The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $\sum_{i \in A}(M_i^*-m^*_i)\Delta\alpha_i+\sum_{i \in B}(M_i^*-m^*_i)\Delta\alpha_i \le \epsilon[\alpha(b) – \alpha(a)]+2K\delta$, I understand that the second term in the right side is made so because $M_i^*-m^*_i\le2K$ for $i\in B$ and $\sum_{i \in B}\Delta\alpha_i<\delta$ as proved in the proof. However, as for the first term in the right side of the inequality, How do we know that $\alpha(b)$ and $-\alpha(a)$ terms are in $\sum_{i \in A} \Delta \alpha_i$? $A$ might not contain $1$ and $n$ so that $\Delta\alpha(x_n)=\alpha(b) – \alpha(x_{n-1})$ and $\Delta\alpha(x_1)=\alpha(x_1)-\alpha(a)$ are not a part of $\sum_{i \in A} \Delta \alpha_i$.
Thank you in advance!
Best Answer
Each $M_i^*-m_i^*\ge0$ and $\Delta\alpha_i\ge0$ as $\alpha_i\ge\alpha_{i-1}$. Also $M_i^*-m_i^*\le\epsilon$ for $i\in A$. Therefore $$\sum_{i\in A}(M_i^*-m_i^*)\Delta\alpha_i\le\epsilon\sum_{i\in A} \Delta\alpha_i.$$ But $$\sum_{i\in A}\Delta\alpha_i\le\sum_{i=1}^n\Delta\alpha_i=\alpha(b)-\alpha(a)$$ as $\Delta\alpha_i\ge0$ for $i\notin A$. Putting this together gives the given bound.