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Here is Theorem 6.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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 $$ \int_a^b f \,d\alpha = \int_a^b f(x) \alpha^\prime(x)\, dx. $$

For terminology, here is the link to my post here on Math SE on Theorem 6.15 in Baby Rudin, 3rd edition.

Here is Rudin's proof of Theorem 6.17:

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). Put $M = \sup \lvert f(x) \rvert$. Since $$ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i = \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( t_i \right) \Delta x_i $$ it follows from (29) that $$ \tag{30} \left\lvert \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i - \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i \right\rvert \leq M \varepsilon. $$ In particular, $$ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i \leq U( P, f \alpha^\prime ) + M \varepsilon, $$ for all choices of $s_i \in \left[ x_{i-1}, x_i \right]$, so that $$ U(P, f, \alpha) \leq U(P, f \alpha^\prime) + M \varepsilon. $$ The same argument leads from (30) to $$ U(P, f \alpha^\prime) \leq U(P, f, \alpha) + M \varepsilon. $$ Thus $$ \tag{31} \left\lvert U(P, f, \alpha) - U(P, f \alpha^\prime ) \right\rvert \leq M \varepsilon. $$

Now note that (28) remains true if $P$ is replaced by any refinement. Hence (31) also remains true. We conclude that $$ \left\lvert \overline{\int}_a^b f \, d \alpha - \overline{\int}_a^b f (x) \alpha^\prime(x) \,dx \right\rvert \leq M \varepsilon. $$ But $\varepsilon$ is arbitrary. Hence $$ \tag{32} \overline{\int}_a^b f \,d \alpha = \overline{\int}_a^b f (x) \alpha^\prime(x) \,dx $$ for any bounded $f$. The equality of the lower integrals follows from (30) in exactly the same way. The theorem follows.

Here is Theorem 6.6 in Baby Rudin, 3rd edition:

$f \in \mathscr{R}(\alpha)$ on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a partition $P$ such that $$ \tag{13} U(P, f, \alpha ) - L(P, f, \alpha ) < \varepsilon. $$

Here is Theorem 6.7:

Theorem 6. 7(a):

If (13) holda for some $P$ and some $\varepsilon$, then (13) holds (with the same $\varepsilon$) for every refinement of $P$.

Theorem 6. 7(b):

If (13) holds for $P = \left\{ \ x_0, \ldots, x_n \ \right\}$ and if $s_i$, $t_i$ are arbitrary points in $\left[ x_{i-1}, x_i \right]$, then $$ \sum_{i=1}^n \left\lvert f \left( s_i \right) - f \left( t_i \right) \right\rvert \Delta \alpha_i < \varepsilon. $$

Theorem 6.7(c):

If $f \in \mathscr{R}(\alpha)$ and the hypotheses of (b) hold, then $$ \left\lvert \sum_{i=1}^n f \left( t_i \right) \Delta \alpha_i - \int_a^b f\, d \alpha \right\rvert < \varepsilon. $$

Now here is my understanding of Rudin's proof of Theorem 6.17:

As $f$ is a bounded real function on $[a, b]$, so there is a positive real number $\varepsilon > 0$ such that $$ \lvert f(x) \rvert < M \tag{0} $$ for all $x \in [a, b]$.

Let $\varepsilon > 0$ be a given real number. As $\alpha^\prime$ is Riemann-integrable on $[a, b]$, so there exists a partition $P = \left\{ \ x_0, \ldots, x_n \ \right\}$ of $[a, b]$ such that $$ U(P, \alpha^\prime) - L(P, \alpha^\prime) < \frac{\varepsilon}{2M}. \tag{1} $$

Now as $\alpha^\prime$ exists on $[a, b]$, so, For each $i \in \{ 1, \ldots, n \}$, the function $\alpha$ is continuous on $\left[ x_{i-1}, x_i \right]$ and differentiable on $\left( x_{i-1}, x_i \right)$, and hence there exists a point $t_i \in \left( x_{i-1}, x_i \right)$ such that $$ \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) = \alpha^\prime \left( t_i \right) \left( x_i - x_{i-1} \right); $$ that is, $$ \Delta \alpha_i = \alpha^\prime \left( t_i \right) \Delta x_i \tag{2} $$ for each $i = 1, \ldots, n$.

And, for each $i \in \{1, \ldots, n \}$, if $s_i \in \left[ x_{i-1}, x_i \right]$, then, as $t_i \in \left[ x_{i-1}, x_i \right]$ also, so we conclude that $$\tag{3a} m_i ( \alpha^\prime ) \leq \alpha^\prime \left( s_i \right) \leq M_i (\alpha^\prime), $$ and
$$\tag{3b} m_i (\alpha^\prime) \leq \alpha^\prime \left( t_i \right) \leq M_i ( \alpha^\prime) , $$ where $$ m_i ( \alpha^\prime ) \colon= \inf \left\{ \ \alpha^\prime (x) \ \colon \ x_{i-1} \leq x \leq x_i \ \right\}, $$ and $$ M_i ( \alpha^\prime) \colon= \sup \left\{ \ \alpha^\prime (x) \ \colon \ x_{i-1} \leq x \leq x_i \ \right\};$$ now (3b) implies that $$ \tag{3c} - M_i ( \alpha^\prime ) \leq - \alpha^\prime \left( t_i \right) \leq - m_i (\alpha^\prime), $$ and upon adding (3c) to (3a), we obtain $$ m_i ( \alpha^\prime ) - M_i( \alpha^\prime) \leq \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \leq M_i ( \alpha^\prime) - m_i( \alpha^\prime), $$ which implies that $$ \left\lvert \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \right\rvert \leq M_i ( \alpha^\prime) - m_i( \alpha^\prime), \tag{3d} $$ and, as $\Delta x_i = x_i - x_{i-1} \geq 0$, so (3d) yields $$ \left\lvert \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i \leq \left[ M_i ( \alpha^\prime) - m_i( \alpha^\prime) \right] \Delta x_i, \tag{3e} $$ for each $i = 1, \ldots, n$.

Now upon adding together all the inequalities in (3e), we get $$ \sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i \leq \sum_{i=1}^n \left[ M_i ( \alpha^\prime) - m_i( \alpha^\prime) \right] \Delta x_i, $$ that is, $$ \sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i \leq U(P, \alpha^\prime) - L(P, \alpha^\prime). \tag{3} $$

From (1) and (3) we get $$ \sum_{i=1}^n \left\lvert \alpha^\prime \left( s_i \right) - \alpha^\prime \left( t_i \right) \right\rvert \Delta x_i < \frac{\varepsilon}{2M}, \tag{4} $$ for any points $s_i \in \left[ x_{i-1}, x_i \right]$, for each $i \in \{1, \ldots, n\}$; the points $t_i \in \left[ x_{i-1}, x_i \right]$ are as given in (2) above.

Now from (2) we see that $$ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i = \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( t_i \right) \Delta x_i, \tag{5a} $$ where the points $s_i$ and $t_i$, for each $i \in \{ 1, \ldots, n \}$, are as in (4) above.

So \begin{align} & \ \ \ \left\lvert \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i - \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i \right\rvert \\ &= \left\lvert \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( t_i \right) \Delta x_i - \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i \right\rvert \\ & \qquad \qquad \mbox{ [ by (5a) ] } \\ &= \left\lvert \sum_{i=1}^n f \left( s_i \right) \left[ \alpha^\prime \left( t_i \right) - \alpha^\prime \left( s_i \right) \right] \Delta x_i \right\rvert \\ &\leq \sum_{i=1}^n \left\lvert f \left( s_i \right) \left[ \alpha^\prime \left( t_i \right) - \alpha^\prime \left( s_i \right) \right] \Delta x_i \right\rvert \\ &= \sum_{i=1}^n \left\lvert f \left( s_i \right) \right\rvert \left\lvert \alpha^\prime \left( t_i \right) - \alpha^\prime \left( s_i \right) \right\rvert \Delta x_i \\ &\leq \sum_{i=1}^n M \left\lvert \alpha^\prime \left( t_i \right) - \alpha^\prime \left( s_i \right) \right\rvert \Delta x_i \\ & \qquad \qquad \mbox{ [ using (0) above ] } \\ &= M \sum_{i=1}^n \left\lvert \alpha^\prime \left( t_i \right) - \alpha^\prime \left( s_i \right) \right\rvert \Delta x_i \\ &< M \frac{\varepsilon}{2M} \qquad \mbox{ [ using (4) above ] } \\ &= \frac{\varepsilon}{2}. \end{align}

Thus we have shown that $$ \left\lvert \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i - \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i \right\rvert < \frac{\varepsilon}{2}, \tag{5} $$ for any points $s_i \in \left[ x_{i-1}, x_i \right]$, for each $i \in \{1, \ldots, n\}$.

From (5) we can conclude that, for any points $s_i \in \left[ x_{i-1}, x_i \right]$, for each $i \in \{ 1, \ldots, n\}$, we have $$ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i - \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i < \frac{\varepsilon}{2}, $$ and so \begin{align} \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i &< \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i + \frac{\varepsilon}{2} \\ &\leq U( P, f \alpha^\prime) + \frac{\varepsilon}{2}, \tag{6a} \end{align} for any point $s_i \in \left[ x_{i-1}, x_i \right]$, for each $i \in \{ 1, \ldots, n \}$.

Now we show that $$ U(P, f, \alpha) = \sup \left\{ \ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i \colon \ s_i \in \left[ x_{i-1}, x_i \right] \mbox{ for each } i = 1, \ldots, n \ \right\}. \tag{6b} $$

For each $i \in \{ 1, \ldots, n \}$, let $$M_i(f) \colon= \sup \left\{ \ f(x) \ \colon \ x_{i-1} \leq x \leq x_i \ \right\}; \tag{6c} $$ now as $ f \left( s_i \right) \leq M_i (f)$ and as $\alpha$ is a monotonically increasing function on $[a, b]$ and hence also on $\left[ x_{i-1}, x_i \right]$, so $$ \Delta \alpha_i = \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) \geq 0,$$ and therefore $ f \left( s_i \right) \Delta \alpha_i \leq M_i(f) \Delta \alpha_i$ for each $i \in \{1, \ldots, n \}$, which implies that $$ \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i \leq \sum_{i=1}^n M_i(f) \Delta \alpha_i = U(P, f, \alpha). $$ Thus $U(P, f, \alpha)$ is an upper bound of the set in (6b).

If $\alpha$ is constant on the interval $[a, b]$, then, for each $i = 1, \ldots, n$, we have $\Delta \alpha_i = 0$ and so all the sums in the set in (6b) are zero, and also $U(P, f, \alpha) = 0$, and so (6b) holds.

So let's assume that $\alpha$ is not constant on the interval $[a, b]$. Then there exists a sub-interval $\left[ x_{k-1}, x_k \right]$ (for some $k = 1, \ldots, n$) such that $$\Delta \alpha_k = \alpha\left( x_k \right) - \alpha\left( x_{k-1} \right) > 0. \tag{*}$$

As $\varepsilon > 0$ and as $\alpha$ is monotonically increasing on $[a, b]$, so, for each $i \in \{ 1, \ldots, n \}$, we have $$ M_i(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } < M_i(f),$$ and thus the real number $ M_i(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } $ is not an upper bound of the set in (6c), which implies that there exists a point $p_i \in \left[ x_{i-1}, x_i \right]$ such that $$ M_i(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } < f\left(p_i\right),$$ and since $\Delta \alpha_i \geq 0$, we have $$ \left[ M_i(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } \right] \Delta \alpha_i \leq f\left(p_i\right) \Delta \alpha_i$$ for each $i = 1, \ldots, n$; but for $i= k$ we have $$ \left[ M_k(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } \right] \Delta \alpha_k < f\left(p_k\right) \Delta \alpha_k;$$ therefore, $$ \sum_{i=1}^n \left[ M_i(f) - \frac{\varepsilon}{ \alpha(b) - \alpha(a) + 1 } \right] \Delta \alpha_i < \sum_{i=1}^n f\left(p_i\right) \Delta \alpha_i,$$ that is, $$ \sum_{i=1}^n M_i(f) \Delta \alpha_i - \frac{\varepsilon }{\alpha(b) - \alpha(a) + 1 } \left[ \alpha(b) - \alpha(a) \right] < \sum_{i=1}^n f\left(p_i\right) \Delta \alpha_i, $$ which is the same as $$ U(P, f, \alpha) - \frac{\varepsilon }{\alpha(b) - \alpha(a) + 1 } \left[ \alpha(b) - \alpha(a) \right] < \sum_{i=1}^n f\left(p_i\right) \Delta \alpha_i. \tag{6d} $$

But, as $a \leq b$ and as $\alpha$ is monotonically increasing, so $$ \alpha(a) \leq \alpha(b), $$ which implies that $$ 0 \leq \alpha(b) - \alpha(a) < \alpha(b) - \alpha(a) + 1, $$ and so $$ 0 \leq \frac{ \alpha(b) - \alpha(a) }{ \alpha(b) - \alpha(a) + 1 } < 1, $$ which then implies that $$ 0 \leq \frac{ \varepsilon }{ \alpha(b) - \alpha(a) +1 } \left[ \alpha(b) - \alpha(a) \right] < \varepsilon, $$ and hence $$ - \varepsilon < - \frac{ \varepsilon }{ \alpha(b) - \alpha(a) +1 } \left[ \alpha(b) - \alpha(a) \right] \leq 0, $$ which implies that $$ U(P, f, \alpha) - \varepsilon < U(P, f, \alpha) - \frac{ \varepsilon }{ \alpha(b) - \alpha(a) +1 } \left[ \alpha(b) - \alpha(a) \right] , \tag{6e} $$

Now from (6d) and (6e) we have $$ U(P, f, \alpha) - \varepsilon < \sum_{i=1}^n f\left(p_i\right) \Delta \alpha_i, $$ which shows that no real number less than $U(P, f, \alpha)$ can be an upper bound of the set in (6b).

Thus the real number $U(P, f, \alpha)$ is an upper bound of the set in (6b), but no real number less than $U(P, f, \alpha)$ is an upper bound of that set. Therefore (6b) holds.

Now from (6a) we see that the real number $U(P, f \alpha^\prime) + \frac{\varepsilon}{2}$ is an upper bound for the set in (6b), and from (6b) we know that $U(P, f, \alpha)$ is the least upper bound of this very set, so from (6a) and (6b) we can conclude that $$ U(P, f, \alpha) \leq U(P, f \alpha^\prime) + \frac{\varepsilon}{2}. \tag{6} $$

Again from (5) we obtain $$ \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i - \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i < \frac{\varepsilon}{2},$$ which implies that \begin{align} \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i &< \sum_{i=1}^n f \left( s_i \right) \Delta \alpha_i + \frac{\varepsilon}{2} \\ &\leq U(P, f, \alpha) + \frac{\varepsilon}{2}, \tag{7a} \end{align} for any point $s_i \in \left[ x_{i-1}, x_i \right]$, for each $i = 1, \ldots, n$.

We can show that $$ U(P, f\alpha^\prime) = \sup \left\{ \ \sum_{i=1}^n f \left( s_i \right) \alpha^\prime \left( s_i \right) \Delta x_i \ \colon \ s_i \in \left[ x_{i-1}, x_i \right] \mbox{ for each } i = 1, \ldots, n \ \right\}. \tag{7b} $$

From (7a) and (7b) we can conclude that $$ U(P, f\alpha^\prime) \leq U(P, f, \alpha) + \frac{\varepsilon}{2}. \tag{7} $$

Now if $Q$ be any partition of $[a, b]$ such that $Q \supset P$, then (by Theorem 6.4 in Baby Rudin, 3rd edition) we have $$ L(P, \alpha^\prime) \leq L(Q, \alpha^\prime) \leq U(Q, \alpha^\prime) \leq U(P, \alpha^\prime), $$ which implies that $$ U(Q, \alpha^\prime) - L(Q, \alpha^\prime) \leq U(P, \alpha^\prime) - L(P, \alpha^\prime), $$ and this together with (1) implies that (1) also holds for $Q$, and therefore both (6) and (7) also hold for $Q$.

That is, if $Q$ is any refinement of $P$, then $$ U(Q, f, \alpha) \leq U(Q, f \alpha^\prime) + \frac{\varepsilon}{2}, \tag{6*} $$ and $$ U( Q, f\alpha^\prime) \leq U(Q, f, \alpha) + \frac{\varepsilon}{2}. \tag{7*} $$

Now let's suppose that $$ \overline{\int}_a^b f \,d \alpha - \overline{\int}_a^b f(x) \alpha^\prime(x) \,d x > \varepsilon. \tag{8a} $$

As $$ \overline{\int}_a^b f(x) \alpha^\prime(x) \,dx + \frac{\varepsilon}{2} > \overline{\int}_a^b f(x) \alpha^\prime(x) \,dx, $$ so there exists a partition $P_1$ of $[a, b]$ such that $$ U\left( P_1, f \alpha^\prime \right) < \overline{\int}_a^b f(x) \alpha^\prime(x) dx + \frac{\varepsilon}{2}, \tag{8b} $$ which implies $$- U\left( P_1, f \alpha^\prime \right) > - \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \frac{\varepsilon}{2},$$ and so by (8a) we obtain \begin{align} \overline{\int}_a^b f d \alpha - U \left( P_1, f \alpha^\prime \right) &> \overline{\int}_a^b f d\alpha - \int_a^b f(x) \alpha^\prime(x) dx - \frac{\varepsilon}{2} \\ &> \varepsilon - \frac{\varepsilon}{2} \\ &= \frac{\varepsilon}{2}, \end{align} which implies that $$ \overline{\int}_a^b f d \alpha - U \left( P_1, f \alpha^\prime \right) > \frac{\varepsilon}{2}. \tag{8c} $$
Now if $Q$ is any partition of $[a, b]$ such that $Q \supset P_1$, then (by Theorem 6.4 in Baby Rudin) $$ U(Q, f\alpha^\prime) \leq U \left( P_1, f\alpha^\prime \right),$$ and this together with (8b) implies that $$ U(Q, f\alpha^\prime) < \overline{\int}_a^b f(x) \alpha^\prime(x) dx + \frac{\varepsilon}{2}, \tag{8b*} $$ from which we obtain $$ \overline{\int}_a^b f d\alpha - U(Q, f \alpha^\prime ) > \frac{\varepsilon}{2}, \tag{8c*} $$ in just the same way as we have obtained (8c) from (8b).

Now let $Q$ be any partition of $[a, b]$ such that $Q \supset P$ and $Q \supset P_1$. Then (8c*) holds for this $Q$ as well.

But as $$ U(Q, f, \alpha) \geq \overline{\int}_a^b f d \alpha, $$ so (8c*) gives $$ U(Q, f, \alpha) - U \left( Q, f \alpha^\prime \right) \geq \overline{\int}_a^b f d \alpha - U \left( Q, f \alpha^\prime \right) > \frac{\varepsilon}{2}, $$
which implies $$ U(Q, f, \alpha) - U \left( Q, f \alpha^\prime \right) > \frac{\varepsilon}{2}, $$
and so $$ U(Q, f, \alpha) > U \left( Q, f \alpha^\prime \right) + \frac{\varepsilon}{2}, $$
which contradicts (6*). Therefore (8a) cannot hold, and so $$ \overline{\int}_a^b f d\alpha - \overline{\int}_a^b f(x) \alpha^\prime(x) dx \leq \varepsilon. \tag{8} $$

Now let's suppose that $$ \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \overline{\int}_a^b f d\alpha > \varepsilon. \tag{9a}$$ As $$ \overline{\int}_a^b f d\alpha + \frac{\varepsilon}{2} > \overline{\int}_a^b f d\alpha, $$ so there exists a partition $P_2$ of $[a, b]$ such that $$ U \left( P_2, f, \alpha \right) < \overline{\int}_a^b f d\alpha + \frac{\varepsilon}{2}, \tag{9b} $$ which implies that $$ - U \left( P_2, f, \alpha \right) > - \overline{\int}_a^b f d\alpha - \frac{\varepsilon}{2}, $$ which together with (9a) gives \begin{align} \overline{\int}_a^b f(x) \alpha^\prime(x) dx - U \left( P_2, f, \alpha \right) &> \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \int_a^b f d\alpha - \frac{\varepsilon}{2} \\ &> \varepsilon - \frac{\varepsilon}{2} \\ &= \frac{\varepsilon}{2}, \end{align} which implies that $$ \overline{\int}_a^b f(x) \alpha^\prime(x) dx - U \left( P_2, f, \alpha \right) > \frac{\varepsilon}{2}. \tag{9c} $$

Now if $Q$ is any partition of $[a, b]$ such that $Q \supset P_2$, then (by Theorem 6.4 in Baby Rudin) $$ U(Q, f, \alpha) \leq U \left( P_2, f, \alpha \right), $$ which together with (9b) yields $$ U(Q, f, \alpha) < \overline{\int}_a^b f d\alpha + \frac{\varepsilon}{2}, \tag{9b*} $$ and then we obtain $$ \overline{\int}_a^b f(x) \alpha^\prime(x) dx - U ( Q, f, \alpha ) > \frac{\varepsilon}{2}, \tag{9c*} $$ in just the same way as we have obtained (9c) from (9b).

Now let $Q$ be any partition of $[a, b]$ such that $Q \supset P$ and $Q \supset P_2$. Then (9c*) holds for this $Q$.

But as $$ U ( Q, f \alpha^\prime) \geq \overline{\int}_a^b f(x) \alpha^\prime(x) dx, $$ so (9c*) gives $$ U(Q, f \alpha^\prime ) - U(Q, f, \alpha) \geq \overline{\int}_a^b f(x) \alpha^\prime (x) dx - U(Q, f, \alpha) > \frac{\varepsilon}{2},$$ which implies $$ U(Q, f \alpha^\prime ) - U(Q, f, \alpha) > \frac{\varepsilon}{2},$$ and so $$ U(Q, f \alpha^\prime ) > U(Q, f, \alpha) + \frac{\varepsilon}{2},$$ which contradicts (7*).

Thus (9a) cannot hold, and so we can conclude that $$ \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \overline{\int}_a^b f d\alpha \leq \varepsilon. \tag{9}$$

Now from (8) and (9) we can conclude that $$\left\lvert \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \overline{\int}_a^b f d\alpha \right\rvert \leq \varepsilon $$ for any real number $\varepsilon > 0$. Hence $$\left\lvert \overline{\int}_a^b f(x) \alpha^\prime(x) dx - \overline{\int}_a^b f d\alpha \right\rvert = 0, $$ which implies that $$ \overline{\int}_a^b f d\alpha = \overline{\int}_a^b f(x) \alpha^\prime(x) dx. \tag{A} $$

An analogous argument gives $$ \underline{\int}_a^b f d\alpha = \underline{\int}_a^b f(x) \alpha^\prime (x) dx. \tag{B} $$

Now suppose $f \in \mathscr{R}(\alpha)$ on $[a, b]$. Then $$ \underline{\int}_a^b f d\alpha = \overline{\int}_a^b f d \alpha, $$ and then (A) and (B) together imply $$ \underline{\int}_a^b f(x) \alpha^\prime(x) dx = \overline{\int}_a^b f(x) \alpha^\prime(x) dx, $$ showing that $f\alpha^\prime \in \mathscr{R}$ on $[a, b]$.

Conversely, suppose $f \alpha^\prime \in \mathscr{R}$ on $[a, b]$. Then $$ \underline{\int}_a^b f(x) \alpha^\prime(x) dx = \overline{\int}_a^b f(x) \alpha^\prime(x) dx, $$ and then (A) and (B) together imply $$ \underline{\int}_a^b f d \alpha = \overline{\int}_a^b f d \alpha , $$ showing that $f \in \mathscr{R}(\alpha)$ on $[a, b]$.

Finally, we assume that $f \in \mathscr{R}(\alpha)$ on $[a, b]$ --- which (as we have just shown) is equivalent to assuming that $f \alpha^\prime \in \mathscr{R}$ on $[a, b]$ --- and then find from (A) and (B) that $$ \int_a^b f d\alpha = \int_a^b f(x) \alpha^\prime(x) dx, $$ as required.

Is my rendering of Rudin's proof correct? If not, then where have I erred?

I admit that my presentation is very very, very lengthy, but it is to demonstrate my reasoning clearly and fully enough.

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  • $\begingroup$ @MichaelHardy thank you for taking the time going through my rather lengthy post. Have I managed to get the proof correctly elaborated? What is it that you've edited in my post? $\endgroup$ Commented Jun 23, 2017 at 11:32
  • $\begingroup$ So far just typesetting issues, such as changing $f d\alpha$ to $f\,d\alpha.$ Possibly I'll go through the math a bit later. $\endgroup$ Commented Jun 23, 2017 at 16:53
  • $\begingroup$ @MichaelHardy I've just editted my post, fixing some more issues. Can you please have a careful look through my post now? $\endgroup$ Commented Jul 14, 2017 at 11:12
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    $\begingroup$ It looks fine. Well done. Math newbie appreciates your teaching. $\endgroup$
    – High GPA
    Commented Aug 1, 2017 at 8:17
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    $\begingroup$ After 9c*, similar argument, which starts after 9a, can be applied to the other upper integral to obtain a similar inequality as 9c*, which then can be added to 9c* to obtain a contradiction. Just need to careful in using common refinements. $\endgroup$
    – Jesse
    Commented Apr 8, 2020 at 11:12

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