Non-measurable set
= Non-measurable set
{title2=$A\notin\Sigma$}
A <subset> $A$ of a <measurable space> $(X,\Sigma)$ is non-measurable when $A\notin\Sigma$. On the <real line>, unless another <sigma-algebra> is specified, this commonly means that $A$ is not a <Lebesgue measurable set>. By <Completeness of Lebesgue measure>, such a set cannot be contained in a <Lebesgue null set>.
For an example, use the <axiom of choice> to select one representative from each equivalence class of $[0,1]$ under $x\sim y$ if $x-y\in\mathbb Q$, and call the resulting set $V$. The translates $V+q$, for $q\in\mathbb Q\cap[-1,1]$, are pairwise disjoint, their union contains $[0,1]$, and their union is contained in $[-1,2]$. If $V$ were a <Lebesgue measurable set>, translation invariance and countable additivity would make that union have measure zero when $m(V)=0$ and infinite measure when $m(V)>0$. Both conclusions contradict these inclusions. Thus $V$ is non-measurable.