Given $c\in (a,b),$ define $\alpha :[a,b]\to\mathbb R$ by: $$\alpha(x) = \begin{cases} -1, & \alpha \le x\le c \\[2ex] 1, & c<x\le b \end{cases}$$ Prove that if $f:[a,b]\to\mathbb R$ satisfies $$\lim_{x\to c^+}f(x)=f(c),$$ then $f\in R(\alpha)$ on $[a,b].$ ($R(\alpha)$ means Riemann-Stieltjes integration respect to $\alpha$ function)
Proof:
We have the continuity of $f$ because of this $\lim_{x\to c^+}f(x)=f(c),$ (I guess..)
So we just need $\alpha $ to be of Bounded Variation, $\alpha\in BV([a,b]).$ (If we have this, then the proof ends)
As $\alpha$ in defined in a compact set, we just need to check that $\alpha $ is continuous to then conclude that $\alpha$ is bounded, and therefore $\alpha\in BV([a,b]).$
Is my idea a good idea to proof the theorem?
Any kind of suggest/help is welcome.