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If the measure $\lambda(E)$ of a set $E$ is finite then $E$ is measurable iff for each $\varepsilon>0$, there exist disjoint finite intervals $I_1,I_2,\dots,I_n$ such that $\lambda(E \triangle \bigcup_{k=1}^n I_k) \leq \varepsilon$ (where $\triangle$ denotes the symmetric difference, and $\lambda$ is the Lebesgue measure).

Can anyone provide me with a counterexample for the case when measure of $E$ is infinite?

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    $\begingroup$ Have you tried $E=\mathbb{R}$? $\endgroup$ – Eric Wofsey Dec 17 '15 at 7:31
  • $\begingroup$ Yes, I had tried but I was not getting how to manage those In's. $\endgroup$ – user268307 Dec 17 '15 at 7:47
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Take $E=\mathbb{R}$. Then, any finite collection of finite intervals will have measure at most n times the largest one, so the symmetric difference will contain the interval $(-\infty, c-1]$ where $c = \inf_{x \in \cup_{i=1}^n I_n} x$.

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  • $\begingroup$ I didn't get how you choose c. $\endgroup$ – user268307 Dec 17 '15 at 7:48
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    $\begingroup$ its just any value thats smaller than the smallest values in any of the intervals. Such a value must exist because there are finitely many finite intervals. $\endgroup$ – Batman Dec 17 '15 at 7:50
  • $\begingroup$ Okie, got it. In your answer in that infimum part you write everything in subscript , that created a confusion. Thanks for your help. $\endgroup$ – user268307 Dec 17 '15 at 7:53

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