Here is part (a) of Theorem 6.12 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $f_1 \in \mathscr{R}(\alpha)$ and $f_2 \in \mathscr{R}(\alpha)$, then $$f_1 + f_2 \in \mathscr{R}(\alpha), $$ and $$ \int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha,$$
And, here is Rudin's proof.
If $f = f_1 + f_2$ and $P$ is any partition of $[a, b]$, we have $$ \tag{20} L\left(P, f_1, \alpha \right) + L\left(P, f_2, \alpha \right) \leq L\left(P, f, \alpha \right) \leq U\left(P, f, \alpha \right) \leq U\left(P, f_1, \alpha \right) + U\left(P, f_2, \alpha \right).$$ If $f_1 \in \mathscr{R}(\alpha)$ and $f_2 \in \mathscr{R}(\alpha)$, let $\varepsilon > 0$ be given. There are partitions $P_j$ $(j = 1, 2)$ such that $$ U \left( P_j, f_j, \alpha \right) - L \left( P_j, f_j, \alpha \right) < \varepsilon. $$ These inequalities persist if $P_1$ and $P_2$ are replaced by their common refinement $P$. Then (20) implies $$ U (P, f, \alpha) - L (P, f, \alpha) < 2 \varepsilon, $$ which proves that $f \in \mathscr{R}(\alpha)$.
With this same $P$ we have $$ U \left( P, f_j , \alpha \right) < \int f_j d \alpha + \varepsilon \qquad (j = 1, 2); $$ hence (20) implies $$ \int f d \alpha \leq U ( P, f, \alpha) < \int f_1 d \alpha + \int f_2 d \alpha + 2 \varepsilon. $$ Since $\varepsilon$ was arbitrary, we conclude that $$ \tag{21} \int f d \alpha \leq \int f_1 d \alpha + \int f_2 d \alpha. $$ If we replace $f_1$ and $f_2$ in (21) by $- f_1$ and $- f_2$, the inequality is reversed, and the equality is proved.
Now here is my reading of Rudin's proof.
Let $f = f_1 + f_2$, and let $P = \left\{ x_0, x_1, \ldots, x_n \right\}$, where $$a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b,$$ be any partition of $[a, b]$.
For each $i = 1, \ldots, n$, if $m_{1i}$, $m_{2i}$, and $m_i$ are the infima of $f_1$, $f_2$, and $f$, respectively, on $\left[ x_{i-1}, x_i \right]$, then we see that $$ m_{1i} \leq f_1(x), \qquad m_{2i} \leq f_2(x) \qquad (x_{i-1} \leq x \leq x_i ),$$ and so $$ m_{1i} + m_{2i} \leq f_1(x) + f_2(x) = f(x) \qquad (x_{i-1} \leq x \leq x_i ),$$ which in turn implies that $$ m_{1i} + m_{2i} \leq m_i \qquad ( i = 1, \ldots, n ),$$ and as $\alpha$ is a monotonically increasing function on $[a, b]$, so $$ m_{1i} \Delta \alpha_i + m_{2i} \Delta \alpha_i \leq m_i \Delta \alpha_i \qquad ( i = 1, \ldots, n ),$$ where $ \Delta \alpha_i = \alpha\left( x_i \right) - \alpha\left( x_{i-1} \right)$. Adding all the inequalities for $i = 1, \ldots, n$, we thus obtain $$ L\left( P, f_1, \alpha \right) + L\left( P, f_2, \alpha \right) \leq L ( P, f, \alpha). \tag{20(1)} $$
Similarly, for each $i = 1, \ldots, n$, if $M_{1i}$, $M_{2i}$, and $M_i$ are the suprema of $f_1$, $f_2$, and $f$, respectively, on $\left[ x_{i-1}, x_i \right]$, then we see that $$ M_{1i} \geq f_1(x), \qquad M_{2i} \geq f_2(x) \qquad (x_{i-1} \leq x \leq x_i ),$$ and so $$ M_{1i} + M_{2i} \geq f_1(x) + f_2(x) = f(x) \qquad (x_{i-1} \leq x \leq x_i ),$$ which in turn implies that $$ M_{1i} + M_{2i} \geq M_i \qquad ( i = 1, \ldots, n ),$$ and as $\alpha$ is a monotonically increasing function on $[a, b]$, so $$ M_{1i} \Delta \alpha_i + M_{2i} \Delta \alpha_i \geq M_i \Delta \alpha_i \qquad ( i = 1, \ldots, n ),$$ and, adding together all these inequalities for $i = 1, \ldots, n$, we thus obtain $$ U\left( P, f_1, \alpha \right) + U\left( P, f_2, \alpha \right) \geq U ( P, f, \alpha). \tag{20(2)} $$ From (20(1)) and (20(2)) we obtain (20) in Rudin's proof.
Let $\varepsilon > 0$ be given. As $f_1$ and $f_2$ are Riemann-integrable with respect to $\alpha$ over $[a, b]$, so (by Theorem 6.6 in Baby Rudin, 3rd edition) we can find partitions $P_1$ and $P_2$ of $[a, b]$ such that $$ U \left( P_1, f_1, \alpha \right) - L \left( P_1, f_1, \alpha \right) < \frac{\varepsilon}{2}, \qquad U \left( P_2, f_2, \alpha \right) - L \left( P_2, f_2, \alpha \right) < \frac{\varepsilon}{2}. \tag{1} $$ Now let $P = P_1 \cup P_2$. Then $P \supset P_1$ and $P \supset P_2$. Then (by Theorem 6.4 in Baby Rudin, 3rd edition) we have $$ L \left( P_1, f_1, \alpha \right) \leq L (P, f_1, \alpha) \leq U(P, f_1, \alpha) \leq U(P_1, f_1, \alpha), $$ and $$ L \left( P_2, f_2, \alpha \right) \leq L (P, f_2, \alpha) \leq U(P, f_2, \alpha) \leq U(P_2, f_2, \alpha), $$ and so we can conclude that $$ U \left(P, f_1, \alpha \right) - L \left(P, f_1, \alpha \right) \leq U ( P_1, f_1, \alpha) - L(P_1, f_1, \alpha) < { \varepsilon \over 2 }, $$ and $$ U \left(P, f_2, \alpha \right) - L \left(P, f_2, \alpha \right) \leq U ( P_2, f_2, \alpha) - L(P_2, f_2, \alpha) < { \varepsilon \over 2 }, $$ which implies $$ U \left(P, f_1, \alpha \right) - L \left(P, f_1, \alpha \right) < { \varepsilon \over 2 }, \qquad \mbox{ and } \qquad U \left(P, f_2, \alpha \right) - L \left(P, f_2, \alpha \right) < { \varepsilon \over 2 }, \tag{2} $$ and this in turn implies that $$ \begin{align} &\ U \left(P, f_1, \alpha \right) + U \left(P, f_2, \alpha \right) - \left[ L \left(P, f_1, \alpha \right) + L \left(P, f_2, \alpha \right) \right] \\ &= \left[ U \left(P, f_1, \alpha \right) - L \left(P, f_1, \alpha \right) \right] + \left[ U \left(P, f_2, \alpha \right) - L \left(P, f_2, \alpha \right) \right] \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align} \tag{3} $$ But (20) implies that $$ \begin{align} & U ( P, f, \alpha) - L(P, f, \alpha) \\ &\leq \left[ U \left(P, f_1, \alpha \right) + U \left(P, f_2, \alpha \right) \right] - \left[ L \left(P, f_1, \alpha \right) + L \left(P, f_2, \alpha \right) \right]. \tag{4} \end{align} $$ Now (3) and (4) together imply that $$ U(P, f, \alpha) - L(P, f, \alpha ) < \varepsilon. \tag{5} $$
Thus for every real number $\varepsilon > 0$ we can find a partition $P$ of $[a, b]$ such that (5) holds. So by Theorem 6.6 in Baby Rudin we can conclude that $f \in \mathscr{R}(\alpha)$.
Thus $$ \underline{\int}_a^b f \ d \alpha = \overline{\int}_a^b f \ d \alpha, $$ that is, $$ \sup \left\{ \ L(Q, f, \alpha) \ \colon \ Q \mbox{ is a partition of } [a, b] \ \right\} = \inf \left\{ \ U(Q, f, \alpha) \ \colon \ Q \mbox{ is a partition of } [a, b] \ \right\}. $$ Moreover, this common value is denoted by $\int_a^b f \ d \alpha$.
Now from (2) we find that $$ U\left( P, f_1, \alpha \right) < L \left( P, f_1, \alpha \right) + { \varepsilon \over 2 } \leq \int_a^b f_1 d \alpha + { \varepsilon \over 2 }, $$ and $$ U\left( P, f_2, \alpha \right) < L \left( P, f_2, \alpha \right) + { \varepsilon \over 2 } \leq \int_a^b f_2 d \alpha + { \varepsilon \over 2 }, $$ and so $$ U\left( P, f_1, \alpha \right) < \int_a^b f_1 d \alpha + { \varepsilon \over 2 }, \qquad \mbox{ and } \qquad U\left( P, f_2, \alpha \right) < \int_a^b f_2 d \alpha + { \varepsilon \over 2 }, $$ and upone adding the last two inequalities we obtain $$ U\left( P, f_1, \alpha \right) + U\left( P, f_2, \alpha \right) < \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha + \varepsilon. \tag{6} $$ Now using (20) we can also conclude $$ \int_a^b f d \alpha \leq U(P, f, \alpha) \leq U\left( P, f_1, \alpha \right) + U\left( P, f_2, \alpha \right), $$ which implies $$ \int_a^b f d \alpha \leq U\left( P, f_1, \alpha \right) + U\left( P, f_2, \alpha \right). \tag{7} $$ Therefore from (6) and (7) we can conclude that for every real number $\varepsilon > 0$, we have $$ \int_a^b f d \alpha < \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha + \varepsilon, $$ which implies that $$ \int_a^b f d \alpha \leq \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha. \tag{A}$$
Again from (2) we see that $$ \int_a^b f_1 d \alpha - { \varepsilon \over 2 } \leq U \left( P, f_1, \alpha \right) - { \varepsilon \over 2 } < L \left( P, f_1, \alpha \right), $$ and $$ \int_a^b f_2 d \alpha - { \varepsilon \over 2 } \leq U \left( P, f_2, \alpha \right) - { \varepsilon \over 2 } < L \left( P, f_2, \alpha \right), $$ which imply $$ \int_a^b f_1 d \alpha - { \varepsilon \over 2 } < L \left( P, f_1, \alpha \right), \qquad \mbox{ and } \qquad \int_a^b f_2 d \alpha - { \varepsilon \over 2 } < L \left( P, f_2, \alpha \right), $$ and upon adding the last two inequalities we obtain $$ \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha - \varepsilon < L \left( P, f_1, \alpha \right) + L \left( P, f_2, \alpha \right). \tag{8}$$ But from (20) we obtain $$ L \left( P, f_1, \alpha \right) + L \left( P, f_2, \alpha \right) \leq L(P, f, \alpha) \leq \int_a^b f d \alpha,$$ which implies $$ L \left( P, f_1, \alpha \right) + L \left( P, f_2, \alpha \right) \leq \int_a^b f d \alpha. \tag{9} $$ From (8) and (9) we have $$ \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha - \varepsilon < \int_a^b f d \alpha, $$ and so $$ \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha < \int_a^b f d \alpha + \varepsilon $$ for every real number $\varepsilon > 0$. Therefore $$ \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha \leq \int_a^b f d \alpha. \tag{B} $$ Now from (A) and (B) we can conclude that $$ \int_a^b f d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha, $$ as required.
Is my rendering of Rudin's proof correct (and as intended by Rudin)?
