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Prove or give a counter example: For every open set $U$ of $\mathbb{R}$, $m^*(\bar{U} - U ) = 0.$

My first impression was that it was true, since if $U$ is an open set in $\mathbb{R}$, then it can be expressed as the disjoint union of a countable collection of open intervals. We also know that the outer measure of an interval is just its length. So, then I get stuck by wondering if $\bar{U} - U$ is always at most countable, which at the moment I cannot think of a case where it's not true. For if this is true, then the outer measure of $\bar{U}-U$ would be zero and I would be done.

What am I missing here?

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See also: – Martin Apr 21 '13 at 22:52
up vote 2 down vote accepted


It is a closed set whose boundary has positive measure. Hence the complement...

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+1: Nicely done. – Cameron Buie Apr 21 '13 at 21:58
The complement obviously is an open set, but I guess I'm not seeing why the boundary of the complement still has positive measure. As a disclaimer, I'm using Royden/Fitzpatrick, 4th Edition. We are up to Section 2.2. – Archie Apr 21 '13 at 22:13
What's the relationship between the boundary of a set and the boundary of it's complement? – Zach L. Apr 21 '13 at 22:14
That the boundary of a set is the intersection of its closure and the closure of its complement? So the measure of the boundary of a set has the same measure of the boundary of its closure? – Archie Apr 21 '13 at 22:22
@Archie: You say that the boundary of a set $C$ is just $$\overline C\cap\overline{\Bbb R\setminus C}.$$ Since $\Bbb R\setminus(\Bbb R\setminus C)=C$ and since $A\cap B=B\cap A$, this means that a set and its complement have the same boundary. – Cameron Buie Apr 21 '13 at 22:33

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