Can $[0,1]$ be partitioned with the following property?

Question

Let $I:= [0,1]$.

Does there exist a partition $\{I_1,I_2\}$ of $I$ such that for all Borel subsets $A$ of $[0,1]$ we have:

$$ \mu (I_1 \cap A) = \frac{1}{2} \mu(A) = \mu (I_2 \cap A)$$

where $\mu$ is the Lebesgue measure?

Many thanks for your help.

Answer 1

A complementary fact due to Kunen: If $m:\mathcal{P}([0, 1]) \to [0, 1]$ is any total extension of Lebesgue measure, then there is some $X \subseteq [0, 1]$ such that for every Borel $B \subseteq [0, 1]$, $m(X \cap B) = m(B)/2$.