Riemann-Stieltjes Integral and the Step Function 
Let $a < c< b$ and let $\alpha (x)$ be defined as $\alpha (x) =\begin{cases} 0 & a \le x \le c \\ 1 &c<x \le b \end{cases}$. Show that $f \in \mathcal{R}(\alpha)$  if and only if $f$ is continuous at $c$ and that the value of the integral is $f(c)$. 

I have been looking for results in in Rudin's Principles of Mathematical Analysis that could be of use to prove this and found that theorem $6.15$ (p.130)
If $a<c<b$, f is bounded on [a,b], f is continuous at c and $\alpha(x)=I(x-c)$ then $\int_{a}^{b}fd\alpha=f(c)$
seem to cover one of the directions (although I could be proved wrong). The other, $f \in \mathcal{R}(\alpha)$, continuous to elude me. Help!
 A: Suppose $f$ is continuous at $c$ with $\alpha$ as given.  Then for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x)-f(c)| < \epsilon$ when $|x-c| < \delta$.
Let $P$ be a partition of $[a,b]$ with $||P||< \delta$. We have some subinterval of $P$  such that $c \in [x_{j-1},x_j)$. 
Note that 
$$S(P,f,\alpha) = \sum_{i=1,i\neq j}^{n}f(\xi_i)[\alpha(x_i)-\alpha(x_{i-1})]+f(\xi_j)[\alpha(x_j)-\alpha(x_{j-1})]$$
The first sum on the RHS vanishes since  $[\alpha(x_i)-\alpha(x_{i-1})] = 0$ if $i \neq j$.
We have $\xi_j,c \in [x_{j-1},x_j]$ and $|\xi_j-c| \leq ||P|| < \delta.$ Hence,
$$|S(P,f,\alpha) -f(c)|= |f(\xi_j)-f(c)| < \epsilon$$
Thus $f \in \mathcal{R}(\alpha)$ with 
$$\int_a^bfd\alpha = f(c).$$
Here is the proof of the converse.
Suppose $f \in \mathcal{R}(\alpha)$ with $\alpha$ as given and
$$\int_a^bfd\alpha = f(c).$$
Then for every $\epsilon > 0$ there exists a partition $P_\epsilon$ such that if $P$ is any refinement and $S(P,f,\alpha)$ is any Riemann-Stieltjes sum then
$$|S(P,f,\alpha)- f(c)|< \epsilon.$$
We have some subinterval of $P$ such that $c \in [x_{j-1},x_j)$. Moreover any $\xi \in  [x_{j-1},x_j]$ can be a tag of the Riemann-Stieltjes sum.
Note that 
$$S(P,f,\alpha) = \sum_{i=1,i\neq j}^{n}f(\xi_i)[\alpha(x_i)-\alpha(x_{i-1})]+f(\xi)[\alpha(x_j)-\alpha(x_{j-1})].$$
The first sum on the RHS vanishes since  $[\alpha(x_i)-\alpha(x_{i-1})] = 0$ if $i \neq j$.
Hence,
$$|S(P,f,\alpha) -f(c)|= |f(\xi)-f(c)| < \epsilon$$
Thus, for any $x$ such that $|x-c| < ||P||$ we have $|f(x)-f(c)|< \epsilon$.
