Real Analysis, Folland Problem 1.5.30 Borel measures 
If $E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$.

Attempted proof/brainstorm - Let $E\in L$ with $m(E) > 0$ and suppose there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$. With out loss of generality, suppose $m(E) < \infty$. By theorem 1.18, there exists an open $U$ such that $E\subset U$ and $$m(E) \leq m(U) \leq m(E) + \epsilon$$
Let $U = \bigcup_{1}^{\infty}I_j$ where $\{I_j\}_{1}^{\infty}$ are open intervals. Then $E\subset \bigcup_{1}^{\infty}I_j$ and we have $$m(E)\leq m(U)\leq \sum_{1}^{\infty}m(I_j)\leq m(E) + \epsilon$$ Note that $m(U) < \infty$. We can write $$m(I_j) = m(I_j\setminus E) + m(E\cap I_j)\leq m(I_j\setminus E) + \alpha m(I_j)$$ then I believe with what we have above we arrive at a contradiction? 
I am not really sure about my approach, any suggestions is greatly appreciated.
 A: Fix $0<\alpha<1$ and let $\epsilon>0$. 
There is a disjoint union of intervals $\cup I_n\supseteq E$ such that $m(E)+\epsilon>\sum m(I_n)$. 
If $m(E\cap I_n) \leq \alpha m(I_n)$, for all $n$, then $E=\bigcup_nE\cap I_n$ so on the one hand
$m(E)=\sum m(E\cap I_n)\leq \alpha \sum m(I_n)$ 
and on the other,
$m(E)>\sum m(I_n)-\epsilon$, 
so if we take $\epsilon$ so small that $\sum m(I_n)-\epsilon>\alpha \sum m(I_n)$
we arrive at a contradiction.
A: Given $\epsilon > 0$, Proposition 1.20 gives a finite union $A = \cup I_j$ of disjoint open intervals
so $m (A - E) +m (E - A) < \epsilon m (E)$ (since $m(E) > 0$). We use this twice. First,
$$m (E) = m (E \cap A) + m(E - A) ≤ \sum m(I_j) + \epsilon m(E)$$
so that $(1 −\epsilon )m(E) ≤ \sum m(I_j)$. Second,
$$\sum m(I_j) = \sum m(I_j \cap E) + \sum m(I_j - E) < \sum m (I_j ∩ E) + \epsilon m(E).$$
Using the first estimate on $m$, this gives
$$\sum m(I_j) < \sum m(I_j \cap E) + 
\frac{\epsilon}{1+\epsilon} 
\sum m(I_j).$$
It follows easily from here if we choose  $\epsilon> 0$ small so $\frac{\epsilon}{1+\epsilon} ≥ \alpha$.
A: Here is a detailed proof following your approach. In particular, showing how to handle the case $\mu(E)=+\infty$. 

If $E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$.

Let $E\in L$ with $m(E) > 0$ and suppose there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$. 
Case 1: Suppose $m(E) <+\infty$. 
By theorem 1.18, for all $\epsilon >0$ there exists an open $U$ such that $E\subset U$ and 
$$m(E) \leq m(U) \leq m(E) + \epsilon$$
Take $\epsilon <(1-\alpha)\mu(E)$. 
Let $U = \bigcup_{1}^{\infty}I_j$ where $\{I_j\}_{1}^{\infty}$ are open intervals. Then $E\subset \bigcup_{1}^{\infty}I_j$ and we have 
$$m(E)\leq m(U)\leq \sum_{1}^{\infty}m(I_j)\leq m(E) + \epsilon <+\infty \tag{1}$$ 
we also have, since the $I_j$'s may not be disjoint, that
$$m(E)\leq \sum_{1}^{\infty}m(E\cap I_j)$$
However, we have that $\mu(E\cap I) \leq \alpha\mu(I_j)$. So we have 
$$m(E)\leq \alpha \sum_{1}^{\infty}m(I_j)$$
So, from $(1)$, we get 
$$ m(E)\leq \sum_{1}^{\infty}m(I_j)\leq \alpha \sum_{1}^{\infty}m(I_j) + \epsilon $$ 
Since $\epsilon <(1-\alpha)\mu(E)$ and $ \alpha \sum_{1}^{\infty}m(I_j) <+\infty$, we have
$$ \sum_{1}^{\infty}m(I_j)\leq \alpha \sum_{1}^{\infty}m(I_j) + \epsilon<  \alpha \sum_{1}^{\infty}m(I_j) +(1-\alpha)\mu(E) \leq \\ \leq \alpha \sum_{1}^{\infty}m(I_j) +(1-\alpha)\sum_{1}^{\infty}m(I_j) =\sum_{1}^{\infty}m(I_j)$$
So
$$ \sum_{1}^{\infty}m(I_j) <\sum_{1}^{\infty}m(I_j)$$
Contradiction. 
Case 2. Suppose $\mu(E)=+\infty$. 
Since $E=\bigcup_{n\in \mathbb{Z}}(E\cap(n,n+1])$, there $k \in \mathbb{Z}$ such that $\mu(E\cap (k,k+1])>0$. Let $E_k= E\cap (k,k+1]$. Clearly we have $0<\mu(E_k)\leq 1$. 
From case 1, we know that for any $\alpha >0$ there is an open interval I such that $\mu(E_k\cap I)>\alpha \mu(I)$.  Let $J=(k,k+1] \cap I$. $J$ is an interval and we have 
$$\mu(E\cap J)=\mu(E\cap (k,k+1] \cap I))=\mu(E_k\cap I)>\alpha \mu(I)\geq \alpha \mu(J)$$
So we prove the result for this second case, which completes the proof. 
Remark: Note that since we are trying to prove that:  for all$E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$, then to work by contradiction we suppose that: there is one $E\in L$ with $m(E) > 0$ such that there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$.
Under this hypothesis, we can not simply replace $E$ by another $E$ with finite measure. In this approach, the best thing is to prove by cases (and case 2 is actually an easy consequence of case 1). 
