I was solving the question if A intersection with Q complément is measurable, then prove or disprove that A is measurable. My logic was as A intersection Q complement is measurable and A intersection Q is measured since set of rationals are measurable, outer measure of A is addition of outer measures of the above two sets which will be zero making A measurable. But then, I considered the possibility of intersection of Vitali set with set of irrationals will be measurable but Vitali set is non measurable. So what is the flaw in my logic?
The intersection of the Vitali set with the irrationals is not measurable.
To see this, let $V$ be the Vitali set, and $\mathbb Q$ the rationals. Since $\mathbb Q$ has measure zero, and Lebesgue measure is complete, any subset of $\mathbb Q$ is measurable, hence $V \cap \mathbb Q$ is measurable.
If $V \cap \mathbb Q^c$ were also measurable, we would have that $V = (V \cap \mathbb Q) \cup (V \cap \mathbb Q^c)$ is measurable (because a union of two measurable sets is measurable), a contradiction.
If $A$ is a Vitali set, then $A\setminus Q$ is still not a measurable set. Any set of rational numbers (or any countable set, or any null set) can be neglected when considering what sets are measurable. Indeed, suppose $A\setminus Q$ were rational. Since $Q$ is measurable, then $A$ is measurable, a contradiction.