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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.

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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?

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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? –  Ron Jeremy 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. –  Ron Jeremy 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

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