My book gave this as an excercise and I did it with a method similar to one given as an example, but I am curious of this alternate method will work. Forgive me if the proof is incoherent or just is wrong, I haven't slept in two days (long story).

Question: Let $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$. Is $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [0,1]$ and if so what is the value of $\displaystyle \int_0^1 f~d\alpha$

Notation: For my sake let $\displaystyle S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}$

Answer: By definition we must show that $\displaystyle \forall \varepsilon>0$ there exists a partition $\displaystyle P$ of $\displaystyle [0,1]$ such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red} (*)}$. Now it is clear that since on any interval $\displaystyle [x_{i-1},x_i]$ there exists values of $\displaystyle x$ such that $\displaystyle x\not\in S$, so consequently $\displaystyle m_i=\inf_{[x_{i-1},x_i]}~f=0$ which in turn implies that for any partiton $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta \alpha_i=0$. This realization enables to prove $\displaystyle \color{red}(*)$ by proving that $\displaystyle U\left(P,f,\alpha\right)<\varepsilon$.

To do this we must merely construct our partition carefully. Before we do this let me establish some "new" notation.

Define $\displaystyle \mathcal{O}_{\varphi}\left(x_i\right)$ as being a neighborhood of radius $\displaystyle \varphi$ around $\displaystyle x_i$

Now construct a partition of $\displaystyle [0,1]$ as follows: Let a partition $\displaystyle P_n=\left\{x_i\right\}~0\leqslant x\leqslant n$ of $\displaystyle [0,1]$ posses the following charcteristics $\displaystyle x_0=a$,$\displaystyle x_n=b$, and $\displaystyle x_i\not\in\mathbb{Q}~~1\leqslant i\leqslant n-1$. Now it is clear that we must have $\displaystyle 2\sum_{i} \mathcal{O}_{\varphi}\left(x_i\right)=1$.

So now it is obvious that we can make the $\displaystyle 2^n$ in $\displaystyle \frac{a}{2^n}$ as big as we want by making $\displaystyle \varphi$ as small as we want, so we can make $\displaystyle M_i=\sup_{[x_{i-1},x_i]}~f$ similarly as small as we want. Finally we can make $\displaystyle \sup_{i}\left(M_i\right)$ as small as we want.

Now let $\displaystyle \delta$ be a number such that $\displaystyle \varphi<\delta\implies \sup_{i}\left(M_i\right)\left(\alpha(b)-\alpha(a)\right)<\varepsilon$. Choose a partition of $\displaystyle [0,1]$ such that $\displaystyle \max\left\{\varphi\right\}<\delta$. Let $\displaystyle \sup_{i}\left(M_i\right)=M$, then

$\displaystyle \begin{aligned}U\left(P,f,\alpha\right)&=\sum_{i=1 }^{n}M_i\cdot\Delta \alpha_i\\

&\leqslant M\sum_{i=1}^{n}\Delta\alpha_i\\

&=M\left(\alpha(b)-\alpha(a)\right)\\

&<\varepsilon\end{aligned}$.

This proves integrability of $\displaystyle f$. Now to find the value just note that since it is integrable that $\displaystyle \int_0^1 f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)=0$

Sorry if that is really far off.