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.
Best Answer
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.