Let $E$ be measurable such that $0 < m(E) < +\infty$. Then for $\varepsilon >0$, there is interval $I$, $m(E \cap I) > (1-\varepsilon)m(I)$. This is what I need to prove:
Let $E$ be a (Lebesgue) measurable subset of $\mathbb R$ such that $0 < m(E) < +\infty$. Show for any $\varepsilon >0$, there is a finite nondegenerate interval $I$ such that $m(E \cap I) > (1-\varepsilon)m(I)$.

Here's my attempt:
First, we notice that since $E$ is measurable, we can write (working backwards) $m (I) = m(I\cap E) + m (I \cap E^c)$. Now if it were the case that $m(I\cap E^c) > (1-\varepsilon) m (I)$ then we would have $\varepsilon m (I) > m (I \cap E^c)$. Conversely if $\varepsilon m (I) > m (I \cap E^c)$, we would have that $m(I\cap E^c) > (1-\varepsilon) m (I)$.
So the problem is equivalent to finding an interval (finite, nondegenerate) such that $\varepsilon m (I) > m (I \cap E^c)$.
Since $m(E) < + \infty$, we have that $m (E^c) = +\infty$.

I couldn't proceed further. Hints would be appreciated.
 A: Let $f(x)=m((-\infty,x)\cap E)$. Then $f$ is increasing and $f(y)-f(x) =m((x,y)\cap E) \leq y-x$ for $x<y$.  So $f$ is a continuous function. $f(x) \to m(E)$ as $x \to \infty$ and $f(x) \to 0$ as $x \to -\infty$. By Intermediate Value Property of continuous functions there exists $x$ such that $m((-\infty,x)\cap E)=f(x) =(1-\epsilon /2) m(E)$.  Can you now show that $m((-M, x) \cap E)> (1-\epsilon ) m(E)$ for lareg enough values of $M$?
A: For each $r>0$,
there is a sequence (finite or infinite) of disjoint open intervals $T_i$
such that $\bigcup T_i\supseteq E$ and also $\sum \text{m}(T_i)=\text{m}(E)+r$.
This is because every open set in $\mathbb{R}$ is a union of disjoint open intervals.
Note: the sets $(E\bigcap T_i)$ are all disjoint, and furthermore $\bigcup(E\bigcap T_i)=E$.
So $\sum \text{m}(E\bigcap T_i)=\text{m}(E)$.
Therefore, it cannot be the case that $\text{m}(T_i)>\big(1+\frac{r}{\text{m}(E)}\big)\times \text{m}(E\bigcap T_i)$ for all $i$. You can show this by summing over $i$.
Therefore, there exists $i$ such that $\text{m}(T_i)\leq\big(1+\frac{r}{m(E)}\big)\times\text{m}(E\bigcap T_i)$.
