# 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 $A$?

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 $A$.

• It looks like we have some misprints in the definition of $\mathcal{B}_0$. And, the complement of a set of infinite measure need not be of measure $0$. – Adam Przeździecki Jul 15 '18 at 8:04
• Adam Przeździecki: Yes, you are right. I have modified my question! – user173856 Jul 15 '18 at 15:31

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.)