 Solution

For each integer $m\ge 1$, letIf $D_m$ had $m+1$ distinct members, finite additivity on their disjoint singleton Borel sets would give total measure at least $(m+1)/m>1$, contradicting $\mu (\mathbb{R})=1$. Thus $|D_m|\le m$.

Every positive singleton measure is at least $1/m$ for some sufficiently large $m$. Consequently the set of point atoms of a measure isA countable union of finite sets is at most countable. This proves the claim, including the possibility that the set is empty. The same argument gives the countability of point atoms of a finite measure, with $|D_m|\le m\mu (\mathbb{R})$ in place of $m$.