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This is a very basic question regarding filtrations. One knows that for a martingale random variable, we need a sequence of nested filtrations, such that $\mathcal F_n \subset \mathcal F_{n+1} \ \forall n \in \mathbb N$.

Now, the so called natural filtration is given by $\mathcal F_n = \sigma(X_1,...,X_n)$, which is the smallest $\sigma$-field such that all $X_1,...,X_n$ are measurable. My question is, how come $\sigma (X_1,...,X_n)\subset \sigma(X_1,...,X_{n+1})$? As I see it, the $\sigma$-field with an added r.v would be “smaller” then the previous one, since it also requires that all the previous r.v’s are measurable. Is there some kind abuse of notation going on?

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    $\begingroup$ Hint: Any $\sigma$-field on which $X_1,\dots,X_n$ are measurable must contain the smallest $\sigma$-field on which $X_1,\dots,X_n$ are measurable. $\endgroup$ Commented Aug 16, 2020 at 14:07

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Denonte by $\mathcal{F} :=\sigma(X_1) \cup \ldots \cup \sigma (X_n)$ and $\mathcal{G} := \sigma(X_1) \cup \ldots \cup \sigma (X_n) \cup \sigma(X_{n+1})$ and observe that

$$\sigma(X_1, \ldots, X_n) = \sigma( \mathcal{F}) \supset \mathcal{F},$$ $$\sigma(X_1, \ldots, X_n, X_{n+1}) = \sigma( \mathcal{G}) \supset \mathcal{G} \supset \mathcal{F}.$$

It is then evident that $\sigma(\mathcal{G})$ contains $\mathcal{F}$, and since $\sigma(\mathcal{F})$ is the smallest sigma-algebra containting $\mathcal{F}$, it follows that $\sigma(\mathcal{F}) \subset \sigma(\mathcal{G})$.

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