# Theorem 6.17 in Baby Rudin, 3rd ed: $\int_a^b f \,d\alpha = \int_a^b f(x) \alpha^\prime(x) \,dx$

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.

• @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? Jun 23, 2017 at 11:32
• 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. Jun 23, 2017 at 16:53
• @MichaelHardy I've just editted my post, fixing some more issues. Can you please have a careful look through my post now? Jul 14, 2017 at 11:12
• It looks fine. Well done. Math newbie appreciates your teaching. Aug 1, 2017 at 8:17
• 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.
– Hiep
Apr 8, 2020 at 11:12