Let $E$ be any Lebesgue measurable set in $\mathbb{R}$ with $0<m(E)< \infty$. I want to show that there exists a finite, nontrivial interval $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$
As $E$ is Lebesgue measurable, then there is an open set $\mathcal{O} \subset \mathbb{R}$ with $E \subset \mathcal{O}$ such that $$m_*(\mathcal{O} - E) < \varepsilon$$
And we remind ourselves that
$m(E) = m_*(E) = \inf \displaystyle\sum_{i=1}^\infty |Q_i|$ for $Q_i$ are closed cubes such that $E \subset \displaystyle\bigcup_{i=1}^\infty Q_i$
Now, take the open subset $\mathcal{O}$ where $E \subset \mathcal{O}$ with $m_*(\mathcal{O} - E) < \varepsilon$. As $\mathcal{O}$ is open, we can write $\mathcal{O}$ as a disjoint countable union of open intervals: $\mathcal{O} = \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$. Now, let $m[(a_i,b_i)] = \ell_i$. Then since the $(a_i,b_i)$'s are disjoint,
\begin{align} m(\mathcal{O}) = m \left( \bigcup_{i=1}^\infty (a_i,b_i) \right) = \sum_{i=1}^\infty m[(a_i,b_i)] = \sum_{i=1}^\infty \ell_i \end{align}
I can see that this approach is falling apart quickly because I need to construct/find a connected closed interval $[a,b]$ whereas by my approach, the $(a_i,b_i)$'s may have a positive distance between each other. If the $(a_i,b_i)$'s were right next to each other, you could simply take the closure of $\cup_i (a_i,b_i)$ and pick a closed interval $[a,b]$ inside such a closure. But the $(a_i,b_i)$'s may be too far apart. What can be done?
