Hey everyone.

The following problem is from Papa Rudin, exercise 2.15; I'm trying to rephrase it so that you won't have to look it up, but there's a chance that the rephrasing will be wrong since I'm not very good at this stuff. I just need a check on this, but I'm not very confident in my answer at all.

Problem: Let $\displaystyle \mu$ be a regular measure; let E be measurable, $\displaystyle \mu(E) < \infty$. Show that $\displaystyle E = N \cup K_1 \cup K_2 \cup \cdots$ for disjoint compact sets $\displaystyle K_i$ and $\displaystyle \mu(N) = 0$.

Attempted Solution Sketch: Choose $\displaystyle K_i$ by using the (inner) regularity of $\displaystyle \mu$ (and finite measure of E):

$\displaystyle \mu(E) \le \mu(K_1) + 2^{-1}$
$\displaystyle \mu(E - K_1) \le \mu(K_2) + 2^{-2}$
$\displaystyle \mu(E - K_1 - \cdots - K_{n - 1}) \le \mu(K_n) + 2^{-n}$
where in all cases $\displaystyle K_i \subset E - K_1 - \cdots - K_{i - 1}$.

Then, $\displaystyle \bigcup K_i \subset E$ and the $\displaystyle K_i$ are disjoint by construction. Thus, $\displaystyle \mu(\bigcup K_i) \le \mu(E)$. Fix $\displaystyle \epsilon > 0$ and find $\displaystyle n$ such that $\displaystyle 2^{-n} < \epsilon$. Then

$\displaystyle \mu(E) = \mu(K_1) + \mu(K_2) + \cdots + \mu(K_{n - 1} + \mu(E - K_1 - \cdots - K_{n - 1}) \le \sum_{i = 1} ^ n \mu(K_i) + 2^{-n} $
\le \sum_{i = 1} ^ \infty \mu(K_i) + \epsilon = \mu(\bigcup K_i) + \epsilon$.

Because $\displaystyle \epsilon > 0$ was arbitrary, $\displaystyle \mu(E) \le \mu(\bigcup K_i)$, which implies equality. Take $\displaystyle N = E - \bigcup K_i$, and the result follows.

Postscript: The details are, possibly, fudged a little bit, but I'm okay with that as long as the main idea is there. I'm not sure this works, as I'm not as comfortable with some of the material here. I'm also not sure if there is a more direct route to this.