Agfdhyk

The theorem and the proof above is from Rudin.
I have a question about the last part of the proof.
For the inequality $sum_{i in A}(M_i^*-m^*_i)Deltaalpha_i+sum_{i in B}(M_i^*-m^*_i)Deltaalpha_i le epsilon[alpha(b) - alpha(a)]+2Kdelta$, I understand that the second term in the right side is made so because $M_i^*-m^*_ile2K$ for $iin B$ and $sum_{i in B}Deltaalpha_i<delta$ 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 $Deltaalpha(x_n)=alpha(b) - alpha(x_{n-1})$ and $Deltaalpha(x_1)=alpha(x_1)-alpha(a)$ are not a part of $sum_{i in A} Delta alpha_i$.

Thank you in advance!

Each $M_i^*-m_i^*ge0$ and $Deltaalpha_ige0$ as $alpha_igealpha_{i-1}$.
Also $M_i^*-m_i^*leepsilon$ for $iin A$. Therefore
$$sum_{iin A}(M_i^*-m_i^*)Deltaalpha_ileepsilonsum_{iin A}
Deltaalpha_i.$$
But
$$sum_{iin A}Deltaalpha_ilesum_{i=1}^nDeltaalpha_i=alpha(b)-alpha(a)$$
as $Deltaalpha_ige0$ for $inotin A$. Putting this together
gives the given bound.

Since $M_i - m_i geqslant 0$ (supremum minus infimum) and $alpha$ is increasing we have

$$sum_{i in A}(M_i-m_i) Delta alpha_i leqslant sum_{i =1}^n(M_i-m_i) Delta alpha_i leqslant epsilon[alpha(b) - alpha(a)]$$