Nonexistence of Lebesgue measurable with certain property Prove that there is no Lebesgue measurable set $A$ in [0,1] with the property that $m(A\cap I) = \frac{1}{4}m(I)$ for every interval $I$
 A: Let $U$ be an open set of $\mathbb{R}$ such that $A \subset U$ and $m(U\setminus A) \leq \epsilon$. Being an open set of $\mathbb{R}$, $U$ can be written as a countable (or finite) disjoint union
$$
U = \bigcup_{n} (a_n,b_n).
$$
Therefore, we obtain
$$
m(A) = m(A \cap U) \leq \sum_n m(A\cap (a_n,b_n)) = \frac{1}{4} \sum_n (b_n-a_n) = \frac{1}{4}m(U) \leq \frac{1}{4}(m(A) + \epsilon),
$$
hence $\forall \epsilon > 0$, $m(A) \leq \epsilon/3$. Finally, $m(A)=0$ and $m(A \cap [0,1]) = 0 \neq 1/4$.
In the exact same way, we have

Lemma. If $A$ is a Borel subset of $\mathbb{R}$ such that there exists
  $\lambda < 1$ with
  $$\forall a,b \in\mathbb{R} ,\quad a < b \implies m(A \cap (a,b)) \leq \lambda(b-a)$$
  then $A$ is negligible.


Alternative proof using Dynkin's lemma.
Suppose that $A$ exists. Then the set of Borel subsets $B$ of $[0,1]$ such that $m(A\cap B) = \frac{1}{4}m(B)$ is a $\lambda$-system which contains the $\pi$-system of all intervals, hence the Borel $\sigma$-algebra of $[0,1]$. In particular, $m(A)=m(A \cap A)=\frac{1}{4}m(A)$ and $m(A)=m(A\cap[0,1])=\frac{1}{4}$, which is absurd. 
A: Hint: Lebesgue's density theorem ( http://en.wikipedia.org/wiki/Lebesgue's_density_theorem ) states that if $A\subset\mathbb{R}$ is Lebesgue-measurable, then for almost every $x\in\mathbb{R}$ we have 
\begin{equation*}
\lim_{r\to 0}\frac{m(A\cap B(x,r))}{m(B(x,r))}=1,
\end{equation*}
where $B(x,r)$ is the open $r$-radius ball centered at $x$. In $\mathbb{R}$ these are open intervals.
A: The following property of Lebesgue measurable sets is useful here: Given a measurable set $A$, there exists sets $F,U$ such that $F$ is closed, $U$ is open, and $F\subset A\subset U$, and most importantly $m(U\setminus F)<\epsilon$ for any $\epsilon$.
From this proposition, you can prove without too much extra work that every Lebesgue measurable set of finite measure greater than zero almost contains a full interval.
