Rudin Theorem $6.17$
Assume $\alpha$ increases monotonically and $\alpha^\prime \in \mathscr{R}$ on $[a, b]$. Let $f$ be a bounded real function on $[a, b]$.
Then $f \in \mathscr{R}(\alpha)$ if and only if $f\alpha^\prime \in \mathscr{R}$. In that case
$$ \int_a^b f \,d\alpha = \int_a^b f(x) \alpha^\prime(x)\, dx. $$
Proof :
Let $\varepsilon > 0$ be given and apply Theorem 6.6 to $\alpha^\prime$: There is a partition $P = \left\{ \ x_0, \ldots, x_n \ \right\}$ of $[a, b]$ such that
$$ \tag{28} U(P, \alpha^\prime) – L(P, \alpha^\prime) < \varepsilon. $$
The mean value theorem furnishes points $t_i \in \left[ x_{i-1}, x_i \right]$ such that
$$ \Delta \alpha_i = \alpha^\prime \left( t_i \right) \Delta x_i $$
for $i = 1, \ldots, n$. If $s_i \in \left[ x_{i-1}, x_i \right]$, then
$$ \tag{29} \sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) – \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i < \varepsilon, $$
by (28) and Theorem 6.7 (b)
Theorem $6.7$ :
$(a).$If $(13)$ hold for some $P$ and some $\epsilon $,then $(13)$ hold (with the same $\epsilon) $for every refinement of $P$
$(b).$ If $(13)$ hold for $P={x_0,x_1,…,x_n}$ and if $s_i ,t_i$ are arbitrary points in $[x_{i-1},x_i]$ ,then $\sum_{i=1}^n|f(s_i)-f(t_i)| \Delta \alpha_i < \epsilon$
My confusion :Im not getting that why $\sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) – \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i < \varepsilon?$
My attempt :Here by $(28)$ , we have $$ U(P, \alpha^\prime) – L(P, \alpha^\prime) < \varepsilon.$$
\begin{align}
U(P, f, \alpha') &= \sum_{i=1}^n M_i \Delta \alpha_i,\tag1 \\
L(P, f, \alpha') &= \sum_{i=1}^n m_i \Delta \alpha_i,\tag2
\end{align}
From $(1)$ and $(2)$ we have $$\sum_{i=1}^n M_i \Delta \alpha_i -\sum_{i=1}^n m_i \Delta \alpha_i < \epsilon$$
It is given that $\Delta \alpha_i = \alpha^\prime \left( t_i \right) \Delta x_i$
$$\implies \sum_{i=1}^n (M_i-m_i) \alpha^\prime \left( t_i \right) \Delta x_i \neq \sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) – \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i < \varepsilon$$
Best Answer
Hint: Given any $\epsilon \gt 0$, note that since $\alpha'$ is Riemann integrable on $[a,b]$, there exists a partition $P=\{a=x_0\lt x_1\lt \dots\lt x_n=b\}$ of $[a,b]$ such that: $$U(P,\alpha')-L(P,\alpha')\lt \epsilon \tag 1$$
This gives $$U(P,\alpha')-L(P,\alpha')=\sum_{i=1}^n(M_i-m_i)\Delta x_i\tag 2$$, where $M_i=\sup \alpha'[x_{i-1},x_i]$ and $m_i=\inf \alpha'[x_{i-1},x_i].$
Now, note that $m_i\leq \alpha' (p_i)\le M_i$ and $M_i\ge\alpha' (q_i)\ge m_i$ for any $p_i,q_i$ is $[x_{i-1},x_i]$.
It follows that for any $p_i,q_i$ in $[x_{i-1},x_i]$, the following holds: $$|\alpha'(p_i)-\alpha'(q_i)|\leq M_i-m_i$$ The above gives: $$|\alpha'(p_i)-\alpha'(q_i)|\Delta x_i\leq (M_i-m_i)\Delta x_i$$ Taking summation on both sides and taking advantage of $(1)$ and $(2)$ gives$$\sum|\alpha'(p_i)-\alpha'(q_i)| \Delta x_i\leq \sum(M_i-m_i)\Delta x_i\lt \epsilon$$
It may be noted here that $t_i$ in OP has been chosen in a certain way. Can you take it from here?