Rudin's RCA, Theorem $2.17$.

Question

There is the $2.14$ theorem: There is the definition which we need for the theorem:

$\bf 2.17\ $ Theorem $\ $ Suppose $X$ is a locally compact, $\sigma$-compact Hausdorff space. If $\frak M$ and $\mu$ are as described in the statement of Theorem $\it 2.14$, then $\frak M$ and $\mu$ have the following properties:

$(a)\ \ $ If $E\in\frak M$ and $\epsilon>0$, there is a closed set $F$ and an open set $V$ such that $F\subset E\subset V$ and $\mu(V-F)<\epsilon$.
$(b)\ \ $ $\mu$ is a regular Borel measure on $X$.
$(c)\ \ $ If $E\in\frak M$, there are sets $A$ and $B$ such that $A$ is an $F_\sigma$, $B$ is a $G_\delta$, $A\subset E\subset B$, and $\mu(B-A)=0.$

There is the proof of the $(b)$:

Let $X$ $=$ $K_1$ $\cup$ $K_2$ $\cup$ $K_3$ $\cup$... where each $K_n$ is compact.

Every closed set $F$ $\subset$ $X$ is $\sigma$-compact, because $F$ $=$ $\cup$($F$ $\cap$ $K_n$). Hence $(a)$ implies that every set $E$ $\in$ $\mathfrak M$ is inner regular. This proves $(b)$.

I don't understand how does the $(a)$ implies that every set $E$ $\in$ $\mathfrak M$ is inner regular by the fact that $F$ is $\sigma$-compact. I also don't understand how all these things proofs the $(b)$. Because in order to $\mu$ be regular Borel measure on $X$ it should be inner regular and outer regular simultaneously and all these things has only proved inner regularity. Rudin has noted in $2.15$ that we already have outer regularity. But where do we use outer regularity in our construction of $\mu$ and $\mathfrak M$?

Any help would be appreciated.

Answer 1

$\sigma$-compactness here is not necessary.

To prove that (a) implies (b), let $E$ be any element of $\mathfrak M$. Clearly $$\mu(E) \le \inf \{ \mu(V) | E \subset V, V \mbox{open} \}$$ because $\mu(E) \le \mu (V)$ by monotonicity of the measure. To prove equality note that if $\mu(E)= \infty$ there is nothing to prove.

Otherwise, if $\mu(E) < \infty$, let $\varepsilon>0$ be arbitrary. Then there exist by condition (a) $$\exists F\subset E \subset V \quad \mbox{such that } \quad \mu(V-F) < \varepsilon$$ Then $$\mu(E) + \varepsilon \ge \mu(F) + \varepsilon \ge \mu (F) + \mu(V-F) = \mu (V) \ge \inf \{ \mu (V) | etc. \}$$ By arbitrarity of $\varepsilon$ we have the opposite inequality $$\mu(E) \ge \inf \{ \mu(V) | E \subset V, V \mbox{open} \}$$

In a similar way you can prove also that $$\mu(E)= \sup \{ \mu(K) | K \mbox{ compact }, K \subset E\}$$

There is only one thing to notice: since $F= \bigcup_n (F \cap K_n)$ , then $$\mu (F) = \sup \{ \mu (F \cap (K_1 \cup \cdots \cup K_n)) | n \ge 1) \}$$ and $F \cap (K_1 \cup \cdots \cup K_n)$ are all compact sets. Now, the measure of $F$ can be appoximated by measure of compact sets, and the measure of $E$ can be approximated by the measure of $F$. Hence you can argue with the usual $\varepsilon$ argument.

Answer 2

Note (like Rudin in $\S$ 2.15) that we already have outer regularity. Hence the proof of (b) reduces to inner regularity. Applying (a) to any $E\in\frak M$, you get$$\mu(E)=\sup\{\mu(F)\mid F\text{ closed}\subset E\}.$$Since any closed $F$ is $\sigma$-compact,$$\mu(F)=\sup\{\mu(K)\mid K \text{ compact}\subset F\}.$$From these two properties you conclude: $$\begin{align}\mu(E)&=\sup_{ F\text{ closed}\subset E}\left(\sup_{ K\text{ compact}\subset F} \mu(K)\right)\\&=\sup_{ K\text{ compact}\subset E} \mu(K),\end{align}$$as required.

As for your new additional question: Rudin has noted in $2.15$ that the $\mu$ and $\mathfrak M$ of Theorem 2.14 already satisfy outer regularity. In Theorem 2.17, he does not construct another $\mu$ and $\mathfrak M$: he takes those constructed in 2.14, and uses outer regularity to say that proving inner regularity is sufficient to prove (b).