Checking a bound on the Stieltjes transform from Terence Tao's notes I'm trying to check a bound on Stieltjes transform of a probability measure, that's given in equation (2.92) on P. 170 in Terence Tao's notes "Topics in Random Matrix Theory". Denote the Stieltjes transform of the probability measure $\mu$ by $s_{\mu}(z)$. Then the bound mentioned in his notes is:
$zs_{\mu}(z)= 1 + o_{\mu}(z)$ as $z= x+iy \to \infty$ so that $|\frac{x}{y}|$ is bounded. N.B. here "$o_{\mu}(z)$" is a notation used to denote $o(z)$ but with highlighting the fact that $z=x+iy\to \infty$ with $|x/y|$ bounded, and that the convergence rate depends on $\mu$.
But all I'm getting, at least under a special case, is: under the same condition of convergence, mentioned just now, $zs_{\mu}(z)= -1 + o_{\mu}(z)$, which I demonstrate below. 
For the special case that I'll treat, just assume that: $\frac{x}{y}=K$. But if you follow my computation below, you'll see that the end result wouldn't change in the limit if you assume $|\frac{x}{y}|\leq K$.
$$zs_{\mu}(z) = \int_{\mathbb{R}} \frac{z}{t-z}d\mu(t)=\int_{\mathbb{R}} \frac{z(t-\bar{z})}{|t-z|^2}d\mu(t)= \int_{\mathbb{R}}\frac{(tx - x^2 - y^2)+i(ty)}  {(t-x)^2 + y^2 }d\mu(t)=  \int_{\mathbb{R}}\frac{(\frac{x}{y}.\frac{t}{y}- (\frac{x}{y})^2-1) + i(\frac{t}{y})}{(\frac{x}{y})^2+1}d\mu(t)= -1 + \int_{\mathbb{R}}\frac{K+1}{K^2 + 1}\frac{t}{y}d\mu(t)= -1 + \frac{K+1}{K^2 + 1}.\frac{\mathbb{E}[\mathbb{I}]}{y}$$, where  $\mathbb{E}[\mathbb{I}]$ is really the expectation of the identity function $\mathbb{I}(t):=t$ w.r.t. $\mu$. Assume it exists for now!
Note that, above, since $z=x+iy \to \infty$ but $|x/y|$ is bounded (actually I assumed that  $|x/y|$ is constant to make things bit easy), we must have $y \to \infty$, yielding: 
$zs_{\mu}(z)= -1 + o_{\mu}(z)$, disproving $zs_{\mu}(z)= 1 + o_{\mu}(z)$. Did I do something wrong in my calculation?
Also note that: if you take: $\mu$ to be the Dirac measure at $0$, i.e. $\mu = \delta_0$, then $s_{\mu}(z)= -1/z$, which does satisfy: $zs_{\mu}(z)= -1 + o_{\mu}(z)$, but not $zs_{\mu}(z)= 1 + o_{\mu}(z)$.
Thanks for taking a look!
 A: I believe there is a minor typo in the definition of the Stieltjes transform on page 169 of those notes, namely instead of
$$
s_\mu(z):=\int_{\mathbb R}\frac{1}{x-z}d\mu(x),\qquad \Im z>0
$$
the intended definition is
$$
s_\mu(z):=\int_{\mathbb R}\frac{1}{z-x}d\mu(x),\qquad \Im z>0.
$$
After fixing the sign mistake caused by swapping $x$ and $z$, this matches the usual definition and also allows a straightforward justification of $(2.92)$ on page 170 as follows. Let $z_n$ be any sequence of complex numbers with $\Im(z_n)>0$ such that $z_n\to\infty$ non-tangentially, and let $f_n(x)=(z_n-x)^{-1}$. Then for all $x\in\mathbb R$
$$
\lim_{n\to\infty}z_nf_n(x)=1,
$$
thus by the dominated convergence theorem it follows that
$$
\lim_{n\to\infty}z_ns_{\mu}(z_n)=\lim_{n\to\infty}\int_{\mathbb R}z_nf_n(x)d\mu(x)=\int_{\mathbb R}\lim_{n\to \infty}z_nf_n(x)d\mu(x)=\int_{\mathbb R}d\mu(x)=1,
$$
which in Tao's notation is equivalent to stating that $zs_{\mu}(z)=1+o_{\mu}(1)$, as claimed.
Note that the applicability of the dominated convergence theorem is justified in the exact same manner as in the observation $(2.91)$ on page 170, since the integrand satisfies
$$
|z_nf_n(x)|\leq \frac{|z_n|}{|\Im(z_n)|}\leq 1.
$$
In other words, we are actually using the special case of the dominated convergence theorem known as the bounded convergence theorem.
A: This may be a little late but maybe this can help:
Though this is not part of the OP, it is mentioned in T. Tao's notes and I will show a short proof of some facts:

*

*Analyticity of the Stieltjes transform:

Let $\mu$ be a complex measure on $(\mathbb{R},\mathscr{B}(\mathbb{R})$, and suppose $\operatorname{sup}(\mu)=\Omega$ so that  $D=\mathbb{C}\setminus\Omega$ is open.
Then, the map
$f:D\rightarrow\mathbb{C}$ given by
$$
\begin{align}
f(z)= \int_\Omega\frac{\mu(d\omega)}{\omega-z}\tag{1}\label{one}
\end{align}
$$
is analytic. Moreover, if the closed ball  $\overline{B}(a;r)\subset D$, then
$$
\begin{align}
f(z)=\sum^\infty_{n=0}c_n(z-a)^n,\qquad z\in B(a;r)\tag{2}\label{two}
\end{align}
$$
where
$$
\begin{align}
c_n=\int_\Omega\frac{\mu(d\omega)}{(\omega-a)^{n+1}},\qquad
|c_n|\leq \frac{\|\mu\|_{TV}}{r^{n+1}},\qquad n\in\mathbb{Z}_+.\tag{3}\label{three}
\end{align}
$$
If $R$ is the radius of convergence of $\eqref{two}$, then $r\leq R$.
Here is a short proof:
If  $\overline{B}(a;r)\subset D$, then
$q:=\inf_{\omega\in\Omega}|\omega-a|>r$, and so
$$
\begin{align}
\frac{|z-a|}{|\omega-a|}\leq \frac{|z-a|}{q}\leq\frac{r}{q}<1,\qquad
\omega\in\Omega,\quad z\in B(a;r).
\end{align}
$$
Hence, for any $z\in B(a;r)$ fixed,  the series
$$
\omega\mapsto
\sum^\infty_{n=0}\frac{(z-a)^n}{(\omega-a)^{n+1}}
=\frac{1}{\omega-z}
$$
converges absolutely and uniformly in $\Omega$. By  dominated convergence (to justify change of order of summation and integration)
$$
f(z)=\int_\Omega \frac{\mu(d\omega)}{\omega-z}=
\int_\Omega \sum^\infty_{n=0}\frac{(z-a)^n}{(\omega-a)^{n+1}}
 \,\mu(d\omega)
= \sum^\infty_{n=0}c_n(z-a)^n,
$$
where the $c_n$ satisfy $\eqref{three}$. The last statement follows from the estimate
$$\begin{align}
\limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}\leq\lim_{n\rightarrow\infty}\frac{1}{r}\sqrt[n]{\frac{\|\mu\|_{TV}}{r}}=\frac{1}{r}
\end{align}
$$

*

*Non tangential asymptotyics. Suppose $|z|\rightarrow\infty$ over a region of the form $T_M=\{z=x+iy\in\mathbb{C}:|x|\leq M|y|\}\setminus\Omega$ for some fixed $M>0$. Then
$$
\Big|\frac{z}{\omega-z}\Big|\leq\frac{|z|}{||z|-|\omega||}\xrightarrow{|z|\rightarrow\infty}1
$$
for each $\omega\in\Omega$, and
$$
\Big|\frac{z}{\omega-z}\Big|\leq\frac{|x|+|y|}{|y|}\leq M+1
$$
for all $\omega\in\Omega$. Then, by dominated convergence, letting $z\rightarrow\infty$ over the region $T_M$
$$
zf(z)=\int_\Omega\frac{z}{\omega-z}\,\mu(d\omega)\xrightarrow{|z|\rightarrow\infty}\mu(\Omega)
$$


*If $\mu$ has compact support, then the asymptotic may be improved considerably . Let $M=\sup\{|\omega|:\omega\in \Omega\}$. Then
$$
\frac{1}{\omega-z}=-\frac{1}{z}\frac{1}{1-\tfrac{\omega}{z}}=-\frac{1}{z}\sum^\infty_{k=0}\frac{\omega^n}{z^n}
$$
By dominated convergence
$$
f(z)=-\frac{1}{z}\sum^\infty_{k=0}z^{-n}\int_\Omega \omega^n\,\mu(d\omega)
$$
Hence
$$zf(z)=\mu(\Omega) +z^{-1}\int_\Omega \omega\,\mu(d\omega) +o_\mu(z^{-1})
$$
