I'm looking for an example of when this theorem fails.

Theorem: Let $E$ be a measurable set of finite measure. For each $\epsilon > 0$, there is a finite disjoint collection of open intervals $I_1,I_2,...$ for which $U = \bigcup_{k=1}^n I_k$ has $m^*(E\setminus U) + m^*(U\setminus E) < \epsilon$.

I've already proved the theorem, but don't see an example of when infinity would make it fail.

  • 1
    $\begingroup$ "Open intervals" include non-bounded intervals like $(1,\infty)$? $\endgroup$ – ajotatxe Oct 5 '17 at 17:34
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    $\begingroup$ Take an infinite set consisting of alternating unit intervals (e.g. the set of reals whose floor is an even integer). How do you approximate this with a finite collection of intervals? $\endgroup$ – Erick Wong Oct 5 '17 at 17:35

From the comments above.

The set $E = \bigcup_{n = -\infty}^\infty (2n,2n+1)$ is a measurable set such that $mE = \infty$. For any choice of $\epsilon > 0$ there is no finite collection $I_1, \dots, I_n$ of open intervals for which $U = \bigcup_{k=1}^n I_k$ has $m^*(E \setminus U) + m^*(U \setminus E) < \epsilon$.


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