Let $\mu$ be a finite positive Borel measure on $\mathbb{T}$ such that $\mu$ is singular with respect to Lebesgue measure. Let $E$ be a closed subset of $\mathbb{T}$ such that $\mu(\{x\})=0$ for every $x\in E$. Prove or disprove the following:
$\mu(E)=0$.
I discussed with some of my friends. So here is an answer:
Consider $[0,1]$ and let $E$ denote the Cantor set. Consider the $\frac{\log2}{\log3}$ dimensional Hausdorff measure on $E.$
Fact: $H(E)=1$ and it is zero on any singleton subset of $E$.
So now we define a measure $\mu$ on $[0,1]$ by $\mu(A)=H(A\cap E),$ then $\mu$ is a finite Borel measure, singular with respect to Lebesgue measure such that $\mu(E)=1$ and $\mu(\{x\})=0$ for every $x\in E.$