$E \subset \mathbb{R}, \mu(E)>0$, prove $\forall \delta > 0$ there is a bdd, open interval $I$ s.t. $\frac{\mu(E \cap I)}{\mu(I)} > 1 - \delta$ For a Lebesgue measurable $E \subset \mathbb{R}, \mu(E)>0$, prove $\forall \delta > 0$ there is a bounded open interval $I$ such that $\frac{\mu(E \cap I)}{\mu(I)$} > 1 - \delta$.
I feel like this question is really easy, which is making me nervous.
We would just need $I \subset E$?  That would make the LHS 1 which is clearly greater than $1 - \delta$.  Am I supposed to show that because $E$ is measurable then there must be some subset of it that is an open, bounded interval?  
My guess is that because $0 < \mu (E) = inf \{\sum_{k=1}^\infty l(I_k): E \subset \bigcup I_k\}$ there is at least one open interval inside $E$ that has positive length and is bounded.  (bounded if E is bounded.  If E is not bounded then take a bounded subset of an unbounded interval).  Then if $I \not \subset E,$ then you can take $I' = int(I \bigcap E) \neq \phi$ since I has positive length.  This $I$ fits the requirements.
Is that correct?
 A: There are measurable sets of positive measure that contain no intervals, for instance take 
$ [0,1]\setminus {\mathbb{Q}}$.
For the questions one can argue as follows. Since $\mu (E) = \inf \{\sum_{k=1}^\infty l(I_k): E \subset \bigcup I_k\}$. There there are $I_n$ such that $E \subset \bigcup I_n$ and $\mu(E)> \sum_{k=1}^\infty \mu(I_k)   -\epsilon $.
So assume that for all $n$ that $\frac{\mu (E \cap I_n)}{\mu (I_n)}\leq 1-\delta  $ or 
$\mu (E \cap I_n)\leq (1-\delta)\mu (I_n)  $.
Summing all these inequalities we obtain
\begin{align*}\sum _n\mu (E \cap I_n)&\leq \sum_n(1-\delta)\mu (I_n)\\
&<(\mu(E)+\epsilon)(1-\delta)\\
&=\mu(E)-\delta\mu(E) +\epsilon -\epsilon \delta
\end{align*}
But, $ \sum _n\mu (E \cap I_n) =\mu(E)  $. So we obtain
$$ 0\leq -\delta\mu(E) +\epsilon -\delta \epsilon \iff \delta \mu(E)\leq \epsilon -\delta \epsilon $$
Which leads to a contradiction when $\epsilon$ is small enough, hence for small enough $\epsilon$ there are $I_n$ where there is $I_{n_0}$ such that 
$$\frac{\mu (E \cap I_{n_0})}{\mu (I_{n_0})}> 1-\delta  $$
