I am trying to find a proof of a result as follows:
Let $\rho(\lambda)$ be a real function. Suppose that $\rho(\lambda)$ is monotone increasing and bounded. Suppose that $f(\lambda)$ is measurable and essentialy bounded with respect $\rho(\lambda)$. Denote by $\chi_{\mu}(\lambda)$ the function defined by $\chi_{\mu}(\lambda)=1$ if $f(\lambda) \leq \mu$ and $\chi_{\mu}(\lambda)=0$ if $f(\lambda)>\mu$. Define $\beta(\mu)=\int_{\mathbb{R}}\chi_{\mu}(\lambda) d\rho(\lambda)$.
Proposition: Let $g(\mu)$ be a real function Lebesgue-Stieltjes integrable over $\mathbb{R}$ with respect $\beta(\mu)$. Then, $g(f(\lambda))$ is Lebesgue-Stieltjes integrable over $\mathbb{R}$ with respect $\rho(\lambda)$ and $\int_{\mathbb{R}}g(f(\lambda))d\rho(\lambda)=\int_{\mathbb{R}}g(\mu)d\beta (\mu).$
Unfortunately, I could not find anything like that.
It's an application of Fubini's theorem, together with the fact that
Now, $\chi_{\mu}(\lambda) g(\mu)=\begin{cases}g(\mu) & f(\lambda)\le \mu \\ 0 & f(\lambda)>\mu\end{cases}$
Namely,
$\int_{\mathbb{R}}g(\mu)d\beta (\mu)=\int_{\mathbb R}g(\mu )\left (\int_{\mathbb{R}}\chi_{\mu}(\lambda) d\rho(\lambda)\right)d\beta(\mu)=\int_{\mathbb{R}}\int_{\{f(\lambda)\le \mu \}}g(\mu)d\beta(\mu)d\rho(\lambda)=$
$\int_{\mathbb{R}}\int_{\{f(\lambda)\le \mu \}}g(\mu)d\beta(\mu)d\rho(\lambda)=\int_{\mathbb{R}}\int^{f(\lambda)}_{-\infty}g(\mu)d\beta(\mu)d\rho(\lambda)=\int_{\mathbb{R}}(g(f(\lambda))-g(-\infty))d\rho(\lambda)=\int_{\mathbb{R}}g(f(\lambda))d\rho(\lambda).$