If $f:[a,b] \to \mathbb R$ is discontinuous in $c \in [a, b]$, c an arbitrary element of the interval then, exist a $g$ function monotone creasing such that $f$ is not $g$-integrable.
I'm trying to prove this but I don't know where to start, I'm completely lost.
We consider the case that $c\in(a,b)$. Define $g:[a,b]\rightarrow\mathbb{R}$ by $$ g(x)=\begin{cases} 0, & \mbox{ if }x<c\\ 1, & \mbox{ if }x\geq c \end{cases}. $$ Clearly $g$ is monotonic increasing. Since $f$ is discontinuous at $c$, there exists $\epsilon_{0}>0$ such that for each $\delta>0$, there exists $x\in(c-\delta,c+\delta)$ such that $|f(x)-f(c)|\geq\epsilon_{0}$. We prove by contradiction. Suppose the contrary that the Riemann-Stieltjes integral $I=\int_{a}^{b}f\,dg$ exists. Then, for $\frac{\epsilon_{0}}{2}$, there exists $\eta>0$ such that for any partition $P=\{x_{0},x_{1},\ldots,x_{n}\}$, where $a=x_{0}<x_{1}<\ldots<x_{n}=b$, and any $\xi_{i}\in[x_{i-1},x_{i}]$, for $i=1,\ldots,n$, $|I-\sum_{i=1}^{n}f(\xi_{i})[g(x_{i})-g(x_{i-1})]|<\frac{\epsilon_{0}}{2}$ whenever $||P||=\max_{1\leq i\leq n}(x_{i}-x_{i-1})<\eta$.
Choose a partition $P=\{x_{0},\ldots,x_{n}\}$ such that $c\in(x_{j-1},x_{j})$ for some $j$ and that $||P||<\eta$. Clearly, such a partition exists. For $i\neq j$, choose $\xi_{i}$ in an arbitrary way. However, consider two different cases for $\xi_{j}$. Case I: $\xi_{j}=c$. Case II: $\xi_{j}\in(x_{j-1},x_{j})$ such that $|f(\xi_{j})-f(c)|\geq\epsilon_{0}$. Note that the $\xi_{j}$ in Case II exists by the discontinuity assumption at the very beginning.
For both cases, the Riemann-Stieltjes sum $\sum_{i=1}^{n}f(\xi_{i})[g(x_{i})-g(x_{i-1})]$ reduces to $f(\xi_{j})$ because $g(x_{i})-g(x_{i-1})=0$ if $i\neq j$ and $g(x_{j})-g(x_{j-1})=1$. It follows that in Case I, we have $|I-f(c)|<\frac{\epsilon_{0}}{2}$, and in Case II, $|I-f(\xi_{j})|<\frac{\epsilon_{0}}{2}$.
Now we have $|f(\xi_{j})-f(c)|\leq|f(\xi_{j})-I|+|I-f(c)|<\epsilon_{0}$, which is a contradiction.