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.
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