Something about Riemann- Stieltjes integral

Question

I need some help with an specific problem about Riemann- Stieltjes integration

Let $\alpha$ be an increasing function on $[a,b]$. Let $f \in R(\alpha)$ in $[a,b]$ and suppose that for some positive number $M$, $|f(x)|>M$ for all $x\in [a,b]$. Prove that $1/f \in R(\alpha)$.

I would be really thankful if someone could bring me a little help

Answer 1

For Riemann Stieltjes integrability of $1/f$ it is necessary that $f\neq0$ in $[a,b]$. Since $\alpha$ is increasing then $f$ have to be bounded variation and by condition $|f(x)|>M $ for all $ x \in [a,b] $ we deduce that $1/f$ is also of bounded variation and so $1/f \in R(\alpha)$.

Answer 2

For the Riemann integral it's easy to see that if $f$ is integrable, the range of $f$ is contained in an interval, and $\phi$ is continuous on that interval then $\phi\circ f$ is integrable.

It's not clear to me whether that holds for Riemann-Stieltjes integrals or not. But it's more or less obvious if $\phi$ satisfies a Lipschitz condition $|\phi(t)-\phi(s)|\le c|t-s|$.

So. The function $1/t$ is Lipschitz on $[M,\infty)$, because its derivative is bounded, hence $1/f$ is integrable wrt $\alpha$.

(The easiest way to see that a Lipschitz condition on $\phi$ is sufficient is probably in terms of upper and lower sums: If $|\phi(s)-\phi(t)|\le c|s-t|$ and $I=[x_{j-1},x_j]$ then $\sup_I\phi\circ f-\inf_I\phi\circ f\le c(\sup_I f-\inf_I f)$.)