Each atom of the measure μ is equivalent to a singleton.

Solution 1:

Let $A=A_0$ be an atom. Use the separability of $X$ to cover $X$ with countably many measurable sets of diameter at most $1$. The intersection of $A$ and one of these sets must have the same measure as $A$, call this intersection $A_1$. Cover $X$ with countably many measurable sets of diameter at most $1/2$. The intersection of $A_1$ and one of these measurable sets must have the same measure as $A_1=A_0=A$. Call this intersection $A_2$- Proceed this way to get a decreasing sequence $\langle A_n\rangle$ of subsets of $A$ that have the same measure as $A$ and whose diameter converges to zero. Since their diameter converges to zero, the intersection contains at most one point. Since measures are downward continuous, the intersection must have the same measure as $A$. So the intersection contains a single point in $A$ with the same measure as $A$.