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 $$[0,a) = \cup_n\left[0,\frac{\lfloor 2^na\rfloor}{2^n}\right)$$
