Show that $\int \frac{1}{x-z}\,\mathrm{d}\mu(x)$ exists on the upper half-plane if $\int (1+|x|)^{-1}\,\mathrm{d}\mu(x)<\infty$ In Harmonic Analysis: A Comprehensive Course in Analysis, Part 3 page 62, he states a definition of Stieltjes transform of a finite (Borel) measure on $\mathbb{R}$:

If $\mu$ is a finite measure on $\mathbb{R}$, its Stieltjes transform, $F_\mu(z)$, is the function on (the upper half-plane) $\mathbb{C}_+$ given by
$$
F_\mu(z)=\int \frac{1}{x-z}\,\mathrm{d}\mu(x).\qquad (*)
$$

It is not difficult to see that this integral is well-defined. Later on page 64, he removed the assumption that $\mu$ is finite;

For $F_\mu$ to exist, one only needs $\int (1+|x|)^{-1}\,\mathrm{d}\mu(x)<\infty$ ...

I want to know the reason behind it. I tried to find the upper bound of the integrand in (*) in different ways to see if somewhere can be at most $(1+|x|)^{-1}$, but no luck! Here is one of my tries: I write $z=u+iv$ with $v>0$, then
$$
\frac{1}{x-z}=\frac{x-u}{(x-u)^2+v^2}+i\frac{v}{(x-u)^2+v^2},
$$
Multiplying both sides by $z=u+iv$, we get
$$
\frac{z}{x-z}=\frac{ux-u^2-v^2}{(x-u)^2+v^2}+i\frac{vx}{(x-u)^2+v^2}\\
=\frac{u(x-u)-v^2}{(x-u)^2+v^2}+i\frac{vx}{(x-u)^2+v^2}\\
=\frac{u(x-u)/v^2-1}{\left ( \frac{x-u}{v} \right )^2+1}+i\frac{x/v}{\left ( \frac{x-u}{v} \right )^2+1}
$$
Then I get stuck ...

For fixed $z=u+iv$ with $ v > 0$ is
$$
 x \mapsto \frac{1+|x|}{|x-z|} = \frac{1+|x|}{\sqrt{(x-u)^2+v^2}}
$$
a continuous function on $\Bbb R$ with limit $1$ for $x \to \pm \infty$. It is therefore bounded, so that
$$
\frac{1}{|x-z|} \le C \cdot \frac{1}{1+|x|}
$$
for some constant $C > 0$ and all $x \in \Bbb R$.
