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Hi I was reading Cohn's book and I have problem with the following exercises (only the return of b is what I don't know), I'd appreciate any help and suggestion, if necessary, for a):

a) Show that a subset $B$ of $\bf{R}$ is Lebesgue measurable iff $\lambda^*(I)=\lambda^*(I\cap B)+\lambda^*(I\cap B^c)$ for any open interval.

b) Let $I$ a bounded subinterval of $\bf{R}$. Show that a subset $B$ of $I$ is Lebesgue measurable iff it satisfies $\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$

a) We only show the sufficiently. Let $A\subset \bf{R}$ and we may assume that $\lambda^* (A)<+\infty$. Let $\{(a_n,b_n)\}$ a sequence of open intervals such that $A\subset \bigcup_n(a_n,b_n)$ and $\sum_nb_n-a_n<\lambda^* (A)+\varepsilon$, where $\varepsilon$ is an arbitrary positive number. Thus

\begin{align}\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \sum _n\lambda^*((a_n,b_n)\cap B)+\sum_n\lambda^*((a_n,b_n)\cap B^c)\\ = \sum _n\lambda^*((a_n,b_n)\cap B)+\lambda^*((a_n,b_n)\cap B^c)\\=\sum _n\lambda^*((a_n,b_n))=\sum_nb_n-a_n\\<\lambda^* (A)+\varepsilon\end{align}

Letting $\varepsilon \downarrow0$, $\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \lambda^* (A)$. Hence $B$ is Lebesgue measurable.

(b) One side is obvious, but the problem is with the return...

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  • $\begingroup$ You made a typo at the ende of proof of (a). You have shown that $B$ is measurable, not $A$. $\endgroup$
    – user251257
    Jul 16, 2015 at 23:56
  • $\begingroup$ @user251257 I don't see the mistake. In the proposition we need to show that $B$ is measurable not $A$. $\endgroup$ Jul 17, 2015 at 0:11
  • $\begingroup$ the last sentence reads: Hence A is Lebesgue measurable $\endgroup$
    – user251257
    Jul 17, 2015 at 0:12
  • $\begingroup$ @user251257 I see, thanks ;) $\endgroup$ Jul 17, 2015 at 0:13

3 Answers 3

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First, show that if $B$ is a subset of a bounded closed interval $I$, then $\lambda_*(B)+\lambda^*(I\backslash B)=\lambda(I)$. Then, by assumption, we have that $\lambda_*(B)+\lambda^*(I \backslash B)=\lambda^*(B)+\lambda^*(I\backslash B)$, hence, $\lambda_*(B)=\lambda^*(B)<\infty$ and $B$ is measurable.

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  • $\begingroup$ I really appreciate your help, but the inner measure is not defined yet in the book. I only know that a set is measurable if divides each subset in such a way that the sizes (as measured by the lebesgue outer measure in this case) of the pieces add properly. $\endgroup$ Jul 18, 2015 at 2:38
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For any bounded open subintervals $J$ we need to show that $\lambda^*(J)\geq \lambda^*(J\cap B)+\lambda^*(J\cap B^c)$ to conclude that $B$ is Lebesgue measurable from (a).

Since subintervals are Lebesgue measurable we can write $\lambda^*(B)=\lambda^*(B\cap J) + \lambda^*(B\cap J^c)$ and $\lambda^*(I\cap B^c)= \lambda^*(I\cap B^c\cap J) + \lambda^*(I\cap B^c\cap J^c).$ Then \begin{align} \lambda^*(I)&=\lambda^*(B)+\lambda^*(I\cap B^c)\\ &=[\lambda^*(B\cap J) + \lambda^*(I\cap B^c\cap J)] + [\lambda^*(B\cap J^c) + \lambda^*(I\cap B^c\cap J^c)] \tag{1}\\ &\geq \lambda^*(I\cap J) + \lambda^*(I\cap J^c)=\lambda^*(I) \end{align} where we use the fact that $(B\cap J)\cup (I\cap B^c\cap J)=I\cap J$ and $(B\cap J^c) \cup (I\cap B^c\cap J^c)=I\cap J^c $ and the subadditivity of outer measure and that $\mu^*$ is a measure on Lebesgue measurable sets.

Denote the first and second term in the square brackets of $(1)$ above as $A$ and $B$. We have $A+B=\lambda^*(I\cap J) + \lambda^*(I\cap J^c)$ and $A\geq \lambda^*(I\cap J)\geq 0$ and $B\geq \lambda^*(I\cap J^c)\geq 0.$ Then $A+B\leq \lambda^*(I\cap J)+B$. Since $B<\infty$ we have $A\leq \lambda^*(I\cap J)$ and we conclude that $A=\lambda^*(B\cap J) + \lambda^*(I\cap B^c\cap J)=\lambda^*(I\cap J).$

Then \begin{align} \lambda^*(J)&=\lambda^*(I\cap J)+\lambda^*(J\cap I^c)\\ &=\lambda^*(B\cap J) + [\lambda^*(I\cap B^c\cap J)+\lambda^*(J\cap I^c)]\\ &\geq \lambda^*(B\cap J) + \lambda^*(B^c\cap J) \end{align} where the last inequality follows from the fact that $(I\cap B^c\cap J)\cup (J\cap I^c)=B^c\cap J.$

The proof is inspired by this argument.

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We prove that the equation holds for each open interval and then use the conclusion of a). Let $L$ be an open interval and $L=(L\cap I)\cup (L\cap I^{c})$. Since $I$ is measurable, we have $$\lambda ^{*}(L)=\lambda ^{*}(L\cap I)+\lambda ^{*}(L\cap I^{c})$$ Also, we have $$\lambda ^{*}(I\cap L)\leq \lambda ^{*}(B\cap I\cap L)+\lambda ^{*}(B^{c}\cap I\cap L)(1)$$ $$\lambda ^{*}(I\cap L^{c})\leq \lambda ^{*}(B\cap I\cap L^{c})+\lambda ^{*}(B^{c}\cap I\cap L^{c})(2)$$ Add these two inequalities and use the fact that $L$ is measurable, we have $$\lambda ^{*}(I)=\lambda ^{*}(L^{c}\cap I)+\lambda ^{*}(L\cap I)\leq \lambda ^{*}(B)+\lambda ^{*}(B^{c}\cap I)=\lambda ^{*}(I)$$

This indicates that the inequalities in $(1)$ and $(2)$ should be strict equalities. Besides, it is easy to check, if an interval $E\cap I=\varnothing $, we have $$\lambda ^{*}(E)=\lambda ^{*}(E\cap B)+\lambda ^{*}(E\cap B^{c})$$Now, we have$$\lambda ^{*}(L)=\lambda ^{*}(L\cap I)+\lambda ^{*}(L\cap I^{c})=\lambda ^{*}(L\cap I\cap B)+\lambda ^{*}(L\cap I\cap B^{c})+\lambda ^{*}(L\cap I^{c}\cap B)+\lambda ^{*}(L\cap I^{c}\cap B^{c})$$ The first term is $\lambda ^{*}(L\cap B)$ and the third term is zero. The sum of second and fourth terms is $\lambda ^{*}(L\cap B^{c})$ since $I$ is measurable.

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