# Subsets $B$ of bounded subinterval $I$ is lebesgue measurable iff $\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$

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...

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

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.

• 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. Jul 18, 2015 at 2:38

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.

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.