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!