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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

$$\sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)\Lambda h_i - |a| A < \sum\limits_{1 \leq i \leq n} (|a|+y_i+\epsilon)(\mu(E_i) + \epsilon/n) - |a|A.$$

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Well, even if. When the strict inequality holds, the weak one holds too. Since at the end one does a limiting operation $\varepsilon \to 0$, the end result would be a (the desired) weak inequality, whether he wrote a strict or weak one before. (Yes, the strict inequality does hold there, but one can always write something weaker than actually holds if convenient.) – Daniel Fischer Jun 29 '13 at 20:56
    
Thank you, I was not sure if this was merely for convenience or not; its my first time studying analysis and I want to make sure I don't miss any subtleties. – Cameron Jun 29 '13 at 21:37
    
Oh, you're doing fine detecting subtleties, it seems. Very promising. – Daniel Fischer Jun 29 '13 at 21:39
    
@DanielFischer Maybe your first comment could be an answer. – Davide Giraudo Jun 30 '13 at 19:22
    
I'm studying this proof and have two very naive questions. 1. Why can we claim that $\Lambda h_i\leq \mu(V_i)$? 2. Do we need to have $y_i\geq 0$, in order to use this in the inequality below? – Lionville Dec 31 '15 at 20:03

Answered in a comment:

Well, even if. When the strict inequality holds, the weak one holds too. Since at the end one does a limiting operation $\varepsilon \to 0$, the end result would be a (the desired) weak inequality, whether he wrote a strict or weak one before. (Yes, the strict inequality does hold there, but one can always write something weaker than actually holds if convenient.) --Daniel Fischer Jun 29 '13 at 20:56

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