6

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?

Saaqib Mahmood
  • 26,762

1 Answers1

2

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.

Wraith1995
  • 1,169