A problem in Riemann Steiltjes integration Let $f:[-1, 1] \rightarrow \mathbb{R}$ be a continuous function. Let $sgn(x)$ be the usual sign function, which is 1 if $x>0$, 0 if $x=0$ and $-1$ if $x<0$. To show that $\int_{[-1, 1]}f\mbox{ }d(sgn)=f(0)$.
My approach: To consider piecewise constant functions $f^{+}$ and $f^{-}$, that respectively majorizes and minorizes $f$ such that 
$p.c.\int_{[-1, 1]}f^{+}\mbox{ }d(sgn)$ and $p.c.\int_{[-1, 1]}f^{-}\mbox{ }d(sgn)$ both are $\epsilon$-close to $f(0)$. This approach is fine as long as one considers partition $\mathbb{P}$ of $[-1, 1]$ containing element of the form  $(-c,0]$ or $[-c,0]$ or $[0, c)$ or $[0, c]$ for some $0<c\leq1$.
The problem: If one considers a partition $\mathbb{P}$ containing the element {$0$},i.e. precisely the case if $c=0$, then both the upper and lower Riemann-Steiltjes integral becomes $0$. How do I get rid of this? Or is it that I cannot use partitions with singleton elements when computing upper or lower Riemann-Steiltjes integral?
P.S. The function $sgn(x)$ is discontinuous at $0$. Is this fact playing any role here? Please explain.
 A: I am not sure of what definition you are using. Take a partition $P$ given by
$-1=x_{0}<x_{1}<\cdots<x_{n}=1$ of $[-1,1]$ and consider the lower sum
$$
L(f,g,P)=\sum_{i=1}^{n}(g(x_{i})-g(x_{i-1}))\inf_{[x_{i-1},x_{i}]}f.
$$
The lower Riemann-Stiltjes integral is given by
$$
\underline{\int_{-1}^{1}}f(x)dg(x)=\sup_{P}L(f,g,P).
$$
By continuity we have that $|f(x)-f(0)|\leq\varepsilon$ for all $|x-0|\leq
\delta$. Given a partition $P$ you can always add more points to the
partition, since if $x_{i-1}<c<x_{i}$, then
\begin{align*}
(g(x_{i})-g(x_{i-1}))\inf_{[x_{i-1},x_{i}]}f  & \leq(g(x_{i}%
)-g(c)+g(c)-g(x_{i-1}))\inf_{[x_{i-1},x_{i}]}f\\
& \leq(g(x_{i})-g(c))\inf_{[x_{i-1},c]}f+(g(c)-g(x_{i-1}))\inf_{[c,x_{i}]}f.
\end{align*}
Thus, without loss of generality you can assume that the points $-\delta$,
$0$, and $\delta$ are in the partition. Let $k$ be such that $x_{k}=0$. Then
$-\delta\leq x_{k-1}<0<x_{k-1}\leq\delta$. Now for $i<k$, we have
$g(x_{i})-g(x_{i-1})=-1-(-1)=0$, while for $i>k+1$ we have $g(x_{i}%
)-g(x_{i-1})=1-(1)=0$, so
\begin{align*}
L(f,g,P)  & =(g(0)-g(x_{k-1}))\inf_{[x_{k-1},0]}f+(g(x_{k+1})-g(0))\inf
_{[0,x_{k+1}]}f\\
& =(0+1)\inf_{[x_{k-1},0]}f+(1-0)\inf_{[0,x_{k+1}]}f=\inf_{[x_{k-1},0]}%
f+\inf_{[0,x_{k+1}]}f\geq2f(0)-2\varepsilon
\end{align*}
since $-\delta\leq x_{k-1}<0<x_{k-1}\leq\delta$. So $\underline{\int_{-1}^{1}%
}f(x)dg(x)\geq2f(0)-2\varepsilon$ and letting $\varepsilon\rightarrow0^{+}$
you get $\underline{\int_{a}^{b}}f(x)dg(x)\geq2f(0)$. Same for the upper
integral. So $\int_{-1}^{1}f(x)\,dg(x)=2f(0)$. Note that
$$
2f(0)=f(0)(g(0)-g_{-}(0))+f(0)(g_{+}(0)-g(0)),
$$
where $g_{-}$ and $g_{+}$ are the left and right limits, so yes, discontinuity
points of $g$ are important.
