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 $\Bbb R$. $\endgroup$ 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.


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