The proof of the converse is slightly clumsy and is independent of the nature of integrator function $\alpha$ and hence we will drop the symbol $\alpha$ and just use $S(f, P)$ in place of $S(f, P, \alpha)$. Moreover in what follows it is necessary to deal with the tags for the partition $P$ and use the notation $T_{P}$ for the a set of tags related to partition $P$. If we have two different tags for partition $P$ we can use notation $T_{P}$ and $T_{P}'$. So the usual notation for Riemann-Stieltjes sums will be $S(f,P, T_{P})$.
Thus we are given that for any $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ such that $$|S(f, P, T_{P}) - S(f, P', T_{P'})| < \epsilon$$ for all $P, P'$ finer than $P_{\epsilon}$. Choosing $\epsilon = 1/n$ for positive integer $n$ we see that there is a partition $P_{n}$ such that $$|S(f, P, T_{P}) - S(f, P', T_{P'})| < \frac{1}{n}$$ for all $P, P'$ finer than $P_{n}$. Clearly we can replace $P_{n + 1}$ by $P_{n}\cup P_{n + 1}$ and hence it is OK to assume that $P_{n}\subseteq P_{n + 1}$.
Consider a sequence $a_{n} = S(f, P_{n}, T_{P_{n}})$ for some choice of tags for $P_{n}$. A different choice of tags will lead to a different sequence $b_{n} = S(f,P_{n}, T_{P_{n}}')$. Clearly if $m > n$ then we have $|a_{m} - a_{n}| < 1/n$ and hence $a_{n}$ is a Cauchy sequence (and so is $b_{n}$). Let $I = \lim_{n \to \infty}a_{n}$. We will prove that $\int_{a}^{b}f\,d\alpha = I$.
Clearly letting $m \to \infty$ in the inequality $|a_{m} - a_{n}| < 1/n$ we get $|a_{n} - I| \leq 1/n$. Now let $\epsilon > 0$ be given and we choose a positive integer $n > 2/\epsilon$. Let $P$ be a partition of $[a, b]$ with $P \supseteq P_{n}$ and $T_{P} $ be any choice of tags associated with partition $P$. Then we have
\begin{align}
|S(f, P, T_{P}) - I| &= |S(f, P, T_{P}) - S(f, P_{n}, T_{P_{n}}) + S(f, P_{n}, T_{P_{n}}) - I|\notag\\
&\leq |S(f, P, T_{P}) - S(f, P_{n}, T_{P_{n}})| + |S(f, P_{n}, T_{P_{n}}) - I|\notag\\
&< \frac{1}{n} + \frac{1}{n}\notag\\
&< \epsilon\notag
\end{align}
It now follows that $f\in R(\alpha)$ and $\int_{a}^{b}f\,d\alpha = I$.
A function $f$ cannot be both unbounded and Riemann-Stieltjes integrable.
This can be shown by producing an $\epsilon > 0$ such that for any real number $A$ and any $\delta > 0$ there is a tagged partition $P$ with $\|P\| < \delta$ and with a Riemann-Stieltjes sum satisfying
$$|S(P,f,\alpha) - A| > \epsilon$$
Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have
$$|S(P,f,\alpha) - A| = \left|f(t_j)(\alpha(x_j) - \alpha(x_{j-1})) + \sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right| \\ \geqslant |f(t_j)|(\alpha(x_j) - \alpha(x_{j-1})) - \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that
$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|}{\alpha(x_j) - \alpha(x_{j-1})},$$
and it follows that no matter how fine the partition $P$ we have
$$|S(P,f, \alpha) - A| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(P,f,\alpha) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.
Best Answer
I know two beautiful direct proofs of this fact. I like a lot the second one!
First proof. If $\alpha$ is monotonically increasing on the set $[a,b]\subsetneq \mathbb R$ , let $\mathcal{R}_\alpha[a, b]$ denote the algebra of Riemann-Stieltjes integrable functions with respect to $\alpha$.
First, if $ f, g\in \mathcal{R}_\alpha[a, b] $, then $ f^2, g^2\in \mathcal{R}_\alpha[a, b] $.
Let us denote $ \lambda:= \displaystyle\sqrt{\int\limits_{a}^{b}f(x)^2\,d\alpha} $ and $\mu:= \displaystyle\sqrt{\int\limits_{a}^{b}g(x)^2\,d\alpha } $ . Define $ h:[a,b]\to \mathbb{R} $ by $ h(x)=\lambda g(x)- \mu f(x)$. Clearly $\lambda,\mu\geq0$ and $ h\in \mathcal{R}_\alpha[a, b] $, so
$ \begin{align}&\\ 0&\leq \int\limits_{a}^{b} h(x)^2\,d\alpha=\int\limits_{a}^{b}\big(\lambda g(x)- \mu f(x)\big)^2\,d\alpha\\ &=\lambda^2\int\limits_{a}^{b}g(x)^2\,d\alpha-2\lambda \mu \int\limits_{a}^{b}f(x) g(x) \,d\alpha+\mu^2 \int\limits_{a}^{b}f(x)^2\,d\alpha \\&=\lambda^2 \mu^2 -2\lambda\mu \int\limits_{a}^{b}f(x)g(x)\,d\alpha+\mu^2 \lambda^2 =2\lambda \mu \left( \lambda \mu- \int\limits_{a}^{b}f(x)g(x)\,d\alpha \right). \end{align} $
Since $f, g\in \mathcal{R}_\alpha[a, b]$ are assumed to be both non-zero (otherwise, the desired inequality is obvious), then $ 2\lambda \mu >0 $, so it has to be that $\displaystyle \lambda \mu- \int\limits_{a}^{b}f(x)g(x)\,d\alpha \geq 0$, that is $$ \int\limits_{a}^{b}f(x)g(x)\,d\alpha \leq\lambda \mu \qquad\qquad (*) $$ On the other hand, $ \displaystyle -\int\limits_{a}^{b}f(x)g(x)\,d\alpha=\int\limits_{a}^{b}\big( (-f)(x) \big) g(x)\,d\alpha $, so, by means of $(*)$, \begin{align} -\int\limits_{a}^{b}f(x)g(x)\,d\alpha&=\int\limits_{a}^{b}\big( (-f)(x) \big) g(x)\,d\alpha\leq \sqrt{\, \int\limits_{a}^{b} \big(-f(x)\big)^2 \,d\alpha}\,\cdot \, \sqrt{\, \int\limits_{a}^{b} g(x)^2\,d\alpha }\\&= \sqrt{\, \int\limits_{a}^{b} f(x)^2 \,d\alpha} \, \cdot \, \sqrt{\, \int\limits_{a}^{b} g(x)^2\,d\alpha }. \end{align} That is, we have proved that $ \displaystyle\left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|\leq \lambda \mu $. Equivalently, $$ \left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|^2\leq \lambda^2 \mu^2= \left(\int\limits_a^b f(x) ^2 \,d\alpha \right) \left(\int\limits_a^b g(x) ^2 \,d\alpha \right), $$ as it was to be shown.
Second proof. We take $\lambda,\mu$ as before, and since $ \displaystyle\left| \int\limits_{a}^{b}f(x)g(x)\,d\alpha\right|\leq \int\limits_{a}^{b}\big|f(x)g(x)\big| \,d\alpha$, the Cauchy-Schwarz inequality can be proved just by showing that $$ \int\limits_{a}^{b}\big|f(x)g(x)\big| \,d\alpha\leq \lambda\mu =\sqrt{\int\limits_{a}^{b}f(x)^2\,d\alpha }\:\cdot \sqrt{\int\limits_{a}^{b}g(x)^2\,d\alpha } . $$ Equivalently (if we asumme that $f\neq 0 \neq g$), $$\int\limits_{a}^{b}\frac{\big|f(x)g(x)\big|}{\lambda \mu }\,d\alpha=\int\limits_{a}^{b}\frac{\big|f(x) \big|}{\lambda } \cdot\frac{\big| g(x)\big|}{ \mu }\,d\alpha \leq 1.$$ Indeed, since for every pair of real numbers $ \zeta, \eta\in\mathbb{R} $ we have that $|\zeta\eta|\leq \frac{\zeta^2+\eta^2}{2}$, it follows that \begin{align*} \int\limits_{a}^{b}\frac{\big|f(x) \big|}{\lambda } \cdot\frac{\big| g(x)\big|}{ \mu }\,d\alpha&\leq \int\limits_{a}^{b}\frac{\frac{f(x)^2}{\lambda^2}+ \frac{g(x)^2}{\mu^2} }{2}\,d\alpha=\frac{1}{2}\left(\, \int\limits_{a}^{b}\frac{f(x)^2}{\lambda^2}\,d\alpha +\int\limits_{a}^{b} \frac{g(x)^2}{\mu^2}\,d\alpha \right)=\frac{1}{2}\left(\frac{\lambda^2}{\lambda^2 }+\frac{ \mu^2}{ \mu^2}\right)=1. \end{align*}