real analysis - Non-measurable set A such that $A cap mathbb{Q}^c$ is measurable

Does there exist a non-(Lebesgue)measurable set $A \subset \mathbb{R}$ such that $B = \{x \in A : x \in \mathbb{R}\setminus\mathbb{Q} \}$ is measurable?

My thoughts: the only non-measurable sets I've seen so far are the Vitali sets, so thinking along this line - are the irrational points in a Vitali set $V$ measurable? It would be enough to show that their complement is measurable - that is, is the set $\{x \in V : x \in \mathbb{Q}\}$ measurable? Since the set $V$ is obtained by taking the quotient group $\mathbb{R} \setminus \mathbb{Q}$, the rational points are all in a single equivalence class represented by, say, $0$, which certainly is measurable as it is a singleton.

Is my thinking correct? If not, does there exist another example?

Answer

No.

Let $\mathbb{I}$ be the set of irrational numbers.

Let $V \subseteq \mathbb{R}$ non measurable.

If $V \cap \mathbb{I}$ is measurable with respect to the Lebesgue measure then $V \cap \mathbb{Q}$ is measurable as a set of Lebesgue measure zero .

Thus $V=(V \cap \mathbb{I}) \cup (V \cap \mathbb{Q})$ would be measurable as a union of measurable sets.

This is a contradiction.

Also note that the Lebesgue measure is complete so a subset of a set of Lebesgue measure zero is measurable.