I'm preparing homework from Cohn's book to have less workload later. The question is in the title, and it's bit of a pity that this guy accepted an answer that was using theory not yet covered in the book, see here. I'll also have you note that he probably used the first edition of the book, as the question is seperated in the second edition.
Let $I$ be a bounded subinterval of $\mathbb{R}$. Show that a subset $B$ of $I$ is Lebesgue-measurable iff $ \lambda^*(I) = \lambda^*(B)+ \lambda^*(I\cap B^c)$ where $\lambda^*(A)$ is defined as $\inf\{\sum_n(b_n-a_n) | A \subseteq \cup_n (a_n,b_n)\}$.
It's sufficiently to show that for all $A$, such that $A\subseteq \mathbb{R}$ and $\lambda^*(A)<+\infty$, that $\lambda^*(A)=\lambda^*(A\cap B)+ \lambda^*(A\cap B^c)$.
I've tried a lot of random stuff, for example: There is a proposition which states that any bounded subinterval is $\lambda^*$-measurable. So \begin{align*} \lambda^*(A)&=\lambda^*(A\cap I)+ \lambda^*(A\cap I^c) \\ &\geq\lambda^*(A\cap B) + (\lambda^*(A\cap I^c) \\ \end{align*} But i don't see how the given assumption can be used.
