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Let $\{X_t\}_{t\in[0,T]}$ be a stochastic process and $(\mathcal{F}_t)_{t\in[0,T]}$ its natural filtration. Show that for any $\mathcal{F}_1$-measurable r.v. $Y$, there exists and countable $S\subset[0,1]$ such that $Y(\omega)=Y(\omega')$ as soon as $X_t(\omega)=X_t(\omega')$, $\forall t\in S$.

I am really stumped and confused by this question, so any help is most appreciated. Thank you in advance.

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By this thread, there exists a countable subset $S$ of $[0,1]$ such that $\sigma(Y)\subset \sigma\left(X_s,s\in S\right)$. Write $S=\{s_n, n\in\mathbb N\}$. By the Doob-Dynkin lemma, there exists a function $f\colon\mathbb R^{\mathbb N}\to\mathbb R$ such that $Y=f\left(X_{s_n}\right)$. If $X_{s_n}(\omega) =X_{s_n}(\omega')$ then $Y(\omega)=Y'(\omega')$.

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  • $\begingroup$ Thank you, I didn't think the Doob-Dynkin lemma would apply as I wasn't sure how you define a measure space around $\mathbb{R}^\mahbb{N}$. Do you consider $\mathbb{R}^\mahbb{N}$ a a topological vector space and take the Borel-algebra to be generated by that topology? Would appreciate if you could help me fill in the missing details there. Thanks. $\endgroup$
    – Ansel B
    Dec 16, 2016 at 9:08
  • $\begingroup$ $\mathbb{R}^{\mathbb{N}}$ to fix the typo. $\endgroup$
    – Ansel B
    Dec 16, 2016 at 12:27

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