Proof of Theorem 6.12(a) in “Principles of Mathematical Analysis 3rd Edition” by Walter Rudin

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 $$\int_a^b c f d \alpha = c \int_a^b f d \alpha$$.

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