Show nonempty interior implies positive exterior measure.

Question

Let $E \subset \mathbb{R}$ I claim if the interior is non empty, then it has positive exterior measure. If $E$ has non empty interior, then

$$E \subset\bigcup_{B \subset E \text{open}}B \neq \emptyset$$

As $ B \subset E \subset \mathbb{R}^n$, open sets are of the form

$$(a,b) \times ... \times (a,b) \space n \text{copies}$$

and so we at least have

$$m_*(E)= \inf \sum_{i=1}^nm_*((a_i,b_i))=\sum_{i=1}^n \vert b_i-a_i \vert > 0.$$

So long as $a_i \neq b_i$ which is guantateed since we have nonempty interior. Is this good enough? Trying to go based strictly off definition of interior and exterior measure. Basic definitions.

Answer 1

Outer measure is monotone. There exist $a_i<b_i, i=1,2..,n$ such that $(a_1,b_1)\times (a_2,b_2)\times....\times (a_n,b_n) \subseteq E$ so $m_{*}(E) \geq m_{*} ((a_1,b_1)\times (a_2,b_2)\times....\times (a_n,b_n))=\prod (b_i-a_i) >0$.