Theorem 2.17 from RCA Rudin 
$\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?
 A: 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.
A: 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 . 
A: $\forall{E\in\mathfrak{M}}$, we can find closed $F$ such
that $\mu\left(E-F\right)<\epsilon$ (using part (a) of the theorem). Also $F$ is $\sigma-$compact
because $F=\cup\left(F\cap K_{n}\right)$. Note that $\left(F\cap K_{n}\right)$
is compact using Theorem 2.5 Corollary (b). Hence, $\mu\left(F\cap K_{n}\right)<\infty$
from Theorem 2.14 (b). Hence $F$ has $\sigma-$finite measure. Hence
${E}$ has $\sigma-$finite measure, since $E=F\cup\left(E-F\right)$,
and hence ${E}$ is inner regular (If $E$ has $\sigma$-finite measure, then $E$ is inner regular).
