First of all, I am searching specifically for a constructive proof. I already have a proof by contradiction.
Here is the question:
Let $E\subset \mathbb{R}$ be a Lebesgue measurable set. Prove that if $m(E)>0$, then for every $\epsilon>0$, there exists interval $[a,b]$ such that $$m(E\cap[a,b]) > (1-\epsilon)\cdot(b-a)$$
Here is what I came up with. I took inspiration from binary search.
First let write $$E=\bigcup_{n\in\mathbb{Z}} E\cap[n,n+1]$$ Then notice that it must be the case that there exists $n\in\mathbb{Z}$ such that $0<m(E\cap[n,n+1])$. Let $B_1 = E\cap[n,n+1]$ such that we have $0<m(B_1) = k_1<\infty$
Consider $a_1 = \inf B_1$, $b_1 =\sup B_1$ and the interval $[a_1 , b_1 ]$. Then $$m(E\cap [a_1 , b_1 ]) = \delta_{k_1} (b_1 - a_1) \quad 0< \delta_{k_1} \leq 1$$
Now, divide $B_1$ in half by considering $$B_- = \left\{ b\in B_1 : b < \frac{b_1-a_1}{2} \right\}$$ $$B_+ = \left\{ b\in B_1 : b > \frac{b_1-a_1}{2} \right\}$$
And take as $B_2 = \begin{cases} B_- & m(B_-) \geq m(B_+) \\ B_+ & otherwise \end{cases}$
Notice that $(b_2 - a_2) = \frac{1}{2}(b_1 - a_1)$, but $m(B_2) \geq \frac{1}{2}m(B_1)$ and hence, for, $$m(E\cap [a_2 , b_2 ]) = \delta_{k_2} (b_2 - a_2) \quad 0< \delta_{k_1} \leq 1$$ we have that $\delta_{k_2} \geq \delta_{k_1}$.
Repeating the process, results in $\delta_{k_n} \to (1-\epsilon)$ as $n\to\infty$.
To see this, suppose by contradiction that this was not the case. Suppose that $\delta_{k_n}$ stays constant after some $n_0$ iterations. First note that this means that the elements of $B_{n_0}$ are evenly spread out over the interval $[a_{n_0},b_{n_0}]$. Note that a union of rectangles of length $b_n-a_n$, touching each other, can be used as a cover of $B_{n_0}$. But it will always be the case that the sum of volumes of such a cover will be equal to $\delta_{k_n}m(B_{n_0})$. Hence, also the infimum is equal to $\delta_{k_n}m(B_{n_0})$, and thus, $\delta_{k_n}$ for $n\geq n_0$ must be equal to $1$ since the infinitum of this particular cover has volume, as most, equal to any other cover of $B_{n_0}$. To see that the last statement is correct, consider any other cover, take, the smallest $R$ in such a cover and then let $n$ be big enough such that $\mathcal{l}(R)>\mathcal{l}([a_n,b_n])$.
Please give me your thoughts on whether you think this proof is valid, incorrect by has potential, or hopeless.