Measure inequality of open interval show that if 0 < r < 1 and E is a measurable set with 0 < m(E) < ∞ , then there exist an open interval I such that m(I ∩ E) > rl(I) ,  where m is measure and l(I) is length of I  
 A: By Lebesgue's density theorem, for almost every $x \in E$, we have that $\displaystyle \frac{ m(E \cap (x - \delta, x + \delta))}{2\delta} \to 1$ as $\delta \to 0$.  In particular, since $m(E) > 0$, the set of points of $E$ with full density is not empty.  So find an $x$ where this happens.   Thus if $0 < r < 1$, for all sufficiently small $\delta$, $\displaystyle \frac{m(E \cap (x - \delta, x + \delta))}{2 \delta}  > r$.  Taking $I = (x - \delta, x + \delta)$ and rearranging that last inequality, we're done.
A: Fix one such $r$ and suppose there is no such interval; this means any interval will satisfy $m(I∩ E) \leq r|I|$. To exploit this, for any epsilon, take a cover of $E$ by open intervals $\{I_k\}_{k=1}^∞ $ such that
\begin{align*}
\sum_{k=1}^{∞} \left|I_k\right| \leq   m(E) + \varepsilon
\end{align*}
which is possible as $m(E)$ is defined as an infimum of such covers. Then as they form a cover, we have
\begin{align*}
m(E) = m\left(E∩ \bigcup_k I_k \right) = m\left(\bigcup_k E∩ I_k \right)  \overset{\tiny \text{subadd.}}{\leq} \sum m(E∩ I_k) \leq r\sum \left|I_k\right| 
 \leq r\left[m(E) + \varepsilon \right] 
\end{align*}
and thus $m(E) \leq r m(E) + \varepsilon_0 $ for any $\varepsilon_0 > 0$; but this can't be true (try $\varepsilon_0 = \frac{1-r}{2} m(E) > 0$.)
