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I'm stumbling over the second step in a proof that $X^{-1}(\mathcal{F}(\mathcal{A})) = \mathcal{F}(X^{-1}(\mathcal{A}))$, where $X: \Omega \to \Theta$ and $\mathcal{A}$ is a collection of subsets of $\Theta$.

(Note: $\mathcal{F}(\cdot)$ generates the smallest $\sigma$-algebra for $\cdot$)

I already have $\mathcal{F}(X^{-1}(\mathcal{A})) \subset X^{-1}(\mathcal{F}(\mathcal{A}))$, but I don't see the other direction. (That step isn't so clear in the text I'm reading.)

zxmkn
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1 Answers1

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Hint:

  1. Show that for any $\sigma$-algebra $\mathcal{M}$ the family $$\Sigma := \{A \in \mathcal{F}(\mathcal{A}); X^{-1}(A) \in \mathcal{M}\}$$ is a $\sigma$-algebra.
  2. Choose $\mathcal{M} := \mathcal{F}(X^{-1}(\mathcal{A}))$. Show that $\Sigma$ contains $\mathcal{A}$.
  3. Conclude that $\mathcal{F}(\mathcal{A}) \subseteq \Sigma$.
saz
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