$g\in C^1[a;b], g\prime(x)\neq0(x\in[a,b])$.Prove that any function from $R_g[a;b]$(Riemann–Stieltjes integral) is bounded. It is given that $g\in C^1[a;b], g\prime(x)\neq0(x\in[a,b])$.Prove that any function from $R_g[a;b]$(Riemann–Stieltjes integral) is bounded.
My work.
So we need to prove that if $f \in R_g[a;b]$ then   $f$ is bounded.
We have $\exists lim_{\lambda\to0} \sum_{i=0}^{n-1}f(\xi_i)[g(x_{i+1})-g(x_i)].$ We can write this as $lim_{\lambda\to0} \sum_{i=0}^{n-1}f(\xi_i)\frac{[g(x_{i+1})-g(x_i)](x_{i+1}-x_i)}{x_{i+1}-x_i}=lim_{\lambda\to0} \sum_{i=0}^{n-1}f(\xi_i)[x_{i+1}-x_i]g\prime(x_i)$. Now I am stuck.How use condition given on $g$.

(This is essentially If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?, reformulated for the Riemann-Stieltjes integral.)
Let $I = \int_a^b f(x) dg(x)$ and choose a partition
$$
 a = x_0 < x_1 < \cdots < x_n = b
$$
such that
$$
 \left|\sum_{i=1}^n f(\xi_i)(g(x_i)-g(x_{i-1}) - I \right| < 1
$$
for all choices of intermediate points $\xi_i \in [x_{i-1}, x_i]$.
For each $1 \le k \le n$ and arbitrary $\xi_k \in [x_{k-1}, x_i]$ it follows, using the triangle inequality, that
$$
 |f(\xi_k)(g(x_k)-g(x_{k-1})| \le |I| + 1 + \sum_{i \ne k} |f(x_i)(g(x_i)-g(x_{i-1})| 
$$
or
$$
 |f(\xi_k)| \le \frac{1}{|g(x_k)-g(x_{k-1})|} \left( 
|I| + 1 + \sum_{i \ne k} |f(x_i)(g(x_i)-g(x_{i-1})| \right)
$$
so that $f$ is bounded on the interval $[x_{k-1}, x_k]$. This is the point where we need that $g$ is strictly monotone, so that $g(x_k)-g(x_{k-1}) \ne 0$.
So $f$ is bounded on each interval $[x_{k-1}, x_k]$, $1 \le k \le n$, and therefore bounded on $[a, b]$.
