# Theorem 6.12 (e) in Baby Rudin: If $f \in \mathscr{R}(\alpha)$ and $c > 0$, then $\ldots$

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:

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 $$L(P, f, c \alpha) = \sum_{i=1}^n \left( \inf_{x_{i-1} \leq x \leq x_i} f(x) \right) \left[ c \alpha\left( x_i \right) - c \alpha\left( x_{i-1} \right) \right] = c \sum_{i=1}^n \left( \inf_{x_{i-1} \leq x \leq x_i} f(x) \right) \left[ \alpha\left( x_i \right) - \alpha\left( x_{i-1} \right) \right] = c L(P, f, \alpha);$$ 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?

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.