Let \((\Omega, \mathscr{A}, \mathbb{P})\) be a probability space. The completion \(\left(\Omega, \mathscr{A}^{\prime}, \mathbb{P}^{\prime}ight)\) is the smallest extension such that \(\mathscr{A}^{\prime}\) contains all subsets of \(\mathscr{A}\) measurable null sets. Show that

\[\begin{aligned}\mathscr{A}^{\prime}:= & \left\{A^{\prime} \subset \Omega: \exists A_{1}, A_{2} \in \mathscr{A}, A_{1} \subset A^{\prime} \subset A_{2}, \mathbb{P}\left(A_{2} \backslash A_{1}ight)=0ight\}, \\& \mathbb{P}^{\prime}\left(A^{\prime}ight):=\mathbb{P}\left(A_{2}ight) \quad\left(A_{2} \text { is as in the definition of } \mathscr{A}^{\prime}ight) .\end{aligned}\]