Let $\alpha \colon [0,1] \to \mathbb{R}$ with $$\alpha (x)=\left\{ \begin{array}{lc} 0, & if & 0 \leq x \leq 1/2 \\ 1, & if & 1/2 < x \leq 1 \end{array} \right. $$
I have not been able to prove that if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$, then $f$ is right-continuous in $[0,1]$.
I don't know how to get continuity with that hypothesis. I thought about proceeding by contradiction but I can not find an argument for the absurd.
I appreciate any help or clue to get continuity.
$f$ need only be right-continuous at $x=1/2:$
let $\epsilon>0$ and choose a partition $P=\{0, x_1,\cdots 1/2,\cdots ,x_{n-1},1\}$ where we have assumed without loss of generality that $x_j=1/2\in P$ for some $1\le j\le n.$ We can also take $P$ fine enough so that
$x\in [x_{j+1}-1/2]\Rightarrow |f(x)-f(1/2)|<\epsilon.\quad \bf(*)$
Then,
$U(f,P)=\sum^n _{i=1}\max \{f(x_i):x\in [x_i-x_{i-1}]\}(\alpha (x_i)-\alpha (x_{i-1}))=$
$\max\{f(x):x\in [1/2,x_{j}]\}$ and similarly, $L(f,P)=\min\{f(x):x\in [1/2,x_{j}]\}.\quad \bf(**)$
It now follows from $\bf(*)$ and $ \bf(**)$ that $U(f,P)-L(f,P)<\epsilon.$