I am having trouble with the beginning of the proof of the Riesz Representation Theorem from Rudin. I will be assuming (for now, I will correct this later) that the notation is familiar to everyone.
I. The Theorem states: Let $X$ be a locally Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra $\mathfrak M$ in $X$ which contains all Borel sets in $X$ and there exists a unique positive measure $\mu$ on $\mathfrak M$ which represents $\Lambda$ in the sense that
(a)$\Lambda f=\int_X f\,d\mu$ for every $f\in C_c(X)$
(b)$\mu(K)<\infty$
(c) For every $E\in\mathfrak M$, we have $$\mu(E)=inf\{\mu(V):E\subset V, V\,\,open\}$$
(d) The relation $$\mu(E)=sup\{\mu(K):K\subset E, K\,\,compact\}$$
II. (Part of) The proof goes:
Let use begin by proving the uniqueness of $\mu$...It suffices to prove that $\mu_1(K)=\mu_2(K)$ for all $K$ whenever both measures ($\mu_1$, $\mu_2$) are measures for which the theorem holds. So fix $K$ and $\epsilon>0$. By (b) and (c), there exists a $K\subset V$ with $\mu_2(V)<\mu_2(K)+\epsilon$; by Urysohn's lemma, there exists an $f$ so that $K\prec f\prec V$; hence $$\mu_1(K)=\int_X \chi_K\,d\mu\leq\int_X f\,d\mu_1=\Lambda f=\int_X f\,d\mu_2\leq\int_X\chi_V=\mu_2(V)<\mu_2(K)+\epsilon$$
Here $K$ is a compact set and $V$ is an open set.
III. What I don't understand
How is $\mu_2(V)<\mu_2(K)+\epsilon$ admissible when $A\subset B\Rightarrow\mu(B)\geq\mu(A)$ and $\epsilon$'s only condition is to be greater than 0?
Also, I don't undestand why $$\int_X f\,d\mu_1=\Lambda f=\int_X f\,d\mu_2$$. How can two different measures be equal to the same linear functional?
$\mu_2(V) < \mu_2(K) + \epsilon$ holds because this is the way you chose $V$! Existence of $V$ comes from the fact that $\mu_2(K) + \epsilon > \mu_2(K) = \inf_{V \supset K} \mu_2(V)$ by (c).
Identity $\int f d\mu_1 = \Lambda f = \int f d \mu_2$ comes from the fact that both $\mu_1$ and $\mu_2$ are supposed to satisfy the conditions of the theorem, in particular Condition (a).