$ f\in\mathcal {R} _ {\alpha} ([a, b])$ and its equivalences Definition: A  partition $ P $ of $ [a, b] $ is a finite set $ \{x_{0}, ..., x_ {n} \} \subseteq {[a, b]} $ such that $ a = x_ {0} <x_ {1} <... <x_ {n-1} <x_ {n} = b $. The norm of a $ P $ partition of $ [a, b] $ is defined by $ || P ||: = \max_ {1 \leq j \leq n} | x_ {j} -x_ {j-1} | $.
The set of all partitions of $ [a, b] $ will be denoted by $ \mathcal {P} ([a, b]) $. If $ P, P '\in \mathcal {P} ([a, b]) $, it is said that $ P' $ is finer than $ P $ if $ P \subseteq {P '} $.
Definition: Let $f,\alpha: [a,b]\rightarrow{\mathbb{R}}$ be bounded. If $P\in\mathcal{P}([a,b])$, $P=\{x_{0}, ..., x_{n}\}$ and $t_{k}\in [x_{k-1}.x_{k}]$, for all $k\in\{1, ..., n\}$. The sum of Riemann - Stieltjes of $ f $ with respect to $ \alpha $ and $ P $ is defined by $$ S (f, \alpha, P): = \sum_ {k = 1} ^ {n } {f (t_ {k}) (\alpha (x_ {k}) - \alpha (x_ {k-1}))} $$.  
Definition: Let $f,\alpha: [a,b]\rightarrow{\mathbb{R}}$ be bounded. We say that $ f $ is Riemann - Stieltjes - Integrable in $ [a, b] $ with respect to $ \alpha $ and we denote $ f \in \mathcal {R} _ {\alpha} ([a, b]) $ if there exists $ A \in \mathbb {R} $ with the following property: Given $ \epsilon> 0 $, there exists a partition $ P _ {\epsilon} \in \mathcal {P} ([a, b]) $ such that $ P \in \mathcal {P} ([a, b] ) $, $ P = \{x_ {0}, ..., x_ {n} \} $, is finner than $ P _ {\epsilon} $ and $ \{t_ {1}, ..., t_ {k} \} $ is any choice of points such that $ t_ {k} \in [x_ {k-1}, x_ {k}] $, for all $ k \in \{1, ..., n \} $ we have $$ | S (f,\alpha, P) - A | <\epsilon $$
Note: If such number $ A $ exists, it is unique, and therefore, we denote it by $ \int_ {a} ^ {b} {fd \alpha} $.
I must demonstrate using the above: Let $f,\alpha: [a, b] \rightarrow{\mathbb {R}}$ be and $\alpha$ monotone increasing in $[a,b]$. show that 
$ f \in \mathcal {R} _ {\alpha} ([a, b]) \Longleftrightarrow{\text{Given}\phantom{a} \epsilon>0, \phantom{a}\text{there exists a partition}\phantom{a} P_{\epsilon} \in \mathcal {P} ([a, b]) \phantom{a}\text{such that if}\phantom{a} P \in \mathcal {P} ([a, b] ) , \phantom{a} P = \{x_ {0}, ..., x_ {n} \} ,\phantom{a}\text{is finner than}\phantom{a}   P _ {\epsilon}\phantom{a}\text{and}\phantom{a} 
 t_{k},t'_{k}\in [x_{k-1},x_{k}]\phantom{a}, \phantom{a}\text{for all}\phantom{a}  k \in \{1, ..., n \}\phantom{a}\text{we have}\phantom{a}    \sum_{k=1}^{n}{|f(t_{k})-f(t'_{k})|(\alpha(x_{k})-\alpha(x_{k-1}))} <\epsilon.}$ 
Does anyone know how to start?
 A: Initially I alluded to Cauchy's principle of convergence in my comment. But Cauchy's principle is more general as it does not require monotone integrator function $\alpha $ and the condition essentially uses two different partitions.
Your result is simpler and an easy consequence of Darboux criterion. Let $P=\{x_0,x_1,x_2,\dots,x_n\}$ be a partition of $[a, b] $ and let $$M_i=\sup\, \{f(x) \mid x\in[x_{i-1},x_i]\} , m_i=\inf\, \{f(x) \mid x\in[x_{i-1},x_i]\} $$ Then we define upper and lower Darboux sums of $f$ with respect to $\alpha$ for partition $P$ via $$U(f, \alpha, P) =\sum_{i=1}^{n}M_i(\alpha(x_i)-\alpha(x_{i-1})),L(f,\alpha,P)=\sum_{i=1}^{n}m_i(\alpha(x_i)-\alpha(x_{i-1}))$$ The Darboux criterion for integrability says that if $f$ is bounded on $[a, b] $ and $\alpha$ is monotone on $[a, b] $ then $f\in\mathcal{R} _{\alpha} ([a, b]) $ if and only if for every $\epsilon >0$ there is a partition $P_{\epsilon} $ of $[a, b] $ such that $$|U(f, \alpha, P) - L(f, \alpha, P) |<\epsilon $$ for all partitions $P$ of $[a, b] $ which are finer than $P_{\epsilon} $.
The result in current question can be obtained from above criterion by observing that $$M_i-m_i=\sup\, \{|f(x) - f(x')|\mid x, x'\in[x_{i-1},x_i]\} $$ Forward direction is easy and I leave it for you. To prove the converse choose $P_{\epsilon/2}$ corresponding to $\epsilon /2$ in the given condition and let $P$ be a finer partition. Choose $t_i, t'_i$ such that $$|f(t_i) - f(t'_i) |>M_i-m_i-\frac{\epsilon} {2|\alpha(b)-\alpha(a)|}$$ and then one can see that $$|U(f, \alpha, P) - L(f, \alpha, P) |<\epsilon $$ and we are done. 
