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 $\lambda^* (A)&lt;+\infty$. Let $\{(a_n,b_n)\}$ a sequence of open intervals such that $A\subset \bigcup_n(a_n,b_n)$ and $\sum_nb_n-a_n&lt;\lambda^* (A)+\varepsilon$, 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\\&lt;\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...
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, $\lambda_*(B)=\lambda^*(B)&lt;\infty$ and $B$ is measurable.
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.