Outer measure defined on sets of $\mathbb R$ - help with showing if outer measure or not I have a long list of definitions of outer measures that I am trying to (a) show IF they are outer measures, and (b) if it is, determine its outer measurable subset of $\mathbb{R}$.
My books requirements for an outer measure are:

  
*
  
*$ \mu^{\ast}(\varnothing) = 0$,
  
*if $A \subset B \subset X$, then $\mu^{\ast}(A) \leq \mu^{\ast}(B)$, and
  
*if $\{A_n\}$ is an infinite sequence of subsets of $X$, then $\mu^{\ast}\left(\bigcup_n\,A_n\right) \leq \sum_n \, \mu^{\ast}(A_n),$
  

For instance, two of them, defined on the power set of $\mathbb{R}$, are (from Cohn's book by the way):
$\mu_3^{\ast}(A) = 0$ if $A$ is bounded, and  $1$ if $A$ is unbounded.
$\mu_4^{\ast}(A) = 0$ if $A$ is empty, $1$ if $A$ is non-empty and bounded, and $+\infty$ if $A$ is unbounded.
$\mu_5^{\ast}(A) = 0$ if $A$ is countable, $b$ if $A$ is uncountable, where $b$ can equal a finite number or $+\infty$.
First, I am only 90% sure the authors means the reals and no the affinely extended reals: $\mathbb{R} \cup \{-\infty\} \cup \{+\infty\}$??
Now for $\mu_3^{\ast}$,
(1) I am assuming the empty set $\varnothing$ is bounded, or else the first criteria of an outer measure fails (and besides I have read different ideas in different place - most seem to think it is both bounded and unbounded?). 
(2) Then for $A \subset B$ (does the author mean proper, as in not $A \subseteq B$??), I can say that the monotonocity inequality is satisfied if they are both the same, bounded or unbounded, and that, since a bounded set (at least on $\mathbb{R}$) cannot contain an unbounded set, the only remaining possibility is if $A$ is bounded and $B$ is unbounded, in which case the inequality is satisfied. 
(3) Lastly, it is certain the power set of $\mathbb{R}$ contains bounded sets, but also it contains $\mathbb{R}$ itself, which is unbounded, so the countably infinite union of elements of the power set of $\mathbb{R}$ is 1 (the left-hand side of the inequality). Since there is only one unbounded (a set missing at least one of either left or right boundaries) subset of $\mathbb{R}$(??) then the right-hand side of the inequality is 1.
For $\mu_4^{\ast}$, 
(1) same as above
(2) Bounded subsets of $\mathbb{R}$ are of the form $(a,b)$, $[a,b]$, $(a,b]$, or $[a,b)$. So case by case, respecting $A \subset B$ (proper) shows this criteria for outer measure is satisfied:
$\hspace{2cm}$ both bounded: $1 \leq 1$
$\hspace{2cm}$ both unbounded: $+\infty \leq +\infty$, as in $\mu^{\ast}((a,+\infty)) \leq \mu^{\ast}(\mathbb{R})$
$\hspace{2cm}$ $A = \varnothing$ and $B$ is bounded and $B \neq \varnothing$ implies $0 \leq 1$
$\hspace{2cm}$ $A = \varnothing$ and $B$ is unbounded implies $0 \leq +\infty$
$\hspace{2cm}$ $A$ is bounded and $B$ is unbounded implies $1 \leq +\infty$
(3) So for the left hand side of the inequality, $\mu^{\ast}(T:=\cup_n\,A_n)$, I know that $T$ is at least as `big' as $\mathbb{R}$, because $\mathbb{R} \in \mathcal{P}(\mathbb{R})$. Therefore, I think it is either $+\infty$ or a higher cardinality - kind of how the natural numbers are infinite but their power set is $\aleph_0$.
For the right-hand side, $\sum_n\,A_n$, I can't see anything more than something like $0 + 0 + \cdots + 1 + 1 + \cdots + {}^+\infty + {}^+\infty + \cdots = {}^+\infty$.
Therefore, the cardinality is the problem I don't understand here(??).
For $\mu_5^{\ast}$ I haven't yet figured out the gist of this with respect to countability, but I do know that countability (not finite) is where there is a bijection between a subset of the naturals ($\aleph_0$), and that the power set of the naturals has cardinality of the continuum, $c = 2^{\aleph_0} = \aleph_1$. If this is the case (??) then maybe any help with the above can help this problem too.
Thank you all for looking at this homework problem - I felt that the only way to accurately describe my questions was in the context of the problem, or else I am afraid there would be a lot of confusion!
 A: The empty set is bounded: it is vacuously true that for every $x\in\varnothing$, $\vert x\vert<1$, for instance. Your author is using $\subset$ to mean $\subseteq$. $2^{\aleph_0}$ is not necessarily equal to $\aleph_1$; the assertion that they are equal is the Continuum Hypothesis, which is consistent with but independent of the usual axioms of set theory.
For $\mu_3^*$ your (1) and (2) are okay, but (3) needs some work. There are two possibilities: either all of the $A_n$ are bounded, or at least one of them is unbounded. If one of the $A_n$ is unbounded, say $A_k$, then $\cup_n A_n$ is unbounded, and we have $$\mu_3^*(\cup_n A_n)=1=\mu_3^*(A_k)\le\sum_n \mu_3^*(A_n).$$ If all of the $A_n$ are bounded, then $\mu_3^*(A_n)=0$ for every $n$, so $\sum_n \mu_3^*(A_n)=0$. That’s fine if $\cup_n A_n$ is bounded, but does it have to be? What if $A_n = [-n,n]$ for each $n$?
For $\mu_4^*$ you’ve gone astray right at the beginning of (2): a bounded subset of $\mathbb{R}$ need not be an interval. For instance, $\{1/n:n\in\mathbb{Z}^+\}$ is bounded, but it doesn’t even contain an interval. (For that matter, every finite set is bounded!) To prove (2) for $\mu_4^*$ you need to show that if $A \subseteq B \subseteq \mathbb{R}$, then $\mu_4^*(A)\le\mu_4^*(B)$. Equivalently, you need to show that if $A$ and $B$ are subsets of $\mathbb{R}$, and $\mu_4^*(A) > \mu_4^*(B)$, then $A \nsubseteq B$. Since $\mu_4^*$ takes on only the three values $0,1$, and $+\infty$, you can do this by looking at just three cases: $$\begin{align*}&\mu_4^*(A)=1 \text{ and } \mu_4^*(B)=0;\\&\mu_4^*(A)=+\infty\text{ and }\mu_4^*(B)=0;\\&\mu_4^*(A)=+\infty\text{ and }\mu_4^*(B)=1.\end{align*}$$
In the first and second cases $A$ is non-empty and $B$ is empty, so it’s certainly true that $A \nsubseteq B$. In the third case $A$ is unbounded and $B$ is bounded. Since $B$ is bounded, there is some positive real number $M$ such that $\vert x\vert < M$ for every $x \in B$, i.e., $B \subseteq [-M,M]$. Since $A$ is unbounded, $A \nsubseteq [-C,C]$ for any positive real number $C$. In particular, $A \nsubseteq [-M,M]$, and therefore $A \nsubseteq B$.
For (3), consider the possibilities for the set $\cup_n A_n$. It could be empty, in which case $\mu_4^*(\cup_n A_n) =$  $0$; $\sum_n \mu_4^*(A_n)$ is a sum of non-negative extended real numbers, so it’s non-negative, and we have $\mu_4^*(\cup_n A_n)\le \sum_n \mu_4^*(A_n)$. $\cup_n A_n$ could be non-empty but bounded, so that $\mu_4^*(\cup_n A_n) = 1$; in that case at least one $A_n$ must be non-empty, so at least one $\mu_4^*(A_n)$ must be $1$, and $\sum_n \mu_4^*(A_n) \ge 1 =$  $\mu_4^*(\cup_n A_n)$. Finally, $\cup_n A_n$ could be unbounded, so that $\mu_4^*(\cup_n A_n) = +\infty$. To get the desired inequality, we need to be sure that $\sum_n \mu_4^*(A_n) = +\infty$.
How could this go wrong? Each $\mu_4^*(A_n)$ is $0,1$, or $+\infty$. We’re in good shape if at least one of the $A_n$ is $+\infty$. We’re also in good shape if infinitely many of them are $1$. We’re in trouble only if all of them are $0$ or $1$, and only finitely many of them are $1$. This happens only if every $A_n$ is bounded, and only finitely many of them are non-empty. In that case $\cup_n A_n$ is essentially the union of finitely many bounded sets. Can you show that the union of finitely many bounded sets is bounded? Once you’ve done that, you’ll have (3) for $\mu_4^*$.
I’m going to leave you to think a bit more about $\mu_5^*$ after you’ve absorbed the foregoing material. The one crucial fact that you’ll need is that the union of a countable number of countable sets is a countable set: if each $A_n$ is countable, so is $\cup_n A_n$.
