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Show that if $m^∗(E) < ∞$ and there exist intervals $I_1, . . . , I_n$ such that $m^∗(E(∆(∪_{i=1}^{n}I_i)))< ∞$, then each of the interval $I_i$ is finite.

I have been asked to prove this and I am thinking along the lines of Littlewood's First Principle, but I haven't made much progress yet. Any help would be appreciated

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  • $\begingroup$ How do you define the outer measure $m^*$? $\endgroup$ – Keen-ameteur 2 days ago
  • $\begingroup$ @Keen-ameteur I have defined $m^*$ as the infimum of the length of the union of the open intervals which contain that set $\endgroup$ – thedumbkid 2 days ago
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There exists an open set $U$ such that $E \subseteq U$ and $m(U) <m^{*}(E)+1$. We can write $U$ as countable disjoint union of open intervals $(a_i,b_i)$. Now $\bigcup_{i=1}^{n} (a_i,b_i)) \subseteq (E\Delta \bigcup_{i=1}^{n} (a_i,b_i)) \cup E$ so $m(\bigcup_{i=1}^{n} (a_i,b_i))<\infty$. This implies that each of the $I_i$'s is a finite interval.

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  • $\begingroup$ does'nt this prove the converse instead? $\endgroup$ – thedumbkid 2 days ago

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