$m(E∩I)≥αm(I)$ for all intervals $I⊂[0,1].$ Prove that $m(E) = 1.$

Question

Let $E$ be a measurable subset of $[0, 1].$ Assume there is a constant $α > 0$ such that $m(E∩I)≥αm(I)$ for all intervals $I⊂[0,1].$

(Here $m(·)$ denotes Lebesgue measure.) Prove that $m(E) = 1.$

I am not sure how to approach this past analysis qual problem. Some help would be awesome. Thanks.

Answer 1

Here's a hint:

Suppose $S\in \mathbb{R}$ is measurable. For every $\epsilon > 0$ there exists some countable collection of intervals $\{I_{\alpha}\}$ that differs from $S$ by a set of measure at most $\epsilon$. (This is an extremely useful theorem).

Try applying this theorem to $E^c$

(let me know if you want another hint, or the answer)

Answer 2

Hints

1. Prove the claim if $E \subseteq [0,1]$ is a closed set.
2. Use that the Lebesgue measure is (inner) regular, i.e. the fact that for any Borel set $E$ there exists a sequence of closed sets $(F_n)_n$ such that $F_n \subseteq E$, $m(F_n) \uparrow m(E)$.