Let $(\Omega, \mathcal{A})$ be a measure space and $A\subset \mathcal{A}$. Show that $\sigma(A) = \sigma(\sigma(A))$.
Do we have to use that the minimal sigma algebra is the intersection of all sigma algebras containing A?
Let $(\Omega, \mathcal{A})$ be a measure space and $A\subset \mathcal{A}$. Show that $\sigma(A) = \sigma(\sigma(A))$.
Do we have to use that the minimal sigma algebra is the intersection of all sigma algebras containing A?
By definition, $\sigma(A) \subset \sigma(\sigma(A))$. Since $\sigma(\sigma(A))$ is the smallest $\sigma$-algebra containing $\sigma(A)$, and $\sigma(A)$ is a $\sigma$-algebra containing $\sigma(A)$, $\sigma(\sigma(A)) \subset \sigma(A)$. Thus $\sigma(A) = \sigma(\sigma(A))$.
Direct from the two following facts: