# Is the measurability of the set E required for this problem to be right or have a solution? [closed]

This is one problem from my text book and since this book is new edition, I have been finding many typos or errors in this book. So I am not sure if this problem has an error that it should have requires the measurability of E somewhere in the problem or maybe I do not know how to solve this problem although the problem does not have any error. The Problem is following:

"Let $E$ have finite outer measure. Show that there is an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F\subseteq E \subseteq G$ and $m_{*}(F)=m*(E)=m^{*}(G)$."

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## closed as too localized by Ｊ. Ｍ., Martin Sleziak, Srivatsan, Matt N., Asaf KaragilaJan 4 '12 at 18:50

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Is $m * (F)$ meant to be $m^*(F)$, the outer measure? You can get the latter symbol by typing m^{*}(F). –  Arturo Magidin Aug 6 '11 at 22:07
In general (possibly non-measurable $E$), you can only get $m(F) = m_*(E)$ and $m(G) = m^*(E)$. –  GEdgar Aug 6 '11 at 23:15
@GEdgar: That would make sense, in which case Emily needs to edit; it seems clear that there was a problem in typesetting. To Emily: to get inner measure, use m_{*}(E). –  Arturo Magidin Aug 6 '11 at 23:37
@Arturo Magidin: Thank you for the comment. I have edited the problem. –  Emily Aug 7 '11 at 2:42
@Emily: You forgot to fix the middle one: $m * (E)$. –  Arturo Magidin Aug 7 '11 at 2:48