Riemann-Stieltjes Integrable If $f$ and $g$ are both Riemann-Stieltjes Integrable with respect to a monotonic function $\alpha$, is it true that $f(g(x))$ is still integrable with respect to $\alpha$? 
 A: No. Take $\alpha(x) = 1/x^3$ and $f(x) = g(x) = x^2$. we have
$$
\int_1^\infty f(x) d\alpha(x) = \int_1^\infty g(x) d\alpha(x) = -3\int_1^\infty \frac 1 {x^2} dx = -3
$$
but
$$
\int_1^\infty f(g(x)) d\alpha(x) = -3\int_1^\infty dx = -\infty
$$
A: This question has never been answered satisfactorily.
The statement is false. However, a counterexample cannot use an infinite interval since the Riemann-Stieltjes integral is defined only for bounded intervals. Furthermore, a counterexample where $f$ or $g$ is unbounded is also invalid, since any Riemann-Stieltjes integrable function must be bounded, as proved here.
An appropriate counterexample is obtained by taking $\alpha(x) = x$ and using a standard counterexample of Riemann integrable functions $f$ and $g$ where $f \circ g$ is not Riemann integrable. For example,$f,g:[0,1] \to \mathbb{R}$ where
$$f(x) = \begin{cases}1, & x \neq 0\\0,&x = 0 \end{cases}, \quad g(x) = \begin{cases}0, & x \not\in \mathbb{Q}\\1,& x = 0\\1/q,& x \in \mathbb{Q}\cap(0,1], \, x = p/q, \, (p,q)=1 \end{cases}$$
and
$$f \circ g(x) = \begin{cases}0, & x \not\in \mathbb{Q}\\ 1,& x \in \mathbb{Q}\end{cases}$$
Here $f$ and $g$ are both Riemann integrable since they are continuous except on sets of measure $0$, but $f\circ g$ is a Dirichlet function which is not Riemann integrable, and, therefore, not Riemann-Stieltjes integrable with respect to $\alpha$.
