a continuous extension to $\partial\Omega$ and for any $z_1,z_2\in\overline{\Omega}$, $$|\phi_\mu(z_1)-\phi_\mu(z_2)|\leq|z_1-z_2|.$$]{}

2.  [For any $z_1,z_2\in\mathbb{C}^+\cup\mathbb{R}$, $$\frac{|z_1-z_2|}{2}\leq|F_\mu(z_1)-F_\mu(z_2)|.$$ Consequently, $G_\mu$ has a continuous extension to $\mathbb{R}$ except one point and the measure $\mu$ has at most one atom.]{}

3.  [The measure $\mu$ has an atom if and only if $0\in\partial\Omega$ and $$\lim_{\epsilon\downarrow0}\frac{F_\mu^{-1}(i\epsilon)-F_\mu^{-1}(0)}{i\epsilon}=m>0,$$ in which case the point $F_\mu^{-1}(0)$ is an atom of $\mu$ with mass $m$.]{}

4.  [The nonatomic part of $\mu$ is absolutely continuous $($with respect to Lebesgue measure$)$.]{}

5.  [The measure $\mu^{\mathrm{ac}}$ is concentrated on the set $\psi(\overline{V^+})$, where $V^+=\{x:f(x)>0\}$.]{}

6.  [At the point $\psi(x)$, $x\in V^+$, the density of $\mu^{\mathrm{ac}}$ is analytic and given by $$\frac{d\mu^{\mathrm{ac}}}{dx}(\psi(x))=\frac{f(x)}{\pi(x^2+f^2(x))}.$$]{}

7.  [The measure $\mu$ is compactly supported if and only if so is $f$.]{}

By letting $p=2$ in Proposition \[prop2.1\] and Theorem \[sub\], we have $H(z)=F_\mu^{-1}(z)=2z-F_{\mathbb{B}^{-1}(\mu)}(z)$, $z\in\mathbb{C}^+$. Then it follows from \[prop2.1\](2) that $$\label{in}
\int_\mathbb{R}\frac{s^2+1}{|s-z|^2}\;d\sigma(s)\leq1,\;\;\;\;\;z\in\overline{\Omega}.$$ where $\sigma$ is the measure in the Nevanlinna representation of $F_{\mathbb{B}^{-1}(\mu)}$. Since $\phi_\mu=E_{\mathbb{B}^{-1}(\mu)}$, the inequality in (1) holds for $z\in\Omega$ by Hölder inequality and (\[in\]), whence (1) holds by continuous extension. The assertion (2) follows from Proposition \[prop2.1\] (3). Observer that $\mu$ has an atom at $\alpha$ if and only if $F_\mu(\alpha)=0$ and the Julia-Carathéodory derivative $F_\mu'(\alpha)<\infty$, which happens 