For set of positive measure $E$, $\alpha \in (0, 1)$, there is interval $I$ such that $m(E \cap I) > \alpha \, m(I)$ I am a graduate student at Iowa State University attempting to return after a five-year hiatus and take the Real/Complex Analysis qualifier on January 8 for potential reinstatement. Since the professors here are on hiatus, I hope you guys don't mind if I bombard the board with a few past questions from our quals here, like this one from Spring 2013: 

Let $m$ denote Lebesgue measure on $\mathbb R$ and $M$ the $\sigma$-algebra of Lebesgue measurable subsets. Given $E \in M$ with $m(E) > 0$ and $\alpha \in (0, 1)$, prove that there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.

I've attempted to force the issue using the definition of outer measure but it doesn't seem to work. I've also attempted using the definition of Lebesgue measurability but that doesn't seem to work either - so I'm at a loss! 
Thanks in advance! 
-Darrin Rasberry 
 A: @AymanHourieh's answer is simple and direct. I will add this slightly longer (yet perhaps more basic) solution that does not rely on Lebesgue's density theorem.
We start with a basic claim on measurable sets (which is sometimes given as the definition for a measurable set):
Claim. Let $E \in M$ of finite measure, then for every $\epsilon>0$ there exists a finite union of disjoint intervals:
$$A_\epsilon := \biguplus_{n=1}^{N}I_n$$
s.t. $m(E \triangle A_\epsilon) < \epsilon$.
Now, let $\epsilon:=(1-\alpha)m(E)$. And let $A_\epsilon$ from the claim. Then
$$
m(E) = m(E\cap A_\epsilon) + m(E\setminus A_\epsilon) \\
\leq m(E\cap A_\epsilon) + m(E \triangle A_\epsilon) \\
< m(E\cap A_\epsilon) + \epsilon \\
= m(E\cap A_\epsilon) + (1-\alpha)m(E)
$$
Then
$$
m(E) < \frac{m(E\cap A_\epsilon)}{\alpha} \tag{1}
$$
Similarly,
$$
m(A_\epsilon) = m(E\cap A_\epsilon) + m(A_\epsilon \setminus E) \\
\leq m(E\cap A_\epsilon) + (1-\alpha)m(E) \\
\overset{\text{from (1)}}{<} m(E\cap A_\epsilon) + \frac{1-\alpha}{\alpha}m(E\cap A_\epsilon) \\
= \frac{m(E\cap A_\epsilon)}{\alpha}
$$
We now note that
$$
m(A_\epsilon ) = \sum_{n=1}^{N}m(I_n) \\
m(E \cap A_\epsilon ) = \sum_{n=1}^{N}m(E \cap I_n).
$$
Therefore,
$$
\sum_{n=1}^{N}m(I_n) < \frac{1}{\alpha}\sum_{n=1}^{N}m(E \cap I_n),
$$
So there must exist $1\leq n \leq N$ s.t.
$$
m(I_n) < \frac{1}{\alpha}m(E \cap I_n),
$$
which completes the proof.
A: The Lebesgue density theorem states that for almost every point $x \in E$ with $m(E) > 0$,
$$
\lim_{\epsilon\to 0} \frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} = 1.
$$
Fix such $x$ and pick $\epsilon$ such that
$$
\frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} > \alpha.
$$
$I = B_\epsilon(x)$ is the desired interval.
A: Here is an (simpler?) solution:
Suppose not. If $m(E) < \infty$. Then for each interval $I$ we have $m(E \cap I) \leq \alpha m(I)$. Let $\varepsilon > 0$. By the definition of the Lebesgue measure, there is some open set $G \supseteq E$ such that $m(G) < m(E) + \varepsilon$. Since every open subset of $\mathbb{R}$ is a (at most) countable union of open, disjoint intervals, we have $G = \bigcup_{k=1}^\infty (a_k,b_k)$. But then
$$
    m(E) = m(E \cap G) =\sum_{k=1}^\infty m(E \cap (a_k,b_k)) \leq \alpha \sum_{k=1}^\infty m((a_k,b_k)) = \alpha m(G) < \alpha (m(E)+\varepsilon) 
$$
Since $\varepsilon > 0$ was arbitrary, we have $m(E) \leq \alpha m(E)$. But then $m(E) = 0$ since $m(E) \geq 0$ and $0 < \alpha < 1$.
If $m(E) = \infty$, then there is disjoint union s.t. $E = \bigcup F_j$ where $m(F_j) < \infty$ since $E$ is $\sigma$-finite. Without loss of generality, suppose $m(F_1) = 0$ and $m(F_j) > 0$ for all $j > 1$. (By taking union operation for null sets, we can obtain as such.)
With $j > 1$, the previous argument implies that there exists an open interval $I_j$ such that $m(F_j \cap I_j) > \alpha m(I_j)$.
Let $I = \bigcup_{j=2}^{\infty} I_j$, then
$$\begin{eqnarray}
m(E \cap I) &\geq& m\left(\bigcup_{j=2}^{\infty} (F_j \cap I_j)\right)\\
&=& \sum_{j=2}^{\infty}m(F_j \cap I_j) \\
&>& \alpha\sum_{j=2}^{\infty}m(I_j) \\
&\geq& \alpha m(I) \\
\end{eqnarray}$$
