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This is somewhat of a stupid question, but I am stuck on coming up with an airtight justification. Let $X_1, \ldots, X_n$ be defined on a probability space $(\Omega, \mathcal{F}, P)$. There are several equivalent definitions for the sigma-field generated by $X_1, \ldots, X_n$. One particular one is:

$$\mathcal{F}_n = \{ B = \{\omega: (X_1(\omega), \ldots, X_n(\omega)) \in A\},\; A \in B(\mathbb{R}^n) \}$$

Where $B(\mathbb{R}^n)$ is the Borel sigma-field on $\mathbb{R}^n$. This is clearly a sigma-algebra since $B(\mathbb{R}^n)$ is a sigma-field, and furthermore is a sub-sigma-field of $\mathcal{F}$ since the transformation is measurable.

My question is: can $\mathcal{F}_n$ be re-written as:

$$\mathcal{F}_n = \{ \cap_{i=1}^n \{\omega: X_i(\omega) \in B_i\}, \;B_i \in B(\mathbb{R}) \}$$

In other words, the set of all sets that can be written as intersections of inverse images of Borel sets under $X_1, \ldots, X_n$. It seems obvious, because $B(\mathbb{R}^n) = B(\mathbb{R}) \times \ldots \times B(\mathbb{R})$, but I feel like I may be missing something. Can anyone confirm or disprove this?

Question answered: Did pointed out (in the comments) where I had gotten confused. Many thanks to everyone. The correct formulation should be: $$\mathcal{F}_n = \sigma \Big(\cap_{i=1}^n \{\omega: X_i(\omega) \in B_i\}, \;B_i \in B(\mathbb{R}) \Big)$$

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    $\begingroup$ Have you checked whether your $\mathcal F_n$ is a $\sigma$-algebra? $\endgroup$ – Davide Giraudo Mar 3 '13 at 16:07
  • $\begingroup$ I thought it was obvious from the above justification, but I haven't rigorously verified it... I will take a look in a bit. Thanks! $\endgroup$ – gogurt Mar 3 '13 at 16:24
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    $\begingroup$ The assertion that $B(\mathbb R^n)$ is $B(\mathbb R)\times\cdots\times B(\mathbb R)$ is either wrong or usually written, to avoid ambiguities, rather as $B(\mathbb R^n)=B(\mathbb R)\otimes\cdots\otimes B(\mathbb R)$. $\endgroup$ – Did Mar 3 '13 at 16:24
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    $\begingroup$ Did you're absolutely right. I got my notation mixed up, which is why I got confused. So essentially, the second "version" of $\mathcal{F}_n$ I gave should be the sigma-algebra generated by what I wrote. Many thanks. $\endgroup$ – gogurt Mar 3 '13 at 16:38
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Either the assertion that $B(\mathbb R^n)$ is $B(\mathbb R)\times\cdots\times B(\mathbb R)$ is wrong or it is more usually (and, to avoid ambiguities, more preferably) written as $B(\mathbb R^n)=B(\mathbb R)\otimes\cdots\otimes B(\mathbb R)$.

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