Prove a set is of full measure

Question

This is a problem in my textbook:

$E\subset[0,1], E$ is Lebesgue measurable, if there exists $\delta\in(0,1)$ such that for any interval $I\subset[0,1]$, $m(E\cap I)\geq \delta|I|$, then $m(E)=1$

A hint of this problem is also given in the textbook: use the lemma below and prove the problem by contradiction

Lemma: If $0<m(E)<\infty$, then $\forall\alpha\in(0,1)$, there exists an interval $I$ such that $m^*(E\cap I)>\alpha|I|$

I am able to prove this lemma. Then I assume $m(E)<1$. Using the lemma, I get $m([0,1]\setminus E\cap I)>\delta|I|, \forall$ interval $I\subset [0,1]$, and I get stuck after this.

Answer 1

First comments: I don't follow your use of the lemma to get the conclusion you state. The lemma provides an interval with some properties and you are getting a conclusion for all intervals. Second, I'm going to write $m(I)$ because I am allergic to using $|I|$ for the Lebesgue measure of an interval.

We have $\delta>0$ such that for any $I\subset [0,1]$, $m(E\cap I)\geq \delta m(I)$.

Suppose $m(E)<1$. Note that this implies $\delta<1$ since otherwise setting $I=[0,1]$ we would get $m(E)\geq \delta\geq 1$.

Now the big hint is to apply the lemma to $F=[0,1]\setminus E$ and $\alpha=1-\delta$. I've hidden the rest.

The lemma provides an interval $I$ so that $m(F\cap I)>\alpha m(I)$.
So we have $m(I)=m(E\cap I)+m(F\cap I)>\delta m(I)+\alpha m(I)=m(I)$.