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:

  1. $ \mu^{\ast}(\varnothing) = 0$,
  2. if $A \subset B \subset X$, then $\mu^{\ast}(A) \leq \mu^{\ast}(B)$, and
  3. 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!

  • $\begingroup$ Hi. This will take me some time to digest but, with respect to your first comment. For your suggestion: $\cup_{\substack{n=1\\n\in \mathbb{N}}}\,A_n = A_1 \cup A_2 \cup \cdots = [1,2] \cup [2,3] \cup \cdots$ it is not bounded because $[n,n+1]$ does not become a limit point in this sequence of closed intervals? This may not use correct notation but I see how it is not bounded. Thank you. $\endgroup$ – nate Sep 26 '11 at 19:36

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 $\cup_n A_n$.

  • $\begingroup$ Thank you for all the work in your explanation! I did miss the fact that an unbounded set could be from the union of bounded or unbounded sets that you and @Davide mentioned. And, as you suggested, I am still contemplating all of the above. :) $\endgroup$ – nate Sep 26 '11 at 22:15
  • $\begingroup$ $\mu_5^{\ast}$: Just to finish this, and possibly get any feedback if I have something to look at again,...(1) is satisfied as $\varnothing$ is countable. (2) is satisfied if both $A$ and $B$ are the same (countable or uncountable), whether or not the inclusion is proper because there are uncountable subsets contained in others $([0,1] \subset \mathbb{R})$. Only other that makes sense is for $A$ to be countable, $B$ not, and this is satisfied. Lastly (3) shows LHS of (3) to be countable or uncountable. If former, all $A_n$ are countable, if latter, at least 1 subset needs to be uncountable. $\endgroup$ – nate Sep 27 '11 at 19:00
  • $\begingroup$ The fact that $b$ can be finite or infinite doesn't stop (3) from being satisfied, just means that $\mu^{\ast}$-measurable subsets need to be uncountable (if $b=+\infty$), (or have at least one). $\endgroup$ – nate Sep 27 '11 at 19:02
  • $\begingroup$ @nate: You’ve got it. $\endgroup$ – Brian M. Scott Sep 27 '11 at 19:04
  • 1
    $\begingroup$ @nate: (My previous comment was a response to your first comment; the second just showed up.) The $\mu^*$-measurable sets include those of measure $0$, so all subsets are $\mu_5^*$-measurable even if $b=+\infty$. $\endgroup$ – Brian M. Scott Sep 27 '11 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.