Hi I'm working with Cohn's book and I have other problem with the necessity condition, I'd appreciate any help.
Let $\lambda^*$ the Lebesgue outer measure on $\bf{R}$, and let $\pi$ be the projection of the first coordinate, that is $\pi (x, y) = x$. Define a function $\mu^∗ : \mathcal{P}({\bf{R}}^2) \to [0, +\infty]$ by $\mu^∗ (A) = \lambda^* ( \pi (A))$.
Show that a subset $B$ of ${\bf{R}}^2$ is measurable for the outer measure $\mu^∗$ iff there are Lebesgue measurable subsets $B_0$ and $B_1$ of $\bf{R}$ such that $B_0 \subset B_1$ , $\lambda^* (B _1 − B_ 0 ) = 0$, and $B_0 \times {\bf{R}} \subset B \subset B_ 1 \times {\bf{R}}$.
Suppose the statement of the sufficiently condition holds, we show that $B$ is $\mu^*$- measurable. Let $A$ be an arbitrary subset of ${\bf{R}}^2$. Thus
\begin{align}\mu^* ( A\cap B)+\mu^* (A\cap B^c)=\lambda^* ( \pi (A\cap B))+\lambda^* ( \pi (A\cap B^c))\le \lambda^* ( \pi (A)\cap B_1))+\lambda^*(\pi(A)\cap B_0^c)\\ =\lambda^* ( \pi (A)\cap B_1))+\lambda^*(\pi(A)\cap B_1^c)\\+\lambda^*(\pi(A)\cap B_1- B_0)\\ \le \lambda^*(\pi(A))=\mu^*(A)\end{align}
Hence $B$ is a $\mu^*$- measurable set.
