# Theorem 6.12(a) Of Baby Rudin. Alternative Proof Of $\int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$

alternative-proofintegrationproof-writingreal-analysissolution-verification

If $$f_1 \in \mathscr{R}(\alpha)$$ and $$f_2 \in \mathscr{R}(\alpha)$$, then
$$\int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$$

I want to prove this theorem using definition of integral(upper and lower Riemann-Stieltjes integral). We’ll make use of $$L(P, f_1, \alpha)+ L(P, f_2, \alpha) \leq L(P, f, \alpha) \leq U(P, f, \alpha) \leq U(P, f_1, \alpha) + U(P, f_2, \alpha )$$ fact.

My attempt: let $$f= f_{1} + f_{2}$$.
$$\inf U(P,f, \alpha) \leq U(P,f, \alpha) \leq U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)$$. Since $$\inf U(P,f, \alpha)$$ is an lower bound of $$U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)$$, we have $$\inf U(P,f, \alpha) \leq \inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\}= \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$$. Similarly, $$\sup L(P,f_{1},\alpha) + \sup L(P,f_{2},\alpha) \leq \sup L(P,f,\alpha)$$.
Thus, $$\sup L(P,f_{1},\alpha) + \sup L(P,f_{2},\alpha) \leq \sup L(P,f,\alpha) \leq \inf U(P,f, \alpha) \leq \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$$. Since, $$\sup L(P,f_{1},\alpha) = \inf U(P,f_{1}, \alpha)$$ and $$\sup L(P,f_{2},\alpha) = \inf U(P,f_{2}, \alpha)$$, we have $$\sup L(P,f,\alpha) = \inf U(P,f, \alpha)= \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$$. Our desired equality $$\int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$$.

Is this proof correct?

You snuck in a major step right here that needs an explanation: $$\inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\}= \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$$ We immediately have $$\inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\} \geq \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$$ but in the context of the proof, this inequality is pointing in the wrong direction. What we really need is the other direction. That is, if we write $$\mathcal{P}$$ for the set of finite partitions of $$[a,b]$$, then for clarity, we need to show $$\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) \leq \inf_{P \in \mathcal{P}} U(P,f_{1}, \alpha) + \inf_{P \in \mathcal{P}} U(P,f_{2}, \alpha).$$
This follows easily from the lemma $$P_1,P_2 \in \mathcal{P} \implies P_1\cup P_2 \in \mathcal{P} \text { and } U(P_1\cup P_2, f, \alpha) \leq U(P_i,f,\alpha) \text{ for } i=1,2$$ as then for any $$P_1,P_2 \in \mathcal{P},$$ \begin{align*}\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) &\leq U(P_1\cup P_2,f_{1}, \alpha) + U(P_1\cup P_2,f_{2}, \alpha) \\ &\leq U(P_1,f_{1}, \alpha) + U(P_2,f_{2}, \alpha)\end{align*} Now by taking the infimum over $$P_1,P_2 \in \mathcal{P}$$, \begin{align*}\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) &\leq \inf_{P_1, P_2\in \mathcal{P}}\Big(U(P_1,f_{1}, \alpha) + U(P_2,f_{2}, \alpha)\Big) \\ &= \inf_{P_1\in \mathcal{P}}U(P_1,f_{1}, \alpha) + \inf_{P_2\in \mathcal{P}}U(P_2,f_{2}, \alpha) \end{align*} [Note that this last equality is the exercise you wanted to use in the comments]
As for the lemma's proof, it should be simple to show $$P_1\cup P_2 \in \mathcal{P}$$. For the inequality, you can induct on the size of $$P_1\cup P_2$$ to assume $$|P_1 \cup P_2| = |P_1| + 1$$, in which case the proof just amounts to showing that $$x_1 < x_2 < x_3$$ implies $$\sup_{x \in [x_1,x_3]}f(x)(\alpha(x_3)-\alpha(x_1)) \geq \sup_{x\in [x_1,x_2]}f(x)(\alpha(x_2)-\alpha(x_1)) + \sup_{x\in [x_2,x_3]}f(x)(\alpha(x_3)-\alpha(x_2))$$