Riemann-Stieltjes integrablity

Question

f and α are bounded functions on [a,b] which both are right discontinuous at c∈(a,b) prove f is not Riemann-Stieltjes integrable with respect to α.

Answer 1

Assume both $f$ and $\alpha$ are not continuous from right. So there is an $\epsilon>0$ such that for any $\delta>0$

$$\exists x,y\in (c, c+\delta)\ s.t.\ |f(x)-f(c)|\geq \epsilon, |\alpha(x)-\alpha(c)|\geq \epsilon$$

since f is $\alpha-$integrable so there exist a partition $P=(x_0,x_1,\cdots, x_n)$ with $\|P\|<\delta$ such that $$U(f,P,\alpha)-L(f,P,\alpha)< \epsilon ^2\tag{*}$$ .

Let we have $x_i\leq c< x_{i+1}$ for $i\geq 0$ . Choose a point $c'\in (c, x_{i+1})$ such that $|\alpha(c')-\alpha(c)|\geq \epsilon$ .

If you put $P'=P\cup\{c, c'\}$ then we have

$$U(f, P',\alpha)-L(f,P',\alpha)\geq (M-m)(\alpha(c')-\alpha (c)\geq \epsilon\times \epsilon=\epsilon^2$$ where $M=\sup_{x\in [c,c']} f(x)$ and $m=\inf_{x\in [c,c']}f(x)$

In the other hand $U(f,P',\alpha)-L(f,P',\alpha)< \epsilon^2$ since $P'\supseteq P$ which is a contradiction with $(*)$