I have some questions regarding step I of Rudin's proof of Riesz Representation Theorem (Real and Complex Analysis). I have included what I've tried. Please scroll down for the attached reference.
Question 1: How does (5) follow from (6)?
$g \prec V_1\cup V_2$ tells us $\Lambda g \le \mu(V_1\cup V_2)$. If I put $g = \chi_{(V_1\cup V_2)}$ in (6), I get (5) – but is this okay to do?
Question 2: Why is $\mu(V) \le \sum_{i=1}^\infty \mu(E_i) + \epsilon$ in the last line?
I was only able to see that if $f\prec V$ (which means $0\le f\le 1$, $f\in C_c(X)$, and $\text{supp}(f) \subset V$), $$\Lambda f = \int_X f\ d\mu = \int_{\text{supp}(f)} f\ d\mu \le \int_{\text{supp}(f)} 1\ d\mu \le \int_V 1\ d\mu = \mu(V)$$ which doesn't really help.
Best Answer
To collect @Gary's comments into an answer:
Q1. $g = \chi_{V_1\cup V_2}$ works, since the support of $g$ is contained in $V_1\cup V_2$, $0 \le g \le 1$, and $$\Lambda g = \int_X g\ d\mu = \mu(V_1\cup V_2)$$
Q2. $$\sum_{i=1}^n \mu(V_i) < \sum_{i=1}^n (\mu(E_i) + \epsilon/2^i) < \sum_{i=1}^\infty \mu(E_i) + \sum_{i=1}^\infty \epsilon/2^i = \sum_{i=1}^\infty \mu(E_i) + \epsilon$$ Taking limits as $n\to\infty$, we get $$\sum_{i=1}^\infty \mu(V_i) \le \sum_{i=1}^\infty\mu(E_i) + \epsilon$$ and by countable subadditivity we have $$\mu(V) \le \sum_{i=1}^\infty \mu(V_i)$$ which gives the result.