> [!theorem] > > Let $E \subset \mathcal L$ and $m$ be the [[Lebesgue Measure|Lebesgue measure]] on $\real$. For any $\eps > 0$, there exists an interval $I \subset \real$ such that > $ > \frac{m(E \cap I)}{m(I)} \ge 1 - \eps > $ > *Proof*. Let $U \supset E$ such that $\frac{m(A)}{m(U)} \ge 1 - \eps$. Write $U = \bigsqcup_{n \in \nat}I_n$ as a disjoint union of open intervals, then > $ > 1 - \eps \le \frac{m(A)}{m(U)} = \frac{1}{m(U)}\sum_{n \in \nat}m(A \cap I_n) = \sum_{n \in \nat}\frac{m(I_n)}{m(U)}\frac{m(A \cap I_n)}{m(I_n)} > $ > so there exists $n \in \nat$ such that $\frac{m(A \cap I_n)}{m(I_n)} \ge 1 - \eps$.