# Additive property of Riemann-Steiltjes integral proof (Rudin)

I am trying to understand the proof of the following theorem from Rudin's Principles of Mathematical Analysis, 3e, Ch 6., pg 128.

$\textbf{6.12 Theorem}$

(a) If $f_1 \in \mathscr{R}(\alpha)$ and $f_2 \in \mathscr{R}(\alpha)$ on $[a,b]$, then $$f_1 + f_2 \in \mathscr{R}(\alpha),$$ $~~~~~cf \in \mathscr{R}(\alpha)$ for every constant $c$, and $$\int_a^b(f_1 + f_2)~d\alpha = \int_a^bf_1 ~d\alpha+\int_a^bf_2~d\alpha,$$ $$\int_a^bcf~d\alpha = c\int_a^bf~d\alpha.$$

$\textbf{Proof}$

In the proof, Rudin argues that if $f = f_1 + f_2$ and $P$ is any partition of $[a,b]$, then $$L(P,f_1,\alpha) + L(P,f_2,\alpha) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq U(P,f_1,\alpha) + U(P,f_2,\alpha)$$

I believe that $L(P,f,\alpha) \leq U(P,f,\alpha),$ but I don't understand why $$L(P,f_1,\alpha) + L(P,f_2,\alpha) \leq L(P,f,\alpha).$$ As I understand, for the lower sum of $f_1$, we select $m_{1i} = inf(f_1(x))$ for each $\Delta x_i$. Similarly we select $m_{2i}$. Since we're just adding things up, we should have $inf(f) = m_{1i} + m_{2i} = m_i,$ correct? Then instead of $\leq$, we should have equality: $$L(P,f_1,\alpha) + L(P,f_2,\alpha) = L(P,f,\alpha)$$.

Could you help me understand where I am going wrong?

Thank you.

In the case of the $L(P,f_i,\alpha)$ you take the infimum independently of the other one. Whereas in the other case, you do it over the sum. In general, you have $$\inf\limits_{x \in [a,b]} \{f(x) + g(x)\} \geqslant \inf\limits_{x \in [a,b]} f(x) + \inf\limits_{x \in [a,b]}g(x),$$ which exactly represents the argument Rudin uses. You can proove that by assuming $$\inf\limits_{x \in [a,b]} \{f(x) + g(x)\} < \inf\limits_{x \in [a,b]} f(x) + \inf\limits_{x \in [a,b]}g(x),$$ which will result in a contradiction.