Let $\mathcal{B}\_{\mathbb{R}}$ be the Borel $\sigma$-algebra on $\mathbb{R}$ and $\mu\_L$ be the Lebesgue measure on $\mathbb{R}$. 


Define a new $\sigma$-algebra $\mathcal{B}\_0$ as follows:
$$\mathcal{B}\_0=\{A\in \mathcal{B}\_{\mathbb{R}}:\mu\_L(A)=0\ \text{or}\ \mu\_L(A^c)=0\}.$$
I want to prove that the family of all locally measurable sets of the measure space $(\mathbb{R},\mathcal{B}\_0,\mu\_L|\_{\mathcal{B}\_0})$, that is, $$\{E\subset \mathbb{R}:E\cap A\in \mathcal{B}\_0\ \text{for all $A\in \mathcal{B}\_0$ such that $\mu\_L(A)<\infty$}\}$$ 
is not the family of all subsets of $\mathbb{R}$.


So I want to ask whether there exists a Lebesgue nonmeasurable set $E$ in $\mathbb{R}$ satisfies that
$E\cap A$ is a Borel null set for every Borel null set $A$.



Does there exist a Lebesgue nonmeasurable set $E$ in $\mathbb{R}$ satisfies that $E\cap A$ is a Borel null set for every Borel null set $A$?
It is consistent that such a set $E$ exists, but I do not know if ZFC proves that such a set exists. 


A set $S$ is called a Sierpiński set if $S$ is uncountable, but $S\cap N$ is countable for every (Borel) null set $N$. 


If $S$ is a Sierpiński set, then $S$ satisfies your requirement: $S$ is not measurable, and each intersection with a Borel null set is countable, so in particular Borel. 


CH implies that Sierpiński sets exist, but their existence is also consistent with the negation of CH. (Add $\aleph\_1$ many random reals to any model, without changing the size of the continuum.)


