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Let $X:(\Omega,\mathcal F) \longrightarrow (S,\mathcal S)$ a r.v. Prove that $X^{-1}(\mathcal A)=\{\{X\in A\}\mid A\in \mathcal A\}$ generate $\sigma (X)=\{\{X\in B\}\mid B\in \mathcal S\}$ if $\mathcal A$ generate $\mathcal S$.

To be honnest, I don't really understand the question. I think that I have to prove that $\sigma (X^{-1}(\mathcal A))=\sigma (X)$, no ? But even, I don't really see how it work. If $\mathcal A$ generate $\mathcal S$, it mean that $\mathcal S=\sigma (\mathcal A)$ ? does it ? But what does it mean ? That all element of $\mathcal S$ is an unuin of ellement of $\mathcal A$ ?

user380364
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  • See http://math.stackexchange.com/questions/2000142/does-the-set-of-inverse-images-of-a-generator-of-a-sigma-algebra-generate-the-si – user160738 Nov 05 '16 at 09:04
  • Yes, that's what you have to prove. It can be shown in general that $\sigma(X^{-1}(\mathcal A))=X^{-1}(\sigma(\mathcal A))$. That will lead to the result. If $\mathcal A$ generates $\mathcal S$ then the RHS will equal $X^{-1}(\mathcal S)=\sigma(X)$. The expression $\mathcal S=\sigma(\mathcal A)$ means that $\mathcal S$ is the smallest $\sigma$-algebra that contains $\mathcal A$ as a subcollection. – drhab Nov 05 '16 at 09:08
  • The answer to this question might be helpful. – drhab Nov 05 '16 at 09:21

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