# 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?

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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 $M_i - m_i \le \epsilon$.