Prove/Disprove $E[1_A | \mathscr{F_t}] = 0 \ \text{or} \ 1 \ \text{a.s.} \ \Rightarrow E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.}$
Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$, let $A \in \mathscr{F}$.
Suppose $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 1 \ \text{a.s.}$$ Does it follow that $$E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.} \ \forall s > t \ ?$$ What about $\forall s < t$?
What if instead $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 0 \ \text{a.s.} \ ?$$ Or what if $$E[1_A | \mathscr{F_t}] = p \ \text{a.s.} \ \text{for some} \ p \in (0,1) \ ?$$
What I tried:
If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$, which is the same as $1_A=1$ (almost surely). In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$.
Likewise, if $\Bbb E[1_A|\mathscr F_t]=0$, then $\Bbb E[1_A]=0$, which is the same as $1_A=0$ (almost surely). In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$.
If $\Bbb E[1_A|\mathscr F_t]=p$, for a constant $p\in(0,1)$, then we have
$\Bbb E[1_A|\mathscr F_s]=E[E[1_A|\mathscr F_t]|\mathscr F_s] = E[p|\mathscr F_s] = p$. This may fail if $s>t$.
Alternatively for $= p$ case:
Let $F$ be a bounded $\mathscr F_t$-measurable random variable.
$$\Bbb E[1_A\cdot F]=\Bbb E[E[1_A\cdot F|\mathscr F_t]]=\Bbb E[F\cdot E[1_A|\mathscr F_t]]$$
$$=\Bbb E[p\cdot F]=p\Bbb E[F]=\Bbb E[1_A]\cdot\Bbb E[F]$$
meaning that $1_A$ and $F$ are independent. In other words, $\sigma(A)$ and $\mathscr F_t$ are independent. So $\sigma(A)$ and $\mathscr F_s$ are also independent if $s<t$ and hence $E[1_A|\mathscr F_s] = E[1_A] = p$ . This may fail if $s>t$.
I guess the idea is that a constant is both independent of $\mathscr F_s$ and $\mathscr F_s$-measurablea constant is both independent of $\mathscr F_s$ and $\mathscr F_s$-measurable.
