Here is Prob. 17, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $\alpha$ increases monotonically on $[a, b]$, $g$ is continuous, and $g(x) = G^\prime (x)$ for $a \leq x \leq b$. Prove that $$ \int_a^b \alpha(x) g(x) \ \mathrm{d} x = G(b) \alpha(b) - G(a) \alpha(a) - \int_a^b G \ \mathrm{d} \alpha. $$
Hint: Take $g$ real, without loss of generality. Given $P = \left\{ \ x_0, x_1, \ldots, x_n \ \right\}$, choose $t_i \in \left( x_{i-1}, x_i \right)$ so that $g \left( t_i \right) \Delta x_i = G \left( x_i \right) - G \left( x_{i-1} \right)$. Show that $$ \sum_{i=1}^n \alpha \left( x_i \right) g \left( t_i \right) \Delta x_i = G(b) \alpha (b) - G(a) \alpha (a) - \sum_{i=1}^n G \left( x_{i-1} \right) \Delta \alpha_i. $$
My Attempt:
Here is the link to one of my Math SE posts where I've included Definitions 6.1 and 6.2 in Rudin:
All the notation to be used below is contained in these definitions.
We first assume that $g$ is real; then so is $G$.
As $g$ is continuous on $[a, b]$, so $g$ is Riemann-integrable, by Theorem 6.8 in Rudin; as $\alpha$ is monotonic on $[a, b]$, so $\alpha$ is Riemann-integrable, by Theorem 6.9 in Rudin. Now as both $\alpha$ and $g$ are Riemann-integrable on $[a, b]$, so is $\alpha g$, by Theorem 6.13 (a) in Rudin. Thus $ \int_a^b \alpha(x) g(x) \ \mathrm{d} x$ exists in $\mathbb{R}$.
As $G$ is differentiable on $[a, b]$, so $G$ is also continuous on $[a, b]$, by Theorem 5.2 in Rudin, and as $G$ is continuous on $[a, b]$ and $\alpha$ is monotonically increasing on $[a, b]$, so $G$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, again by Theorem 6.8 in Rudin. Thus $\int_a^b G \ \mathrm{d} \alpha$ exists in $\mathbb{R}$.
Let $P = \left\{ \ x_0, x_1, \ldots, x_n \ \right\}$ be a partition of $[a, b]$.
For each $i = 1, \ldots, n$, as $G$ is continuous on $\left[ x_{i-1}, x_i \right]$ and differentiable on $\left( x_{i-1}, x_i \right)$, so by the mean-value theorem we can find a point $t_i \in \left( x_{i-1}, x_i \right)$ such that $$ G \left( x_i \right) - G \left( x_{i-1} \right) = G^\prime \left( t_i \right) \cdot \left( x_i - x_{i-1} \right) = g \left( t_i \right) \Delta x_i. \tag{0} $$ So $$ \begin{align} \sum_{i=1}^n \alpha \left( x_i \right) g \left( t_i \right) \Delta x_i &= \sum_{i=1}^n \alpha \left( x_i \right) \left[ G \left( x_i \right) - G \left( x_{i-1} \right) \right] \qquad \mbox{ [ using (0) ] } \\ &= \sum_{i=1}^n \alpha \left( x_i \right) G \left( x_i \right) - \sum_{i=1}^n \alpha \left( x_i \right) G \left( x_{i-1} \right) \\ &= \sum_{i=1}^n \alpha \left( x_i \right) G \left( x_i \right) - \sum_{i=0}^{n-1} \alpha \left( x_{i+1} \right) G \left( x_i \right) \\ &= \alpha \left(x_n \right) G \left( x_n \right) + \sum_{i=1}^{n-1} \alpha \left( x_i \right) G \left( x_i \right) - \sum_{i=0}^{n-1} \alpha \left( x_{i+1} \right) G \left( x_i \right) \\ &= \alpha(b) G(b) - \alpha (a) G(a) + \sum_{i=0}^{n-1} \alpha \left( x_i \right) G \left( x_i \right) - \sum_{i=0}^{n-1} \alpha \left( x_{i+1} \right) G \left( x_i \right) \\ &= \alpha(b) G(b) - \alpha (a) G(a) + \sum_{i=0}^{n-1} \left[ \alpha \left( x_i \right) - \alpha \left( x_{i+1} \right) \right] G \left( x_i \right) \\ &= \alpha(b) G(b) - \alpha(a) G(a) - \sum_{i=0}^{n-1} \left[ \alpha \left( x_{i+1} \right) - \alpha \left( x_i \right) \right] G \left( x_i \right) \\ &= \alpha(b) G(b) - \alpha(a) G(a) - \sum_{i=1}^{n} \left[ \alpha \left( x_{i} \right) - \alpha \left( x_{i-1} \right) \right] G \left( x_{i-1} \right) \\ &= \alpha(b) G(b) - \alpha(a) G(a) - \sum_{i=1}^{n} G \left( x_{i-1} \right) \Delta \alpha_i. \end{align} $$
What next? How to proceed from here?
Now we assume that $g$ is a complex function on $[a, b]$, that $g$ is continuous, and $g(x) = G^\prime (x)$. Then $g = \Re g + \iota \Im g$, and $\Re g$ and $\Im g$ are both real continuous functions on $[a, b]$. Moreover, the function $G$ too is complex, and $G = \Re G + \iota \Im G$, where $\Re G$ and $\Im G$ are real functions on $[a, b]$ such that $ \left( \Re G \right)^\prime = \Re g$ and $ \left( \Im G \right)^\prime = \Im g$. The function $\alpha$ is still real of course.
So $$ \begin{align} \int_a^b \alpha(x) g(x) \ \mathrm{d} x &= \int_a^b \alpha(x) \Re g(x) \ \mathrm{d} x + \iota \int_a^b \alpha(x) \Im g(x) \ \mathrm{d} x \\ &= \left[ \alpha (b) \Re G(b) - \alpha(a) \Re G(a) - \int_a^b \Re G \ \mathrm{d} \alpha \right] \\ & \qquad + \iota \left[ \alpha (b) \Im G(b) - \alpha(a) \Im G(a) - \int_a^b \Im G \ \mathrm{d} \alpha \right] \\ &= \alpha (b) G(b) - \alpha(a) G(a) - \int_a^b G \ \mathrm{d} \alpha, \end{align} $$ as required.
Is this proof for complex functions correct? If so, then is it rigorous enough too? If not, then where have I erred?
