Let $X_1, ..., X_n$ be a sequence of random variables. Show that
$\hspace{60pt}$ $\mathcal{F}_n$ = $\sigma(X_1, ..., X_n)$
is the smallest filtration such that the sequence $X_1, ..., X_n$ is adapted to $\mathcal{F}_1, ..., \mathcal{F}_n$.
Because the sequence $X_1, ..., X_n$ is adapted to the filtration $\mathcal{G}_1, ..., \mathcal{G}_n$, it follows that $X_n$ is $\mathcal{G}_n$-measurable for each $n$. But
$\hspace{60pt}$ $\mathcal{G}_1 \subset \mathcal{G}_2 \subset ...$
so $X_1, ..., X_n$ are $\mathcal{G}_n$-measurable for each $n$. Why?
For example, $X_2$ is $\mathcal{G}_2$-measurable and from $\mathcal{G}_1 \subset \mathcal{G}_2 \subset ...$ it follows that $X_2$ is $\mathcal{G}_n$-measurable for each $n \geq 2$, i.e $\{ X_2 \in B \} \in \mathcal{G}_n$ for each $n \geq 2$ and each $B \in \mathcal{B}(\mathbb{R})$. Why $X_2$ is $\mathcal{G}_1$-measurable?
Thank you!
First of all, your second and third paragraphs. Recall that if $\xi$ is $F$-measurable and $F\subseteq G$ then $\xi$ is $G$-measurable: that follows from the definition of measurability. Now, if $\Bbb F = (F_n)_{n\geq 0}$ is a filtration and $\xi = (\xi_n)_{n\geq 0}$ is $\Bbb F$-adapted, then for each $n$ we have $\xi_n$ is $F_n$-measurable. Since $F_n\subseteq F_m$ for $m\geq n$ we obtain that $\xi_n$ is $F_m$-measurable. Equivalently, $\xi_i$ is $F_n$-measurable for all $i\leq n$ so that $\xi_1,\dots,\xi_n$ are all $F_n$-measurable. Now, regarding your last paragraph: of course $\xi_2$ does not have to be $F_1$-measurable: the index of the $\sigma$-algebra in the filtration must be at least as a big as the index of the random variable for measurability to be guaranteed by adaptness.
Now, by definition $\xi$ is $\Bbb G$-adapted iff $\xi_1,\dots,\xi_n$ are $G_n$-measurable for all $n\geq 0$. Again, by definition $F_n = \sigma(\xi_1,\dots,\xi_n)$ is the smallest $\sigma$-algebra with such property, so $F_n\subseteq G_n$. Since $\xi$ is $\Bbb F$-adapted, this is the smallest filtration.