Measure and the intersection with an open interval 
Let $E \subset \Re$ be measurable with $\mu(E) > 0$. Show that for
  every $0 < \epsilon < 1$ there is an open interval $I$ such that
  $\mu(E \bigcap I) > (1-\epsilon)\mu(I)$

Let $0 < \epsilon < 1$, $\mu(E)>0$. How would I start this?
 A: Lemma. Let $F\subset[a,b]$ be measurable set of positive measure, then for all $\varepsilon>0$ there exist open interval $I$ such that $\mu(F\cap I)>(1-\varepsilon)\mu(I)$
Proof. Assume there exist $\varepsilon>0$ such that for all open interavals $I$ holds $\mu(F\cap I)\leq (1-\varepsilon)\mu(I)$. From definition of measure it follows that there exist a family of open intervals $\{J_k:k=1,\ldots,m\}$ such that 
$$
F\subset\bigcup_{k=1}^n J_k\quad\text{and}\quad\sum\limits_{k=1}^n\mu(J_k)<\mu(F)(1+\varepsilon)
$$
From assumption we get
$$
\mu(F)=\mu\left(F\cap\bigcup_{k=1}^n J_k\right)\leq\sum\limits_{k=1}^n\mu(F\cap J_k)\leq\sum\limits_{k=1}^n(1-\varepsilon)\mu(J_k)=(1-\varepsilon^2)\mu(F)
$$
Contradiction.
Theorem. Let $E\subset\mathbb{R}$ be measurable set of positive measure, then for all $\varepsilon>0$ there exist open interval $I$ such that $\mu(F\cap I)>(1-\varepsilon)\mu(I)$
Proof. Fix $\varepsilon>0$. For each $n\in\mathbb{N}$, denote $F_n=E\cap[n,n+1)$. Then $\mu(E)=\sum\limits_{n=1}^\infty\mu(F_n)$. Since $\mu(E)>0$, there exist some  $n\in\mathbb{N}$ such that $\mu(F_n)>0$. By previous lemma there exist open interval $I$ such that $\mu(F_n\cap I)>(1-\varepsilon)\mu(I)$. Hence
$$
\mu(E\cap I)\geq\mu(F_n\cap I)>(1-\varepsilon)\mu(I)
$$
