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(1) If $E$ is a subset of $\mathbb R$ with finite outer measure, i.e. $m^{*}(E) <\infty$; and (2) $E$ is not Lebesgue measurable, i.e. there exists $F$ such that $m^{*} (F) < m^{*}(EF) + m^{*}(FE^{c}).$

[Claim] There exists an open set $O\supset E$ with finite outer measure, such that $$m^{*}(O E^{c}) > m^{*} (O) - m^{*} (E).$$

My questions are the following.

(1) Is the above claim correct? I've seen this in Royden's book, but have a bit concern. Note that a set $E$ is Leb measurable if it satisfies caratheodory condition: $m^{*} (F) = m^{*}(EF) + m^{*}(FE^{c})$ for all subset $F$. If the claim is yes, it seems that measurability condition can be reduced from all subset $F$ to Borel set $F$.

(2) If the claim is correct, a proof is needed.

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    $\begingroup$ Originally, Lebesgue had a condition for measurability (for Lebesgue measure) where the sets $F$ were only intervals. Caratheodory wanted to define measurability in other settings where that does not work (perhaps because all intervals have infinite measure). So he observed that the Lebesgue criterion is still correct if you allow all sets for $F$. Then he used that as his definition. $\endgroup$ – GEdgar Sep 23 '15 at 17:04
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    $\begingroup$ @GEdgar Thanks for the explanation on the equivalence. $\endgroup$ – user79963 Sep 23 '15 at 18:17
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We always have $$ m^\ast (O) \leq m^\ast (E) + m^\ast (O \cap E^c). $$ Thus, assume towards a contradiction that there is no open set as you want it. Hence, for every open set $O \supset E$, we have $$ m^\ast (O) = m^\ast (E) + m^\ast (O \cap E^c). $$

More or less by definition of $m^\ast$ (I don't know your precise definition), there is a sequence of open sets $O_n \supset E$ with $m^\ast (O_n) \to m^\ast (E)$. Hence, $$ m^\ast (O_n \setminus E) = m^\ast (O_n) - m^\ast (E) \to 0. $$ Set $A := \bigcap_n O_n \supset E$. Note that $A$ is a Borel set. By the above, we get $m^\ast (A \setminus E) \leq m^\ast (O_n \setminus E) \to 0$, so that $A\setminus E$ is measurable (since it is a null set).

Finally, we see that $E = A \setminus (A \setminus E)$ is measurable.

This contradiction shows that there exists an open set $O$ with the properties you desire.

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  • $\begingroup$ Thanks. It confirms my question. $\endgroup$ – user79963 Sep 23 '15 at 18:03

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