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 $x \in \Bbb R$.