Theorem: Integration by Substitution/Riemann-Stieltjes Integral/Increasing

Let $g$ be a real function that is continuous and strictly increasing on $\closedint a b$.
Let $f, \alpha$ be real functions that are bounded on $\closedint {\map g a} {\map g b}$.
Further suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint {\map g a} {\map g b}$.
Let $h, \beta : \closedint a b \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
:$\map \beta x = \map \alpha {\map g x}$
Then, $h$ is Riemann-Stieltjes integrable with respect to $\beta$ on $\closedint a b$ and:
:$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$

Proof:
For any $x \in \closedint a b$, $\map g x < \map g a$ implies $x < a$ by definition of strictly increasing, so:
:$\forall x \in \closedint a b: \map g x \ge \map g a$
Likewise, $\map g x > \map g b$ implies $x > b$, so:
:$\forall x \in \closedint a b: \map g x \le \map g b$
Together:
:$\Img g \subseteq \closedint {\map g a} {\map g b}$
But by the Intermediate Value Theorem:
:$\closedint {\map g a} {\map g b} \subseteq \Img g$
Therefore:
:$\Img g = \closedint {\map g a} {\map g b}$
Then, by Inverse of Strictly Monotone Function, $g$ has an inverse $g^{-1} : \closedint {\map g a} {\map g b} \to \closedint a b$.
Furthermore, $g^{-1}$ is strictly increasing.
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, there exists a subdivision $P'_\epsilon$ of $\closedint {\map g a} {\map g b}$ such that:
:For each $P'$ finer than $P'_\epsilon$, $\ds \size {\map S {P', f, \alpha} - \int_{\map g a}^{\map g b} f \rd \alpha} < \epsilon$.
Suppose $P'_\epsilon = \set {y_0, y_1, \dotsc, y_{n - 1}, y_n}$.
Then, define $P_\epsilon = g^{-1} \sqbrk {P'_\epsilon}$.
Since $g^{-1}$ is strictly increasing:
:$a = \map {g^{-1}} {\map g a} = \map {g^{-1}} {y_0} < \map {g^{-1}} {y_1} < \dotso < \map {g^{-1}} {y_{n - 1}} < \map {g^{-1}} {y_n} = \map {g^{-1}} {\map g b} = b$
we have that $P_\epsilon$ is a subdivision of $\closedint a b$.
Now, let $P = \set {x_0, \dotsc, x_m}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.
Define $P' = g \sqbrk P$.
As above, since $g$ is strictly increasing, $P'$ is a subdivision of $\closedint {\map g a} {\map g b}$.
Furthermore, for every $k$:
:$y_k = \map g {\map {g^{-1}} {y_k}} \in g \sqbrk {P_\epsilon} \subseteq g \sqbrk P = P'$
Thus, $P'$ is finer than $P'_\epsilon$.
Hence:
{{begin-eqn}}
{{eqn | l = \map S {P, h, \beta}
      | r = \sum_{k \mathop = 1}^m \map h {t_k} \paren {\map \beta {x_k} - \map \beta {x_{k - 1} } }
      | c = {{Defof|Riemann-Stieltjes Sum}}
}}
{{eqn | r = \sum_{k \mathop = 1}^m \map f {\map g {t_k} } \paren {\map \alpha {\map g {x_k} } - \map \alpha {\map g {x_{k - 1} } } }
      | c = Definitions of $h$ and $\beta$
}}
{{eqn | r = \map S {P', f, \alpha}
      | c = {{Defof|Riemann-Stieltjes Sum}}, since $\map g {x_{k - 1} } \le \map g {t_k} \le \map g {x_k}$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\map S {P, h, \beta} - \int_{\map g a}^{\map g b} f \rd \alpha}
      | r = \size {\map S {P', f, \alpha} - \int_{\map g a}^{\map g b} f \rd \alpha}
      | c = Above
}}
{{eqn | o = <
      | r = \epsilon
      | c = Definition of $P'_\epsilon$
}}
{{end-eqn}}
Since $\epsilon > 0$ was arbitrary, by definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b h \rd \beta = \int_{\map g a}^{\map g b} f \rd \alpha$
{{qed}}