TITLE: The (non)existence of a measurable set
QUESTION [1 upvotes]: Does there exist a measurable set $E \subset [0,1]$, such that for any open interval $I \subset [0,1]$, $$m(E~\cap~I)=\frac{1}{2}m(I)~?$$

Here $m$ denotes the Lebesgue measure.

REPLY [2 votes]: First, every open set in $[0,1]$ is a countable disjoint union of open intervals.
That means that your propery extend to the open sets as well by sigma-additivity :
If $O = \bigcup_{j\in \Bbb N} I_j$ then 
$$m(O) = \sum_{j=0}^\infty m(I_j) = \sum_{j=0}^\infty \frac{1}{2}m(I_j \cap E) = \frac{1}{2}m(\bigcup_{j\in \Bbb N} (I_j \cap E) ) = \frac{1}{2}m(O \cap E )  $$
Then you need to use the fact that the Lebesgue measure is regular. You have that 
$$\forall E \text{ measurable }, m(E) = \inf_{E\subset O, O \text{ open }} m(O)$$
Hence
$$m(E) = \inf_{E\subset O, O \text{ open }} m(O) = \inf_{E\subset O, O \text{ open }} 2m(O\cap E) = \inf_{E\subset O, O \text{ open }} 2m( E) = 2m(E)$$
So  this would imply $m(E) = 0$, but this is impossible as $m(E) = m(E\cap [0,1]) = \frac{1}{2} m([0,1]) = \frac{1}{2}$
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