TITLE: Theorem 6.12 (e) in Baby Rudin: If $f \in \mathscr{R}(\alpha)$ and $c > 0$, then $\ldots$
QUESTION [6 upvotes]: If $f \in \mathscr{R}(\alpha)$ on $[a, b]$ and if $c$ is a positive constant, then $f \in \mathscr{R}(c \alpha)$, and 
  $$ \int_a^b f d (c \alpha) = c \int_a^b f d \alpha.$$

This is part of Theorem 6.12 (e) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 
My proof: 
Here is my Math Stack Exchange post on the definition of integral as used in Baby Rudin.

As $c > 0$ and as $\alpha$ is a monotonically increasing function on $[a, b]$, so $c \alpha$ is also  monotonically increasing on $[a, b]$. 
And, for any partition $P = \left\{ \ x_0, x_1, \ldots, x_n \ \right\}$ of $[a, b]$, we also have 
  $$ 
\begin{align}
L(P, f, c \alpha) &= \sum_{i=1}^n \left[ \left(  \inf_{x_{i-1} \leq x \leq x_i} f(x) \right) \left[ c \alpha\left( x_i \right) - c \alpha\left( x_{i-1} \right) \right] \right] \\ 
&=  c \sum_{i=1}^n \left(  \inf_{x_{i-1} \leq x \leq x_i} f(x) \right) \left[  \alpha\left( x_i \right) -  \alpha\left( x_{i-1} \right) \right]  \\
 &= c L(P, f, \alpha);
\end{align}
$$ 
  that is, 
  $$  L(P, f, c \alpha) = c L(P, f, \alpha ) \ \mbox{ and similarly} \   U(P, f, c \alpha) = c U(P, f, \alpha). \tag{1} $$
Now as $f \in \mathscr{R}(\alpha)$ on $[a, b]$, so for every real number $\varepsilon > 0$ we can find a partition $P$ of $[a, b]$ such that 
  $$ U(P, f, \alpha) - L(P, f, \alpha) < { \varepsilon \over c }, $$
  and so from (1) we conclude that for this same partition $P$ of $[a, b]$, we have 
  $$ U(P, f, c \alpha) - L(P, f, c \alpha) < \varepsilon, \tag{2} $$
  from which it follows that $f \in \mathscr{R}( c \alpha)$. 
Also from (1) and  (2) we see that 
  \begin{align}
\int_a^b f d (c \alpha) &\leq U(P, f, c \alpha) \\
&<  L(P, f, c \alpha ) + \varepsilon \\
&= c L(P, f, \alpha) + \varepsilon \\ 
&\leq c \int_a^b f d \alpha + \varepsilon
\end{align}
  for every real number $\varepsilon > 0$, which implies that 
  $$ \int_a^b f d (c \alpha) \leq c \int_a^b f d \alpha. \tag{A}$$
And, again from (1) and  (2) we also have
\begin{align}
c \int_a^b f d \alpha & \leq c U(P, f, \alpha) \\
&= U(P, f, c \alpha) \\
&< L(P, f, c \alpha) + \varepsilon \\
&\leq \int_a^b f d (c \alpha) + \varepsilon 
\end{align}
  for every real number $\varepsilon > 0$, which implies that 
  $$ c \int_a^b f d \alpha \leq \int_a^b f d (c \alpha). \tag{B}$$ 
From (A) and (B) the required result follows. 

Is my proof satisfactory enough?

REPLY [2 votes]: The proof is correct, but you could consider shortening the second half of the proof. In particular, you could use that $\sup L(P,f,c\alpha) = c\sup L(P,f,\alpha)$ to reduce the second part of the proof down to a line or two. Since you already have $L(P,f,c\alpha) = cL(P,f,\alpha)$, you just need to quickly note a certain property of the $\sup$ and then use the definition of the integral.
Edit: if you want more feedback, you can use this to cut out even more of the proof. If you use this fact about the $\sup$ and $\inf$, then you do not need to use the $P-\epsilon$ formulation of the integral.