# 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 $A$. – user251257 Jul 16 '15 at 23:56
• @user251257 I don't see the mistake. In the proposition we need to show that $B$ is measurable not $A$. – Jose Antonio Jul 17 '15 at 0:11
• the last sentence reads: Hence A is Lebesgue measurable – user251257 Jul 17 '15 at 0:12
• @user251257 I see, thanks ;) – Jose Antonio Jul 17 '15 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.