# Bounded subset with positive outer measure

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.

• That would indeed be a good approach. Commented Jun 30, 2013 at 5:18
• 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. Commented Jun 30, 2013 at 5:32
• Have you yet proved that a countable union of sets of measure zero has measure zero? Commented Jun 30, 2013 at 5:38
• It would be easy if you go for a proof by contradiction. Suppose that every one of them is of zero outer measure... Commented Jun 30, 2013 at 5:39
• @PJMiller If you have found the solution, you might want to post it so the question does not go unanswered. Commented Jul 2, 2013 at 21:33

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.