Let $\mu$ be a complex, regular, Borel measure on the circle group $\mathbb{T}$. Define its distribution function $F_\mu : [0, 2\pi) \rightarrow \mathbb{C}$, where $F_\mu(x)=\mu([0,x))$ if $x \in (0, 2\pi)$ and $F_\mu(0)=\mu(\{0\})$. Say $F_\mu$ satisfies Lipschitz condition of order $\alpha>0$ iff there is some $C \geq 0 $ such that for any $x, y \in [0, 2\pi)$,
$|F_\mu(x) - F_\mu(y)| \leq C |x - y| ^\alpha$
How do I show that this implies that there is some $K \geq 0 $ such that for any interval $I \subseteq [0, 2\pi)$ of length $h$, $|\mu(I)| \leq K h^\alpha$ ?
I've noted that if I let the endpoints of such an itnerval $I$ be $x < y$ I have that
$|F_\mu(x) - F_\mu(y)| = |\mu([x,y))|$
From an article I am reading it appears that $\mu(\{z\})=0$ for any $z$. I thought about using the Lipschitz condition for $F_\mu$ to go about this, and I saw that
$|\mu([z,z+\varepsilon))| \leq C \varepsilon^\alpha$
But how would I compare $|\mu(\{z\})|$ and $|\mu([z,z+\varepsilon))|$, since $\mu$ is a complex measure?
~~~EDIT~~~
solution (to a corrected version of the question) is in comments below