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$\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 $G_b$, $A\subset E\subset B$, and $\mu(B-A)=0.$

$\rm P\scriptstyle{\rm ROOF}$

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

I understood the proof of points $(a)$ and $(c)$. But I can't understand the proof of $(b)$. It's obvious that every closed set is $\sigma$-compact. But how Rudin applies $(a)$ here?

We have to show that if $\alpha>0$ then exists compact set $K\subset E$ such that $\mu(K)>\alpha$.

Can anyone explain it to me please?

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Given $E$ of infinite measure and $\epsilon > 0$, let $V$ and $F$ be as in (a). Then $\mu(E-F) < \epsilon$. For each $n$, put $$ F_n = F\cap \left(\bigcup_{i=1}^{n}K_i\right), $$ $F_n$, as a closed subset of a compact set, is itself compact. We have $$ \lim_{n\to \infty} \mu(F_n) = \mu(F) = +\infty, $$ from which we conclude that $\mu$ is regular.

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  • $\begingroup$ Limit equality is true since $F_n$ is increasing sets and their union is $F$ right? $\endgroup$ – ZFR Jun 25 '16 at 6:03
  • $\begingroup$ Also where did you use that $\mu(E-F)<\epsilon$? $\endgroup$ – ZFR Jun 25 '16 at 6:05
  • $\begingroup$ To show that $\mu(F) = +\infty$. $\endgroup$ – Qiyu Wen Jun 25 '16 at 6:06
  • $\begingroup$ You mean that if $\mu(E-F)<\epsilon$ then $\mu(E)<\mu(F)+\epsilon$. Right? Note that $\mu(E-F)$ is not equal to $\mu(E)-\mu(F)$ in general. It's true when $\mu(F)<\infty$. Why it's true in our case? $\endgroup$ – ZFR Jun 25 '16 at 6:11
  • $\begingroup$ The reasoning is that, if $\mu(F) < +\infty$, then $\mu(E) < +\infty$, in contradiction to our assumption. $\endgroup$ – Qiyu Wen Jun 25 '16 at 6:14
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note that regular as in Rudin book mean that every borel set is inner and outer regular, it mean : $$ \mu(F)= \sup\{\mu(K) \; K \; \textrm{compact of } F\}\\ \mu(F)=\inf\{\mu(O) \; O \; \textrm{open which contient } F\}\\ $$ it's clear that $\mu(K)\leq \mu(F)$ to proof that it will be the sup we can use the proprety of supremum bounded $$ \forall \epsilon >0 \; \exists K_\epsilon \subset F \; \textrm{compact} \; ; \; \mu(F)-\epsilon\leq \mu(K_\epsilon) $$ that mean $\mu(F-K_\epsilon)\leq \epsilon $ but the condition (a) give such $K_\epsilon$ since $\mu(F-K)\leq \mu(O-K)<\epsilon$ the same technique can be used to prove the other one .

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