I have the following filtration on $\Omega = [0,1]$. $$\mathcal{F}_n = \sigma([(k-1)2^{-n},k2^{-n}), 1\leq k \leq 2^n)$$ I also define $\mathcal{F}_{\infty} = \sigma\left(\cup_n\mathcal{F}_n\right)$. Is it then true that $\mathcal{F}_{\infty} = \mathcal{B}([0,1])$?

The smallest generator for $\mathcal{B}([0,1])$ that I can think of is the collection $\{[0,q), q \in [0,1]\cap \mathbb{Q}\}$. But would dyadic rationals do the job as well? Can I just say the dyadic rationals in $[0,1]$ are dense $[0,1]$, therefore I can approximate any element of $a \in [0,1]$ by dyadics and create intervals of the form $[0,a)$? Sorry for the terrible handwaving.

I realize now that I also need to be able to approach any real on $[0,1]$ either from right or from left with a dyadic sequence so that I can either take intersections or unions of the intervals to generate $[0,a)$ for any $a \in [0,1]$. Am I thinking in the right direction?

This took me a while to figure out but I think the following argument is all I need $$[0,a) = \cup_n\left[0,\frac{\lfloor 2^na\rfloor}{2^n}\right)$$

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    $\begingroup$ Any open interval of [0,1] is the union of countably many closed intervals with dyadic-rational endpoints. $\endgroup$ Mar 18, 2016 at 23:21
  • $\begingroup$ @user254665 Could you please take a look at my latest edit and let me know if the last argument is OK? $\endgroup$
    – Calculon
    Mar 19, 2016 at 14:06
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    $\begingroup$ Your argument is correct. You could also argue that any real number can be approximated by dyadic numbers, which is given by the dyadic expansion. $\endgroup$ Mar 19, 2016 at 15:34
  • $\begingroup$ It is OK. . If $D$ is any dense subset of $[0,1$] and $0\in D$ then $\{[a,b] :a,b\in D\}$ generates $B([0,1]). $ $\endgroup$ Mar 19, 2016 at 19:12
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    $\begingroup$ @aduh I don't remember why I needed dyadic rationals in particular. But I guess you can choose any other positive integer as well. $\endgroup$
    – Calculon
    Jul 11, 2017 at 17:01


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