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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?

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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 $\sigma(A) = \sigma(\sigma(A))$.

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Direct from the two following facts:

  • $\sigma(A)$ is a sigma-algebra.
  • For every sigma-algebra $\mathcal B$, $\sigma(\mathcal B)=\mathcal B$.
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