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Why can every set $E$ in the real numbers with $\mu^{*}(E)=\infty$ be realized as the disjoint union of countably many measurable sets, each of which has finite outer measure? I'm trying to see this without the approximation properties by open, closed sets.

Edit: Perhaps one can cover by countably many intervals, intersect each with $E$, and then inductively define the $k$th set by subtracting off the union of the previous $k-1$?

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I didn't think this claim would be true. But would appreciate it if someone would correct me if I am wrong. See a proposed counterexample below:

Define $E$ the union of the following two sets:

1) a Vitali set on $[0,1)$

2) $[1,\infty)$

The set in (1) is non-measureable, so and thus prevents this part of $E$ as being expressed by a countable union of measureable sets.

The set in (2) ensures that $E$ has infinite outer measure.

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  • $\begingroup$ You're right. I didn't see the "countably many measurable sets" and focused in the "finite outer measure" part. $\endgroup$ – leo Oct 2 '14 at 1:18
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You can write $E$ as a disjoint union of countably many sets with finite outer measure, because you can divide the entire space as such union. Namely, for each $n\in \Bbb Z$, let $I_n = [n,n+1)$. Then $$\Bbb R = \bigcup_{n\in \Bbb Z} I_n.$$

Now given such $E$, define $E_n=E\cap I_n$. Then $E$ is the disjoint union of those $E_n$ and each $E_n$ has finite outer measure because $E_n\subseteq I_n$.

Bit in general we can't ensure measurability of the subsets involved in the decomposition. See Ats's answer.

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