set theory - Non Lebesgue measurable subsets with "large" outer measure

Yes, I believe so - since subsets of a null set $A$ (i.e., $m(A)=0$) are not necessarily measurable, but will obviously still have outer measure 0, given any measurable set $E$ you "should" (i.e., I think so, but not sure) be able to find a non-measurable subset $S$ of a null set $A$ inside $E$, remove $S$, and since $m(E)=m(E-S)+m(S)$ and $m(S)$ isn't anything, we must have that $m(E-S)$ isn't anything, while we still have $m^{ast}(E)=m^{ast}(E-S)+m^{ast}(S)=m^{ast}(E-S)$.

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