I am wanting to try to prove the question below, but there is a step that I can't get pass. I know that the proof is worthless if I assume incorrectly, and should have stopped proving from there, but I feel that I am close and possibly just missing a theorem or something that might be able to salvage the proof. But if ther is no way then I will just try a different attempt altogether.
Book: here
page: 166
I would really appreciate any help\insight you can offer.
Question
Let $\alpha$ be an increasing function on $[a,b]$ and suppose $\alpha \in R(\alpha )$ on $[a,b]$. Show that $\int^a_b\alpha d \alpha = \frac{1}{2}[\alpha (b)^2 - \alpha(a)^2]$
Note: $\alpha \in R(\alpha )$ this is showing that $\alpha$ is Riemann-integrable
My attempt
Let $P$ be a partition on $[a,b]$
Let as $\alpha$ is increasing, thus $\alpha(x) \leq \alpha(y)$ where $x<y$ for $x,y\in[a,b]$
Let $M_k = sup\{\alpha(x) | x_{k-1} \leq x \leq x_k\} = \alpha(x_k)$
Let $m_k = inf\{\alpha(x) | x_{k-1} \leq x \leq x_k\} = \alpha(x_{k-1})$
let $\Delta\alpha_k = \alpha(x_k) - \alpha(x_{k-1})$
Now the upper Stieltjies integral: $U(P,\alpha,\alpha) = \sum\limits_{k=1}^n M_k\Delta\alpha_k = \sum\limits_{k=1}^n\alpha(x_k)\Delta\alpha_k$
and the lower Stieltjies integral: $L(P,\alpha,\alpha) = \sum\limits_{k=1}^n m_k\Delta\alpha_k = \sum\limits_{k=1}^n\alpha(x_{k-1})\Delta\alpha_k$
As $\alpha$ is Riemann-integrable thus the upper Stieltjies integral $=$ lower Stieltjies integral,
thus $\inf\{U(P,\alpha,\alpha)|$ where is P is a partition on $[a,b]\}$
$ = \sup\{L(P,\alpha,\alpha)|$ where is P is a partition on $[a,b]\}$
$ = \int^a_b\alpha d \alpha$
Let $\int^a_b\alpha d \alpha = \frac{1}{2}[U(P,\alpha,\alpha) + L(P,\alpha,\alpha)]$ <<< this is the problem step
$= \frac{1}{2}[\sum\limits_{k=1}^n\alpha(x_k)\Delta\alpha_k + \sum\limits_{k=1}^n\alpha(x_{k-1})\Delta\alpha_k]$
$= \frac{1}{2}\sum\limits_{k=1}^n[\alpha(x_k)+\alpha(x_{k-1})]\Delta\alpha_k$
$= \frac{1}{2}[(\alpha(x_1) + \alpha(x_0))(\alpha(x_1)- \alpha(x_0))+ (\alpha(x_2) + \alpha(x_1))(\alpha(x_2)- \alpha(x_1))+\cdots]$
$= \frac{1}{2}[\alpha(x_1)^2 - \alpha(x_0)^2+ \alpha(x_2)^2 - \alpha(x_1)^2+\cdots]$
$= \frac{1}{2}[\alpha(x_{last})^2 - \alpha(x_0)^2] = \frac{1}{2}[\alpha(b)^2 - \alpha(a)^2]$
the problem
this $\int^a_b\alpha d \alpha = \frac{1}{2}[U(P,\alpha,\alpha) + L(P,\alpha,\alpha)]$
should be $\int^a_b\alpha d \alpha = \frac{1}{2}[\inf\{U(P,\alpha,\alpha)|$ for $P$ on $[a,b]\} + \sup\{L(P,\alpha,\alpha)| $for $P$ on $[a,b]\}]$
but I can't get rid on the $\inf$ and $\sup$. Is there away to do this?