TITLE: If $E$ has $\sigma$-finite measure, then $E$ is inner regular
QUESTION [9 upvotes]: In Rudin's Real & Complex Analysis, page 47:

It is easy to see that if $E \in \mathfrak{M}$ and E has $\sigma$-finite measure, then $E$ is inner regular.

$\mathfrak{M}$ in this context is the $\sigma$-algebra constructed to prove the Riesz representation theorem.
I can't for the life of me figure out how this is easy. Can anyone explain it to me?

REPLY [10 votes]: If $\mu(E) < \infty$, $E$ is inner regular.
Hence we can assume that $\mu(E) = \infty$.
Since $E$ is $\sigma$-finite, $E$ can be written as $E = \bigcup E_n$, where each $E_n$ has a finite measure and they are pair-wise disjoint.
Let $\epsilon > 0$ be a positive real number.
Since each $E_n$ is inner regular, there exists a compact set $K_n$ such that 
$K_n \subset E_n$, $\mu(E_n) < \mu(K_n) + \epsilon/2^n$.
Then $\mu(E) = \sum \mu(E_n) \le \sum \mu(K_n) + \epsilon$.
Since $\mu(E) = \infty$, $\sum \mu(K_n) = \infty$.
Hence sup$\{\mu(K_1 \cup\dots\cup K_m)\colon m = 1,2,\dots\} = \infty$.
Thus $E$ is inner regular.