Integral with respect to a CDF $F(x)$ mutiplied by $x$ I recently saw a mathematical expression of the following
$$\int^\infty_xdF(z)z.$$
It looks like a Riemann–Stieltjes integration of $1$ with respect to a function $F(z)z$. Here, $F$ is a CDF and it is assumed to have a support $\underline z$ to infinity.
What confuses me is if $F$ is differentiable, we may express the above integration as
$$\int^\infty_xf(z)z+F(z)dz$$
which is equivalent to $$\int^\infty_xf(z)zdz+\int^\infty_xF(z)dz.$$
Applying the integration by parts on the second term, we end up with the following expression:
$$\int^\infty_xf(z)zdz+F(z)z|^\infty_{z=x}-\int^\infty_xf(z)zdz=F(z)z|^\infty_{z=x}$$
However, we can't evaluate this term because of the infinite component.
My question is: Can we say the function $\int^\infty_xdF(z)z$ is well defined for such a CDF that has an unbounded support? Or is there any problem in my reasoning?
 A: If $F$ is a CDF, then $0 \leq F(z) \leq 1$ for all $z \in \mathbb{R}$. So the upper bound on your first integral shouldn't be $\infty$, and indeed shouldn't be greater than $1$, since you're integrating with respect to the differential $dF$, and the values of $F$ don't exceed 1. The support of $F$ may be unbounded, but the range of $F$ is not. But instead of
$\int_x^\infty zdF$,
we could consider
$\int_x^1 zdF$,
for example.
Then, as @Kavi Rama Murthy said, if the probability density function is $f$ and the CDF is $F$, this means that
$\frac{dF}{dz} = f$.
So then it's just a variable substitution in the integral. If we assume that $F: \mathbb{R} \rightarrow (0, 1)$ is strictly increasing (which is the case for many CDFs, for example the CDF of the normal distribution), then it has an inverse $F^{-1}: (0, 1) \rightarrow \mathbb{R}$. Because $F$ is a CDF, we also know that $\lim_{z \rightarrow \infty}F(z) = 1$, and thus $\lim_{x \rightarrow 1}F^{-1}(x) = \infty$. Therefore we have
$\int_x^1 zdF = \int_{F^{-1}(x)}^{\infty} z(\frac{dF}{dz})dz = \int_{F^{-1}(x)}^\infty zf(z)dz$.
