0
$\begingroup$

In probability theory one often seeks to construct the $\sigma$-field $ \sigma(X_{1},X_{2},\ldots) $ for a sequence of R.V.'s $\{X_n\}_{n\in \mathbb{N}}$ (Assumption: $X_n:(\Omega, \mathcal{F}) \rightarrow (\Omega',\mathcal{F'}) )$.

I realized that I'm still not 100% sure what $\sigma(X_1,X_{2},\ldots)$ means. My recollection is the following definition: $\sigma(X_1,X_{2},\ldots) = \sigma\left\{ \bigcap_{n=1}^{\infty} X_n^{-1}(A_n') \mid A_1',A_2',\ldots \in \mathcal{F}' \right\}$. Is this correct? If not, what is the correct formal definition?

$\endgroup$
1
  • 1
    $\begingroup$ Another convention. When $\Omega' = \mathbb R$ and no sigma-algebra $\mathcal F'$ is mentioned, we assume it is the Borel sets. This is true even if Lebesgue measure has been mentioned. The notion of Lebesgue-to-Lebesgue measurability is hardly ever useful. $\endgroup$
    – GEdgar
    Nov 20, 2020 at 12:42

1 Answer 1

1
$\begingroup$

The definition of $\sigma(X_1,X_2,..)$ is it is the smallest sigma algebra which makes each $X_i$ measurable. It is the intersection of all sigma algebras which make each $X_i$ measurable. It is equal to the smallest sigma algebra which contains sets of the form $X_1^{-1}(A_1)\cap X_2^{-1}(A_2)\cap...\cap X_n^{-1}(A_n)$ where $n$ is a positive integer and $A_i$'s are in $\mathcal F'$. It is possible replace the finite intersections here by infinite intersections $X_1^{-1}(A_1)\cap X_2^{-1}(A_2)\cap...$ so your description is also correct.

$\endgroup$
2
  • $\begingroup$ It is also the smallest sigma-algebra which contains sets of the form $X_n^{-1}(A)$, where $n \in \mathbb N$ and $A \in \mathcal F'$. Instead of all $\mathcal F'$, you could use only a generating set for it. $\endgroup$
    – GEdgar
    Nov 20, 2020 at 12:39
  • $\begingroup$ Thank you. I wonder if the following identity is also correct: $\sigma(X_1,X_2,\dotsc)= \sigma(\cup_{n=1}^{+\infty} \sigma(X_1,\dotsc,X_n)) $. Is it true? $\endgroup$ Jul 7, 2023 at 11:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .