If each $(B^n)_{t\in[0,\:T]}$ is $\mathcal F$-adapted, then $\sigma\left(\bigcup_{n\in\mathbb N}\sigma\left(B^n_s:s\le t\right)\right)=\mathcal F_t$

Question

Let

• $T>0$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration on $(\Omega,\mathcal A)$
• $(B^n)_{t\in[0,\:T]}$ be an $\mathcal F$-adapted stochastic process on $(\Omega,\mathcal A,\operatorname P)$, for all $n\in\mathbb N$

Can we show that $$\mathcal G_t:=\sigma\left(\bigcup_{n\in\mathbb N}\sigma\left(B^n_s:s\le t\right)\right)=\mathcal F_t$$ for all $t\in[0,T]$?

Clearly, $$\sigma(B_s^n)\subseteq\mathcal F_s\;\;\;\text{for all }s\in[0,T]\tag 1$$ and hence (since $\mathcal F_s\subseteq\mathcal F_t$ for all $0\le s\le t\le T$) $$\sigma\left(B_s^n:s\le t\right)=\sigma\left(\bigcup_{s\le t}\sigma(B_s^n)\right)\subseteq\sigma\left(\bigcup_{s\le t}\mathcal F_s\right)=\sigma(\mathcal F_t)=\mathcal F_t\;\;\;\text{for all }t\in[0,T]\;.\tag 2$$ Thus, $$\mathcal G_t\subseteq\mathcal F_t\;.\tag 3$$

Answer 1

Of course not…

Given any non-trivial filtration take $B^n \equiv n$ for all $n\in\mathbb{N}$. Then $\mathcal{G}_t$ is trivial for each t but $\mathcal{F}_t$ is not, so they are not equal but totally satisfy your settings…