Royce's introductory definition of outer measure. Let $A\subset R$. Then:

$$m^*(A)=\text{inf}\left\{\sum_{n=1}^{\infty}l(I_n):\,A\subset \cup_{n=1}^{\infty}I_n\right\},$$

where $I_n=(a_n,b_n)$.

However, in reading this, I suddenly question whether it is a fact that any subset of $R$ can be covered by a countable collection of open, bounded intervals of the form $(a,b)$.

  • 1
    $\begingroup$ Take $I_n=(n-1,n+1)$. Then you have a cover of all of $\Bbb R$. $\endgroup$ – David Mitra Jun 24 '13 at 19:56
  • $\begingroup$ Ah, of course, and then you cover any subset of R. $\endgroup$ – David Jun 24 '13 at 20:29

For $\mathbb{R}$, this is definitly possible. For example, for every integer $i$ take the interval $I_i = (i-1,i+1)$. For ever $i$ the interval is open and bounded. The countable (since $\mathbb{Z}$ is countable) union gives: $$\bigcup_{i \in \mathbb{Z}} I_i = \mathbb{R} \ .$$ This fullfills your requirements, because $\mathbb{R}$ covers every subset of itself.


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.