Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that for sets $A,B$ bounded in $\mathbb{R}$:

If there exists $\alpha > 0$ such that $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$.

So these sets aren't necessarily measurable. This comes out of section 2.2 of Royden's Real Analysis. I'm really having trouble with this one for some reason. The only theorem that I can see that might be of some help is that outer measure is preserved under set translation. But I would have to translate each point of one of these sets a different amount, so that seems hopeless.

Because I have so few theorems to work with my hunch is that I need to go back to the very definition of outer measure and do something clever with it, but so far I haven't had any luck. Can anyone help me? Thanks.

share|cite|improve this question
up vote 4 down vote accepted

HINT: Let $$U=\bigcup_{a\in A}\left(a-\frac{\alpha}2,a+\frac{\alpha}2\right)$$ and $$V=\bigcup_{b\in B}\left(b-\frac{\alpha}2,b+\frac{\alpha}2\right)\;.$$ Suppose that $x\in U\cap V$; then there are $a\in A$ and $b\in B$ such that $$|x-a|,|x-b|<\frac{\alpha}2\;.$$ Is this actually possible?

share|cite|improve this answer
I can see it's not, and I feel like your getting at compactness here... I thought about this one all last night though and I still can't figure it out. You think you could help me out a bit more? – Thoth Oct 13 '12 at 20:12
@IntegrateDisNutsSucka: Not compactness, but simply the fact that $U\cap V=\varnothing$. Thus, $U$ and $V$ are disjoint open sets such that $A\subseteq U$ and $B\subseteq V$. Thus, when you cover $A$ with a countable collection of intervals, you might as well assume that these intervals are contained in $U$, since you’re eventually going to take an infimum anyway. Similarly, the intervals that you use to cover $B$ can be assumed to be contained within $V$. But then the total length of the intervals covering $A\cup B$ is just the sum of the total length of those covering $A$ and the total ... – Brian M. Scott Oct 13 '12 at 20:19
... length of those covering $B$. There are a number of details to sort out when you write it up, but that’s the basic idea. – Brian M. Scott Oct 13 '12 at 20:20
Ok ya I kinda went down that path too, but there aren't any theorems about the outer measure of the countable union of disjoint intervals being equal to the sum of there lengths. – Thoth Oct 13 '12 at 20:23
@IntegrateDisNutsSucka: You don’t need such a theorem, since you’re not looking at a countable union of pairwise disjoint intervals. In covering $A\cup B$ you’re looking at a countable family $\mathscr{I}$ of intervals that are generally not disjoint but that can be split into two subfamilies, $\mathscr{A}$ covering $A$ and $\mathscr{B}$ covering $B$, in such a way that each interval in $\mathscr{A}$ is disjoint from each interval in $\mathscr{B}$. Thus, $s(\mathscr{I})=s(\mathscr{A})+s(\mathscr{B})$, where $s(\cdot)$ is the sum of the lengths of the intervals in a given family. – Brian M. Scott Oct 13 '12 at 20:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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