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Show that if a set $E$ has positive outer measure, then there is a bounded subset of $E$ that also has positive outer measure.

Suppose $E$ has outer measure $a$. That means for any positive integer $n$, there exists a countable collection of open intervals that cover $E$ that has total length in $[a, a+\frac1n)$.

Then I have to prove that there exists a bounded subset of $E$ with posiitve outer measure. Maybe I should look at $E\cap(-1,1), E\cap(-2,2), \ldots,E\cap(-n,n),\ldots$ and prove that one of them must have positive outer measure.

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  • $\begingroup$ That would indeed be a good approach. $\endgroup$ – Brian M. Scott Jun 30 '13 at 5:18
  • $\begingroup$ Hmm.. a positive outer measure means there there exists $\epsilon$ such that $E\cap(-n,n)$ cannot be covered with open intervals of length less than $\epsilon$. I'm not sure how to proceed to there. $\endgroup$ – PJ Miller Jun 30 '13 at 5:32
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    $\begingroup$ Have you yet proved that a countable union of sets of measure zero has measure zero? $\endgroup$ – Brian M. Scott Jun 30 '13 at 5:38
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    $\begingroup$ It would be easy if you go for a proof by contradiction. Suppose that every one of them is of zero outer measure... $\endgroup$ – i707107 Jun 30 '13 at 5:39
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    $\begingroup$ @PJMiller If you have found the solution, you might want to post it so the question does not go unanswered. $\endgroup$ – Michael Greinecker Jul 2 '13 at 21:33
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Since PJ Miller never posted an answer to his problem, which I venture to guess comes from Royden's Real Analysis book, I will do so for the sake of completeness.

First, observe that $E = \bigcup_{k \in \Bbb{Z}} E \cap [k,k+1)$. Thus, if each set $E \cap [k,k+1)$ were of measure zero, then

$$m^*(E) = m^* \left(\bigcup_{k \in \Bbb{Z}} E \cap [k,k+1) \right) \le \sum_{k=1}^\infty m^*(E \cap [k,k+1))=0,$$

forcing $m^*(E) = 0$, which is contrary to our hypothesis. Hence there must be some $k \in \Bbb{Z}$ such that $m^*(E \cap [k,k+1))>0$, where $E \cap [k,k+1)$ is a bounded subset of $E$.

The reason for choosing this decomposition of $E$, instead of PJ Miller's decomposition, is that it will be useful in one's further study of measure theory, so it is important to get such a decomposition ingrained into one's head early on.

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