Cohn Measure Theory exercise question 9 chapter 1.3 This exercise can be found in Cohn's book about measure theory (exercise 9, chapter 1.3, p. 22):

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
  $B_0 \times \mathbb{R} \subseteq B \subseteq B_1 \times \mathbb{R}$.

Somehow I am stuck proving the direction $\Leftarrow$. My attempt: Let $A \subseteq \mathbb{R}^2$. Then
\begin{align}
\mu^*(A \cap B) + \mu^*(A \cap B^c) &\leq \mu^*(A \cap (B_1 \times \mathbb{R})) + \mu^*(A \cap (B_0 \times \mathbb{R})^c)\\
&= \lambda^*(\pi(A \cap (B_1 \times \mathbb{R}))) + \lambda^*(\pi(A \cap (B_0 \times \mathbb{R})^c))\\
&\leq \lambda^*(\pi(A) \cap \pi(B_1 \times \mathbb{R}))) + \lambda^*(\pi(A) \cap \pi((B_0 \times \mathbb{R})^c)))\\
&= \lambda^*(\pi(A) \cap B_1) + \lambda^*(\pi(A) \cap B_0^c)
\end{align}
Maybe this is completely wrong. Has someone another idea or an idea how to proceed?
 A: You are almost there!
Hint: From your last inequality, write $B_1 = B_0 \cup ( B_1 \setminus B_0)$ and use subadditivity.
Also, your third line should be an inequality. We always have the inclusion $\pi(A \cap (B_1 \times \mathbb{R})) \subset \pi(A) \cap \pi( B_1 \times \mathbb{R})$, but not equality in general. So you need to use monotonicity here.
A: Using the hint we get
\begin{align}
\mu^*(A \cap B) + \mu^*(A \cap B^c) &\leq \mu^*(A \cap (B_1 \times \mathbb{R})) + \mu^*(A \cap (B_0 \times \mathbb{R})^c)\\
&= \lambda^*(\pi(A \cap (B_1 \times \mathbb{R}))) + \lambda^*(\pi(A \cap (B_0 \times \mathbb{R})^c))\\
&\leq \lambda^*(\pi(A) \cap \pi(B_1 \times \mathbb{R}))) + \lambda^*(\pi(A) \cap \pi((B_0 \times \mathbb{R})^c)))\\
&= \lambda^*(\pi(A) \cap B_1) + \lambda^*(\pi(A) \cap B_0^c)\\ &= \lambda^*(\pi(A) \cap (B_0 \cup (B_1\setminus B_0))) + \lambda^*(\pi(A) \cap B_0^c)\\
&= \lambda^*((\pi(A) \cap B_0) \cup (\pi(A) \cap (B_1\setminus B_0))) + \lambda^*(\pi(A) \cap B_0^c)\\
&\leq \lambda^*(\pi(A) \cap B_0) + \lambda^*(\pi(A) \cap (B_1\setminus B_0)) + \lambda^*(\pi(A) \cap B_0^c)\\
&= \lambda^*(\pi(A) \cap B_0) + \lambda^*(\pi(A) \cap B_0^c) \\
&= \lambda^*(\pi(A))\end{align}
For any $A \subseteq \mathbb{R}^2$.
