# Thread: Exterior measure - show there exists an open interval...

1. ## Exterior measure - show there exists an open interval...

( $m$ is the exterior/outer measure). Let $E \subset \mathbb{R}$ with $m(E)>0$. Prove that for each $0 < \alpha < 1$, there exists an open interval $I$ such that $m(E \cap I) \geq \alpha m(I)$.

Im not sure how to proceed....Ive tried starting with the definition of the outer measure and considering a cover of open intervals but im not making any progress. Ive also looked at contradiction, but not sure if im getting anywhere with that. Any help will be appreciated
Thanks

2. Originally Posted by ramdayal9
( $m$ is the exterior/outer measure). Let $E \subset \mathbb{R}$ with $m(E)>0$. Prove that for each $0 < \alpha < 1$, there exists an open interval $I$ such that $m(E \cap I) \geq \alpha m(I)$.

Im not sure how to proceed....Ive tried starting with the definition of the outer measure and considering a cover of open intervals but im not making any progress. Ive also looked at contradiction, but not sure if im getting anywhere with that. Any help will be appreciated
Thanks

Hmm, have you learned the concept of Lebesgue density? There's a theorem that states if E is measurable , then for a.e. x in E

d(x, E) = $lim_{ (B)\rightarrow(x)}\ m(B \cap E)/m(B)$ = 1 almost everywhere where B is a ball that contains x.

Fix an alpha, $0 < \alpha < 1$ and suppose that no such interval, I, exists.

Thus, for all x in E, we have:

$lim_{ (B)\rightarrow(x)}\ m(B \cap E)/m(B) \leq \alpha < 1$, which obviously contradicts the above statement.

EDIT: Oops, you never said E was measurable, but I think it would still work with outer measures, let me confirm.