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.

  • $\begingroup$ That would indeed be a good approach. $\endgroup$ 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
  • 2
    $\begingroup$ Have you yet proved that a countable union of sets of measure zero has measure zero? $\endgroup$ Jun 30 '13 at 5:38
  • 1
    $\begingroup$ It would be easy if you go for a proof by contradiction. Suppose that every one of them is of zero outer measure... $\endgroup$ Jun 30 '13 at 5:39
  • 2
    $\begingroup$ @PJMiller If you have found the solution, you might want to post it so the question does not go unanswered. $\endgroup$ Jul 2 '13 at 21:33

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.


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.