A Cauchy Transform characterization for certain analytic functions on the upper plane I'm trying to prove the following proposition taken from [Bercovici, Hari & Voiculescu, Dan] Free Convolution of Measures with Unbounded Support paper (Section 5, Prop. 5.1). Before enunciate the proposition let's define the $Upper\,cone$ with $\alpha>0$ width as
$$
\Gamma_\alpha=
\Big\{z\in\mathbb{C}\,:\,Im(z)>0\quad\text{&}\quad|Re(z)|<\alpha\,Im(z)\Big\}.
$$
And recall the Cauchy transform given a probability measure $\mu$ on $\mathbb{R}$ being
$$
G_\mu(z)=\int_\mathbb{R}\,\frac{1}{z-t}\,d\,\mu(t),\quad\forall z\notin\mathbb{R}.
$$

Proposition 5.1: Let $G:\mathbb{C}^+\to\mathbb{C}^-$ be an analytic function. Then the following assertions are equivalent.
  $$
(i)\,\text{There exists a probability measure}\,\mu\text{ on }\mathbb{R}\text{ such that }G_\mu=G\,\text{ in }\,\mathbb{C}^+
$$
  $$
(ii)\,\text{For each }\alpha>0 \text{ we have }\lim_{|z|\to\infty,\,\,z\in\Gamma_\alpha}zG(z)=1
$$
  $$
(iii)\text{ We have }
\lim_{y\to\infty}iyG(iy)=1
$$

I have already proved $(i)\Rightarrow\,(ii)\Rightarrow(iii)$. I got stucked trying $(iii)\Rightarrow(i)$.
First I define $\tilde{G}=-G$. So $\tilde{G}:\mathbb{C^+}\to\mathbb{C^+}$ and is still analytic. Next I'm using a representation for Nevanlinna functions that says that for every analytic function $f:\mathbb{C}^+\to\mathbb{C^+}$ we have the following representation
$$
f(z)=az+b+\int_\mathbb{R}\,\frac{1+uz}{u-z}\,d\tau(u)\quad\forall z\in\mathbb{C^+},
$$
where $a>0,b\in\mathbb{R}$ are constants and $\tau(u)\in BV(\mathbb{R})$ is non-decreasing. So, since $\tilde{G}$ happens to be a Nevanlinna function we get that
$$
G(z)=-az-b+\int_\mathbb{R}\,\frac{1+uz}{z-u}\,d\,\tau(u)\quad z\in\mathbb{C^+}.
$$
Now I'm supposed to give a probability measure on $\mathbb{R}$, which I'm trying by denoting $d\mu(t)$ specifically, where a $d\tau(u)$ is involved. I have tried a series of variable changes trying $-az-b+\int_\mathbb{R}\,\frac{1+uz}{z-u}\,d\,\tau(u)=G_\mu(z)$ to hold, but haven't succeeded yet. Another thing I'm also missing is how $(iii)$ could help, since I've already examine $iyG(iy)$ and seems to doesn't give something clear about finding a probability measure. Any help would be appreciated.
UPDATE:
Found that
$$
\frac{1}{u-z}-\frac{u}{1+u^2}=\frac{1+uz}{(u-z)(1+u^2)}
$$
so
$$
\Big(\frac{1}{u-z}-\frac{u}{1+u^2}\Big)(1+u^2)=\frac{1+uz}{(u-z)}
$$
so
$$
\int_\mathbb{R}\frac{1+uz}{(u-z)}\,d\,\tau (u)
=
\int_\mathbb{R}\Big(\frac{1}{u-z}-\frac{u}{1+u^2}\Big)(1+u^2)\,d\,\tau (u)
$$
So it seems that $d\mu(u)=(1+u^2)\,d\,\tau (u)$ should be the probability measure I'm looking for. In this way
$$
G(z)=-az-b+G_\mu (z)-\int_\mathbb{R}\,u\,d\,\tau (u)
$$
So now it suffices to show that $az+b+\int_\mathbb{R}\,u\,d\tau (u)=0$, which happens iff $a=0$ and $b=-\int_\mathbb{R}\,u\,d\tau (u)$. So now I think here is where the fact that $iyG(iy)\to 1$ helps. What do you think? Any help would be appreciated.
 A: I got it. First
$$
iyG(iy)=iy\Big(-iay-b+\int_\mathbb{R}\frac{1}{iy-u}d\mu(u)-\int_\mathbb{R}ud\tau(u)\Big)
$$
$$
=iy\Big(-iay-b+\int_\mathbb{R}\frac{u}{y^2+u^2}d\mu(u)+i\int_\mathbb{R}\frac{y}{y^2+u^2}d\mu(u)-\int_\mathbb{R}ud\tau(u)\Big)
$$
$$
=ay^2-iby+i\int_\mathbb{R}\frac{yu}{y^2+u^2}d\mu(u)-\int_\mathbb{R}\frac{y^2}{y^2+u^2}d\mu(u)-i\int_\mathbb{R}yu\,d\tau(u)
$$
$$
=A+iB,
$$
where
$$
A_y=y^2\Big(a-\int_\mathbb{R}\frac{1}{y^2+u^2}\,d\tau(u)\Big)
$$
$$
B_y=-y\Big(b+\int_\mathbb{R}u\,d\tau(u)-\int_\mathbb{R}\frac{u}{y^2+u^2}\,d\tau(u)\Big).
$$
And since $|iyG(iy)|$ is bounded for all $y\ge 0$, so do are $|A_y|,|B_y|$, for all $y\ge 0$. Thus
$$
\lim_{y\to\infty}\frac{1}{|y|^2}\,|A_y|=0
$$
$$
\lim_{y\to\infty}\frac{1}{|y|}\,|B_y|=0
$$
iff
$$
a-\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\tau(u)=0
$$
$$
b+\int_\mathbb{R}\,u\,d\tau(u)-\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\tau(u)=0,
$$
where $\frac{u}{1+(u/y)^2}$ and $\frac{1}{1+(u/y)^2}$ are bounded, thus dominated, for every $y\ge 0$, so 
$$
\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\tau(u)=0
$$
$$
\lim_{y\to\infty}\frac{1}{y^2}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\tau(u)=0.
$$
Thereby $a=0$ and $b=-\int_\mathbb{R}\,u\,d\tau(u)$. So $G(z)=G_\mu(z)$, for all $z\in\mathbb{C}^+$.
That $\mu(u)$ is a probability measure follows from the fact that $\lim_{y\to\infty}|iyG(iy)|=1$, because
$$
\lim_{y\to\infty}|iyG(iy)|=1
$$
iff
$$
\lim_{y\to\infty}\Big|-\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\mu(u)+i\frac{1}{y}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\mu(u)\Big|=1
$$
iff
$$
\lim_{y\to\infty}\int_\mathbb{R}\frac{1}{1+(u/y)^2}\,d\mu(u)
+
\lim_{y\to\infty}\frac{1}{y}\int_\mathbb{R}\frac{u}{1+(u/y)^2}\,d\mu(u)
=1
$$
iff
$$
\int_\mathbb{R}\,d\mu(u)=1
$$
