[proofplan] Realize $\sigma(\mathcal A)$ as the intersection of all sigma-algebras on $\Omega$ that contain $\mathcal A$. Intersections of sigma-algebras are sigma-algebras. [/proofplan] [step:Define the generated sigma-algebra] Let $\mathscr S$ be the collection of sigma-algebras $\mathcal G$ on $\Omega$ such that $\mathcal A\subset\mathcal G$. This collection is nonempty because $\mathcal P(\Omega)\in\mathscr S$. Define \begin{align*} \sigma(\mathcal A)=\bigcap_{\mathcal G\in\mathscr S}\mathcal G. \end{align*} [/step] [step:Check it is a sigma-algebra containing the generators] The intersection of any family of sigma-algebras on $\Omega$ is again a sigma-algebra: $\Omega$ belongs to every member of the family, complements remain in every member, and countable unions remain in every member. Therefore $\sigma(\mathcal A)$ is a sigma-algebra. Since each $\mathcal G\in\mathscr S$ contains $\mathcal A$, their intersection contains $\mathcal A$. [/step] [step:Prove minimality] If $\mathcal G$ is a sigma-algebra with $\mathcal A\subset\mathcal G$, then $\mathcal G\in\mathscr S$. Since $\sigma(\mathcal A)$ is the intersection over all members of $\mathscr S$, it is contained in $\mathcal G$. This proves the minimality property. [/step]