Counter example on Riemann-Stieltjes theorem

Question

Suppose $f$ is boounded on $[a,b]$, $f$ has only finitely many points of discontinuity on $[a,b]$ and $ \alpha $ is continuous at every point of discontinuity. Then $f \in \Re(\alpha)$

Is there any example that if $f$ is bounded on $[a,b]$ and discontinuous at $ x=c \in $[a,b], $ \alpha(x) $ is discontinuous at $ x=c $ as well, but $ f \in \Re(\alpha)$?

I've edited it, I'm pointing at the moon and you are looking at my finger — Alberto, Nov 26 '20 at 16:18
If integrand and integrator are both discontinuous from the right or the left then the RS integral does not exist, as shown in the link. However if one is right-continuous and the other is left-continuous you get your example. — RRL, Nov 26 '20 at 16:39

score 0 · Accepted Answer · answered Nov 26 '20 at 17:22

An example where both the integrand and integrator are discontinuous but the Riemann-Stieltjes integral exists is$$f(x) = \begin{cases}0, & a \leqslant x < c \\ 1, & c \leqslant x \leqslant b \end{cases}\quad \alpha(x) = \begin{cases}0, & a \leqslant x \leqslant c \\ 1, & c < x \leqslant b \end{cases}$$

For a partition with subinterval $I_c =[c,c+\delta]$ we have both the upper and lower Darboux-Stieltjes sums equal to $1$ since $\sup_{x\in I_c} f(x) = \inf_{x \in I_c} f(x) = 1$ and $\alpha(c+\delta) - \alpha(c) = 1$. This proves that $f \in \mathcal{R}(\alpha)$ since for any $\epsilon > 0$ there is a partition such that $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.

Counter example on Riemann-Stieltjes theorem

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