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I am able to prove the first principle. Let $E$ be a measurable set of finite outer measure. Then for each $\epsilon > 0$, there is a finite disjoint collection of open intervals $I_k$ for which if $\mathcal{O} = \cup_{k=1}^n I_k$, then $m(E \sim O) < \epsilon$.

Can you give an $E$ such that as $\epsilon \to 0$, we have $n \to \infty$?

Any help is greatly appreciated!

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  • $\begingroup$ Are you asking for a specific measurable set such that, as $\epsilon$ goes to zero, the number of intervals in such a collection is forced to run off to infinity? $\endgroup$ – Josh Keneda Sep 24 '14 at 6:35
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Hint: Take Lebesgue measure on the real line, and let $B(x, r)$ be the ball of radius $r$ about the point $x$, and consider $$E=\bigcup_{n=1}^\infty B\left(n, \frac{1}{2^n}\right).$$

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