I am having trouble understanding step III in Rudin's proof of the Riesz Representation theorem:
I'm not going to define any of the symbols that occur in the picture or provided any of the background details, so don't complain about there not being enough details for you to answer the question and acquire more points. This question is for those who have a copy of Rudin's Real and Complex Analysis.
Obviously if $K$ is compact set contained in the open set $V$, then $\mu(K) \le \mu(V)$ for every $K \subseteq V$, and so $\mu(V)$ is an upper bound for $\{\mu(K) \mid K \subseteq V, K \mbox{ is compact } \}$. Hence, $\sup \{\mu(K) \mid K \subseteq V, K \mbox{ is compact } \} \le \mu(V)$....However, I don't know how to get the reverse inequality...Perhaps it's due to the fact that I've already spent an entire (on and off) working through the proof, so that my brain is fried...What I don't understand is, which inequality is Rudin proving? I presume it's the one I'm having trouble with. I don't understand how considering any open set $W$ that contains $\mbox{supp} (f)$ helps show the other inequality
EDIT:
For those who find my post unclear, here is the cliffsnotes version:
Rudin is proving that $ \mu(V) = \sup \{\mu(K) \mid K \subseteq V, K \mbox{ is compact } \}$; showing $\sup \{\mu(K) \mid K \subseteq V, K \mbox{ is compact } \} \le \mu(V)$ is trivial; and I'm wondering how to get the other inequality.
Yes it is unclear. My copy of the book is missing - I had no idea what was being proved until your edit. Having clarified that, what step do you have a problem with?
(i) If $\alpha < \mu(V)$ there exists $K\subset V$ with $\mu(K)>\alpha$.
(ii) It follows that $\sup_K\mu(K)\ge\mu(V)$.