Trying to understand this very long theorem of which I think good understanding is very educational. I am going through all the steps specifically now there's a subtlety in the conclusion of step II which I cannot get.
I am sure is a silly thing. I'll write the proof for reference
Step II : If $K$ is compact, then $K \in \mathcal{M}_F$ and $$ \mu(K) = \inf\left\{\Lambda f : K \prec f \right\} \;\;\;\; (7) $$ This implies assertion (b) of the theorem.
Proof: If $K \prec f$ and $0 < \alpha < 1$, let $V_\alpha = \left\{x : f(x) > \alpha \right\}$. Then $K \subset V_\alpha$ and $\alpha g \leq f$ whenever $g \prec V_\alpha$. Hence $$ \mu(K) \leq \mu(V_\alpha) = \sup \left\{ \Lambda g : g \prec V_\alpha \right\} \leq \alpha^{-1} \Lambda f $$ Let $\alpha \to 1$ to conclude that $$ \mu(K) \leq \Lambda f \;\;\;\; (8) $$ Thus $\mu(K) < \infty$. Since $K$ evidently satisfies (3), $K \in \mathcal{M}_F$. If $\epsilon > 0$, there's $V, K \subset V$ with $\mu(V) < \mu(K) + \epsilon$. By Urysohn's lemma $K \prec f \prec V$ for some $f$. Thus $$ \Lambda f \leq \mu(V) < \mu(K) + \epsilon $$ which, combined with (8), gives (7)
Here's the question or maybe clarification... combining (7) with (8) provides to me that
$$ \mu(K) \leq \Lambda f \leq \mu(K) + \epsilon \;\;\;\; (9) $$
and bedause $\epsilon > 0$ is arbitrary we end up with
$$ \mu(K) = \Lambda f $$
However in (8) the relationship $\mu(K) \leq \Lambda f$ holds for any $f$ with $K \prec f$ and specifically we have
$$ \mu(K) \leq \inf \left\{ \Lambda f : K \prec f \right\} $$
On the other hand for $9$ the $f$ used depends on the open $V$ chosen therefore to me the combination of (8) and (9) to get (7) should be done as
$$ \mu(K) \leq \inf \left\{\Lambda f : K \prec f \right\} \leq \Lambda f \leq \mu(K) + \epsilon \Rightarrow \mu(K) \leq \inf \left\{\Lambda f : K \prec f \right\} \leq \mu(K) + \epsilon $$
and as $\epsilon \to 0$ we have
$$ \mu(K) = \inf \left\{\Lambda f : K \prec f \right\} $$
Is this the right conclusion?
Your conclusion is correct but perhaps overly complicated.
The first part of the proof says that for every $f$ such that $K \prec f$, we have $\mu(K) \leq \Lambda f$. In other words, $\mu(K)$ is a lower bound for the set $\{\Lambda f\colon K \prec f\}$.
The second part of the proof says that for every $\varepsilon > 0$, there exists an $f$ such that $K \prec f$ and $\Lambda f < \mu(K) + \varepsilon$. In other words, $\mu(K) + \varepsilon$ is not a lower bound for the set $\{\Lambda f\colon K \prec f\}$.
So $\mu(K)$ is a lower bound, and everything that’s greater than $\mu(K)$ is not a lower bound. This is precisely what it means to be the greatest lower bound. Hence
$$\mu(K) = \inf{\{\Lambda f\colon K \prec f\}}.$$