, such simpler proof is similar to your proof.
Proof: Let $(\Omega, \mathcal F, P)$ be a probability space. Assume that singletons are measurable. Suppose $A \in \mathcal F$ is atomless,  $P(A)>0$ and $\omega \in A$.
Let $B_1= A$. Then $B_1$ is atomless and $P(B_1)=P(A) >0$. So,  there is $C \subseteq B_1$ such that $0< P(C) < P(B_1)$. If  $ P(C) \leq \frac{P(B_1)}{2}$, take $B_2=C$. If  $ P(C) > \frac{P(B_1)}{2}$, then $P(B_1 \setminus C)  < \frac{P(B_1)}{2}$ and take $B_2=B_1 \setminus C$.
By repeating the argument above, we can build a sequence of sets $\{B_n\}_n$ such that $A \supseteq B_1 \supseteq B_2  \supseteq ...$ and, for all $n$, $P(B_{n+1})\leq \frac{P(B_n)}{2}$. Then $P(\bigcap_n B_n) =0$. Let $D= \bigcap_n B_n$ and let, for each $n$, $A_n= (B_n \setminus D) \cup \{\omega\}$.
Clearly, we have $A \supseteq A_1 \supseteq A_2  \supseteq ...$, for all $n$, $P(A_n) > 0$,  and $\bigcap_n A_n =\{\omega\}$.
$\square$
