Let $f:[a,b] \to \mathbb{R}$ be bounded, $a < s < b$ and $\alpha :[a,b] \to \mathbb{R}$ be $\alpha(x) = I(x-s)$. Prove $f$ is integrable with respect to $\alpha$,$ f \in \mathcal{R}(\alpha) \iff \lim_x \to {s^+} f(x) = f(s)$ and in this case $\int_a^b fd\alpha = f(s)$.
Edit: Here's what I have so far... any criticism is appreciated.
Let $f \in \mathcal{R}(\alpha)$. Let $(P_n)$ be the sequence of partitions of $[a,b]$ with $2n+1$ elements, i.e. $P_1 = \{ a,s,b\}$, $P_2 = \{a, x_1, s, x_3, b\}$, $P_3 = \{a, x_1, x_2, s, x_4, x_5, b\}$, etc. where $s$ is the $n^{\text{th}}$ element.
We see for every $P_n$, $\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1}) = 1$ if $i = n$ and $0$ otherwise. Thus $$ U(P_n) = \sum_{i=0}^{2n+1} M_i \Delta \alpha_i = M_n(\alpha(x_{n+1}) - \alpha(s)) = M_n(1 - 0) = M_n. $$ Claim: $\lim_{n \to \infty} U(P_n) = f(s)$. Note that as $n \to \infty$, $x_{n+1} \to s$. \begin{align*} \lim_{n \to \infty} U(P_n) & = \lim_{n \to \infty} \sup \{ f(x) | x \in [s, x_{n+1}]\} = sup \{ \lim_{n \to \infty} \{f(x) | x \in [s, x_{n+1}]\}\} \\ & = sup \{f(x) | x \in \{s\} \} = sup f(s) = f(s). \end{align*} Conversely, if $\lim_{x \to s^+} f(x) = f(s)$ and $\int_a^b f d\alpha = f(s)$ then $f \in \mathcal{R}(\alpha)$.