TITLE: Rudin Real And Complex Definition 2.16
QUESTION [5 upvotes]: I'm having some difficulties in the remark Rudin makes in definition 2.16, when he says that , if we consider the situation described in theorem 2.14(Riesz representation theorem), if $E$ is in the $\sigma$-algebra and has $\sigma$-finite measure then $E$ is inner regular. I don't understand why this is true.
And i think it's key to prove theorem 2.18 item (b), so the trouble is really to prove theorem 2.18.
Sorry for my bad english.

REPLY [5 votes]: If $E$ has finite measure, then the Riesz representation theorem (item (d) in theorem 2.14) applies and it follows that $E$ is inner regular.
Otherwise, put $E = \bigcup_{n=1}^\infty E_n$ where $\mu(E_n) < \infty$. The sets $E_n$ can be made disjoint by a construction similar to step VII in theorem 2.14.
Fix $M > 0$. What we need to do is find a compact set $K \subset E$, so that $\mu(K) > M$.
Again, by (d) in theorem 2.14, we can find compact sets $K_n \subset E_n$ so that:
$$
\mu(E_n) \le \mu(K_n) + 2^{-n} \epsilon
$$
We have:
$$
\mu(E) = \sum_{n=1}^\infty \mu(E_n) \le \sum_{n=1}^\infty \mu(K_n) + \epsilon
$$
Since $\mu(E) = \infty$, we also have $\sum_{n=1}^\infty \mu(K_n) = \infty$ and we can find $N$ so that:
$$
\sum_{n=1}^N \mu(K_n) > M
$$
Put:
$$
K = \bigcup_{n=1}^N K_n
$$
Clearly $K \subset E$. Furthermore, $K$ is the union of a finite collection of compact sets. Hence it's compact. The desired result follows.