Hint on Cohn, Section 1.3, Problem 8 (Measure Theory) Here's the question:

Suppose $I \subseteq \mathbf{R}$ is a bounded interval, and $B\subseteq I$ satisfies $$\lambda^\ast(I) = \lambda^\ast(B) + \lambda^\ast(B^c \cap I).$$
  Show that $B$ is $\lambda^\ast$-measurable.

I'm hoping to get a hint (not a solution) or maybe a suggestion on a simple case to start with. 
Note that measurability of $B$ with respect to $\lambda^\ast$ is defined as
$$
\lambda^\ast(A) = \lambda^\ast(A \cap B) + \lambda^\ast(A \cap B^c), 
$$
for all $A \subseteq \mathbf{R}$. 
 A: I have a solution (outlined below), but it is more equation manipulation than anything else. If someone can give intuition, I'd be grateful. 
$\newcommand{\R}{\mathbf{R}}$
I will assume the following fact. 

$E \subseteq \R$ is $\lambda^\ast$-measurable if and only if $\lambda^\ast(J) \geq \lambda^\ast(J \cap E) + \lambda^\ast(J \cap E^c)$ for all open intervals $J \subseteq \R$. 

Let $J \subseteq \R$ be an open interval. Since $J$ is $\lambda^\ast$ measurable, it follows that for any $S \subseteq \R$, 
\begin{equation}\label{eon:meas}
\lambda^\ast(S) = \lambda^\ast(J \cap S) + \lambda^\ast(J^c \cap S). \qquad (1)
\end{equation}
By assumption, $B\subseteq \R$ satisfies
\begin{align*}
\lambda^\ast(I) &= \lambda^\ast(B) + \lambda^\ast(I \cap B^c) & \\ 
&= (\lambda^\ast(J \cap B) + \lambda^\ast(J^c \cap B)) + (\lambda^\ast(J \cap I \cap B^c) + \lambda^\ast(J^c \cap I \cap B^c)) & \text{eq. (1)}\\
&= (\lambda^\ast(J \cap B) +  \lambda^\ast(J \cap I \cap B^c)) + ( \lambda^\ast(J^c \cap B)) +\lambda^\ast(J^c \cap I \cap B^c)) & \text{arrange by $J, J^c$}\\
&\overset{(*)}{\geq}  \lambda^\ast(J \cap I) + \lambda^\ast(J^c \cap I)  & \text{cntble subadd.}\\
&= \lambda^\ast(I) & \text{$\lambda^\ast$-meas. of $J$}
\end{align*}
Thus the inequality above is equality. Notice that if $a + b = c + d$, and $a \geq c$ and $b \geq d$, then 
$$
0 \leq a - c = d - b \leq 0,
$$
so $a = c$ and $b = d$. Applying this to ineq. $(*)$, we've shown for any interval 
$J \subseteq \R$,
$$
\lambda(J \cap I) = \lambda^\ast(J \cap B) +  \lambda^\ast(J \cap I \cap B^c). \qquad \text{(2)}
$$ 
Suppose $J \subseteq I$. Then the equation above reduces to 
$$
\lambda^\ast(J) = \lambda^\ast(J \cap B) +  \lambda^\ast(J \cap B^c). 
$$
On the other hand, suppose that $J \not \subseteq I$, then since $I$ is $\lambda^\ast$-measurable,
\begin{align*}
\lambda^\ast(J) &= \lambda^\ast(J \cap I) + \lambda^\ast(J \cap I^c) & \text{$\lambda^\ast$-meas of $I$}\\
&= \lambda^\ast(J \cap B) + (\lambda^\ast(J \cap I \cap B^c) + \lambda^\ast(J \cap I^c)) & \text{eq. (2)} \\ 
&\geq \lambda^\ast(J \cap B) + \lambda^\ast(J \cap B^c) & 
\text{subadd., $B^c = (I \cap B^c) \cup I^c$}.
\end{align*}
Thus, we've shown that $B \subseteq \R$ satisfies 
$$
\lambda^\ast(J) \geq \lambda^\ast(J \cap B) + \lambda^\ast(J \cap B^c),
$$
where $J \subseteq \R$ was an arbitrary open interval. The claim follows by the fact. 
