Non-measurable set ($A\notin \Sigma$)

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.