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I aim to show that $\mathcal{F}_n = \sigma(X_1, X_2, \dots X_n)$ is the smallest filtration such that $X_1, X_2, \dots X_n$ is adapted to $\mathcal{F}_1, \mathcal{F}_2, \dots \mathcal{F}_n$. Is this proof convincing?

Proof:

$\mathcal{F}_n = \sigma(X_1, X_2, \dots X_n) = \sigma(\cup_{i=1}^{n} \sigma(X_i)) = \sigma(\cup_{i=1}^n \{X_i \in B\})$, $\forall B \in \mathcal{B}(\mathcal(R))$.

Consider some other filtration $G_n$ where $X_1, X_2, \dots X_n$ is adapted to $\mathcal{G}_1, \mathcal{G}_2, \dots \mathcal{G}_n$. If some event $A \in \mathcal{F}_n$ then $A = \cup_{i \in I} \{X_i \in B \} \implies A \in \mathcal{G}_n$ since $G_n$ is an adapted filtration.

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