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Some probably silly questions about such theorem, the proof is quite long and i don't understand some of the steps of this theorem. (i will probably update this post because of the length...).

The theorem state that

"Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_{c}(X)$. Then there exists a $\sigma$-algebra $\mathcal{M}$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathcal{M}$ which represents $\Lambda$ in the sense that:

  1. $\Lambda f = \int_X f d\mu \;\; \forall f \in C_c(X)$;
  2. $\mu(K) < \infty \forall K \subset X$, $K$ compact set;
  3. $\forall E \in \mathcal{M}$, we have $$\mu(E) = inf \left\{ \mu(V) : E \subset V, V \text{ open} \right\}$$
  4. The relation $$ \mu(E) = sup \left\{ \mu(K) : K \subset E, K \text{ compact} \right\}$$ Holds for every open set $E$, and for every $E \in \mathcal{M}$ with $\mu(E) < \infty$
  5. If $E \in \mathcal{M}$, $A subset E$ and $\mu(E) = 0$, then $A \in \mathcal{M}$. "

Here there are the questions:

  1. The theorem states that $X$ is a locally compact Hausdorff space but i don't understand where it is used such hypothesis, since it is never mentioned esplicitly.
  2. The step two of the theorem (i can report all the proof if you want) states that, at some point that the relation

$$\Lambda f \leq \mu(V) < \mu(K) + \epsilon$$ combined with $$\mu(K) \leq \Lambda f$$ gives $$\mu(K) = \inf \left\{ \Lambda f : K \prec f \right\}$$

where $K$ is a compact set, $V$ is open and $\epsilon > 0$. I don't understand the implication...

Update. I probably had a clue by my self. The statement is : If $K$ is compact, then $K \in \mathcal{M}$ and $$\mu(K) = \text{inf} \left\{ \Lambda f : K \prec f \right\}$$

Proof:

  1. Let's take an arbitrary $K \prec f$ 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 $$\mu(K) \leq \mu(V_\alpha) = \text{sup} \left\{ \Lambda g : g \prec \Lambda f \right\}.$$ Thus $\mu(K) < \infty$. $K$ Satisfies the condition for belonging to $\mathcal{M}_F$.
  2. If $\epsilon > 0$ there exists $V$ such that $K \subset V$ 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 the inequality derived in part 1 gives the thesis.

(As you can see i reported more or less the proof of the book). I think the key is that the $f$ taken in step 1 is arbitrary while the $f$ taken at the part 2 is not arbitrary but depends on the $\epsilon$ choosen, since the inequality is not strict then we have the thesis. However i still struggle in formalize this logic step.


This question can also be useful, so mapping my own question in the other basically what here is $f$ there is $s$ and what here is $\Lambda$ there is $f$, the $\alpha$'s role is unchanged.

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  • $\begingroup$ Urysohn's lemma is used somewhere in the proof, so the condition on $X$ is used there. $\endgroup$
    – user99914
    Nov 2, 2015 at 9:17
  • $\begingroup$ You're right... any clue on the point 2 instead? $\endgroup$ Nov 2, 2015 at 10:13

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