My question has to do with the inequalities at the end. However, I will summarize the step to give some context to my question.
We want to show that for some complex functional $\Lambda $on $C_c(X) $ , that $\Lambda f \le \int_X f d \mu$ when $f$ is real.
Let $K$ be the support of $f$. Choose $[a, b]$ to be an interval containing the range of $f$. choose $\epsilon > 0$ and choose $y_i$ such that $y_i – y_{i-1} < \epsilon$ and $$y_0 < a < y_1 < \cdots < y_n = b.$$
Let $E_i = \{ x : y_{i-1} < f(x) \le y_i \} \cap K$. Then there are open sets $E_i \subset V_i$ such that $$\mu (V_i) < \mu (E_i) + \frac{\epsilon}{n}$$ and $f(x) < y_i + \epsilon$ for $x \in V_i$. Then using previous steps we find functions $h_i \prec V_i$ such that $\sum h_i = 1$. Then $\mu (K) \le \sum \Lambda h_i$, and $h_i f \le (y_i + \epsilon)h_i$. Now we get a long string of equalities and inequalities.
$$\hspace{-.5 in}\Lambda f = \sum^n \Lambda (h_if) \le \sum^n (y_i + \epsilon) \Lambda h_i \\
\hspace{.15in}= \sum ^n(|a| + y_i + \epsilon)\Lambda h_i – |a|\sum^n \Lambda h_i \\
\hspace{.55in}\le \sum^n(|a| + y_i + \epsilon)[\mu (E_i) + \frac{\epsilon}{n}] – |a| \mu(K) \\
\hspace{1.16in}= \sum^n(y_i – \epsilon)\mu(E_i) + 2\epsilon \mu(K) + \frac{\epsilon}{n}\sum^n(|a| + y_i + \epsilon) \\
\hspace{.15in}\le \int_X f d \mu + \epsilon[2 \mu(K) + |a| + b + \epsilon].$$
since this is true for any $\epsilon > 0$, we have $$\Lambda f \le \int_x f d \mu.$$
My question is why do we inject $|a|$ into the inequality? What purpose does it serve? It appears to me that if we never introduced $|a|$, second sum on line four would be $\frac{\epsilon}{n}\sum(y_i + \epsilon)$ which is less than $\frac{\epsilon}{n} \sum(b + \epsilon) = \epsilon(b + \epsilon.)$
Best Answer
Rudin introduces $|a|$ so that $|a| + y_i + \varepsilon \ge 0$. This allows Rudin to use the fact that $\Lambda h_i \le \mu(E_i) + \frac{\varepsilon}{n}$ in going from line 2 to 3.