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I am reading "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

There is the following theorem on p.128 without a proof.

Is the following proof ok?

Theorem 6.12(a):

If $f \in \Re (\alpha)$, then $c f \in \Re (\alpha)$ and $\int_a^b c f d \alpha = c \int_a^b f d \alpha$ for any real number $c \in \mathbb{R}$.

Proof:
(1)
If $c = 0$, then $0 = \int_a^b 0 d \alpha = \int_a^b c f d \alpha$. So $c f = 0 \in \Re (\alpha)$ and $0 = \int_a^b c f d \alpha = c \int_a^b f d \alpha$.

(2)
Assume that $c > 0$.
Then, it is easy to show that the following equalities hold
$c \sup L(P, f, \alpha) = \sup L(P, c f, \alpha)$,
$c \inf U(P, f, \alpha) = \inf U(P, c f, \alpha)$.

And
$\sup L(P, f, \alpha) = \inf U(P, f, \alpha)$ since $f \in \Re (\alpha)$.

So, $\sup L(P, c f, \alpha) = c \sup L(P, f, \alpha) = c \inf U(P, f, \alpha) = \inf U(P, c f, \alpha)$.

$\therefore c f \in \Re (\alpha)$ and $\int_a^b c f d \alpha = c \int_a^b f d \alpha$.

(3)
Assume that $c < 0$.
Then, it is easy to show that the following equalities hold
$c \inf U(P, f, \alpha) = \sup L(P, c f, \alpha)$,
$c \sup L(P, f, \alpha) = \inf U(P, c f, \alpha)$.

And
$\sup L(P, f, \alpha) = \inf U(P, f, \alpha)$ since $f \in \Re (\alpha)$.

So, $\sup L(P, c f, \alpha) = c \inf U(P, f, \alpha) = c \sup L(P, f, \alpha) = \inf U(P, c f, \alpha)$.

$\therefore c f \in \Re (\alpha)$ and $\int_a^b c f d \alpha = c \int_a^b f d \alpha$.

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    $\begingroup$ Looks fine to me. $\endgroup$ – Kavi Rama Murthy Feb 24 at 5:28
  • $\begingroup$ @KaviRamaMurthy Thank you very much! $\endgroup$ – tchappy ha Feb 24 at 6:57

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