Let $(B^1,B^2)$ be independent Brownian motions with corresponding filtration $\mathcal{F}_t$. Let $\mathcal{F}^2_t$ be the filtration generated by $B^2$. How does one prove that for any $s<t$ and any $A$ measurable with respect to $\mathcal{F}_s$, we have $\mathbb{E}\left[A\mid\mathcal{F}^2_s\right]=\mathbb{E}\left[A\mid\mathcal{F}^2_t\right]$?
This seems "intuitively" obvious, but I'm not sure to prove it formally. I think the issue is I don't know how to relate the disjoint $\sigma$-algebras $\mathcal{F}_s$ and $\mathcal{F}^2_t$.
Intuitively, the idea seems to be that $A$ can in principal be written in terms of $(B^1,B^2)$ up to time $s$. Because $B^2$ is a Brownian motion, observing its value over $(s,t]$ tells us nothing about $A$.
Define $\mathcal I:=\left\{A_1\cap A_2,A_1\in\mathcal F_s^1,A_2\in\mathcal F_s^2\right\}$.
Then $\mathcal I$ is a $\pi$-system. Since the collection $$\left\{A\in\mathcal F_s,\mathbb E\left[A\mid\mathcal F_s^2\right]=\mathbb E\left[A\mid\mathcal F_t^2\right]\right\} $$ is a $\lambda$-system which contains $\mathcal I$, we can conclude by the $\pi$-$\lambda$ theorem.