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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?

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  • $\begingroup$ Why do you say that the intersection of a vitali set with the irrationals is measurable? $\endgroup$ – littleO Sep 7 '17 at 5:50
  • $\begingroup$ That was the doubt as irrationals in R are measurable, I got confused. $\endgroup$ – Vaishnavi Avachat Sep 7 '17 at 8:47
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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.

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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.

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