For a real constant $\gamma$ and $G(y)$ is a bounded, non-decreasing and left-continuous function. If we define
$\Lambda(x) = \int_{-\infty}^{x}(1 - \frac{\sin y}{y})\frac{1 + y^2}{y^2}d\mu_{G}$,
here $\mu_{G}$ denotes the Lebesgue-Stieltjes measure that deduced by function $G(y)$. The questions that I have are listed below.
Let $\mu_{\Lambda}$ denotes the Lebesgue-Stieltjes measure that is deduced by $\Lambda(x)$, then whether $\mu_{\Lambda} \ll \mu_{G}$?
If function $\Lambda(x)$ is predetermined, then can we solve a unique $G(x)$ from the expression of $\Lambda(x)$ above? And how?
The answer to 1. is yes and follows directly from definition of the integral. We have for each $\mu_G$-null set A and each $f \ge 0$ $$\int_A f d\mu_G = 0$$ so it follows $$\mu_{\Lambda}(A) = \int_A(1 - \frac{\sin y}{y})\frac{1 + y^2}{y^2}d\mu_{G} = 0$$ hence $\mu_{\Lambda} \ll \mu_{G}$
The answer to 2. is no because if the measure $\mu_G$ is induced by $G$ then it is also induced by $G + c$ for $c \in \Bbb R$. So if you have found a solution $G$ then there are infinite many solutions $G_c = G+c$ as well.
If we assume $\Lambda$ and $G$ to be differentiable then we can deduce $$G'(x) = \frac{x^3}{(1+x^2)(x - \sin(x))} \Lambda'(x)$$ by the fundamental theorem of calculus, so again $G$ is determined unless an additive constant.