Existence of $\int_a^bf\,dg$ when $f,g \not\in BV([a,b])$ This question without satisfactory answer asks about necessary and sufficient conditions for existence of the  Riemann-Stieltjes integral $\int_a^b fdg$ when both $f$ and $g$ are continuous. Related question is the existence of integral when both $f$ and $g$ have unbounded variation on $[a,b]$.  
I have proved that if $g$ is continuous and $g \not\in BV([a,b])$ and $f 
\in BV([a,b])$ then $\int_a^bfdg$ exists. I use integration by parts argument that $f \in \mathcal{R_g([a,b])} \iff g \in \mathcal{R_f([a,b])}$ and $\int_a^g gdf$ always exists under these conditions. 
I tried to find an example of (distinct) continuous functions $f,g \not\in BV([a,b])$ where $\int_a^b fdg$ exists but am unable. Is it possible the integral never exists in this case?
 A: If both integrand and integrator have unbounded variation on an interval, the Riemann-Stieltjes integral can exist.
Consider $$f(x) =\begin{cases}x \cos(\pi/x), &0 < x \leqslant 1\\0 , & x= 0 \end{cases}$$
and $g(x) = f(1-x)$. Both $f$ and $g$ are continuous but of unbounded variation on $[0,1]$.
Since $g$ has bounded variation on $[0,1/2]$ and $f$ has bounded variation on $[1/2,1]$ we have existence of the integrals,
$$\int_0^{1/2}f \, dg, \,\,\,\int_{1/2}^1g \, df $$
Using integration by parts, we also have the existence of
$$\int_{1/2}^1f \,dg = f(1)g(1) - f(1/2)g(1/2) - \int_{1/2}^1 g \, df,$$
and, therefore, existence of
$$\int_{0}^1f \,dg = \int_{0}^{1/2}f \,dg + \int_{1/2}^1f \,dg $$
A: I am not sure if a necessary and sufficient condition for the existence of the integral, even restricted to continuous functions, has been known. But here are some known results that may provide some idea on the existence of Riemann-Stieltjes integral. First, for $f : [a, b] \to \mathbb{R}$, its $p$-variation ($p>0$) is defined as
$$ V^p(f,[a,b])=\sup_{\Pi} \sum_{i=1}^{n} |f(x_i) - f(x_{i-1})|^{p}, $$
where the supremum is taken over all partitions $\Pi=\{a=x_0<\cdots<x_n=b\}$ of $[a, b]$. We easily note the followings:


*

*$1$-variation is the usual variation discussed in the context of $BV$.

*If $f$ has finite $p$-variation, then it has finite $q$-variation for any $q > p$.

*If $f$ is $\alpha$-Hölder continuous for $\alpha \in (0, 1]$, then $f$ has finite $\frac{1}{\alpha}$-variation. In particular, for each $p > 1$ there exists $f \in C([a, b])$ such that $V^1(f,[a, b]) = +\infty$ but $V^p(f,[a,b])<+\infty$.


Now the following theorem provides a large family of non-trivial examples.

Theorem. (L. C. Young, 1936) Let $p, q > 0$ be such that $\frac{1}{p}+\frac{1}{q} > 1$. Suppose that $f, g : [a, b] \to \mathbb{R}$ have no common discontinuity, $f$ has finite $p$-variation, and $g$ has finite $q$-variation. Then $\int_{a}^{b} f \, \mathrm{d}g$ exists.

This type of integral is called Young integral. There are some extensions to more general types of variations.
On the other hand, the above theorem is sharp in the sense that $1/p+1/q = 1$ case bears counter-examples. Young provided an example of $f$ and $g$ with finite $2$-variations such that $\int_{a}^{b}f\,\mathrm{d}g$ does not exist. Counter-examples for any other values of $p, q$ with $1/p+1/q = 1$ also exist.
