Alternative Proof to Theorem 6.12c) of Baby Rudin

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Theorem 6.12 (c) of Principles of Mathematical Analysis by Walter Rudin, 3rd edition is as follows:

If $f\in \mathscr{R}(\alpha)$ on [a,b] and if $a<c<b$, then $f\in \mathscr{R}(\alpha)$ on [a,c] and on[c,b], and

$\int_{a}^{b} f \,d\alpha = \int_{a}^{c} f \,d\alpha + \int_{c}^{b} f \,d\alpha$

I was wondering if I can construct an alternative proof of the last part theorem above using the following lemma:

Let A and B be nonempty sets bounded above and let $$A+B = \{a+b| a \in A , b \in B\}$$ Then $$sup(A+B) = sup A + sup B$$

Proof:

We let $$A = \{L(P,f,\alpha)| P\text{ is a partition of [a,c]}\}$$ and let $$B = \{L(P,f,\alpha)| Q\text{ is a partition of [c,b]}\}$$

then $$A+B = \{L(P \cup Q, f ,\alpha )|P \cup Q\text{ is a partition of [a,b]}\}$$

then by the above lemma and since f is integrable on [a,b], [a,c], and [c,b] : $$\int_{a}^{b} f \,d\alpha = \underline \int_a^b f \,d\alpha = \underline\int_{a}^{c} f \,d\alpha + \underline\int_{c}^{b} f \,d\alpha = \int_{a}^{c} f \,d\alpha + \int_{c}^{b} f \,d\alpha$$

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Besides what I pointed out in the comments, your reasoning is good.