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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?

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Answers above may be incorrect. E=∪En,μEn<∞, but En may not be measureable. –  user103953 Oct 29 '13 at 3:43
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2 Answers

up vote 2 down vote accepted

Let $E = \cup_{n=1}^\infty E_n$, where the $E_n$ can be taken, without loss of generality, to be pairwise disjoint and $\mu E_n < \infty $. Then by Theorem 2.14, $\mu E_n = \sup \{ \mu K | K \subset E_n, K \text{compact} \}$.

Let $\epsilon> 0$ and choose $K_n \subset E_n$ compact so that $\mu E_n \leq \mu K_n + \frac{\epsilon}{2\cdot 2^n}$. Then we have $\mu E = \sum_{n=1}^\infty \mu E_n \leq \sum_{n=1}^\infty \mu K_n+ \frac{\epsilon}{2}$. (Note that since $K_n \subset E_n$, we have $ \sum_{n=1}^\infty \mu K_n \leq \mu E$.)

Suppose $\sum_{n=1}^\infty \mu K_n = \infty$. Then let $C_N = \cup_{n=1}^N K_n \subset E$. $C_N$ is compact, $\mu C_n = \sum_{n=1}^N K_n$, and $\mu C_N \to \infty$ which gives $\mu E = \sup \{ \mu K | K \subset E, K \text{compact} \}$.

If $\sum_{n=1}^\infty \mu K_n < \infty$, we have $\sum_{n=1}^N \mu K_n \to \sum_{n=1}^\infty \mu K_n$, we can choose $N$ sufficiently large so that $\mu E \leq \sum_{n=1}^N \mu K_n+ \epsilon$. As above, let $C_N = \cup_{n=1}^N K_n \subset E$. Then $\mu E \leq \mu C_N + \epsilon$. Again, this gives $\mu E = \sup \{ \mu K | K \subset E, K \text{compact} \}$.

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Makes perfect sense. Thank you. –  PeterM Sep 20 '12 at 22:02
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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.

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