# Why does this theorem fail when $m(E) = \infty$?

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.

• "Open intervals" include non-bounded intervals like $(1,\infty)$? – ajotatxe Oct 5 '17 at 17:34
• 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? – Erick Wong Oct 5 '17 at 17:35

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 $m^*(E \setminus U) + m^*(U \setminus E) < \epsilon$.