Cohn Measure Theory exercise question 9 chapter 1.3 other direction This exercise can be found in Cohn's book about measure theory (exercise 9, chapter 1.3, p. 22) and I have already asked for help for the direction $\Leftarrow$ here: 

Define $\mu^*(A) := \lambda^*(\pi(A))$ for $A \subseteq \mathbb{R}^2$,
  $\pi(x,y) := x$ and $\lambda^*$ Lebesgue outer measure on
  $\mathbb{R}$. Then $\mu^*$ is an outer measure. Show, that $B
 \subseteq \mathbb{R}^2$ is measurable with respect to $\mu^*$ if and
  only if there exists $B_0, B_1 \subseteq \mathbb{R}$ Lebesgue
  measurable, $B_0 \subseteq B_1$, $\lambda^*(B_1\setminus B_0) = 0$ and
  $B_0 \times \mathbb{R} \subseteq B \subseteq B_1 \times \mathbb{R}$.

However, I do not quite get the direction $\Rightarrow$. My idea was to define some sets $B_0$ and $B_1$ and then to show that they fulfill the given properties. I think a good choice of $B_1$ would be $\pi(B)$ since then $B \subseteq B_1 \times \mathbb{R}$. I am not sure how to prove that the so defined $B_1$ is Lebesgue measurable and I am not sure how to define $B_0$ aswell.
 A: Your choice of $B_1$ is right. To see it is Lebesgue measurable, take any set $A \subset \mathbb{R}$ and notice that $A= \pi(A \times \mathbb{R})$. Then:
\begin{align}
\lambda^*(A \cap B_1) + \lambda^*(A \cap (B_1)^c)  
&= \lambda^*( \pi(A \times \mathbb{R}) \cap \pi(B) ) + \lambda^*(\pi(A\times \mathbb{R}) \cap \pi(B)^c) \\
&\leq \lambda^*( \pi( (A\times \mathbb{R}) \cap B))+ \lambda^*(\pi((A\times \mathbb{R}) \cap B^c))
\end{align}
(I'll let you fill in the details of why this last step is correct.)
The choice of $B_0$ is the tricky part, but once we get the idea it is not so hard. Define $B_0 = \pi(B) \cap \pi(B^c)^c$. Here's a drawing to see why this is a good choice (but beware, the set $B$ in the drawing will not be measurable):

I am not so good at drawing on the computer but I hope this gives an idea. Now, clearly $B_0 \subset B_1$, and $B_0 \times \mathbb{R} \subset B$. What remains is to see that $\lambda^*(B_1 \setminus B_0) =0$. 
Hint: To do this, remember that $B$ is $\mu^*$-measurable, so for any set $A \subset \mathbb{R}^2$ we have:
$$
\mu^*(A) = \mu^*(A \cap B) + \mu^*(A \cap B^c)
$$
Now, if the result were to be true, $B$ should be very close (in measure) to the rectangle $\pi(B) \times \mathbb{R}$. This gives us a clue that we should try to put $A= \pi(B) \times \mathbb{R}$ in this equality and see where this gets us. 
