Proof of Riemann Stieltjes Integral 

Does anyone help me to understand why the last inequality holds?  
2M$\epsilon$ comes from that $M_i-m_i \le 2M$ and delta($\alpha$)< $\epsilon$ but why the other term appeared?

Let $ a_j $ be such numbers that $ x_{a_j} = u_j $ and $ b_j $ s. t. $ x_{b_j} = v_j $. Let $ k $ be the number for which $ [u_k, v_k] $ is the last of the intervals covering $ E $. Now we can break down the main sum this way:
$ \sum_1^n(M_i-m_i)\Delta\alpha_i = \sum_1^{a_1}(M_i-m_i)\Delta\alpha_i + \sum_{a_1+1}^{b_1}(M_i-m_i)\Delta\alpha_i + \sum_{b_1+1}^{a_2}(M_i-m_i)\Delta\alpha_i + \sum_{a_2+1}^{b_2}(M_i-m_i)\Delta\alpha_i + ... + \sum_{a_k+1}^{b_k}(M_i-m_i)\Delta\alpha_i + \sum_{b_k+1}^{n}(M_i-m_i)\Delta\alpha_i$
In the sums of the form $ \sum_{a_j+1}^{b_j}(M_i-m_i)\Delta\alpha_i $ we have $ M_i - m_i \le 2M $ and $ \alpha(x_{b_j}) - \alpha(x_{a_j}) = \alpha(v_j) - \alpha(u_j) \le \epsilon $. In the others, we can use $ M_i - m_i \le \epsilon $.
Try manipulating the sums an you should get the desired inequality.
