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Let $X_t : \Omega \rightarrow S$ define a stochastic process $Z:=(X_t)_{t \in [0,T]}$ and $\pi_t : S^{[0,T]} \rightarrow S, \pi_t(Z):=X_t$ be the canonical projection.

Let $S$ be equipped with the sigma algebra $B.$

Then we defined the sigma-algebra $B^{[0,T]}$ on $S^{[0,T]}$ as

$B^{[0,T]}:=\sigma \left( \{ \pi_t^{-1}(A);t \in T,A \in B \} \right).$

Now, I was wondering how $A:=\bigcup_{I \subset [0,T];|I| < \infty} \pi_I^{-1}(B^I)$ is related to the sigma algebra $B^{[0,T]}$. Here $B^I$ denotes elements of the product sigma algebra (since $I$ is finite) and $\pi_{\{i_1,...,i_n\} } = (\pi_{i_1},...,\pi_{i_n})$

Our professor said something like (he did not write it down, so I was wondering whether I remember this correctly): $A$ is an algebra and stable under intersections, moreover $A$ generates the sigma algebra $B^{[0,T]}$ in the sense $B^{[0,T]} = \sigma(A).$ Is this true?

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Yes, that's correct.

Since, for any set $C \in \mathcal{B}$, we have

$$\pi_t^{-1}(C) \in \mathcal{A} \subseteq \sigma(A)$$

(just choose $I = \{t\}$), we find that $\pi_t$ is measurable with respect to $\sigma(\mathcal{A})$. As $t$ is arbitrary and $\mathcal{B}^{[0,T]}$ is the smallest $\sigma$-algebra (on $S^{[0,T]}$) such that $\pi_t$ is measurable for any $t$, we get

$$\mathcal{B}^{[0,T]} \subseteq \sigma(\mathcal{A}).$$

On the other hand, if $B^I = B_{i_1} \times \dots \times B_{i_n}$ is an element of the product $\sigma$-algebra and $I = \{i_1,\ldots,i_n\}$, then

$$\pi_I^{-1}(B^I) = \bigcap_{i=1}^n \underbrace{\pi_{t_i}^{-1}(B_i)}_{\in \mathcal{B}^{[0,T]}} \in \mathcal{B}^{[0,T]}.$$

This proves $\mathcal{A} \subseteq \mathcal{B}^{[0,T]}$. Hence,$$\sigma(\mathcal{A}) \subseteq \sigma(\mathcal{B}^{[0,T]}) = \mathcal{B}^{[0,T]}.$$

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