Let $\alpha$ be an increasing function on $[a,b]$. Show that $\int^a_b\alpha d \alpha = \frac{1}{2}[\alpha (b)^2 - \alpha(a)^2]$

Question

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?

Answer 1

If you can establish that the Riemann-Stieltjes integral $\int_a^b\alpha \, d\alpha$ exists by hypothesis or otherwise, then it is straightforward to show that

$$I= \int_a^b \alpha \, d\alpha = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$

Note that for any partition $a = x_0 < x_1 < \ldots < x_n = b$ we have

$$\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]=\frac{1}{2}\sum_{j=1}^n[\alpha^2(x_j) - \alpha^2(x_{j-1})] =\frac{1}{2}\sum_{j=1}^n[\alpha(x_j) + \alpha(x_{j-1})][\alpha(x_j) - \alpha(x_{j-1})] \\ = \frac{1}{2}\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})] +\frac{1}{2}\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})] $$

Thus,

$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|\leqslant \\ \frac{1}{2}\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right| +\frac{1}{2}\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right|$$

Note that the sums on the RHS are both Riemann-Stieltjes sums. Since the Riemann-Stieltjes integral, $I$, exists, for any $\epsilon > 0$ there exists a partition such that

$$\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right|< \epsilon, \\\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right| < \epsilon$$

Hence, for any $\epsilon > 0$ we have

$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|< \epsilon,$$

and it follows that

$$I = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$

---

If $\alpha$ is increasing then it might have jump discontinuities in which case the RS integral of $\alpha $ with respect to $\alpha$ cannot exist. To carry on we must assume that $\alpha$ is continuous. Then it is not difficult to show that $\int_a^b\alpha \, d\alpha$ exists using both continuity and the fact that $\alpha$ is increasing (or more generally because it is of bounded variation).

Answer 2

If you have derived the product rule (or integration by parts) for Riemann–Stieltjes integrals like this one, then you can apply that to get:

$$ \mathrm{d}\alpha^2 = 2\alpha \mathrm{d}\alpha $$

integration gives

$$ \int_a^b \alpha \mathrm{d}\alpha = \frac{1}{2} \int_a^b \mathrm{d}\alpha^2 = \frac{1}{2} (\alpha(b)^2 - \alpha(a)^2). $$

If I were you, I would just copy the proof of the product rule from somewhere and go like this, not making my hands dirty with going back to definition directly.

Or here just plug in the $\alpha$ inside the integration by parts formula: https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral