Suppose $\mu(E\cap I)\geq C\mu(I)$ for every interval $I\subseteq [0,1]$. Show that $\mu(E)=1$. Suppose $E\subseteq [0,1].$
$C$ is some positive number and $\mu$ is Lebesgue measure.
My approach:
Assume $\mu(E)<\frac{q}{p}<1$.
We know $E$ can be approximated by finite summation of intervals. Intervals can be approximated by summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$. Therefore, $E$ can be approximated by finite summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$.
Choose a $N$ such that there is a $\delta$-approximation.
Suppose there are $n$ intervals used in the approximation. Since $n\cdot 2^{-N}\leq \frac{q}{p}+\delta$, we have $2^N-n\geq \frac{p-q-p\delta}{p}2^N$, which means there are at least about $\frac{p-q-p\delta}{p}2^N$ intervals remained blank.
Then, since it's a $\delta$-approximation, there is some interval on which $\mu(E\cap I)\leq \frac{\delta p}{p-q-p\delta}2^{-N}$.
Then, we have $\mu(E\cap I)/\mu(I)\leq \frac{\delta p}{p-q-p\delta}\to 0$ as $\delta\to 0$, which is a contradiction.
I think this is a correct proof, but is too ugly. Is there an elegant one?
 A: There's this nice proof using Lebesgue's differentiation theorem, at least assuming $E$ is measurable:
Let $I_{\varepsilon, x} := [x - \varepsilon, x + \varepsilon]$.  We know that for almost every $x \not\in E$ we have $$\lim_{\varepsilon \rightarrow 0} \frac{\mu(E \cap I_{\varepsilon, x})}{\mu(I_{\varepsilon, x})} = 0,$$
but
$$\lim_{\varepsilon \rightarrow 0} \frac{\mu(E \cap I_{\varepsilon, x})}{\mu(I_{\varepsilon, x})} \ge C > 0$$
so every $x \in E^c$ is contained in the exceptional set (which has measure $0$).  Hence we must have that $\mu(E^c) = 0$ and therefore $\mu(E) = 1$.
A: The premises imply that $E$ is measurable because $\mu (E)=\mu(E\cap I)$ exists when $I=[0,1].$
Let $E'=E\setminus \{0,1\}.$ I found it easier to show that $\mu(E')=1.$ Note that $\mu(E'\cap J)\ge C\mu(J)$ for every interval $J\subseteq (0,1).$
By contradiction, suppose $\mu(E')=1-r<1.$ Lebesgue measure is inner-regular: $\mu(E')=\sup \{\mu (B): B=\overline B\subseteq E'\}.$ So for any $d>0$ there exists $B_d=\overline B_d\subseteq E'$ with $$\mu(B_d)>\mu(E')-d=1-r-d.$$
Let $F_d= (0,1)\setminus  B_d.$ Then $F_d\supseteq (0,1)\setminus E'$ so $$\mu(F_d)\ge 1-\mu(E')=r.$$
Now $F_d$ is open in $\Bbb R$ and is a subset of $(0,1)$ so $F_d=\cup G_d$ where $G_d $ is a countable pairwise-disjoint family of open intervals, with $J\subseteq (0,1)$ for every $J\in G_d.$ Hence $$\mu (E'\cap  F_d)=\sum_{J\in G_d}\mu (E'\cap J)\ge$$ $$\ge\sum_{J\in G_d}C\mu(J)=$$ $$=C\sum_{J\in G_d}\mu(J)=C\mu(\cup G_d)=$$ $$=C\mu(F_d)\ge Cr.$$ Now since $B_d, F_d$ are disjoint and $B_d\cup F_d=(0,1),$ and since $B_d\subseteq E'$ we have $$1-r=\mu(E')= \mu (E'\cap B_d)+\mu (E'\cap F_d)=$$ $$=\mu(B_d)+\mu(E'\cap F_d)>$$ $$>(1-r-d)+Cr.$$  This holds for any $d>0$ so $$\forall d>0\,(\,1-r>(1-r-d)+Cr\,)$$ which implies $1-r\ge 1-r+Cr,$ which is absurd if $C$ and $r$ are positive.
