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Suppose $E\subseteq \mathbb{R}$ and $m^*(E)<\infty$. Prove that $E$ is measurable if and only if for any subset $A \subseteq E$, we have: $$m^*(E)=m^*(A)+m^*(E-A)$$

If $E$ is measurable and $A$ is any subset of $E$ (not necessarily measurable), since $E=A \cup (E-A)$ and $A\cap (E-A) = \emptyset$ we have: $m^*(E) = m^*(A) + m^*(E-A)$ (a proof can be found here)

For the other direction, I think one possible strategy is to prove that the additivity property for two disjoint sets still holds and then use it to prove $m^*(X)=m^*(X\cap E)+m^*(X \cap E^c)$ for any $X \subseteq \mathbb{R}$. However, I haven't succeeded yet.

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  • $\begingroup$ If $E$ is measurable and $A$ is ANY subset of $E$, then $A$ and $E-A$ may be non-measurable sets, and then you cannot use the result presented in math.stackexchange.com/questions/469027/… In fact, it may not be true that $m^*(E) = m^*(A) + m^*(E-A)$. $\endgroup$ – Ramiro May 11 '16 at 5:12
  • $\begingroup$ @Ramiro: Read the statement of the problem in that question again. It proves it for any $A$ and nowhere in the proof given by martini measurability of $A$ has been assumed. $\endgroup$ – user246836 May 11 '16 at 6:30
  • $\begingroup$ @H.Z. In the problem in that question (math.stackexchange.com/questions/469027/… ) we have two sets, namely $A$ and $E$, and one of them is measurable (namely $E$) then we can conclude that $$m^*(E\cup A)+m^*(E\cap A) = m^*(E)+m^*(A)$$ Here you are incorrectly applying that result to $A$ and $E-A$, where both may be non-measurable. $\endgroup$ – Ramiro May 11 '16 at 12:43
  • $\begingroup$ @H.Z. You seem to be confusing your $E$ here with the $E$ there. But to use the result in that question, the way you did, it is your $E-A$ (which may be non-measurable) that would play the role of $E$ there. $\endgroup$ – Ramiro May 11 '16 at 12:49
  • $\begingroup$ @Ramiro OK. Now I understand it. Thanks. $\endgroup$ – user246836 May 11 '16 at 13:04

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