First of all is the proof legitimate? If so, then are there other methods to achieve the same result that are neater than mine?
Let $\alpha : [a,b] \to \mathbb{R}$ be a step function with discontinuities at $c_{1} < \dots < c_{n}$ where $a \le c_{1}$ and $c_{n} \le b$. Suppose $f : [a,b] \to \mathbb{R}$ is continuous at each $c_{j}, 1 \le j \le n$. Then $f \in \mathcal{R}(\alpha)$ on $[a,b]$ and \begin{equation} \int_{a}^{b}f d \alpha \sum_{j=1}^{n}f(c_{j})[\alpha(c_{j}+)-\alpha(c_{j}-)]\end{equation}
Proof:
Let $\epsilon >0$ and choose a partition $P_{\epsilon}=\{a=\tilde{x}_{0}< \tilde{x}_{1} < \dots < \tilde{x}_{m}=b \}$ such that $\{ c_{1}, \dots, c_{n} \} \subset P_{\epsilon}$ and $\|P_{\epsilon}\|=\max_{1 \le i \le m}|\tilde{x}_{i}-\tilde{x}_{i-1}|< \delta$ with $\delta =\min (|c_{2}-c_{1}|,\dots,c_{n}-c_{n-1},\delta_{0})$ with $\delta_{0} =\min (\delta_{1}, \dots, \delta_{n})$ where $\delta_{j}>0$ exists due to the fact that $ƒ$ is continuous at $c_{j}$ implies
\begin{equation}
|f(t)-f(c_{j})| < \frac{\epsilon}{\sum_{k=1}^{n}\Big( |\alpha(c_{k})-\alpha(c_{k}-)|+|\alpha(c_{k}+)-\alpha(c_{k})|\Big)} \tag{*}
\end{equation}
for all $t$ with $|t-c_{j}|<\delta_{j}$.
Now, let $P=\{a=x_{0},x_{1},\dots,x_{p}=b\}$. be a refinement of $P_{\epsilon}$ and $T=\{t_{1},\dots,t_{p}\}$ be any choice for $P$ and consider each term in \begin{equation} S(P;T,f,\alpha)=\sum_{i=1}^{p}f(t_{i})[\alpha(x_{i})-\alpha(x_{i-1})] \end{equation}
We have three different types of intervals for each $1\le j \le p$:
1) Neither $x_{i}$ nor $x_{i-1}$ is in $\{c_{1},\dots,c_{n}\}$ and so both $x_{i}$ and $x_{i-1}$ lie in a subinterval of $[a,b]$- this could be $[a,c_{1}), (c_{j-1},c_{j}),$ or $(c_{n},b]$ for $2\le j \le n$ - on which $\alpha $ is constant. As such $\alpha (x_{i})-\alpha(x_{i-1})=0$.
2) There is a $1 \le j \le n$ such that $x_{i}=c_{j}$. This means $\alpha(x_{i})-\alpha(x_{i-1})=\alpha(c_{j})-\alpha(c_{j}-)$.
3) There is a $1 \le j \le n$ such that $x_{i-1}=c_{j}$. This means $\alpha(x_{i})-\alpha(x_{i-1})=\alpha(c_{j}+)-\alpha(c_{j})$.
By reason of this, we have: \begin{align*} S(P,T,f,\alpha)&=(\text{case 2})+(\text{case 3}) \\ &=\sum_{j=1}^{n}f(t_{i_{j}}) [\alpha( c_{j})-\alpha(c_{j}-)]+\sum_{j=1}^{n}f(t_{i_{j}^*}) [\alpha( c_{j}+)-\alpha(c_{j})] \\ &=\sum_{j=1}^{n}(f(t_{i_{j}}) [\alpha( c_{j})-\alpha(c_{j}-)]+f(t_{i_{j}^*}) [\alpha( c_{j}+)-\alpha(c_{j})]) \tag{**} \end{align*} where we let $t_{i_{j}}$ denote points that lie in the subinterval of $P$ whose right hand end point is $c_{j}$ and $t_{i_{j}^{*}}$ denote points that lie in the subinterval of $P$ whose left hand end point is $c_{j}$. Since $\|P \|<\delta$, we have $c_{j}-\delta < t_{i_{j}} \leq c_{j}$ and $c_{j} \leq t_{i_{j}^{*}}<c_{j}+\delta$. We
\begin{equation} \sum_{j=1}^{n}f(c_{j})[\alpha(c_{j}+)-\alpha(c_{j}-)]=\sum_{j=1}^{n}\left(f(c_{j})[\alpha(c_{j})-\alpha(c_{j}-)]+f(c_{j})[\alpha(c_{j}+)-\alpha(c_{j})] \right) \tag{***} \end{equation} Thus $(**)$ and $(***)$, together with the linearity of finite sums, and the triangle inequality gives us the following chain of inequalities: \begin{align*} \left| S(P,T,f,\alpha)-\sum_{j=1}^{n}f(c_{j})[\alpha (c_{j}+)-\alpha(c_{j}-)] \right| &\le \sum_{j=1}^{n} \Big(|f(t_{i_{j}})-f(c_{j})||\alpha (c_{j})-\alpha (c_{j}-)| \\ &+ |f(t_{i_{j}^{*}})-f(c_{j})||\alpha (c_{j}+)-\alpha(c_{j})| \Big) \\ &< \epsilon \end{align*} where the last inequality follows from $(*)$. Therefore, the desired result holds.