In Rudin's proof of the Riesz Representation Theorem (step ten), he proves that
$$\Lambda h_i \leq \mu(V_i) < \mu(E_i) + \epsilon/n , \quad \mu(K) \leq \sum\limits_{1 \leq i \leq n} \Lambda h_i.$$
Writing $A = \sum\limits_{1 \leq i \leq n} \Lambda h_i$, he then asserts that
$$\sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)\Lambda h_i - |a| A \leq \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)(\mu(E_i) + \epsilon/n) - |a|\mu(K).$$
My question is, how does he avoid strict inequality? As $\Lambda h_i < \mu(E_i) + \epsilon/n$, it seems to follow that
$$ \begin{align*} \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)\Lambda h_i - |a| A &< \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)(\mu(E_i) + \epsilon/n) - |a|A\\ &\leq \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)(\mu(E_i) + \epsilon/n) - |a|\mu(K) \end{align*} $$ and therefore that
$$\sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)\Lambda h_i - |a| A < \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)(\mu(E_i) + \epsilon/n) - |a|A.$$