Please help me with the following part of the proof of Tauberian theorem I cannot overcome.
Let us make the following assumptions: Suppose that $U:[0,\infty)\to [0,\infty)$ is a non-decreasing, right continuous function. Denote by $U(dx), x\geq 0$ its associated measure, with the convention that there is an atom of size $U(0)$ at $x=0$. Furthermore assume that $L$ is slowly varying function at infinity in the sense that (for any $\lambda>0$): $$\lim_{x\to\infty}\frac{L(\lambda x)}{L(x)}=1$$
Our main assumption is:
$$U(x)\approx \frac{x^{\rho}L(x)}{\Gamma(1+\rho)}\mbox{ as }x\to\infty.$$
The author states that this assumption expressed in terms of vague convergence implies that, on a bounded intervals of $[0,\infty)$
$$\lim_{x\to\infty}\frac{U(xdy)}{U(x)}=\rho y^{\rho-1}dy$$
This statement is not clear for me. Especially how to understand $U(xdy)$.
My guess is that $U(x\mathrm{d} y)$ refers to the measure associated with the (nondecreasing right-continuous) function $y \mapsto U(xy)$. In particular, the distribution function of $\frac{U(x \mathrm{d}y)}{U(x)}$ is given by $$\int_0^\lambda \frac{U(x\mathrm{d}y)}{U(x)} = \frac{U(\lambda x)}{U(x)} \to \lambda^\rho = \int_0^\lambda \rho y^{\rho-1}\, \mathrm{d}y \tag{1}$$ as $x \to \infty$, where the RHS is the distribution function of $\rho y^{\rho-1}\mathrm{d}y$. This holds for every $\lambda >0$.
For $\lambda = 0$, notice that $$\int_{\lbrace 0\rbrace} \frac{U(x\mathrm{d}y)}{U(x)} = \frac{U(0)}{U(x)} \to 0 = \int_{\lbrace 0 \rbrace} \rho y^{\rho-1}\, \mathrm{d}y \tag{2}$$ since $\lim_{x \to \infty} U(x)= \lim_{x\to \infty} \frac{x^\rho L(x)}{\Gamma(1+\rho)}= \infty$, assuming that $\rho >0$. To conclude, you have to argue that the family of measures $\left(\frac{U(x\mathrm{d}y)}{U(x)}\right)_{x\geq 0}$ is bounded on every bounded interval of $[0,\infty)$, but this follows from $(1)$.