Define the filtration $(\mathcal{F}_n)_{n \in \mathbb{N}}$ on $[0,1]$ via $$ \mathcal{F}_n := \sigma\Big(\big[k 2^{-n},(k+1)2^{-n}\big) \mid k = 0, 1, \dotsc, 2^{n}-1 \Big). $$ Is it true that the "limit" of this filtration is the Borel $\sigma$-algebra, i.e., $$ \bigcup_{n \in \mathbb{N}} \mathcal{F}_n = \mathcal{B}([0,1])~~? $$
The direction $\subseteq$ seems obvious to me, but I cannot show the other direction.
No. Take $A=[0,1/3)$ : we have $A\notin \mathcal{F}_n$ for all $n$, but $A\in \mathcal{B}([0,1])$.
A true statement is : $$\mathcal{B}([0,1])=\sigma\bigg(\bigcup_{n\in\mathbb{N}} \left\{[k2^{-n},(k+1)2^{-n}) \mid k=0,1,\dotsc,2^n-1\right\}\bigg).$$