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...
 A: 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. 
A: 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.
