How does one determine the $\sigma$-algebra of $\mu^{*}$-measurable subsets for the following $\mu^{*}$? This is a homework question. 
Define $\mu^{*} : \mathscr{P}(\mathbb{R}) \to [0, \infty ]$ by
$$\mu^{*}(A) = 
     \begin{cases}
       0 &\quad\text{if } A = \emptyset \\
       1 &\quad\text{if } A \text{ is bounded (and } A \neq \emptyset ) \\
       \infty &\quad\text{if } A \text{ is unbounded and (} A \neq \emptyset ) 
        \ 
     \end{cases}
$$
First, we (ought to) have shown that $\mu^{*}$ is an outer measure. I think that went alright. 
Now, the next question is: determine the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $\mathbb{R}$. How does one do that? 
I suspect it may help that we know that $\mu^{*}$ is an outer measure. For example, there is a theorem from Carathéordory in the book "Measure and Integration Theory" by H. Bauer, which we use for the course, that states: 
"Let $\mu^{*}$ be an outer measure on a set $\Omega$. Then the system $\mathscr{A}^{*}$ of all $\mu^{*}$-measurable sets $A \subset \Omega$ is a $\sigma$-algebra in $\Omega$. Moreover, the restriction of $\mu^{*}$ to $\mathscr{A}^{*}$ is a measure." 
In the proof of this theorem, the author notes that a requirement for a subset $A$ of $\Omega$ to lie in $\mathscr{A}^{*}$ is $$ \mu^{*} (Q) = \mu^{*} (Q \cap A) + \mu^{*} (Q \setminus A) \qquad \text{for all } Q \in \mathscr{P}(\Omega)  .$$
Can we use this fact to find all subsets $A$ of $\Omega$ that lie in the system $\mathscr{A}^{*}$ (which, apparently, is a $\sigma$-algebra)? If so, how? 
 A: Hint: As you know the restriction is a measure, you see that at most one bounded set (beside the empty set) is in the $\sigma$-algebra, because else it wouldn't be additive. As $\sigma$-algebras are closed under taking the complement at most one unbounded set (beside $\mathbb{R}$) may be in it. 
For a very formal answer, look at $Q_n=[-n,n]$, then
$\mu^\ast (Q_n)=1$, but on the right hand side for some $n$ the sum should be $2$.
A: Check the condition.
Since you have an outer measure, the inequality "$\leq$" always holds, so that you only have to check "$\geq$". This is trivial for $\mu^\ast (Q) = \infty$, so that we can assume $\mu^\ast (Q) <\infty$.
So you have to check for which sets $A$ the following is true: For arbitrary $Q$ with $\mu^\ast (Q) < \infty$ (which is equivalent to $Q$ being bounded) we have
$$
\mu^\ast (Q) \geq \mu^\ast (Q \cap A)  + \mu^\ast (Q \setminus A).
$$
For $Q = \emptyset$, this trivially holds, since the sets on the right-hand side are also empty.
So, we only need to check it for $Q\neq \emptyset$ bounded, i.e. we have to check for which sets $A$, we have
$$
1 \geq \mu^\ast (Q \cap A) + \mu^\ast (Q \setminus A)
$$
for all bounded sets $Q \neq \emptyset$.
Using the definition, you should be able to show that this only holds for $A = \emptyset$ or $A = \Bbb{R}$.
