We let $\mathcal{F}_\infty=\bigvee_n\mathcal{F}_n$. (What does $\bigvee$ usually denote? Is $\bigvee$ different from $\cup$? Can't find it in the text.)

The theorem tells us when we have $\lim_n E(Y|\mathcal{F}_n)=E(Y|\mathcal{F}_\infty)$.

I'd like to know:

  • Is $\mathcal{F}_\infty$ a $\sigma$-algebra?
  • Is it true that $A\in\mathcal{F}_\infty$ implies $A\in\mathcal{F}_n$ for some $n$?

It seems the answer to the former (assuming the latter is true) is no: If $A\in \mathcal{F}_1$ and $B\in\mathcal{F}_2$, then $A\cup B$ need not be an element of $\mathcal{F_n}$ for any $n$.

But then $E(Y|\mathcal{F}_\infty)$ is nonsensical, since conditional expectation (as a random variable) is defined using $\sigma$-algebras.


$\bigvee_n \mathcal{F}_n$ is the $\sigma$-algebra generated by $\bigcup_n \mathcal{F}_n$ (i.e., smallest $\sigma$-algebra containing $\bigcup_n \mathcal{F}_n$).

This is needed because in general, $\bigcup_n \mathcal{F}_n$ might not be a $\sigma$-algebra. See here for an easy example. See here for an example where the sequence is nested, i.e. $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \cdots$.

So, by definition $\mathcal{F}_\infty$ is a $\sigma$-algebra. For your second bullet-point, I think the answer is no.

  • $\begingroup$ Bummer. I'd like to see why we have $\int_A Y=\int_A X_n=\int_A X_\infty$, where $A\in\mathcal{F}_\infty$. This is clear when $A\in\mathcal{F}_\infty$ implies $A\in\mathcal{F}_n$ for some $n$ (i.e. second bullet point is true). Not sure otherwise. Any pointers? $\endgroup$ – manofbear Apr 28 '17 at 17:40
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    $\begingroup$ @manofbear: Your comment seems to involve a lot of additional context, so you should ask it as a new question. $\endgroup$ – Nate Eldredge Apr 28 '17 at 17:46
  • $\begingroup$ @NateEldredge Thanks for the tip, will do $\endgroup$ – manofbear Apr 28 '17 at 17:48

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