Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $T>0$
- $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration of $\mathcal A$
Let $$\mathcal R_a:=\bigcup_{F\in\mathcal F_a}F\times\left\{a\right\}\cup\bigcup_{a\le s<t\le T}\bigcup_{F\in\mathcal F_s}F\times(s,t]$$ and $$\mathcal P_a:=\sigma(\mathcal R_a)$$ for $a\in[0,T]$. Are we able to show that $$\left.\mathcal P_a\right|_{\Omega\times[b,\:T]}:=\left\{A\cap(\Omega\times[b,T]:A\in\mathcal P_a\right\}=\mathcal P_b\tag1$$ for all $0\le a\le b\le T$?
It's easy to see that $$\left.\mathcal R_a\right|_{\Omega\times[b,\:T]}\subseteq\mathcal P_b\tag2$$ (where the system on the left-hand side is defined in the obvious way). Now, I've hoped that I could show $\mathcal R_b\subseteq\sigma\left(\left.\mathcal R_a\right|_{\Omega\times[b,\:T]}\right)$. For this purpose, let $B\in\mathcal R_b$. If $$B=F\times(s,t]$$ for some $b\le s<t\le T$ and $F\in\mathcal F_s$, then (since $a\le b$) $$B\in\mathcal R_a$$ and hence $$B=B\cap(\Omega\times[b,T])\in\left.\mathcal R_a\right|_{\Omega\times[b,\:T]}\;.$$
Otherwise, $$B=F\times\left\{b\right\}$$ for some $F\in\mathcal F_b$ and I think that we've encountered a problem here: $B$ doesn't belong to $\mathcal R_a$.
We've only got $$F\times\left\{t\right\}=F\times\bigcap_{n\in\mathbb N}\left(t-\frac1n,t\right]\in\mathcal P_b\;\;\;\text{for all }t\in(b,T]\;.\tag3$$
Can we solve that problem or is $(1)$ wrong?