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It is known that for any set( not necessarily measurable) $E$, there exists a measurable set(Lebesgue measurable) $F$, such that $E \subset F$ and outer measure $m^*(E)$ is equal to the measure $m(F)$. This is easy to show, by approximating $E$ by some open set $U_\epsilon$ covering $E$, such that $$m^*(E)+\epsilon \ge m(U_\epsilon) $$. Taking $\epsilon $ smaller and smaller and taking intersection of all such open sets, we obtain the result. My question is that if there are two arbitrary sets $E_1, E_2$ and $E_1 \subset E_2$, are there two measurable sets $F_1, F_2$ such that $E_1 \subset F_1$, $E_2 \subset F_2$, $m^*(E_1)=m(F_1)$, $m^*(E_2)=m(F_2)$, $F_1 \subset F_2$. Note that without the last condition, this problem is trivial. If there are not such $F_1, F_2$ with $F_1 \subset F_2$, can we instead make it satisfy that $F_1 \cap {F_2}^c$ is null.

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    $\begingroup$ With regards to the set $E,$ we take open $U_n\supset E$ such that $m^*(E)>1/n+m(U_n)$ and let $F=\cap_{n\in \mathbb N}U_n$ so $F$ is measurable as it is a countable intersection of measurable sets. An uncountable intersection of measurable sets need not be measurable. $\endgroup$ – DanielWainfleet Aug 6 '17 at 17:48
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Yes, there are such sets.

Apply the same construction as before to obtain $F_2$. Now, in order to obtain $F_1$, you can use the same process, but this time instead of using $U_\varepsilon$, you use $U_\varepsilon'=U_\varepsilon\cap F_2$. It is still a set that contains $E_1$ and it is true that $m^*(E_2)+\varepsilon\geqslant m(U_\varepsilon')$. Of course, $U_\varepsilon'$ may not be open, but it is still measurable. So, the intersection $F_1$ of all these sets is a:

  • measurable set;
  • subset of $F_2$;
  • set whose measure is $m^*(E_1).$
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