TITLE: caratheodory measurability condition
QUESTION [6 upvotes]: (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.

REPLY [2 votes]: 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.