$\newcommand{\sgn}{\operatorname{sgn}}$Let $\sgn:\mathbb R \rightarrow\mathbb R$ be the signum function
$$\left\{ \begin{array}{rcl} 1& \mbox{when} & x>0 \\ 0 & \text{when} & x=0 \\ -1 & \text{for} & x<0 \end{array}\right.$$
let $f:[-1,1] \rightarrow\mathbb R$ be a countinuous function. Show that $f$ is Riemann Stieltjes integrable with respect to sgn, and that
$$\int_{[-1,1]}f \, d\sgn = 2f(0).$$
I have the hint that says find for every $\varepsilon>0$ a piecewise continuous function majorizing and minorizing $f$ whose Riemann-Stieltjes integral is $\varepsilon$-close to $2f(0).$
I am having continuous function on a closed inteval $[-1,1],$ so $f$ is uniformaly continuous and hence by the theorem it is RSI. Also, $\sgn(x)$ is continuous and uniformly continuous since its derivative is bounded, and hence I conclude that $\int_{[-1,1]}f \,d \sgn$ is Riemann-integrable. But the problem I am getting stock on is showing that $$\int_{[-1,1]}f \, d \sgn=2f(0)$$
So I would appreaciate any help with that. Thank you.
Start writing the sum corresponding to the integral $$ \sum_{i=0}^nf(x_*)\left(\alpha(x_{i+1})-\alpha(x_i)\right) $$ with $\alpha(x)=\sgn(x)$ and $x_{i+1}\geq x_*\geq x_{i}$. The only value of this sum which contributes is the term including $$ \alpha(x_{m+1})-\alpha(x_m) $$ where $$ x_{m+1}>0>x_m $$ which then has value $\alpha(x_{m+1})-\alpha(x_m)=2$. In the limit, $f(x_*)\to f(0)$, which is finite since $f$ is continuous on a compact domain. Such a term exists thanks to $\sgn(0)=0$.