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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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  • 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 http://math.stackexchange.com/questions/469027/for-an-outer-measure-m-does-me-cup-ame-cap-a-mema-alw?lq=1

    In fact, it may not be true that $m^(E) = m^(A) + m^*(E-A)$.

     – Ramiro May 11 '16 at 05:12
  • @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. – user246836 May 11 '16 at 06:30
  • @H.Z. In the problem in that question (http://math.stackexchange.com/questions/469027/for-an-outer-measure-m-does-me-cup-ame-cap-a-mema-alw?lq=1 ) 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. – Ramiro May 11 '16 at 12:43
  • @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. – Ramiro May 11 '16 at 12:49
  • @Ramiro OK. Now I understand it. Thanks. – user246836 May 11 '16 at 13:04
  • @H.Z. Here is an example where $E\subseteq \mathbb{R}$ is measurable, $m^(E)<\infty$ and there is a subset $A \subseteq E$, such that $$m^(E)<m^(A)+m^(E-A)$$ Take $E=[0,1]$ and $A$ to be a Vitali set of outer measure 1 (they exist, see http://math.stackexchange.com/questions/157532/vitali-set-of-outer-measure-exactly-1?rq=1 ). Since $A$ is not Lebesgue measurable, $[0,1]-A$ is not Lebesgue measurable and so $m^([0,1]-A)>0$. So we have $$m^([0,1])=1<1+m^([0,1]-A)=m^(A)+m^*([0,1]-A)$$ – Ramiro May 11 '16 at 13:17
  • @H.Z. Here is a simpler example where $E\subseteq \mathbb{R}$ is measurable, $m^(E)<\infty$ and there is a subset $A\subseteq E$ such that $$ m^(E)<m^(A)+m^(E-A)$$ Take $E=[0,1]$ and $A$ be a Vitali set. Since $A$ is not Lebesgue measurable, we have that $m^(A)>0$. On the other hand, we know that the inner measure of $A$ is zero, that is $m_(A)=0$ (see, for inatance, http://math.stackexchange.com/questions/739089/how-to-get-the-inner-measure-of-the-vitali-set ), and so we have $m^([0,1]-A)=1$. Then $$m^([0,1])=1<1 +m^(A) = m^(A) + m^*([0,1]-A)$$ – Ramiro May 12 '16 at 02:59

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