If $E\subseteq \mathbb{R}$ is measurable and $\delta>0$ then there exists open set $U$ s.t. $\delta \mu(U)<\mu(E)$ In part of the proof of a problem I am trying to solve I need the following fact (assume that $\mu$ is the Lebesgue measure): 
If $E\subseteq \mathbb{R}$ is measurable and $\delta>0$ then there exists open set $U\subseteq \mathbb{R}$, such that $E\subseteq U$ and $\,$ $\delta \mu(U)<\mu(E)$.
I know and have proven the following fact:
Suppose $E \subseteq \mathbb{R}$. Then for each $\epsilon>0$ there exists an open set $U\subseteq \mathbb{R}$ such that $E\subseteq U$ and $\mu(U)< \mu(E)+\epsilon$.
I am pretty sure I can use the second fact to prove the first fact, but I keep getting a value of $\epsilon$ that is in terms of $\mu(U)$, which isn't good because $U$ should depend on $\epsilon$, not the other way around. Some help?

Let me expand Ian's comment.
First, the inequality can only hold for $\delta<1$.
For example, if $E=[0,1]$ and $\delta\geq1$, there never is such a $U$.
But the claim is indeed true for all $\delta\in (0,1)$.
Take any such $\delta$ and work backwards:
You want $\delta \mu(U)< \mu(E)$.
That is, you want
$$\mu(U) < \delta^{-1} \mu(E) = \mu(E) + (\delta^{-1}-1)\mu(E). $$
Now take $\epsilon=(\delta^{-1}-1) \mu(E)$ in your lemma.
Then  $\mu (U)<\mu (E)+\epsilon $ for some open $U \supset E$.
Since this is the estimate you want, $U $ is the set you want.
