Constructing a Borel set

Question

I have to make a Borel set $E \subset R $ such that $ 0 < m(E \cap I) < m(E)$, for all real segment $I$, where $ m $ is the Lebesgue measure. I also have to say if it is possible for E to have $m(E)< \infty$

I know that I have to take a dense sequence $(r_j)$, such as $Q$, $a_j > 0$ such that $a_k > \sum_{j=k+1} ^{\infty} {a_j}$ , such as $ a_j = 1/3^j$, and then define $I_j = (r_j-a_j,r_j+a_j)$ , $E_k= I_k \setminus \bigcup _{j=k+1}^{\infty} {I_j} $, and finaly $E = \bigcup E_k$.

I just don't know how to calculate its measure, nor how to prove that $m(E)$ can or can't be finite.

Answer 1

Let $J_n$ be an enumeration of all intervals with rational endpoints.

We construct two disjoint Cantor like sets(or ''fat'' Cantor sets) $A_1,B_1 \subseteq J_1$ and $0<m(A_1),m(B_1)<\frac{1}{2}$

Inductively we choose disjoint Cantor like sets $A_n,B_n \subseteq J_n\setminus \bigcup_{m<n}(A_m \cup B_m)$ and $0<m(A_n),m(B_n)<\frac{1}{2^n}$

The desired set is $C:=\bigcup_{n=1}^{\infty}A_n$

Note that $C$ is Borel as a countable union of Borel sets.