of the form $\Omega = \bigcup_{\alpha > 0} \Gamma_{\alpha, \beta_\alpha}$ for some $\beta_{\alpha} > 0$, where $\Gamma_{\alpha, \beta}=\{z \in {\mathbb{C}_{-}};  |{\text{\normalfont Re~}}z| < \alpha |{\text{\normalfont Im~}}z|,~{\text{\normalfont Im~}}z < - \beta \}$.

Let $\mu$ be a probability measure in $\mathcal{P}_2$, $a = {\text{\normalfont Re~}}\!\big( m_1(\mu)\big)$ and $b = {\text{\normalfont Im~}}\!\big( m_1 (\mu)\big)$. The probability measures $\mu_{N}^T \equiv (D_{\frac{1}{N}}\mu)^{\ast N}$, $\mu_{N}^F \equiv (D_{\frac{1}{N}}\mu)^{\boxplus N}$ and $\mu_{N}^B \equiv (D_{\frac{1}{N}}\mu)^{\uplus N}$ all converge weakly to the Cauchy distribution $\mu_{a,b}$. If $b = 0$, we understand that $\mu_{a,0}= \delta_a$.

As usual we take a decomposition $\mu = \nu + \lambda$ in (\[decomp\]). It is well known that $\int_{\operatorname*{\mathbb{R}}} e^{ixz}\nu(dx) = \nu(\operatorname*{\mathbb{R}}) + im_1(\nu)z + o(z)$ as $z \to 0$. Therefore we have $$\notag
\begin{split}
\mathcal{F}_{\mu_N^T}(z) &= \Big(\mathcal{F}_{\mu}\Big(\frac{z}{N}\Big)\Big) ^{N} \\
              &= \Big(1 + \frac{m_1(\mu)}{N}(iz) + o(N^{-1}) \Big) ^N \\
             & \to e^{im_1(\mu)z}~~\text{as~}N \to \infty \\
              & = e^{i az - bz}
              \end{split}$$ for $z \geq 0$. We notice that $m_1(\mu)$ has a non-negative imaginary part from Proposition \[prop11\]. Therefore the limit is the Fourier transform of the Cauchy distribution $\mu_{a,b}$ with $a = {\text{\normalfont Re~}}\!(m_1(\mu))$ and $b= {\text{\normalfont Im~}}\!(m_1(\mu))$. Since $\mathcal{F}_{\rho}(z) = \overline{\mathcal{F}_\rho(-z)}$ for $z < 0$, $\mathcal{F}_{\mu_N^T}(z) $ converges to $\mathcal{F}_{\mu_{a,b}}(z)$ for all $z \in 